{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0001","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c3cd5b7a39fde475440347de2df15fe66069e89503e7419009e2195685a1577c","family_id":"qf-mod28-commring-qualification","seed":202610281001,"semantic_template":"qf-mod28-commring-qualification/{\"addition\":[[0,1,2,3,4,5,6,7],[1,0,3,2,5,4,7,6],[2,3,0,1,6,7,4,5],[3,2,1,0,7,6,5,4],[4,5,6,7,0,1,2,3],[5,4,7,6,1,0,3,2],[6,7,4,5,2,3,0,1],[7,6,5,4,3,2,1,0]],\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,0,1,0,1,0,1],[0,2,0,2,0,2,0,2],[0,3,0,3,0,3,0,3],[0,0,2,2,4,4,6,6],[0,1,2,3,4,5,6,7],[0,2,2,0,4,6,6,4],[0,3,2,1,4,7,6,5]],\"n\":8}"},"fields":{"title":"Multiplicative Commutativity Is Additional to Complete Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 0, 3, 2, 5, 4, 7, 6], [2, 3, 0, 1, 6, 7, 4, 5], [3, 2, 1, 0, 7, 6, 5, 4], [4, 5, 6, 7, 0, 1, 2, 3], [5, 4, 7, 6, 1, 0, 3, 2], [6, 7, 4, 5, 2, 3, 0, 1], [7, 6, 5, 4, 3, 2, 1, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 0, 1, 0, 1, 0, 1], [0, 2, 0, 2, 0, 2, 0, 2], [0, 3, 0, 3, 0, 3, 0, 3], [0, 0, 2, 2, 4, 4, 6, 6], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 2, 0, 4, 6, 6, 4], [0, 3, 2, 1, 4, 7, 6, 5]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.","answer":{"kind":"expression","value":"(8, 0, 5, [[1, 2, 0, 2], [1, 3, 1, 3], [1, 6, 0, 2], [1, 7, 1, 3], [2, 3, 2, 0], [2, 4, 0, 2], [2, 6, 0, 2], [3, 4, 0, 2], [3, 7, 3, 1], [4, 6, 6, 4], [4, 7, 6, 4], [6, 7, 4, 6]], No, unital_ring_not_commutative)","explanation":"Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 5, [[1, 2, 0, 2], [1, 3, 1, 3], [1, 6, 0, 2], [1, 7, 1, 3], [2, 3, 2, 0], [2, 4, 0, 2], [2, 6, 0, 2], [3, 4, 0, 2], [3, 7, 3, 1], [4, 6, 6, 4], [4, 7, 6, 4], [6, 7, 4, 6]], No, unital_ring_not_commutative)."},"skills":["Multiplicative Commutativity Is Additional to Complete Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0002","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b31e699a2100c5d56457b8d7f97de8c87aedd0a990669d652543cfa408f7852b","family_id":"qf-mod28-commring-qualification","seed":202610281002,"semantic_template":"qf-mod28-commring-qualification/{\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,4,5,7,0,6],[2,3,4,5,7,6,1,0],[3,4,5,7,6,0,2,1],[4,5,7,6,0,1,3,2],[5,7,6,0,1,2,4,3],[6,0,1,2,3,4,7,5],[7,6,0,1,2,3,5,4]],\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,7,0,2,7,4],[0,3,7,1,4,6,5,2],[0,4,0,4,0,4,4,0],[0,5,2,6,4,1,3,7],[0,6,7,5,4,3,1,2],[0,7,4,2,0,7,2,4]],\"n\":8}"},"fields":{"title":"Multiplicative Commutativity Is Additional to Complete Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 7, 0, 6], [2, 3, 4, 5, 7, 6, 1, 0], [3, 4, 5, 7, 6, 0, 2, 1], [4, 5, 7, 6, 0, 1, 3, 2], [5, 7, 6, 0, 1, 2, 4, 3], [6, 0, 1, 2, 3, 4, 7, 5], [7, 6, 0, 1, 2, 3, 5, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 7, 0, 2, 7, 4], [0, 3, 7, 1, 4, 6, 5, 2], [0, 4, 0, 4, 0, 4, 4, 0], [0, 5, 2, 6, 4, 1, 3, 7], [0, 6, 7, 5, 4, 3, 1, 2], [0, 7, 4, 2, 0, 7, 2, 4]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.","answer":{"kind":"expression","value":"(8, 0, 1, [], Yes, commutative_ring)","explanation":"Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring)."},"skills":["Multiplicative Commutativity Is Additional to Complete Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0003","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3546b16cb0a0df026275990546ad59c798d5c5d776de68a10d695a0abbf8501b","family_id":"qf-mod28-commring-qualification","seed":202610281003,"semantic_template":"qf-mod28-commring-qualification/{\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,4,6,7,5,0],[2,3,4,6,5,0,7,1],[3,4,6,5,7,1,0,2],[4,6,5,7,0,2,1,3],[5,7,0,1,2,4,3,6],[6,5,7,0,1,3,2,4],[7,0,1,2,3,6,4,5]],\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,5,0,4,2,5],[0,3,5,1,4,2,7,6],[0,4,0,4,0,0,4,4],[0,5,4,2,0,4,5,2],[0,6,2,7,4,5,1,3],[0,7,5,6,4,2,3,1]],\"n\":8}"},"fields":{"title":"Multiplicative Commutativity Is Additional to Complete Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 6, 7, 5, 0], [2, 3, 4, 6, 5, 0, 7, 1], [3, 4, 6, 5, 7, 1, 0, 2], [4, 6, 5, 7, 0, 2, 1, 3], [5, 7, 0, 1, 2, 4, 3, 6], [6, 5, 7, 0, 1, 3, 2, 4], [7, 0, 1, 2, 3, 6, 4, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 5, 0, 4, 2, 5], [0, 3, 5, 1, 4, 2, 7, 6], [0, 4, 0, 4, 0, 0, 4, 4], [0, 5, 4, 2, 0, 4, 5, 2], [0, 6, 2, 7, 4, 5, 1, 3], [0, 7, 5, 6, 4, 2, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.","answer":{"kind":"expression","value":"(8, 0, 1, [], Yes, commutative_ring)","explanation":"Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring)."},"skills":["Multiplicative Commutativity Is Additional to Complete Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0004","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"35e736d65b7e5a9fd74dee64db43e34fb396b3edcf5677de5abf1ee9fd4711be","family_id":"qf-mod28-commring-qualification","seed":202610281004,"semantic_template":"qf-mod28-commring-qualification/{\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,4,6,0,7,5],[2,3,4,6,7,1,5,0],[3,4,6,7,5,2,0,1],[4,6,7,5,0,3,1,2],[5,0,1,2,3,7,4,6],[6,7,5,0,1,4,2,3],[7,5,0,1,2,6,3,4]],\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,7,0,7,2,4],[0,3,7,1,4,6,5,2],[0,4,0,4,0,4,4,0],[0,5,7,6,4,1,3,2],[0,6,2,5,4,3,1,7],[0,7,4,2,0,2,7,4]],\"n\":8}"},"fields":{"title":"Multiplicative Commutativity Is Additional to Complete Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 6, 0, 7, 5], [2, 3, 4, 6, 7, 1, 5, 0], [3, 4, 6, 7, 5, 2, 0, 1], [4, 6, 7, 5, 0, 3, 1, 2], [5, 0, 1, 2, 3, 7, 4, 6], [6, 7, 5, 0, 1, 4, 2, 3], [7, 5, 0, 1, 2, 6, 3, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 7, 0, 7, 2, 4], [0, 3, 7, 1, 4, 6, 5, 2], [0, 4, 0, 4, 0, 4, 4, 0], [0, 5, 7, 6, 4, 1, 3, 2], [0, 6, 2, 5, 4, 3, 1, 7], [0, 7, 4, 2, 0, 2, 7, 4]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.","answer":{"kind":"expression","value":"(8, 0, 1, [], Yes, commutative_ring)","explanation":"Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring)."},"skills":["Multiplicative Commutativity Is Additional to Complete Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0005","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a56a8030fa3a11b7e21fd2bbfb035fffe0cd4b48d9c816e4d1fa93ce980dcfb8","family_id":"qf-mod28-commring-qualification","seed":202610281005,"semantic_template":"qf-mod28-commring-qualification/{\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,4,7,6,0,5],[2,3,4,7,5,0,1,6],[3,4,7,5,6,1,2,0],[4,7,5,6,0,2,3,1],[5,6,0,1,2,4,7,3],[6,0,1,2,3,7,5,4],[7,5,6,0,1,3,4,2]],\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,5,0,4,5,2],[0,3,5,1,4,2,7,6],[0,4,0,4,0,0,4,4],[0,5,4,2,0,4,2,5],[0,6,5,7,4,2,1,3],[0,7,2,6,4,5,3,1]],\"n\":8}"},"fields":{"title":"Multiplicative Commutativity Is Additional to Complete Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 7, 6, 0, 5], [2, 3, 4, 7, 5, 0, 1, 6], [3, 4, 7, 5, 6, 1, 2, 0], [4, 7, 5, 6, 0, 2, 3, 1], [5, 6, 0, 1, 2, 4, 7, 3], [6, 0, 1, 2, 3, 7, 5, 4], [7, 5, 6, 0, 1, 3, 4, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 5, 0, 4, 5, 2], [0, 3, 5, 1, 4, 2, 7, 6], [0, 4, 0, 4, 0, 0, 4, 4], [0, 5, 4, 2, 0, 4, 2, 5], [0, 6, 5, 7, 4, 2, 1, 3], [0, 7, 2, 6, 4, 5, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.","answer":{"kind":"expression","value":"(8, 0, 1, [], Yes, commutative_ring)","explanation":"Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring)."},"skills":["Multiplicative Commutativity Is Additional to Complete Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0006","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e630e9d36b23f5cce2b5f02334e8611c039dece1f94f6a5b3d6b85a14d2df1a3","family_id":"qf-mod28-commring-qualification","seed":202610281006,"semantic_template":"qf-mod28-commring-qualification/{\"addition\":[[0,1,2,3,4,5,6,7],[1,0,3,2,7,6,5,4],[2,3,0,1,6,7,4,5],[3,2,1,0,5,4,7,6],[4,7,6,5,0,3,2,1],[5,6,7,4,3,0,1,2],[6,5,4,7,2,1,0,3],[7,4,5,6,1,2,3,0]],\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,0,1,0,1,0,1],[0,2,0,2,0,2,0,2],[0,3,0,3,0,3,0,3],[0,0,2,2,4,6,6,4],[0,3,2,1,4,7,6,5],[0,2,2,0,4,4,6,6],[0,1,2,3,4,5,6,7]],\"n\":8}"},"fields":{"title":"Multiplicative Commutativity Is Additional to Complete Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 0, 3, 2, 7, 6, 5, 4], [2, 3, 0, 1, 6, 7, 4, 5], [3, 2, 1, 0, 5, 4, 7, 6], [4, 7, 6, 5, 0, 3, 2, 1], [5, 6, 7, 4, 3, 0, 1, 2], [6, 5, 4, 7, 2, 1, 0, 3], [7, 4, 5, 6, 1, 2, 3, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 0, 1, 0, 1, 0, 1], [0, 2, 0, 2, 0, 2, 0, 2], [0, 3, 0, 3, 0, 3, 0, 3], [0, 0, 2, 2, 4, 6, 6, 4], [0, 3, 2, 1, 4, 7, 6, 5], [0, 2, 2, 0, 4, 4, 6, 6], [0, 1, 2, 3, 4, 5, 6, 7]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.","answer":{"kind":"expression","value":"(8, 0, 7, [[1, 2, 0, 2], [1, 3, 1, 3], [1, 5, 1, 3], [1, 6, 0, 2], [2, 3, 2, 0], [2, 4, 0, 2], [2, 6, 0, 2], [3, 4, 0, 2], [3, 5, 3, 1], [4, 5, 6, 4], [4, 6, 6, 4], [5, 6, 6, 4]], No, unital_ring_not_commutative)","explanation":"Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 7, [[1, 2, 0, 2], [1, 3, 1, 3], [1, 5, 1, 3], [1, 6, 0, 2], [2, 3, 2, 0], [2, 4, 0, 2], [2, 6, 0, 2], [3, 4, 0, 2], [3, 5, 3, 1], [4, 5, 6, 4], [4, 6, 6, 4], [5, 6, 6, 4]], No, unital_ring_not_commutative)."},"skills":["Multiplicative Commutativity Is Additional to Complete Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0007","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"838c9c0eb5e11fcf8ad5b47537a3056b9078c2a9a76e97c29af1eccaef6fd050","family_id":"qf-mod28-commring-qualification","seed":202610281007,"semantic_template":"qf-mod28-commring-qualification/{\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,5,6,4,7,0],[2,3,5,4,7,6,0,1],[3,5,4,6,0,7,1,2],[4,6,7,0,2,1,3,5],[5,4,6,7,1,0,2,3],[6,7,0,1,3,2,5,4],[7,0,1,2,5,3,4,6]],\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,6,2,0,5,6],[0,3,6,1,7,5,2,4],[0,4,2,7,1,5,6,3],[0,5,0,5,5,0,0,5],[0,6,5,2,6,0,5,2],[0,7,6,4,3,5,2,1]],\"n\":8}"},"fields":{"title":"Multiplicative Commutativity Is Additional to Complete Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 6, 4, 7, 0], [2, 3, 5, 4, 7, 6, 0, 1], [3, 5, 4, 6, 0, 7, 1, 2], [4, 6, 7, 0, 2, 1, 3, 5], [5, 4, 6, 7, 1, 0, 2, 3], [6, 7, 0, 1, 3, 2, 5, 4], [7, 0, 1, 2, 5, 3, 4, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 6, 2, 0, 5, 6], [0, 3, 6, 1, 7, 5, 2, 4], [0, 4, 2, 7, 1, 5, 6, 3], [0, 5, 0, 5, 5, 0, 0, 5], [0, 6, 5, 2, 6, 0, 5, 2], [0, 7, 6, 4, 3, 5, 2, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.","answer":{"kind":"expression","value":"(8, 0, 1, [], Yes, commutative_ring)","explanation":"Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring)."},"skills":["Multiplicative Commutativity Is Additional to Complete Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0008","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e67dd92204e2e5a2ea8421fa8c2cd2e52952825d5e559c655247d68b8cbde102","family_id":"qf-mod28-commring-qualification","seed":202610281008,"semantic_template":"qf-mod28-commring-qualification/{\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,5,7,4,0,6],[2,3,5,4,6,7,1,0],[3,5,4,7,0,6,2,1],[4,7,6,0,2,1,5,3],[5,4,7,6,1,0,3,2],[6,0,1,2,5,3,7,4],[7,6,0,1,3,2,4,5]],\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,7,2,0,7,5],[0,3,7,1,6,5,4,2],[0,4,2,6,1,5,3,7],[0,5,0,5,5,0,5,0],[0,6,7,4,3,5,1,2],[0,7,5,2,7,0,2,5]],\"n\":8}"},"fields":{"title":"Multiplicative Commutativity Is Additional to Complete Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 7, 4, 0, 6], [2, 3, 5, 4, 6, 7, 1, 0], [3, 5, 4, 7, 0, 6, 2, 1], [4, 7, 6, 0, 2, 1, 5, 3], [5, 4, 7, 6, 1, 0, 3, 2], [6, 0, 1, 2, 5, 3, 7, 4], [7, 6, 0, 1, 3, 2, 4, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 7, 2, 0, 7, 5], [0, 3, 7, 1, 6, 5, 4, 2], [0, 4, 2, 6, 1, 5, 3, 7], [0, 5, 0, 5, 5, 0, 5, 0], [0, 6, 7, 4, 3, 5, 1, 2], [0, 7, 5, 2, 7, 0, 2, 5]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.","answer":{"kind":"expression","value":"(8, 0, 1, [], Yes, commutative_ring)","explanation":"Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring)."},"skills":["Multiplicative Commutativity Is Additional to Complete Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0009","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"14d62b7b20f92b17dca8165b996c316d783457da2d1ff5d77924d68f7fb726fa","family_id":"qf-mod28-commring-qualification","seed":202610281009,"semantic_template":"qf-mod28-commring-qualification/{\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,5,7,6,4,0],[2,3,5,6,0,4,7,1],[3,5,6,4,1,7,0,2],[4,7,0,1,5,2,3,6],[5,6,4,7,2,0,1,3],[6,4,7,0,3,1,2,5],[7,0,1,2,6,3,5,4]],\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,4,5,0,2,4],[0,3,4,1,2,5,7,6],[0,4,5,2,5,0,4,2],[0,5,0,5,0,0,5,5],[0,6,2,7,4,5,1,3],[0,7,4,6,2,5,3,1]],\"n\":8}"},"fields":{"title":"Multiplicative Commutativity Is Additional to Complete Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 7, 6, 4, 0], [2, 3, 5, 6, 0, 4, 7, 1], [3, 5, 6, 4, 1, 7, 0, 2], [4, 7, 0, 1, 5, 2, 3, 6], [5, 6, 4, 7, 2, 0, 1, 3], [6, 4, 7, 0, 3, 1, 2, 5], [7, 0, 1, 2, 6, 3, 5, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 4, 5, 0, 2, 4], [0, 3, 4, 1, 2, 5, 7, 6], [0, 4, 5, 2, 5, 0, 4, 2], [0, 5, 0, 5, 0, 0, 5, 5], [0, 6, 2, 7, 4, 5, 1, 3], [0, 7, 4, 6, 2, 5, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.","answer":{"kind":"expression","value":"(8, 0, 1, [], Yes, commutative_ring)","explanation":"Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring)."},"skills":["Multiplicative Commutativity Is Additional to Complete Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0010","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"83219141f02edcb6260532ba11a6f30a255c65abdcff6a7540f9f43073b3f1b0","family_id":"qf-mod28-commring-qualification","seed":202610281010,"semantic_template":"qf-mod28-commring-qualification/{\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,5,0,6,7,4],[2,3,5,6,1,7,4,0],[3,5,6,7,2,4,0,1],[4,0,1,2,7,3,5,6],[5,6,7,4,3,0,1,2],[6,7,4,0,5,1,2,3],[7,4,0,1,6,2,3,5]],\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,7,7,0,2,5],[0,3,7,1,6,5,4,2],[0,4,7,6,1,5,3,2],[0,5,0,5,5,0,5,0],[0,6,2,4,3,5,1,7],[0,7,5,2,2,0,7,5]],\"n\":8}"},"fields":{"title":"Multiplicative Commutativity Is Additional to Complete Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 0, 6, 7, 4], [2, 3, 5, 6, 1, 7, 4, 0], [3, 5, 6, 7, 2, 4, 0, 1], [4, 0, 1, 2, 7, 3, 5, 6], [5, 6, 7, 4, 3, 0, 1, 2], [6, 7, 4, 0, 5, 1, 2, 3], [7, 4, 0, 1, 6, 2, 3, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 7, 7, 0, 2, 5], [0, 3, 7, 1, 6, 5, 4, 2], [0, 4, 7, 6, 1, 5, 3, 2], [0, 5, 0, 5, 5, 0, 5, 0], [0, 6, 2, 4, 3, 5, 1, 7], [0, 7, 5, 2, 2, 0, 7, 5]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.","answer":{"kind":"expression","value":"(8, 0, 1, [], Yes, commutative_ring)","explanation":"Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring)."},"skills":["Multiplicative Commutativity Is Additional to Complete Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0011","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"077b145a8553e51754b424da83c7d19d1d1b1edcc42efc2708d16f46798b79da","family_id":"qf-mod28-commring-qualification","seed":202610281011,"semantic_template":"qf-mod28-commring-qualification/{\"addition\":[[0,1,2,3,4,5,6,7],[1,0,3,2,6,7,4,5],[2,3,0,1,5,4,7,6],[3,2,1,0,7,6,5,4],[4,6,5,7,0,2,1,3],[5,7,4,6,2,0,3,1],[6,4,7,5,1,3,0,2],[7,5,6,4,3,1,2,0]],\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,0,1,0,0,1,1],[0,2,0,2,0,0,2,2],[0,3,0,3,0,0,3,3],[0,2,2,0,4,5,5,4],[0,0,2,2,4,5,4,5],[0,3,2,1,4,5,7,6],[0,1,2,3,4,5,6,7]],\"n\":8}"},"fields":{"title":"Multiplicative Commutativity Is Additional to Complete Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 0, 3, 2, 6, 7, 4, 5], [2, 3, 0, 1, 5, 4, 7, 6], [3, 2, 1, 0, 7, 6, 5, 4], [4, 6, 5, 7, 0, 2, 1, 3], [5, 7, 4, 6, 2, 0, 3, 1], [6, 4, 7, 5, 1, 3, 0, 2], [7, 5, 6, 4, 3, 1, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 0, 1, 0, 0, 1, 1], [0, 2, 0, 2, 0, 0, 2, 2], [0, 3, 0, 3, 0, 0, 3, 3], [0, 2, 2, 0, 4, 5, 5, 4], [0, 0, 2, 2, 4, 5, 4, 5], [0, 3, 2, 1, 4, 5, 7, 6], [0, 1, 2, 3, 4, 5, 6, 7]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.","answer":{"kind":"expression","value":"(8, 0, 7, [[1, 2, 0, 2], [1, 3, 1, 3], [1, 4, 0, 2], [1, 6, 1, 3], [2, 3, 2, 0], [2, 4, 0, 2], [2, 5, 0, 2], [3, 5, 0, 2], [3, 6, 3, 1], [4, 5, 5, 4], [4, 6, 5, 4], [5, 6, 4, 5]], No, unital_ring_not_commutative)","explanation":"Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 7, [[1, 2, 0, 2], [1, 3, 1, 3], [1, 4, 0, 2], [1, 6, 1, 3], [2, 3, 2, 0], [2, 4, 0, 2], [2, 5, 0, 2], [3, 5, 0, 2], [3, 6, 3, 1], [4, 5, 5, 4], [4, 6, 5, 4], [5, 6, 4, 5]], No, unital_ring_not_commutative)."},"skills":["Multiplicative Commutativity Is Additional to Complete Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0012","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c3336bb68babede04bb051cca8cc03a84365c02835531034bf4dd0010ed5e484","family_id":"qf-mod28-commring-qualification","seed":202610281012,"semantic_template":"qf-mod28-commring-qualification/{\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,5,0,7,4,6],[2,3,5,7,1,6,0,4],[3,5,7,6,2,4,1,0],[4,0,1,2,6,3,7,5],[5,7,6,4,3,0,2,1],[6,4,0,1,7,2,5,3],[7,6,4,0,5,1,3,2]],\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,6,6,0,5,2],[0,3,6,1,7,5,2,4],[0,4,6,7,1,5,2,3],[0,5,0,5,5,0,0,5],[0,6,5,2,2,0,5,6],[0,7,2,4,3,5,6,1]],\"n\":8}"},"fields":{"title":"Multiplicative Commutativity Is Additional to Complete Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 0, 7, 4, 6], [2, 3, 5, 7, 1, 6, 0, 4], [3, 5, 7, 6, 2, 4, 1, 0], [4, 0, 1, 2, 6, 3, 7, 5], [5, 7, 6, 4, 3, 0, 2, 1], [6, 4, 0, 1, 7, 2, 5, 3], [7, 6, 4, 0, 5, 1, 3, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 6, 6, 0, 5, 2], [0, 3, 6, 1, 7, 5, 2, 4], [0, 4, 6, 7, 1, 5, 2, 3], [0, 5, 0, 5, 5, 0, 0, 5], [0, 6, 5, 2, 2, 0, 5, 6], [0, 7, 2, 4, 3, 5, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.","answer":{"kind":"expression","value":"(8, 0, 1, [], Yes, commutative_ring)","explanation":"Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring)."},"skills":["Multiplicative Commutativity Is Additional to Complete Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0013","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"961350efa140591fa2356e6c7d95df1d245c98d14131e012e6fafee522195fbc","family_id":"qf-mod28-commring-qualification","seed":202610281013,"semantic_template":"qf-mod28-commring-qualification/{\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,6,5,7,4,0],[2,3,6,4,7,0,5,1],[3,6,4,5,0,1,7,2],[4,5,7,0,2,3,1,6],[5,7,0,1,3,6,2,4],[6,4,5,7,1,2,0,3],[7,0,1,2,6,4,3,5]],\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,5,2,6,0,5],[0,3,5,1,7,2,6,4],[0,4,2,7,1,5,6,3],[0,5,6,2,5,6,0,2],[0,6,0,6,6,0,0,6],[0,7,5,4,3,2,6,1]],\"n\":8}"},"fields":{"title":"Multiplicative Commutativity Is Additional to Complete Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 5, 7, 4, 0], [2, 3, 6, 4, 7, 0, 5, 1], [3, 6, 4, 5, 0, 1, 7, 2], [4, 5, 7, 0, 2, 3, 1, 6], [5, 7, 0, 1, 3, 6, 2, 4], [6, 4, 5, 7, 1, 2, 0, 3], [7, 0, 1, 2, 6, 4, 3, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 5, 2, 6, 0, 5], [0, 3, 5, 1, 7, 2, 6, 4], [0, 4, 2, 7, 1, 5, 6, 3], [0, 5, 6, 2, 5, 6, 0, 2], [0, 6, 0, 6, 6, 0, 0, 6], [0, 7, 5, 4, 3, 2, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.","answer":{"kind":"expression","value":"(8, 0, 1, [], Yes, commutative_ring)","explanation":"Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring)."},"skills":["Multiplicative Commutativity Is Additional to Complete Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0014","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1e795952cfb6f342f78f288660184ec5eed4eedc53816fa0e7aa48045395fd3a","family_id":"qf-mod28-commring-qualification","seed":202610281014,"semantic_template":"qf-mod28-commring-qualification/{\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,6,7,0,4,5],[2,3,6,4,5,1,7,0],[3,6,4,7,0,2,5,1],[4,7,5,0,2,6,1,3],[5,0,1,2,6,7,3,4],[6,4,7,5,1,3,0,2],[7,5,0,1,3,4,2,6]],\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,7,2,7,0,6],[0,3,7,1,5,4,6,2],[0,4,2,5,1,3,6,7],[0,5,7,4,3,1,6,2],[0,6,0,6,6,6,0,0],[0,7,6,2,7,2,0,6]],\"n\":8}"},"fields":{"title":"Multiplicative Commutativity Is Additional to Complete Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 7, 0, 4, 5], [2, 3, 6, 4, 5, 1, 7, 0], [3, 6, 4, 7, 0, 2, 5, 1], [4, 7, 5, 0, 2, 6, 1, 3], [5, 0, 1, 2, 6, 7, 3, 4], [6, 4, 7, 5, 1, 3, 0, 2], [7, 5, 0, 1, 3, 4, 2, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 7, 2, 7, 0, 6], [0, 3, 7, 1, 5, 4, 6, 2], [0, 4, 2, 5, 1, 3, 6, 7], [0, 5, 7, 4, 3, 1, 6, 2], [0, 6, 0, 6, 6, 6, 0, 0], [0, 7, 6, 2, 7, 2, 0, 6]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.","answer":{"kind":"expression","value":"(8, 0, 1, [], Yes, commutative_ring)","explanation":"Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring)."},"skills":["Multiplicative Commutativity Is Additional to Complete Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0015","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"09b5f869faacf0c511d6fd9c06ea8f25e082258b669a39f0f30bd720d6e0231b","family_id":"qf-mod28-commring-qualification","seed":202610281015,"semantic_template":"qf-mod28-commring-qualification/{\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,6,7,4,5,0],[2,3,6,5,0,7,4,1],[3,6,5,4,1,0,7,2],[4,7,0,1,6,3,2,5],[5,4,7,0,3,2,1,6],[6,5,4,7,2,1,0,3],[7,0,1,2,5,6,3,4]],\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,4,6,2,0,4],[0,3,4,1,2,7,6,5],[0,4,6,2,6,4,0,2],[0,5,2,7,4,1,6,3],[0,6,0,6,0,6,0,6],[0,7,4,5,2,3,6,1]],\"n\":8}"},"fields":{"title":"Multiplicative Commutativity Is Additional to Complete Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 7, 4, 5, 0], [2, 3, 6, 5, 0, 7, 4, 1], [3, 6, 5, 4, 1, 0, 7, 2], [4, 7, 0, 1, 6, 3, 2, 5], [5, 4, 7, 0, 3, 2, 1, 6], [6, 5, 4, 7, 2, 1, 0, 3], [7, 0, 1, 2, 5, 6, 3, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 4, 6, 2, 0, 4], [0, 3, 4, 1, 2, 7, 6, 5], [0, 4, 6, 2, 6, 4, 0, 2], [0, 5, 2, 7, 4, 1, 6, 3], [0, 6, 0, 6, 0, 6, 0, 6], [0, 7, 4, 5, 2, 3, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.","answer":{"kind":"expression","value":"(8, 0, 1, [], Yes, commutative_ring)","explanation":"Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring)."},"skills":["Multiplicative Commutativity Is Additional to Complete Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0016","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b5eec7dede2b7d9bf16fac492a263ac184942aa1c04cead442857dca79456bbb","family_id":"qf-mod28-commring-qualification","seed":202610281016,"semantic_template":"qf-mod28-commring-qualification/{\"addition\":[[0,1,2,3,4,5,6,7],[1,0,3,2,7,6,5,4],[2,3,0,1,5,4,7,6],[3,2,1,0,6,7,4,5],[4,7,5,6,0,2,3,1],[5,6,4,7,2,0,1,3],[6,5,7,4,3,1,0,2],[7,4,6,5,1,3,2,0]],\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,0,1,1,1,0,0],[0,2,0,2,2,2,0,0],[0,3,0,3,3,3,0,0],[0,3,2,1,5,4,6,7],[0,1,2,3,4,5,6,7],[0,0,2,2,7,6,6,7],[0,2,2,0,6,7,6,7]],\"n\":8}"},"fields":{"title":"Multiplicative Commutativity Is Additional to Complete Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 0, 3, 2, 7, 6, 5, 4], [2, 3, 0, 1, 5, 4, 7, 6], [3, 2, 1, 0, 6, 7, 4, 5], [4, 7, 5, 6, 0, 2, 3, 1], [5, 6, 4, 7, 2, 0, 1, 3], [6, 5, 7, 4, 3, 1, 0, 2], [7, 4, 6, 5, 1, 3, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 0, 1, 1, 1, 0, 0], [0, 2, 0, 2, 2, 2, 0, 0], [0, 3, 0, 3, 3, 3, 0, 0], [0, 3, 2, 1, 5, 4, 6, 7], [0, 1, 2, 3, 4, 5, 6, 7], [0, 0, 2, 2, 7, 6, 6, 7], [0, 2, 2, 0, 6, 7, 6, 7]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.","answer":{"kind":"expression","value":"(8, 0, 5, [[1, 2, 0, 2], [1, 3, 1, 3], [1, 4, 1, 3], [1, 7, 0, 2], [2, 3, 2, 0], [2, 6, 0, 2], [2, 7, 0, 2], [3, 4, 3, 1], [3, 6, 0, 2], [4, 6, 6, 7], [4, 7, 7, 6], [6, 7, 7, 6]], No, unital_ring_not_commutative)","explanation":"Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 5, [[1, 2, 0, 2], [1, 3, 1, 3], [1, 4, 1, 3], [1, 7, 0, 2], [2, 3, 2, 0], [2, 6, 0, 2], [2, 7, 0, 2], [3, 4, 3, 1], [3, 6, 0, 2], [4, 6, 6, 7], [4, 7, 7, 6], [6, 7, 7, 6]], No, unital_ring_not_commutative)."},"skills":["Multiplicative Commutativity Is Additional to Complete Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0017","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7def6addf7d8c6a9f77796ca73741d96ec3c84a78ce0b33163e3dd1f1957bfef","family_id":"qf-mod28-commring-qualification","seed":202610281017,"semantic_template":"qf-mod28-commring-qualification/{\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,6,5,0,7,4],[2,3,6,7,0,1,4,5],[3,6,7,4,1,2,5,0],[4,5,0,1,6,7,2,3],[5,0,1,2,7,4,3,6],[6,7,4,5,2,3,0,1],[7,4,5,0,3,6,1,2]],\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,4,6,4,0,2],[0,3,4,1,2,7,6,5],[0,4,6,2,6,2,0,4],[0,5,4,7,2,1,6,3],[0,6,0,6,0,6,0,6],[0,7,2,5,4,3,6,1]],\"n\":8}"},"fields":{"title":"Multiplicative Commutativity Is Additional to Complete Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 5, 0, 7, 4], [2, 3, 6, 7, 0, 1, 4, 5], [3, 6, 7, 4, 1, 2, 5, 0], [4, 5, 0, 1, 6, 7, 2, 3], [5, 0, 1, 2, 7, 4, 3, 6], [6, 7, 4, 5, 2, 3, 0, 1], [7, 4, 5, 0, 3, 6, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 4, 6, 4, 0, 2], [0, 3, 4, 1, 2, 7, 6, 5], [0, 4, 6, 2, 6, 2, 0, 4], [0, 5, 4, 7, 2, 1, 6, 3], [0, 6, 0, 6, 0, 6, 0, 6], [0, 7, 2, 5, 4, 3, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.","answer":{"kind":"expression","value":"(8, 0, 1, [], Yes, commutative_ring)","explanation":"Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring)."},"skills":["Multiplicative Commutativity Is Additional to Complete Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0018","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"85ecd37e835b2b5af360273e342863f7ce64391ba45669167de97f3f7e4541ab","family_id":"qf-mod28-commring-qualification","seed":202610281018,"semantic_template":"qf-mod28-commring-qualification/{\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,6,0,4,7,5],[2,3,6,7,1,0,5,4],[3,6,7,5,2,1,4,0],[4,0,1,2,5,7,3,6],[5,4,0,1,7,6,2,3],[6,7,5,4,3,2,0,1],[7,5,4,0,6,3,1,2]],\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,5,5,6,0,2],[0,3,5,1,7,2,6,4],[0,4,5,7,1,2,6,3],[0,5,6,2,2,6,0,5],[0,6,0,6,6,0,0,6],[0,7,2,4,3,5,6,1]],\"n\":8}"},"fields":{"title":"Multiplicative Commutativity Is Additional to Complete Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 0, 4, 7, 5], [2, 3, 6, 7, 1, 0, 5, 4], [3, 6, 7, 5, 2, 1, 4, 0], [4, 0, 1, 2, 5, 7, 3, 6], [5, 4, 0, 1, 7, 6, 2, 3], [6, 7, 5, 4, 3, 2, 0, 1], [7, 5, 4, 0, 6, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 5, 5, 6, 0, 2], [0, 3, 5, 1, 7, 2, 6, 4], [0, 4, 5, 7, 1, 2, 6, 3], [0, 5, 6, 2, 2, 6, 0, 5], [0, 6, 0, 6, 6, 0, 0, 6], [0, 7, 2, 4, 3, 5, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.","answer":{"kind":"expression","value":"(8, 0, 1, [], Yes, commutative_ring)","explanation":"Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring)."},"skills":["Multiplicative Commutativity Is Additional to Complete Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0019","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1b6e03d112c5b569d7426a0759e12221e40fcd67cfdbe9c4573e1763f1c5c96c","family_id":"qf-mod28-commring-qualification","seed":202610281019,"semantic_template":"qf-mod28-commring-qualification/{\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,7,5,6,0,4],[2,3,7,4,6,0,1,5],[3,7,4,5,0,1,2,6],[4,5,6,0,2,3,7,1],[5,6,0,1,3,7,4,2],[6,0,1,2,7,4,5,3],[7,4,5,6,1,2,3,0]],\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,5,2,7,5,0],[0,3,5,1,6,2,4,7],[0,4,2,6,1,5,3,7],[0,5,7,2,5,7,2,0],[0,6,5,4,3,2,1,7],[0,7,0,7,7,0,7,0]],\"n\":8}"},"fields":{"title":"Multiplicative Commutativity Is Additional to Complete Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 5, 6, 0, 4], [2, 3, 7, 4, 6, 0, 1, 5], [3, 7, 4, 5, 0, 1, 2, 6], [4, 5, 6, 0, 2, 3, 7, 1], [5, 6, 0, 1, 3, 7, 4, 2], [6, 0, 1, 2, 7, 4, 5, 3], [7, 4, 5, 6, 1, 2, 3, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 5, 2, 7, 5, 0], [0, 3, 5, 1, 6, 2, 4, 7], [0, 4, 2, 6, 1, 5, 3, 7], [0, 5, 7, 2, 5, 7, 2, 0], [0, 6, 5, 4, 3, 2, 1, 7], [0, 7, 0, 7, 7, 0, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.","answer":{"kind":"expression","value":"(8, 0, 1, [], Yes, commutative_ring)","explanation":"Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring)."},"skills":["Multiplicative Commutativity Is Additional to Complete Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0020","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"da0eaead4d1eddf432a69e0022da4f564353254d1e0183503289211863921801","family_id":"qf-mod28-commring-qualification","seed":202610281020,"semantic_template":"qf-mod28-commring-qualification/{\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,7,6,0,5,4],[2,3,7,4,5,1,0,6],[3,7,4,6,0,2,1,5],[4,6,5,0,2,7,3,1],[5,0,1,2,7,6,4,3],[6,5,0,1,3,4,7,2],[7,4,6,5,1,3,2,0]],\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,6,2,6,7,0],[0,3,6,1,5,4,2,7],[0,4,2,5,1,3,6,7],[0,5,6,4,3,1,2,7],[0,6,7,2,6,2,7,0],[0,7,0,7,7,7,0,0]],\"n\":8}"},"fields":{"title":"Multiplicative Commutativity Is Additional to Complete Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 6, 0, 5, 4], [2, 3, 7, 4, 5, 1, 0, 6], [3, 7, 4, 6, 0, 2, 1, 5], [4, 6, 5, 0, 2, 7, 3, 1], [5, 0, 1, 2, 7, 6, 4, 3], [6, 5, 0, 1, 3, 4, 7, 2], [7, 4, 6, 5, 1, 3, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 6, 2, 6, 7, 0], [0, 3, 6, 1, 5, 4, 2, 7], [0, 4, 2, 5, 1, 3, 6, 7], [0, 5, 6, 4, 3, 1, 2, 7], [0, 6, 7, 2, 6, 2, 7, 0], [0, 7, 0, 7, 7, 7, 0, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.","answer":{"kind":"expression","value":"(8, 0, 1, [], Yes, commutative_ring)","explanation":"Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring)."},"skills":["Multiplicative Commutativity Is Additional to Complete Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0021","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f9f97b14bf2eeb2be8ccfe6c2140e98195be074a2fce2bccbc30b54cd0db8105","family_id":"qf-mod28-commring-qualification","seed":202610281021,"semantic_template":"qf-mod28-commring-qualification/{\"addition\":[[0,1,2,3,4,5,6,7],[1,0,3,2,6,7,4,5],[2,3,0,1,7,6,5,4],[3,2,1,0,5,4,7,6],[4,6,7,5,0,3,1,2],[5,7,6,4,3,0,2,1],[6,4,5,7,1,2,0,3],[7,5,4,6,2,1,3,0]],\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,0,1,0,1,1,0],[0,2,0,2,0,2,2,0],[0,3,0,3,0,3,3,0],[0,2,2,0,4,4,7,7],[0,1,2,3,4,5,6,7],[0,3,2,1,4,6,5,7],[0,0,2,2,4,7,4,7]],\"n\":8}"},"fields":{"title":"Multiplicative Commutativity Is Additional to Complete Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 0, 3, 2, 6, 7, 4, 5], [2, 3, 0, 1, 7, 6, 5, 4], [3, 2, 1, 0, 5, 4, 7, 6], [4, 6, 7, 5, 0, 3, 1, 2], [5, 7, 6, 4, 3, 0, 2, 1], [6, 4, 5, 7, 1, 2, 0, 3], [7, 5, 4, 6, 2, 1, 3, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 0, 1, 0, 1, 1, 0], [0, 2, 0, 2, 0, 2, 2, 0], [0, 3, 0, 3, 0, 3, 3, 0], [0, 2, 2, 0, 4, 4, 7, 7], [0, 1, 2, 3, 4, 5, 6, 7], [0, 3, 2, 1, 4, 6, 5, 7], [0, 0, 2, 2, 4, 7, 4, 7]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.","answer":{"kind":"expression","value":"(8, 0, 5, [[1, 2, 0, 2], [1, 3, 1, 3], [1, 4, 0, 2], [1, 6, 1, 3], [2, 3, 2, 0], [2, 4, 0, 2], [2, 7, 0, 2], [3, 6, 3, 1], [3, 7, 0, 2], [4, 6, 7, 4], [4, 7, 7, 4], [6, 7, 7, 4]], No, unital_ring_not_commutative)","explanation":"Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 5, [[1, 2, 0, 2], [1, 3, 1, 3], [1, 4, 0, 2], [1, 6, 1, 3], [2, 3, 2, 0], [2, 4, 0, 2], [2, 7, 0, 2], [3, 6, 3, 1], [3, 7, 0, 2], [4, 6, 7, 4], [4, 7, 7, 4], [6, 7, 7, 4]], No, unital_ring_not_commutative)."},"skills":["Multiplicative Commutativity Is Additional to Complete Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0022","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c68fe1832155370a7f5b291e64ab32ffb52803054169720eaff9e77d030426c8","family_id":"qf-mod28-commring-qualification","seed":202610281022,"semantic_template":"qf-mod28-commring-qualification/{\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,7,0,6,4,5],[2,3,7,5,1,4,0,6],[3,7,5,6,2,0,1,4],[4,0,1,2,6,7,5,3],[5,6,4,0,7,2,3,1],[6,4,0,1,5,3,7,2],[7,5,6,4,3,1,2,0]],\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,6,6,2,7,0],[0,3,6,1,5,4,2,7],[0,4,6,5,1,3,2,7],[0,5,2,4,3,1,6,7],[0,6,7,2,2,6,7,0],[0,7,0,7,7,7,0,0]],\"n\":8}"},"fields":{"title":"Multiplicative Commutativity Is Additional to Complete Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 0, 6, 4, 5], [2, 3, 7, 5, 1, 4, 0, 6], [3, 7, 5, 6, 2, 0, 1, 4], [4, 0, 1, 2, 6, 7, 5, 3], [5, 6, 4, 0, 7, 2, 3, 1], [6, 4, 0, 1, 5, 3, 7, 2], [7, 5, 6, 4, 3, 1, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 6, 6, 2, 7, 0], [0, 3, 6, 1, 5, 4, 2, 7], [0, 4, 6, 5, 1, 3, 2, 7], [0, 5, 2, 4, 3, 1, 6, 7], [0, 6, 7, 2, 2, 6, 7, 0], [0, 7, 0, 7, 7, 7, 0, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.","answer":{"kind":"expression","value":"(8, 0, 1, [], Yes, commutative_ring)","explanation":"Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring)."},"skills":["Multiplicative Commutativity Is Additional to Complete Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0023","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8058b88622d7bc5e8af6d92aca0ac7b1fed1a04a94b589722162a56a072fbd70","family_id":"qf-mod28-commring-qualification","seed":202610281023,"semantic_template":"qf-mod28-commring-qualification/{\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,7,5,0,4,6],[2,3,7,6,0,1,5,4],[3,7,6,4,1,2,0,5],[4,5,0,1,7,6,3,2],[5,0,1,2,6,4,7,3],[6,4,5,0,3,7,2,1],[7,6,4,5,2,3,1,0]],\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,4,7,4,2,0],[0,3,4,1,2,6,5,7],[0,4,7,2,7,2,4,0],[0,5,4,6,2,1,3,7],[0,6,2,5,4,3,1,7],[0,7,0,7,0,7,7,0]],\"n\":8}"},"fields":{"title":"Multiplicative Commutativity Is Additional to Complete Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 5, 0, 4, 6], [2, 3, 7, 6, 0, 1, 5, 4], [3, 7, 6, 4, 1, 2, 0, 5], [4, 5, 0, 1, 7, 6, 3, 2], [5, 0, 1, 2, 6, 4, 7, 3], [6, 4, 5, 0, 3, 7, 2, 1], [7, 6, 4, 5, 2, 3, 1, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 4, 7, 4, 2, 0], [0, 3, 4, 1, 2, 6, 5, 7], [0, 4, 7, 2, 7, 2, 4, 0], [0, 5, 4, 6, 2, 1, 3, 7], [0, 6, 2, 5, 4, 3, 1, 7], [0, 7, 0, 7, 0, 7, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.","answer":{"kind":"expression","value":"(8, 0, 1, [], Yes, commutative_ring)","explanation":"Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring)."},"skills":["Multiplicative Commutativity Is Additional to Complete Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0024","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"81754845ea1979d694f60537e86276dc53495cbdb439a38a42534e1e2f843ae5","family_id":"qf-mod28-commring-qualification","seed":202610281024,"semantic_template":"qf-mod28-commring-qualification/{\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,7,0,4,5,6],[2,3,7,6,1,0,4,5],[3,7,6,5,2,1,0,4],[4,0,1,2,5,6,7,3],[5,4,0,1,6,7,3,2],[6,5,4,0,7,3,2,1],[7,6,5,4,3,2,1,0]],\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,5,5,7,2,0],[0,3,5,1,6,2,4,7],[0,4,5,6,1,2,3,7],[0,5,7,2,2,7,5,0],[0,6,2,4,3,5,1,7],[0,7,0,7,7,0,7,0]],\"n\":8}"},"fields":{"title":"Multiplicative Commutativity Is Additional to Complete Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 0, 4, 5, 6], [2, 3, 7, 6, 1, 0, 4, 5], [3, 7, 6, 5, 2, 1, 0, 4], [4, 0, 1, 2, 5, 6, 7, 3], [5, 4, 0, 1, 6, 7, 3, 2], [6, 5, 4, 0, 7, 3, 2, 1], [7, 6, 5, 4, 3, 2, 1, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 5, 5, 7, 2, 0], [0, 3, 5, 1, 6, 2, 4, 7], [0, 4, 5, 6, 1, 2, 3, 7], [0, 5, 7, 2, 2, 7, 5, 0], [0, 6, 2, 4, 3, 5, 1, 7], [0, 7, 0, 7, 7, 0, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.","answer":{"kind":"expression","value":"(8, 0, 1, [], Yes, commutative_ring)","explanation":"Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring)."},"skills":["Multiplicative Commutativity Is Additional to Complete Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0025","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8c579dca8005fea00d288efe40dc7a83f6295723f6a272f7d6f8e6626d96efdf","family_id":"qf-mod28-commring-qualification","seed":202610281025,"semantic_template":"qf-mod28-commring-qualification/{\"addition\":[[0,1,2,3,4,5,6,7],[1,2,4,5,3,6,7,0],[2,4,3,6,5,7,0,1],[3,5,6,0,7,1,2,4],[4,3,5,7,6,0,1,2],[5,6,7,1,0,2,4,3],[6,7,0,2,1,4,3,5],[7,0,1,4,2,3,5,6]],\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,3,0,6,2,3,6],[0,3,0,0,3,3,0,3],[0,4,6,3,1,7,2,5],[0,5,2,3,7,1,6,4],[0,6,3,0,2,6,3,2],[0,7,6,3,5,4,2,1]],\"n\":8}"},"fields":{"title":"Multiplicative Commutativity Is Additional to Complete Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 4, 5, 3, 6, 7, 0], [2, 4, 3, 6, 5, 7, 0, 1], [3, 5, 6, 0, 7, 1, 2, 4], [4, 3, 5, 7, 6, 0, 1, 2], [5, 6, 7, 1, 0, 2, 4, 3], [6, 7, 0, 2, 1, 4, 3, 5], [7, 0, 1, 4, 2, 3, 5, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 3, 0, 6, 2, 3, 6], [0, 3, 0, 0, 3, 3, 0, 3], [0, 4, 6, 3, 1, 7, 2, 5], [0, 5, 2, 3, 7, 1, 6, 4], [0, 6, 3, 0, 2, 6, 3, 2], [0, 7, 6, 3, 5, 4, 2, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Give (n,zero,one,all [a,b,ab,ba] failures with a<b in label order,commutative?,classification). Classification is commutative_ring or unital_ring_not_commutative.","answer":{"kind":"expression","value":"(8, 0, 1, [], Yes, commutative_ring)","explanation":"Compare both multiplication orders for every distinct unordered pair. Off-diagonal failures disprove commutativity; the full ring qualification is a prior premise rather than something repaired by this audit. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (8, 0, 1, [], Yes, commutative_ring)."},"skills":["Multiplicative Commutativity Is Additional to Complete Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0026","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"031d000333b7b88790e0074bd6fbeed3100b2e87cd2fd63078977e121e360647","family_id":"qf-mod28-commring-product-swap","seed":202610281026,"semantic_template":"qf-mod28-commring-product-swap/{\"a\":1,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2],[4,5,6,7,0,1,2,3],[5,6,7,0,1,2,3,4],[6,7,0,1,2,3,4,5],[7,0,1,2,3,4,5,6]],\"b\":2,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,6,0,2,4,6],[0,3,6,1,4,7,2,5],[0,4,0,4,0,4,0,4],[0,5,2,7,4,1,6,3],[0,6,4,2,0,6,4,2],[0,7,6,5,4,3,2,1]],\"n\":8}"},"fields":{"title":"Swap Two Factors Only after Global Commutative-Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 6, 0, 2, 4, 6], [0, 3, 6, 1, 4, 7, 2, 5], [0, 4, 0, 4, 0, 4, 0, 4], [0, 5, 2, 7, 4, 1, 6, 3], [0, 6, 4, 2, 0, 6, 4, 2], [0, 7, 6, 5, 4, 3, 2, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=1, b=2. Give (a,b,ab,ba,equal?).","answer":{"kind":"expression","value":"(1, 2, 2, 2, Yes)","explanation":"Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (1, 2, 2, 2, Yes)."},"skills":["Swap Two Factors Only after Global Commutative-Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0027","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"40a9b71570d5e7eb6adc967245bb5632e8f840093b64abac3097f54743089f9f","family_id":"qf-mod28-commring-product-swap","seed":202610281027,"semantic_template":"qf-mod28-commring-product-swap/{\"a\":2,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,4,5,7,0,6],[2,3,4,5,7,6,1,0],[3,4,5,7,6,0,2,1],[4,5,7,6,0,1,3,2],[5,7,6,0,1,2,4,3],[6,0,1,2,3,4,7,5],[7,6,0,1,2,3,5,4]],\"b\":5,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,7,0,2,7,4],[0,3,7,1,4,6,5,2],[0,4,0,4,0,4,4,0],[0,5,2,6,4,1,3,7],[0,6,7,5,4,3,1,2],[0,7,4,2,0,7,2,4]],\"n\":8}"},"fields":{"title":"Swap Two Factors Only after Global Commutative-Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 7, 0, 6], [2, 3, 4, 5, 7, 6, 1, 0], [3, 4, 5, 7, 6, 0, 2, 1], [4, 5, 7, 6, 0, 1, 3, 2], [5, 7, 6, 0, 1, 2, 4, 3], [6, 0, 1, 2, 3, 4, 7, 5], [7, 6, 0, 1, 2, 3, 5, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 7, 0, 2, 7, 4], [0, 3, 7, 1, 4, 6, 5, 2], [0, 4, 0, 4, 0, 4, 4, 0], [0, 5, 2, 6, 4, 1, 3, 7], [0, 6, 7, 5, 4, 3, 1, 2], [0, 7, 4, 2, 0, 7, 2, 4]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=2, b=5. Give (a,b,ab,ba,equal?).","answer":{"kind":"expression","value":"(2, 5, 2, 2, Yes)","explanation":"Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (2, 5, 2, 2, Yes)."},"skills":["Swap Two Factors Only after Global Commutative-Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0028","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"98ad56539a33a48da324765d6bcd67b15734fb2a15dd4db25ef2b726dcfce74f","family_id":"qf-mod28-commring-product-swap","seed":202610281028,"semantic_template":"qf-mod28-commring-product-swap/{\"a\":3,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,4,6,7,5,0],[2,3,4,6,5,0,7,1],[3,4,6,5,7,1,0,2],[4,6,5,7,0,2,1,3],[5,7,0,1,2,4,3,6],[6,5,7,0,1,3,2,4],[7,0,1,2,3,6,4,5]],\"b\":0,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,5,0,4,2,5],[0,3,5,1,4,2,7,6],[0,4,0,4,0,0,4,4],[0,5,4,2,0,4,5,2],[0,6,2,7,4,5,1,3],[0,7,5,6,4,2,3,1]],\"n\":8}"},"fields":{"title":"Swap Two Factors Only after Global Commutative-Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 6, 7, 5, 0], [2, 3, 4, 6, 5, 0, 7, 1], [3, 4, 6, 5, 7, 1, 0, 2], [4, 6, 5, 7, 0, 2, 1, 3], [5, 7, 0, 1, 2, 4, 3, 6], [6, 5, 7, 0, 1, 3, 2, 4], [7, 0, 1, 2, 3, 6, 4, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 5, 0, 4, 2, 5], [0, 3, 5, 1, 4, 2, 7, 6], [0, 4, 0, 4, 0, 0, 4, 4], [0, 5, 4, 2, 0, 4, 5, 2], [0, 6, 2, 7, 4, 5, 1, 3], [0, 7, 5, 6, 4, 2, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=3, b=0. Give (a,b,ab,ba,equal?).","answer":{"kind":"expression","value":"(3, 0, 0, 0, Yes)","explanation":"Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (3, 0, 0, 0, Yes)."},"skills":["Swap Two Factors Only after Global Commutative-Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0029","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f1d78a2aa2ab85675014c1c854406ce55bf13b32d9d0cfae5b8813b1c40da440","family_id":"qf-mod28-commring-product-swap","seed":202610281029,"semantic_template":"qf-mod28-commring-product-swap/{\"a\":4,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,4,6,0,7,5],[2,3,4,6,7,1,5,0],[3,4,6,7,5,2,0,1],[4,6,7,5,0,3,1,2],[5,0,1,2,3,7,4,6],[6,7,5,0,1,4,2,3],[7,5,0,1,2,6,3,4]],\"b\":3,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,7,0,7,2,4],[0,3,7,1,4,6,5,2],[0,4,0,4,0,4,4,0],[0,5,7,6,4,1,3,2],[0,6,2,5,4,3,1,7],[0,7,4,2,0,2,7,4]],\"n\":8}"},"fields":{"title":"Swap Two Factors Only after Global Commutative-Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 6, 0, 7, 5], [2, 3, 4, 6, 7, 1, 5, 0], [3, 4, 6, 7, 5, 2, 0, 1], [4, 6, 7, 5, 0, 3, 1, 2], [5, 0, 1, 2, 3, 7, 4, 6], [6, 7, 5, 0, 1, 4, 2, 3], [7, 5, 0, 1, 2, 6, 3, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 7, 0, 7, 2, 4], [0, 3, 7, 1, 4, 6, 5, 2], [0, 4, 0, 4, 0, 4, 4, 0], [0, 5, 7, 6, 4, 1, 3, 2], [0, 6, 2, 5, 4, 3, 1, 7], [0, 7, 4, 2, 0, 2, 7, 4]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=4, b=3. Give (a,b,ab,ba,equal?).","answer":{"kind":"expression","value":"(4, 3, 4, 4, Yes)","explanation":"Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (4, 3, 4, 4, Yes)."},"skills":["Swap Two Factors Only after Global Commutative-Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0030","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b99e7ccd7335e7b7b6117af41f7f011208ab2a05e77b1fd419e4c82ccbab1e8f","family_id":"qf-mod28-commring-product-swap","seed":202610281030,"semantic_template":"qf-mod28-commring-product-swap/{\"a\":7,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,4,7,6,0,5],[2,3,4,7,5,0,1,6],[3,4,7,5,6,1,2,0],[4,7,5,6,0,2,3,1],[5,6,0,1,2,4,7,3],[6,0,1,2,3,7,5,4],[7,5,6,0,1,3,4,2]],\"b\":5,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,5,0,4,5,2],[0,3,5,1,4,2,7,6],[0,4,0,4,0,0,4,4],[0,5,4,2,0,4,2,5],[0,6,5,7,4,2,1,3],[0,7,2,6,4,5,3,1]],\"n\":8}"},"fields":{"title":"Swap Two Factors Only after Global Commutative-Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 7, 6, 0, 5], [2, 3, 4, 7, 5, 0, 1, 6], [3, 4, 7, 5, 6, 1, 2, 0], [4, 7, 5, 6, 0, 2, 3, 1], [5, 6, 0, 1, 2, 4, 7, 3], [6, 0, 1, 2, 3, 7, 5, 4], [7, 5, 6, 0, 1, 3, 4, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 5, 0, 4, 5, 2], [0, 3, 5, 1, 4, 2, 7, 6], [0, 4, 0, 4, 0, 0, 4, 4], [0, 5, 4, 2, 0, 4, 2, 5], [0, 6, 5, 7, 4, 2, 1, 3], [0, 7, 2, 6, 4, 5, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=7, b=5. Give (a,b,ab,ba,equal?).","answer":{"kind":"expression","value":"(7, 5, 5, 5, Yes)","explanation":"Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (7, 5, 5, 5, Yes)."},"skills":["Swap Two Factors Only after Global Commutative-Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0031","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bb59d4818a8e6dd7f742d362d4c3c27244f785285ae1b649a5bfb802e110150f","family_id":"qf-mod28-commring-product-swap","seed":202610281031,"semantic_template":"qf-mod28-commring-product-swap/{\"a\":6,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,4,7,0,5,6],[2,3,4,7,6,1,0,5],[3,4,7,6,5,2,1,0],[4,7,6,5,0,3,2,1],[5,0,1,2,3,6,7,4],[6,5,0,1,2,7,4,3],[7,6,5,0,1,4,3,2]],\"b\":1,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,6,0,6,4,2],[0,3,6,1,4,7,2,5],[0,4,0,4,0,4,0,4],[0,5,6,7,4,1,2,3],[0,6,4,2,0,2,4,6],[0,7,2,5,4,3,6,1]],\"n\":8}"},"fields":{"title":"Swap Two Factors Only after Global Commutative-Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 7, 0, 5, 6], [2, 3, 4, 7, 6, 1, 0, 5], [3, 4, 7, 6, 5, 2, 1, 0], [4, 7, 6, 5, 0, 3, 2, 1], [5, 0, 1, 2, 3, 6, 7, 4], [6, 5, 0, 1, 2, 7, 4, 3], [7, 6, 5, 0, 1, 4, 3, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 6, 0, 6, 4, 2], [0, 3, 6, 1, 4, 7, 2, 5], [0, 4, 0, 4, 0, 4, 0, 4], [0, 5, 6, 7, 4, 1, 2, 3], [0, 6, 4, 2, 0, 2, 4, 6], [0, 7, 2, 5, 4, 3, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=6, b=1. Give (a,b,ab,ba,equal?).","answer":{"kind":"expression","value":"(6, 1, 6, 6, Yes)","explanation":"Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (6, 1, 6, 6, Yes)."},"skills":["Swap Two Factors Only after Global Commutative-Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0032","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6a7b33b43fb25c672d2f0555e0d164000f8f0ae1e3e3f8e59a3084d719df5e75","family_id":"qf-mod28-commring-product-swap","seed":202610281032,"semantic_template":"qf-mod28-commring-product-swap/{\"a\":7,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,5,6,4,7,0],[2,3,5,4,7,6,0,1],[3,5,4,6,0,7,1,2],[4,6,7,0,2,1,3,5],[5,4,6,7,1,0,2,3],[6,7,0,1,3,2,5,4],[7,0,1,2,5,3,4,6]],\"b\":5,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,6,2,0,5,6],[0,3,6,1,7,5,2,4],[0,4,2,7,1,5,6,3],[0,5,0,5,5,0,0,5],[0,6,5,2,6,0,5,2],[0,7,6,4,3,5,2,1]],\"n\":8}"},"fields":{"title":"Swap Two Factors Only after Global Commutative-Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 6, 4, 7, 0], [2, 3, 5, 4, 7, 6, 0, 1], [3, 5, 4, 6, 0, 7, 1, 2], [4, 6, 7, 0, 2, 1, 3, 5], [5, 4, 6, 7, 1, 0, 2, 3], [6, 7, 0, 1, 3, 2, 5, 4], [7, 0, 1, 2, 5, 3, 4, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 6, 2, 0, 5, 6], [0, 3, 6, 1, 7, 5, 2, 4], [0, 4, 2, 7, 1, 5, 6, 3], [0, 5, 0, 5, 5, 0, 0, 5], [0, 6, 5, 2, 6, 0, 5, 2], [0, 7, 6, 4, 3, 5, 2, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=7, b=5. Give (a,b,ab,ba,equal?).","answer":{"kind":"expression","value":"(7, 5, 5, 5, Yes)","explanation":"Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (7, 5, 5, 5, Yes)."},"skills":["Swap Two Factors Only after Global Commutative-Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0033","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e209a10447f7e96e8a7560d3e9f8af1df7a8451ee754da7ace464682adfc5304","family_id":"qf-mod28-commring-product-swap","seed":202610281033,"semantic_template":"qf-mod28-commring-product-swap/{\"a\":0,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,5,7,4,0,6],[2,3,5,4,6,7,1,0],[3,5,4,7,0,6,2,1],[4,7,6,0,2,1,5,3],[5,4,7,6,1,0,3,2],[6,0,1,2,5,3,7,4],[7,6,0,1,3,2,4,5]],\"b\":6,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,7,2,0,7,5],[0,3,7,1,6,5,4,2],[0,4,2,6,1,5,3,7],[0,5,0,5,5,0,5,0],[0,6,7,4,3,5,1,2],[0,7,5,2,7,0,2,5]],\"n\":8}"},"fields":{"title":"Swap Two Factors Only after Global Commutative-Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 7, 4, 0, 6], [2, 3, 5, 4, 6, 7, 1, 0], [3, 5, 4, 7, 0, 6, 2, 1], [4, 7, 6, 0, 2, 1, 5, 3], [5, 4, 7, 6, 1, 0, 3, 2], [6, 0, 1, 2, 5, 3, 7, 4], [7, 6, 0, 1, 3, 2, 4, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 7, 2, 0, 7, 5], [0, 3, 7, 1, 6, 5, 4, 2], [0, 4, 2, 6, 1, 5, 3, 7], [0, 5, 0, 5, 5, 0, 5, 0], [0, 6, 7, 4, 3, 5, 1, 2], [0, 7, 5, 2, 7, 0, 2, 5]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=0, b=6. Give (a,b,ab,ba,equal?).","answer":{"kind":"expression","value":"(0, 6, 0, 0, Yes)","explanation":"Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (0, 6, 0, 0, Yes)."},"skills":["Swap Two Factors Only after Global Commutative-Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0034","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"def3c4a5c5b95d1623945ac1e2f2fef4d403dbcb723010497b514d8c34217589","family_id":"qf-mod28-commring-product-swap","seed":202610281034,"semantic_template":"qf-mod28-commring-product-swap/{\"a\":1,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,5,7,6,4,0],[2,3,5,6,0,4,7,1],[3,5,6,4,1,7,0,2],[4,7,0,1,5,2,3,6],[5,6,4,7,2,0,1,3],[6,4,7,0,3,1,2,5],[7,0,1,2,6,3,5,4]],\"b\":2,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,4,5,0,2,4],[0,3,4,1,2,5,7,6],[0,4,5,2,5,0,4,2],[0,5,0,5,0,0,5,5],[0,6,2,7,4,5,1,3],[0,7,4,6,2,5,3,1]],\"n\":8}"},"fields":{"title":"Swap Two Factors Only after Global Commutative-Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 7, 6, 4, 0], [2, 3, 5, 6, 0, 4, 7, 1], [3, 5, 6, 4, 1, 7, 0, 2], [4, 7, 0, 1, 5, 2, 3, 6], [5, 6, 4, 7, 2, 0, 1, 3], [6, 4, 7, 0, 3, 1, 2, 5], [7, 0, 1, 2, 6, 3, 5, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 4, 5, 0, 2, 4], [0, 3, 4, 1, 2, 5, 7, 6], [0, 4, 5, 2, 5, 0, 4, 2], [0, 5, 0, 5, 0, 0, 5, 5], [0, 6, 2, 7, 4, 5, 1, 3], [0, 7, 4, 6, 2, 5, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=1, b=2. Give (a,b,ab,ba,equal?).","answer":{"kind":"expression","value":"(1, 2, 2, 2, Yes)","explanation":"Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (1, 2, 2, 2, Yes)."},"skills":["Swap Two Factors Only after Global Commutative-Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0035","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"31f6d3cb64171c0c4643799d8f56ef215473abf63191dfb383082cf049eb727c","family_id":"qf-mod28-commring-product-swap","seed":202610281035,"semantic_template":"qf-mod28-commring-product-swap/{\"a\":2,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,5,0,6,7,4],[2,3,5,6,1,7,4,0],[3,5,6,7,2,4,0,1],[4,0,1,2,7,3,5,6],[5,6,7,4,3,0,1,2],[6,7,4,0,5,1,2,3],[7,4,0,1,6,2,3,5]],\"b\":6,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,7,7,0,2,5],[0,3,7,1,6,5,4,2],[0,4,7,6,1,5,3,2],[0,5,0,5,5,0,5,0],[0,6,2,4,3,5,1,7],[0,7,5,2,2,0,7,5]],\"n\":8}"},"fields":{"title":"Swap Two Factors Only after Global Commutative-Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 0, 6, 7, 4], [2, 3, 5, 6, 1, 7, 4, 0], [3, 5, 6, 7, 2, 4, 0, 1], [4, 0, 1, 2, 7, 3, 5, 6], [5, 6, 7, 4, 3, 0, 1, 2], [6, 7, 4, 0, 5, 1, 2, 3], [7, 4, 0, 1, 6, 2, 3, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 7, 7, 0, 2, 5], [0, 3, 7, 1, 6, 5, 4, 2], [0, 4, 7, 6, 1, 5, 3, 2], [0, 5, 0, 5, 5, 0, 5, 0], [0, 6, 2, 4, 3, 5, 1, 7], [0, 7, 5, 2, 2, 0, 7, 5]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=2, b=6. Give (a,b,ab,ba,equal?).","answer":{"kind":"expression","value":"(2, 6, 2, 2, Yes)","explanation":"Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (2, 6, 2, 2, Yes)."},"skills":["Swap Two Factors Only after Global Commutative-Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0036","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ef5d44f072e3330498dbcedd109ba91ef9fa40b30b14eeef5a1207bae228f36c","family_id":"qf-mod28-commring-product-swap","seed":202610281036,"semantic_template":"qf-mod28-commring-product-swap/{\"a\":3,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,5,6,7,0,4],[2,3,5,7,0,4,1,6],[3,5,7,4,1,6,2,0],[4,6,0,1,5,2,7,3],[5,7,4,6,2,0,3,1],[6,0,1,2,7,3,4,5],[7,4,6,0,3,1,5,2]],\"b\":0,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,4,5,0,4,2],[0,3,4,1,2,5,7,6],[0,4,5,2,5,0,2,4],[0,5,0,5,0,0,5,5],[0,6,4,7,2,5,1,3],[0,7,2,6,4,5,3,1]],\"n\":8}"},"fields":{"title":"Swap Two Factors Only after Global Commutative-Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 6, 7, 0, 4], [2, 3, 5, 7, 0, 4, 1, 6], [3, 5, 7, 4, 1, 6, 2, 0], [4, 6, 0, 1, 5, 2, 7, 3], [5, 7, 4, 6, 2, 0, 3, 1], [6, 0, 1, 2, 7, 3, 4, 5], [7, 4, 6, 0, 3, 1, 5, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 4, 5, 0, 4, 2], [0, 3, 4, 1, 2, 5, 7, 6], [0, 4, 5, 2, 5, 0, 2, 4], [0, 5, 0, 5, 0, 0, 5, 5], [0, 6, 4, 7, 2, 5, 1, 3], [0, 7, 2, 6, 4, 5, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=3, b=0. Give (a,b,ab,ba,equal?).","answer":{"kind":"expression","value":"(3, 0, 0, 0, Yes)","explanation":"Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (3, 0, 0, 0, Yes)."},"skills":["Swap Two Factors Only after Global Commutative-Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0037","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"55c30f87765d1802cf1d42cf1292917fb627f21a261f26a125dfd551ded0b901","family_id":"qf-mod28-commring-product-swap","seed":202610281037,"semantic_template":"qf-mod28-commring-product-swap/{\"a\":5,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,5,0,7,4,6],[2,3,5,7,1,6,0,4],[3,5,7,6,2,4,1,0],[4,0,1,2,6,3,7,5],[5,7,6,4,3,0,2,1],[6,4,0,1,7,2,5,3],[7,6,4,0,5,1,3,2]],\"b\":3,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,6,6,0,5,2],[0,3,6,1,7,5,2,4],[0,4,6,7,1,5,2,3],[0,5,0,5,5,0,0,5],[0,6,5,2,2,0,5,6],[0,7,2,4,3,5,6,1]],\"n\":8}"},"fields":{"title":"Swap Two Factors Only after Global Commutative-Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 0, 7, 4, 6], [2, 3, 5, 7, 1, 6, 0, 4], [3, 5, 7, 6, 2, 4, 1, 0], [4, 0, 1, 2, 6, 3, 7, 5], [5, 7, 6, 4, 3, 0, 2, 1], [6, 4, 0, 1, 7, 2, 5, 3], [7, 6, 4, 0, 5, 1, 3, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 6, 6, 0, 5, 2], [0, 3, 6, 1, 7, 5, 2, 4], [0, 4, 6, 7, 1, 5, 2, 3], [0, 5, 0, 5, 5, 0, 0, 5], [0, 6, 5, 2, 2, 0, 5, 6], [0, 7, 2, 4, 3, 5, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=5, b=3. Give (a,b,ab,ba,equal?).","answer":{"kind":"expression","value":"(5, 3, 5, 5, Yes)","explanation":"Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (5, 3, 5, 5, Yes)."},"skills":["Swap Two Factors Only after Global Commutative-Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0038","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"321739456b6f47c96ef2b0da8ee3deaa98db1e404a32a3ac1a0dae63c360c575","family_id":"qf-mod28-commring-product-swap","seed":202610281038,"semantic_template":"qf-mod28-commring-product-swap/{\"a\":4,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,6,5,7,4,0],[2,3,6,4,7,0,5,1],[3,6,4,5,0,1,7,2],[4,5,7,0,2,3,1,6],[5,7,0,1,3,6,2,4],[6,4,5,7,1,2,0,3],[7,0,1,2,6,4,3,5]],\"b\":5,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,5,2,6,0,5],[0,3,5,1,7,2,6,4],[0,4,2,7,1,5,6,3],[0,5,6,2,5,6,0,2],[0,6,0,6,6,0,0,6],[0,7,5,4,3,2,6,1]],\"n\":8}"},"fields":{"title":"Swap Two Factors Only after Global Commutative-Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 5, 7, 4, 0], [2, 3, 6, 4, 7, 0, 5, 1], [3, 6, 4, 5, 0, 1, 7, 2], [4, 5, 7, 0, 2, 3, 1, 6], [5, 7, 0, 1, 3, 6, 2, 4], [6, 4, 5, 7, 1, 2, 0, 3], [7, 0, 1, 2, 6, 4, 3, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 5, 2, 6, 0, 5], [0, 3, 5, 1, 7, 2, 6, 4], [0, 4, 2, 7, 1, 5, 6, 3], [0, 5, 6, 2, 5, 6, 0, 2], [0, 6, 0, 6, 6, 0, 0, 6], [0, 7, 5, 4, 3, 2, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=4, b=5. Give (a,b,ab,ba,equal?).","answer":{"kind":"expression","value":"(4, 5, 5, 5, Yes)","explanation":"Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (4, 5, 5, 5, Yes)."},"skills":["Swap Two Factors Only after Global Commutative-Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0039","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2d89dd84d4d4de34b0568ca5dab757c312de8a1ec31a93f625af1113dd755e8e","family_id":"qf-mod28-commring-product-swap","seed":202610281039,"semantic_template":"qf-mod28-commring-product-swap/{\"a\":7,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,6,7,0,4,5],[2,3,6,4,5,1,7,0],[3,6,4,7,0,2,5,1],[4,7,5,0,2,6,1,3],[5,0,1,2,6,7,3,4],[6,4,7,5,1,3,0,2],[7,5,0,1,3,4,2,6]],\"b\":1,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,7,2,7,0,6],[0,3,7,1,5,4,6,2],[0,4,2,5,1,3,6,7],[0,5,7,4,3,1,6,2],[0,6,0,6,6,6,0,0],[0,7,6,2,7,2,0,6]],\"n\":8}"},"fields":{"title":"Swap Two Factors Only after Global Commutative-Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 7, 0, 4, 5], [2, 3, 6, 4, 5, 1, 7, 0], [3, 6, 4, 7, 0, 2, 5, 1], [4, 7, 5, 0, 2, 6, 1, 3], [5, 0, 1, 2, 6, 7, 3, 4], [6, 4, 7, 5, 1, 3, 0, 2], [7, 5, 0, 1, 3, 4, 2, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 7, 2, 7, 0, 6], [0, 3, 7, 1, 5, 4, 6, 2], [0, 4, 2, 5, 1, 3, 6, 7], [0, 5, 7, 4, 3, 1, 6, 2], [0, 6, 0, 6, 6, 6, 0, 0], [0, 7, 6, 2, 7, 2, 0, 6]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=7, b=1. Give (a,b,ab,ba,equal?).","answer":{"kind":"expression","value":"(7, 1, 7, 7, Yes)","explanation":"Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (7, 1, 7, 7, Yes)."},"skills":["Swap Two Factors Only after Global Commutative-Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0040","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"aa9352c85d1552c40523208275947a464171f94346ef512987f743878b75b215","family_id":"qf-mod28-commring-product-swap","seed":202610281040,"semantic_template":"qf-mod28-commring-product-swap/{\"a\":7,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,6,7,4,5,0],[2,3,6,5,0,7,4,1],[3,6,5,4,1,0,7,2],[4,7,0,1,6,3,2,5],[5,4,7,0,3,2,1,6],[6,5,4,7,2,1,0,3],[7,0,1,2,5,6,3,4]],\"b\":6,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,4,6,2,0,4],[0,3,4,1,2,7,6,5],[0,4,6,2,6,4,0,2],[0,5,2,7,4,1,6,3],[0,6,0,6,0,6,0,6],[0,7,4,5,2,3,6,1]],\"n\":8}"},"fields":{"title":"Swap Two Factors Only after Global Commutative-Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 7, 4, 5, 0], [2, 3, 6, 5, 0, 7, 4, 1], [3, 6, 5, 4, 1, 0, 7, 2], [4, 7, 0, 1, 6, 3, 2, 5], [5, 4, 7, 0, 3, 2, 1, 6], [6, 5, 4, 7, 2, 1, 0, 3], [7, 0, 1, 2, 5, 6, 3, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 4, 6, 2, 0, 4], [0, 3, 4, 1, 2, 7, 6, 5], [0, 4, 6, 2, 6, 4, 0, 2], [0, 5, 2, 7, 4, 1, 6, 3], [0, 6, 0, 6, 0, 6, 0, 6], [0, 7, 4, 5, 2, 3, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=7, b=6. Give (a,b,ab,ba,equal?).","answer":{"kind":"expression","value":"(7, 6, 6, 6, Yes)","explanation":"Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (7, 6, 6, 6, Yes)."},"skills":["Swap Two Factors Only after Global Commutative-Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0041","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1c282c1a420dd0af954466b9a34995177b5fe423c3991f15132a6bc84d955b18","family_id":"qf-mod28-commring-product-swap","seed":202610281041,"semantic_template":"qf-mod28-commring-product-swap/{\"a\":0,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,6,0,7,5,4],[2,3,6,5,1,4,7,0],[3,6,5,7,2,0,4,1],[4,0,1,2,7,6,3,5],[5,7,4,0,6,2,1,3],[6,5,7,4,3,1,0,2],[7,4,0,1,5,3,2,6]],\"b\":4,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,7,7,2,0,6],[0,3,7,1,5,4,6,2],[0,4,7,5,1,3,6,2],[0,5,2,4,3,1,6,7],[0,6,0,6,6,6,0,0],[0,7,6,2,2,7,0,6]],\"n\":8}"},"fields":{"title":"Swap Two Factors Only after Global Commutative-Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 0, 7, 5, 4], [2, 3, 6, 5, 1, 4, 7, 0], [3, 6, 5, 7, 2, 0, 4, 1], [4, 0, 1, 2, 7, 6, 3, 5], [5, 7, 4, 0, 6, 2, 1, 3], [6, 5, 7, 4, 3, 1, 0, 2], [7, 4, 0, 1, 5, 3, 2, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 7, 7, 2, 0, 6], [0, 3, 7, 1, 5, 4, 6, 2], [0, 4, 7, 5, 1, 3, 6, 2], [0, 5, 2, 4, 3, 1, 6, 7], [0, 6, 0, 6, 6, 6, 0, 0], [0, 7, 6, 2, 2, 7, 0, 6]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=0, b=4. Give (a,b,ab,ba,equal?).","answer":{"kind":"expression","value":"(0, 4, 0, 0, Yes)","explanation":"Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (0, 4, 0, 0, Yes)."},"skills":["Swap Two Factors Only after Global Commutative-Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0042","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0c85701072f62b2ac60c76acd466ea1025f3abb13caff3d81add06400c1133fe","family_id":"qf-mod28-commring-product-swap","seed":202610281042,"semantic_template":"qf-mod28-commring-product-swap/{\"a\":1,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,6,5,0,7,4],[2,3,6,7,0,1,4,5],[3,6,7,4,1,2,5,0],[4,5,0,1,6,7,2,3],[5,0,1,2,7,4,3,6],[6,7,4,5,2,3,0,1],[7,4,5,0,3,6,1,2]],\"b\":2,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,4,6,4,0,2],[0,3,4,1,2,7,6,5],[0,4,6,2,6,2,0,4],[0,5,4,7,2,1,6,3],[0,6,0,6,0,6,0,6],[0,7,2,5,4,3,6,1]],\"n\":8}"},"fields":{"title":"Swap Two Factors Only after Global Commutative-Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 5, 0, 7, 4], [2, 3, 6, 7, 0, 1, 4, 5], [3, 6, 7, 4, 1, 2, 5, 0], [4, 5, 0, 1, 6, 7, 2, 3], [5, 0, 1, 2, 7, 4, 3, 6], [6, 7, 4, 5, 2, 3, 0, 1], [7, 4, 5, 0, 3, 6, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 4, 6, 4, 0, 2], [0, 3, 4, 1, 2, 7, 6, 5], [0, 4, 6, 2, 6, 2, 0, 4], [0, 5, 4, 7, 2, 1, 6, 3], [0, 6, 0, 6, 0, 6, 0, 6], [0, 7, 2, 5, 4, 3, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=1, b=2. Give (a,b,ab,ba,equal?).","answer":{"kind":"expression","value":"(1, 2, 2, 2, Yes)","explanation":"Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (1, 2, 2, 2, Yes)."},"skills":["Swap Two Factors Only after Global Commutative-Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0043","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"85bf31f95cbfbe972d1db01ed693937a0db821564b910146ca6c7659a1a74575","family_id":"qf-mod28-commring-product-swap","seed":202610281043,"semantic_template":"qf-mod28-commring-product-swap/{\"a\":2,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,6,0,4,7,5],[2,3,6,7,1,0,5,4],[3,6,7,5,2,1,4,0],[4,0,1,2,5,7,3,6],[5,4,0,1,7,6,2,3],[6,7,5,4,3,2,0,1],[7,5,4,0,6,3,1,2]],\"b\":7,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,5,5,6,0,2],[0,3,5,1,7,2,6,4],[0,4,5,7,1,2,6,3],[0,5,6,2,2,6,0,5],[0,6,0,6,6,0,0,6],[0,7,2,4,3,5,6,1]],\"n\":8}"},"fields":{"title":"Swap Two Factors Only after Global Commutative-Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 0, 4, 7, 5], [2, 3, 6, 7, 1, 0, 5, 4], [3, 6, 7, 5, 2, 1, 4, 0], [4, 0, 1, 2, 5, 7, 3, 6], [5, 4, 0, 1, 7, 6, 2, 3], [6, 7, 5, 4, 3, 2, 0, 1], [7, 5, 4, 0, 6, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 5, 5, 6, 0, 2], [0, 3, 5, 1, 7, 2, 6, 4], [0, 4, 5, 7, 1, 2, 6, 3], [0, 5, 6, 2, 2, 6, 0, 5], [0, 6, 0, 6, 6, 0, 0, 6], [0, 7, 2, 4, 3, 5, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=2, b=7. Give (a,b,ab,ba,equal?).","answer":{"kind":"expression","value":"(2, 7, 2, 2, Yes)","explanation":"Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (2, 7, 2, 2, Yes)."},"skills":["Swap Two Factors Only after Global Commutative-Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0044","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b8cab29f55a8351c267a559d9b25587576c2d4450d420408bb8aad8b96f1d294","family_id":"qf-mod28-commring-product-swap","seed":202610281044,"semantic_template":"qf-mod28-commring-product-swap/{\"a\":3,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,7,5,6,0,4],[2,3,7,4,6,0,1,5],[3,7,4,5,0,1,2,6],[4,5,6,0,2,3,7,1],[5,6,0,1,3,7,4,2],[6,0,1,2,7,4,5,3],[7,4,5,6,1,2,3,0]],\"b\":0,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,5,2,7,5,0],[0,3,5,1,6,2,4,7],[0,4,2,6,1,5,3,7],[0,5,7,2,5,7,2,0],[0,6,5,4,3,2,1,7],[0,7,0,7,7,0,7,0]],\"n\":8}"},"fields":{"title":"Swap Two Factors Only after Global Commutative-Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 5, 6, 0, 4], [2, 3, 7, 4, 6, 0, 1, 5], [3, 7, 4, 5, 0, 1, 2, 6], [4, 5, 6, 0, 2, 3, 7, 1], [5, 6, 0, 1, 3, 7, 4, 2], [6, 0, 1, 2, 7, 4, 5, 3], [7, 4, 5, 6, 1, 2, 3, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 5, 2, 7, 5, 0], [0, 3, 5, 1, 6, 2, 4, 7], [0, 4, 2, 6, 1, 5, 3, 7], [0, 5, 7, 2, 5, 7, 2, 0], [0, 6, 5, 4, 3, 2, 1, 7], [0, 7, 0, 7, 7, 0, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=3, b=0. Give (a,b,ab,ba,equal?).","answer":{"kind":"expression","value":"(3, 0, 0, 0, Yes)","explanation":"Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (3, 0, 0, 0, Yes)."},"skills":["Swap Two Factors Only after Global Commutative-Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0045","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"96d33ed82a5b9eca49b23372df20fc71123b4181a4e4d19c3fd26f61eb73e34e","family_id":"qf-mod28-commring-product-swap","seed":202610281045,"semantic_template":"qf-mod28-commring-product-swap/{\"a\":7,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,7,6,0,5,4],[2,3,7,4,5,1,0,6],[3,7,4,6,0,2,1,5],[4,6,5,0,2,7,3,1],[5,0,1,2,7,6,4,3],[6,5,0,1,3,4,7,2],[7,4,6,5,1,3,2,0]],\"b\":3,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,6,2,6,7,0],[0,3,6,1,5,4,2,7],[0,4,2,5,1,3,6,7],[0,5,6,4,3,1,2,7],[0,6,7,2,6,2,7,0],[0,7,0,7,7,7,0,0]],\"n\":8}"},"fields":{"title":"Swap Two Factors Only after Global Commutative-Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 6, 0, 5, 4], [2, 3, 7, 4, 5, 1, 0, 6], [3, 7, 4, 6, 0, 2, 1, 5], [4, 6, 5, 0, 2, 7, 3, 1], [5, 0, 1, 2, 7, 6, 4, 3], [6, 5, 0, 1, 3, 4, 7, 2], [7, 4, 6, 5, 1, 3, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 6, 2, 6, 7, 0], [0, 3, 6, 1, 5, 4, 2, 7], [0, 4, 2, 5, 1, 3, 6, 7], [0, 5, 6, 4, 3, 1, 2, 7], [0, 6, 7, 2, 6, 2, 7, 0], [0, 7, 0, 7, 7, 7, 0, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=7, b=3. Give (a,b,ab,ba,equal?).","answer":{"kind":"expression","value":"(7, 3, 7, 7, Yes)","explanation":"Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (7, 3, 7, 7, Yes)."},"skills":["Swap Two Factors Only after Global Commutative-Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0046","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0f71d2a9598c5f6b98bc9e1e1167af4cad062ccb8a4e883914f397a50bf304d1","family_id":"qf-mod28-commring-product-swap","seed":202610281046,"semantic_template":"qf-mod28-commring-product-swap/{\"a\":5,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,7,6,4,0,5],[2,3,7,5,0,6,1,4],[3,7,5,4,1,0,2,6],[4,6,0,1,7,3,5,2],[5,4,6,0,3,2,7,1],[6,0,1,2,5,7,4,3],[7,5,4,6,2,1,3,0]],\"b\":4,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,4,7,2,4,0],[0,3,4,1,2,6,5,7],[0,4,7,2,7,4,2,0],[0,5,2,6,4,1,3,7],[0,6,4,5,2,3,1,7],[0,7,0,7,0,7,7,0]],\"n\":8}"},"fields":{"title":"Swap Two Factors Only after Global Commutative-Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 6, 4, 0, 5], [2, 3, 7, 5, 0, 6, 1, 4], [3, 7, 5, 4, 1, 0, 2, 6], [4, 6, 0, 1, 7, 3, 5, 2], [5, 4, 6, 0, 3, 2, 7, 1], [6, 0, 1, 2, 5, 7, 4, 3], [7, 5, 4, 6, 2, 1, 3, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 4, 7, 2, 4, 0], [0, 3, 4, 1, 2, 6, 5, 7], [0, 4, 7, 2, 7, 4, 2, 0], [0, 5, 2, 6, 4, 1, 3, 7], [0, 6, 4, 5, 2, 3, 1, 7], [0, 7, 0, 7, 0, 7, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=5, b=4. Give (a,b,ab,ba,equal?).","answer":{"kind":"expression","value":"(5, 4, 4, 4, Yes)","explanation":"Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (5, 4, 4, 4, Yes)."},"skills":["Swap Two Factors Only after Global Commutative-Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0047","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a8903a73fcc07c663e5e6dbafd8c0f5169b79371f26b89dc81041889417db96f","family_id":"qf-mod28-commring-product-swap","seed":202610281047,"semantic_template":"qf-mod28-commring-product-swap/{\"a\":6,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,7,0,6,4,5],[2,3,7,5,1,4,0,6],[3,7,5,6,2,0,1,4],[4,0,1,2,6,7,5,3],[5,6,4,0,7,2,3,1],[6,4,0,1,5,3,7,2],[7,5,6,4,3,1,2,0]],\"b\":1,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,6,6,2,7,0],[0,3,6,1,5,4,2,7],[0,4,6,5,1,3,2,7],[0,5,2,4,3,1,6,7],[0,6,7,2,2,6,7,0],[0,7,0,7,7,7,0,0]],\"n\":8}"},"fields":{"title":"Swap Two Factors Only after Global Commutative-Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 0, 6, 4, 5], [2, 3, 7, 5, 1, 4, 0, 6], [3, 7, 5, 6, 2, 0, 1, 4], [4, 0, 1, 2, 6, 7, 5, 3], [5, 6, 4, 0, 7, 2, 3, 1], [6, 4, 0, 1, 5, 3, 7, 2], [7, 5, 6, 4, 3, 1, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 6, 6, 2, 7, 0], [0, 3, 6, 1, 5, 4, 2, 7], [0, 4, 6, 5, 1, 3, 2, 7], [0, 5, 2, 4, 3, 1, 6, 7], [0, 6, 7, 2, 2, 6, 7, 0], [0, 7, 0, 7, 7, 7, 0, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=6, b=1. Give (a,b,ab,ba,equal?).","answer":{"kind":"expression","value":"(6, 1, 6, 6, Yes)","explanation":"Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (6, 1, 6, 6, Yes)."},"skills":["Swap Two Factors Only after Global Commutative-Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0048","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c36b6b152365604a4b7e137c081bc80e7dea32ed81ad0b6dc2110c302e482e00","family_id":"qf-mod28-commring-product-swap","seed":202610281048,"semantic_template":"qf-mod28-commring-product-swap/{\"a\":5,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,7,5,0,4,6],[2,3,7,6,0,1,5,4],[3,7,6,4,1,2,0,5],[4,5,0,1,7,6,3,2],[5,0,1,2,6,4,7,3],[6,4,5,0,3,7,2,1],[7,6,4,5,2,3,1,0]],\"b\":7,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,4,7,4,2,0],[0,3,4,1,2,6,5,7],[0,4,7,2,7,2,4,0],[0,5,4,6,2,1,3,7],[0,6,2,5,4,3,1,7],[0,7,0,7,0,7,7,0]],\"n\":8}"},"fields":{"title":"Swap Two Factors Only after Global Commutative-Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 5, 0, 4, 6], [2, 3, 7, 6, 0, 1, 5, 4], [3, 7, 6, 4, 1, 2, 0, 5], [4, 5, 0, 1, 7, 6, 3, 2], [5, 0, 1, 2, 6, 4, 7, 3], [6, 4, 5, 0, 3, 7, 2, 1], [7, 6, 4, 5, 2, 3, 1, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 4, 7, 4, 2, 0], [0, 3, 4, 1, 2, 6, 5, 7], [0, 4, 7, 2, 7, 2, 4, 0], [0, 5, 4, 6, 2, 1, 3, 7], [0, 6, 2, 5, 4, 3, 1, 7], [0, 7, 0, 7, 0, 7, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=5, b=7. Give (a,b,ab,ba,equal?).","answer":{"kind":"expression","value":"(5, 7, 7, 7, Yes)","explanation":"Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (5, 7, 7, 7, Yes)."},"skills":["Swap Two Factors Only after Global Commutative-Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0049","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"daee49b6e84d798de39fcf7c1ef90ec55e0eee49e7b036027f679657e1c04b82","family_id":"qf-mod28-commring-product-swap","seed":202610281049,"semantic_template":"qf-mod28-commring-product-swap/{\"a\":0,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,7,0,4,5,6],[2,3,7,6,1,0,4,5],[3,7,6,5,2,1,0,4],[4,0,1,2,5,6,7,3],[5,4,0,1,6,7,3,2],[6,5,4,0,7,3,2,1],[7,6,5,4,3,2,1,0]],\"b\":4,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,5,5,7,2,0],[0,3,5,1,6,2,4,7],[0,4,5,6,1,2,3,7],[0,5,7,2,2,7,5,0],[0,6,2,4,3,5,1,7],[0,7,0,7,7,0,7,0]],\"n\":8}"},"fields":{"title":"Swap Two Factors Only after Global Commutative-Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 0, 4, 5, 6], [2, 3, 7, 6, 1, 0, 4, 5], [3, 7, 6, 5, 2, 1, 0, 4], [4, 0, 1, 2, 5, 6, 7, 3], [5, 4, 0, 1, 6, 7, 3, 2], [6, 5, 4, 0, 7, 3, 2, 1], [7, 6, 5, 4, 3, 2, 1, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 5, 5, 7, 2, 0], [0, 3, 5, 1, 6, 2, 4, 7], [0, 4, 5, 6, 1, 2, 3, 7], [0, 5, 7, 2, 2, 7, 5, 0], [0, 6, 2, 4, 3, 5, 1, 7], [0, 7, 0, 7, 7, 0, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=0, b=4. Give (a,b,ab,ba,equal?).","answer":{"kind":"expression","value":"(0, 4, 0, 0, Yes)","explanation":"Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (0, 4, 0, 0, Yes)."},"skills":["Swap Two Factors Only after Global Commutative-Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0050","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0dac098e42102338f64cceb1c4a5e4c0361577a6f9f1be202c8923e3c40502c8","family_id":"qf-mod28-commring-product-swap","seed":202610281050,"semantic_template":"qf-mod28-commring-product-swap/{\"a\":1,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,4,5,3,6,7,0],[2,4,3,6,5,7,0,1],[3,5,6,0,7,1,2,4],[4,3,5,7,6,0,1,2],[5,6,7,1,0,2,4,3],[6,7,0,2,1,4,3,5],[7,0,1,4,2,3,5,6]],\"b\":2,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,3,0,6,2,3,6],[0,3,0,0,3,3,0,3],[0,4,6,3,1,7,2,5],[0,5,2,3,7,1,6,4],[0,6,3,0,2,6,3,2],[0,7,6,3,5,4,2,1]],\"n\":8}"},"fields":{"title":"Swap Two Factors Only after Global Commutative-Ring Qualification","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 4, 5, 3, 6, 7, 0], [2, 4, 3, 6, 5, 7, 0, 1], [3, 5, 6, 0, 7, 1, 2, 4], [4, 3, 5, 7, 6, 0, 1, 2], [5, 6, 7, 1, 0, 2, 4, 3], [6, 7, 0, 2, 1, 4, 3, 5], [7, 0, 1, 4, 2, 3, 5, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 3, 0, 6, 2, 3, 6], [0, 3, 0, 0, 3, 3, 0, 3], [0, 4, 6, 3, 1, 7, 2, 5], [0, 5, 2, 3, 7, 1, 6, 4], [0, 6, 3, 0, 2, 6, 3, 2], [0, 7, 6, 3, 5, 4, 2, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require the complete ring also to be commutative. a=1, b=2. Give (a,b,ab,ba,equal?).","answer":{"kind":"expression","value":"(1, 2, 2, 2, Yes)","explanation":"Look up both ordered products in the same multiplication table. One equal pair alone would not establish whole-ring commutativity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (1, 2, 2, 2, Yes)."},"skills":["Swap Two Factors Only after Global Commutative-Ring Qualification"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0051","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c6f475fe0ca0a6d1bdfcdae8fd984e3ab0ad42adb3cb9af4ee4a6da87c882b44","family_id":"qf-mod28-commring-power-product","seed":202610281051,"semantic_template":"qf-mod28-commring-power-product/{\"a\":1,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2],[4,5,6,7,0,1,2,3],[5,6,7,0,1,2,3,4],[6,7,0,1,2,3,4,5],[7,0,1,2,3,4,5,6]],\"b\":2,\"exponent\":1,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,6,0,2,4,6],[0,3,6,1,4,7,2,5],[0,4,0,4,0,4,0,4],[0,5,2,7,4,1,6,3],[0,6,4,2,0,6,4,2],[0,7,6,5,4,3,2,1]],\"n\":8}"},"fields":{"title":"A Power-of-Product Identity in a Qualified Commutative Ring","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 6, 0, 2, 4, 6], [0, 3, 6, 1, 4, 7, 2, 5], [0, 4, 0, 4, 0, 4, 0, 4], [0, 5, 2, 7, 4, 1, 6, 3], [0, 6, 4, 2, 0, 6, 4, 2], [0, 7, 6, 5, 4, 3, 2, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=1, b=2, nonnegative exponent=1. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).","answer":{"kind":"expression","value":"(1, 2, 1, 2, 2, 2, Yes)","explanation":"Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (1, 2, 1, 2, 2, 2, Yes)."},"skills":["A Power-of-Product Identity in a Qualified Commutative Ring"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0052","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1a8cea20c750ff3dd3ddad3ba374c7f42959f3c81d36469c88a72649d2871580","family_id":"qf-mod28-commring-power-product","seed":202610281052,"semantic_template":"qf-mod28-commring-power-product/{\"a\":2,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,4,5,7,0,6],[2,3,4,5,7,6,1,0],[3,4,5,7,6,0,2,1],[4,5,7,6,0,1,3,2],[5,7,6,0,1,2,4,3],[6,0,1,2,3,4,7,5],[7,6,0,1,2,3,5,4]],\"b\":5,\"exponent\":2,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,7,0,2,7,4],[0,3,7,1,4,6,5,2],[0,4,0,4,0,4,4,0],[0,5,2,6,4,1,3,7],[0,6,7,5,4,3,1,2],[0,7,4,2,0,7,2,4]],\"n\":8}"},"fields":{"title":"A Power-of-Product Identity in a Qualified Commutative Ring","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 7, 0, 6], [2, 3, 4, 5, 7, 6, 1, 0], [3, 4, 5, 7, 6, 0, 2, 1], [4, 5, 7, 6, 0, 1, 3, 2], [5, 7, 6, 0, 1, 2, 4, 3], [6, 0, 1, 2, 3, 4, 7, 5], [7, 6, 0, 1, 2, 3, 5, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 7, 0, 2, 7, 4], [0, 3, 7, 1, 4, 6, 5, 2], [0, 4, 0, 4, 0, 4, 4, 0], [0, 5, 2, 6, 4, 1, 3, 7], [0, 6, 7, 5, 4, 3, 1, 2], [0, 7, 4, 2, 0, 7, 2, 4]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=2, b=5, nonnegative exponent=2. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).","answer":{"kind":"expression","value":"(2, 5, 2, 2, 4, 4, Yes)","explanation":"Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (2, 5, 2, 2, 4, 4, Yes)."},"skills":["A Power-of-Product Identity in a Qualified Commutative Ring"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0053","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3ef86811aa5ea6c89e51eb4ab83b7668159653ede4d56225537d7e25957f0c8c","family_id":"qf-mod28-commring-power-product","seed":202610281053,"semantic_template":"qf-mod28-commring-power-product/{\"a\":3,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,4,6,7,5,0],[2,3,4,6,5,0,7,1],[3,4,6,5,7,1,0,2],[4,6,5,7,0,2,1,3],[5,7,0,1,2,4,3,6],[6,5,7,0,1,3,2,4],[7,0,1,2,3,6,4,5]],\"b\":0,\"exponent\":3,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,5,0,4,2,5],[0,3,5,1,4,2,7,6],[0,4,0,4,0,0,4,4],[0,5,4,2,0,4,5,2],[0,6,2,7,4,5,1,3],[0,7,5,6,4,2,3,1]],\"n\":8}"},"fields":{"title":"A Power-of-Product Identity in a Qualified Commutative Ring","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 6, 7, 5, 0], [2, 3, 4, 6, 5, 0, 7, 1], [3, 4, 6, 5, 7, 1, 0, 2], [4, 6, 5, 7, 0, 2, 1, 3], [5, 7, 0, 1, 2, 4, 3, 6], [6, 5, 7, 0, 1, 3, 2, 4], [7, 0, 1, 2, 3, 6, 4, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 5, 0, 4, 2, 5], [0, 3, 5, 1, 4, 2, 7, 6], [0, 4, 0, 4, 0, 0, 4, 4], [0, 5, 4, 2, 0, 4, 5, 2], [0, 6, 2, 7, 4, 5, 1, 3], [0, 7, 5, 6, 4, 2, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=3, b=0, nonnegative exponent=3. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).","answer":{"kind":"expression","value":"(3, 0, 3, 0, 0, 0, Yes)","explanation":"Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (3, 0, 3, 0, 0, 0, Yes)."},"skills":["A Power-of-Product Identity in a Qualified Commutative Ring"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0054","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ae47bf061a667083d66890085f40bd6560c713ff6bc9dee68942c8065457197b","family_id":"qf-mod28-commring-power-product","seed":202610281054,"semantic_template":"qf-mod28-commring-power-product/{\"a\":4,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,4,6,0,7,5],[2,3,4,6,7,1,5,0],[3,4,6,7,5,2,0,1],[4,6,7,5,0,3,1,2],[5,0,1,2,3,7,4,6],[6,7,5,0,1,4,2,3],[7,5,0,1,2,6,3,4]],\"b\":3,\"exponent\":4,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,7,0,7,2,4],[0,3,7,1,4,6,5,2],[0,4,0,4,0,4,4,0],[0,5,7,6,4,1,3,2],[0,6,2,5,4,3,1,7],[0,7,4,2,0,2,7,4]],\"n\":8}"},"fields":{"title":"A Power-of-Product Identity in a Qualified Commutative Ring","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 6, 0, 7, 5], [2, 3, 4, 6, 7, 1, 5, 0], [3, 4, 6, 7, 5, 2, 0, 1], [4, 6, 7, 5, 0, 3, 1, 2], [5, 0, 1, 2, 3, 7, 4, 6], [6, 7, 5, 0, 1, 4, 2, 3], [7, 5, 0, 1, 2, 6, 3, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 7, 0, 7, 2, 4], [0, 3, 7, 1, 4, 6, 5, 2], [0, 4, 0, 4, 0, 4, 4, 0], [0, 5, 7, 6, 4, 1, 3, 2], [0, 6, 2, 5, 4, 3, 1, 7], [0, 7, 4, 2, 0, 2, 7, 4]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=4, b=3, nonnegative exponent=4. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).","answer":{"kind":"expression","value":"(4, 3, 4, 4, 0, 0, Yes)","explanation":"Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (4, 3, 4, 4, 0, 0, Yes)."},"skills":["A Power-of-Product Identity in a Qualified Commutative Ring"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0055","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fb249ca3933d9740d678af89e86dccd87ad26f4dfaac73d7c7061d9667ee862e","family_id":"qf-mod28-commring-power-product","seed":202610281055,"semantic_template":"qf-mod28-commring-power-product/{\"a\":7,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,4,7,6,0,5],[2,3,4,7,5,0,1,6],[3,4,7,5,6,1,2,0],[4,7,5,6,0,2,3,1],[5,6,0,1,2,4,7,3],[6,0,1,2,3,7,5,4],[7,5,6,0,1,3,4,2]],\"b\":5,\"exponent\":5,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,5,0,4,5,2],[0,3,5,1,4,2,7,6],[0,4,0,4,0,0,4,4],[0,5,4,2,0,4,2,5],[0,6,5,7,4,2,1,3],[0,7,2,6,4,5,3,1]],\"n\":8}"},"fields":{"title":"A Power-of-Product Identity in a Qualified Commutative Ring","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 7, 6, 0, 5], [2, 3, 4, 7, 5, 0, 1, 6], [3, 4, 7, 5, 6, 1, 2, 0], [4, 7, 5, 6, 0, 2, 3, 1], [5, 6, 0, 1, 2, 4, 7, 3], [6, 0, 1, 2, 3, 7, 5, 4], [7, 5, 6, 0, 1, 3, 4, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 5, 0, 4, 5, 2], [0, 3, 5, 1, 4, 2, 7, 6], [0, 4, 0, 4, 0, 0, 4, 4], [0, 5, 4, 2, 0, 4, 2, 5], [0, 6, 5, 7, 4, 2, 1, 3], [0, 7, 2, 6, 4, 5, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=7, b=5, nonnegative exponent=5. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).","answer":{"kind":"expression","value":"(7, 5, 5, 5, 0, 0, Yes)","explanation":"Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (7, 5, 5, 5, 0, 0, Yes)."},"skills":["A Power-of-Product Identity in a Qualified Commutative Ring"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0056","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dee458e664b68fec4366ec1000a3bc126bf8a72897eb21c63fd4c049a82b99a1","family_id":"qf-mod28-commring-power-product","seed":202610281056,"semantic_template":"qf-mod28-commring-power-product/{\"a\":6,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,4,7,0,5,6],[2,3,4,7,6,1,0,5],[3,4,7,6,5,2,1,0],[4,7,6,5,0,3,2,1],[5,0,1,2,3,6,7,4],[6,5,0,1,2,7,4,3],[7,6,5,0,1,4,3,2]],\"b\":1,\"exponent\":6,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,6,0,6,4,2],[0,3,6,1,4,7,2,5],[0,4,0,4,0,4,0,4],[0,5,6,7,4,1,2,3],[0,6,4,2,0,2,4,6],[0,7,2,5,4,3,6,1]],\"n\":8}"},"fields":{"title":"A Power-of-Product Identity in a Qualified Commutative Ring","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 7, 0, 5, 6], [2, 3, 4, 7, 6, 1, 0, 5], [3, 4, 7, 6, 5, 2, 1, 0], [4, 7, 6, 5, 0, 3, 2, 1], [5, 0, 1, 2, 3, 6, 7, 4], [6, 5, 0, 1, 2, 7, 4, 3], [7, 6, 5, 0, 1, 4, 3, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 6, 0, 6, 4, 2], [0, 3, 6, 1, 4, 7, 2, 5], [0, 4, 0, 4, 0, 4, 0, 4], [0, 5, 6, 7, 4, 1, 2, 3], [0, 6, 4, 2, 0, 2, 4, 6], [0, 7, 2, 5, 4, 3, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=6, b=1, nonnegative exponent=6. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).","answer":{"kind":"expression","value":"(6, 1, 6, 6, 0, 0, Yes)","explanation":"Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (6, 1, 6, 6, 0, 0, Yes)."},"skills":["A Power-of-Product Identity in a Qualified Commutative Ring"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0057","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7594f7afbfaa0724c691e18d0859803de16c98f0548dbb7f19afe6c06988b7d6","family_id":"qf-mod28-commring-power-product","seed":202610281057,"semantic_template":"qf-mod28-commring-power-product/{\"a\":7,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,5,6,4,7,0],[2,3,5,4,7,6,0,1],[3,5,4,6,0,7,1,2],[4,6,7,0,2,1,3,5],[5,4,6,7,1,0,2,3],[6,7,0,1,3,2,5,4],[7,0,1,2,5,3,4,6]],\"b\":5,\"exponent\":7,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,6,2,0,5,6],[0,3,6,1,7,5,2,4],[0,4,2,7,1,5,6,3],[0,5,0,5,5,0,0,5],[0,6,5,2,6,0,5,2],[0,7,6,4,3,5,2,1]],\"n\":8}"},"fields":{"title":"A Power-of-Product Identity in a Qualified Commutative Ring","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 6, 4, 7, 0], [2, 3, 5, 4, 7, 6, 0, 1], [3, 5, 4, 6, 0, 7, 1, 2], [4, 6, 7, 0, 2, 1, 3, 5], [5, 4, 6, 7, 1, 0, 2, 3], [6, 7, 0, 1, 3, 2, 5, 4], [7, 0, 1, 2, 5, 3, 4, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 6, 2, 0, 5, 6], [0, 3, 6, 1, 7, 5, 2, 4], [0, 4, 2, 7, 1, 5, 6, 3], [0, 5, 0, 5, 5, 0, 0, 5], [0, 6, 5, 2, 6, 0, 5, 2], [0, 7, 6, 4, 3, 5, 2, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=7, b=5, nonnegative exponent=7. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).","answer":{"kind":"expression","value":"(7, 5, 7, 5, 0, 0, Yes)","explanation":"Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (7, 5, 7, 5, 0, 0, Yes)."},"skills":["A Power-of-Product Identity in a Qualified Commutative Ring"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0058","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1851801f7fce50e697a07487050d032b8d82defb01fe6ad81a26e0ab73b1692a","family_id":"qf-mod28-commring-power-product","seed":202610281058,"semantic_template":"qf-mod28-commring-power-product/{\"a\":0,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,5,7,4,0,6],[2,3,5,4,6,7,1,0],[3,5,4,7,0,6,2,1],[4,7,6,0,2,1,5,3],[5,4,7,6,1,0,3,2],[6,0,1,2,5,3,7,4],[7,6,0,1,3,2,4,5]],\"b\":6,\"exponent\":1,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,7,2,0,7,5],[0,3,7,1,6,5,4,2],[0,4,2,6,1,5,3,7],[0,5,0,5,5,0,5,0],[0,6,7,4,3,5,1,2],[0,7,5,2,7,0,2,5]],\"n\":8}"},"fields":{"title":"A Power-of-Product Identity in a Qualified Commutative Ring","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 7, 4, 0, 6], [2, 3, 5, 4, 6, 7, 1, 0], [3, 5, 4, 7, 0, 6, 2, 1], [4, 7, 6, 0, 2, 1, 5, 3], [5, 4, 7, 6, 1, 0, 3, 2], [6, 0, 1, 2, 5, 3, 7, 4], [7, 6, 0, 1, 3, 2, 4, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 7, 2, 0, 7, 5], [0, 3, 7, 1, 6, 5, 4, 2], [0, 4, 2, 6, 1, 5, 3, 7], [0, 5, 0, 5, 5, 0, 5, 0], [0, 6, 7, 4, 3, 5, 1, 2], [0, 7, 5, 2, 7, 0, 2, 5]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=0, b=6, nonnegative exponent=1. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).","answer":{"kind":"expression","value":"(0, 6, 1, 0, 0, 0, Yes)","explanation":"Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (0, 6, 1, 0, 0, 0, Yes)."},"skills":["A Power-of-Product Identity in a Qualified Commutative Ring"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0059","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5dfe1d64b1cb02a6c2b78100f0485effb4752a521ee6f646696c0f7042ecaf6c","family_id":"qf-mod28-commring-power-product","seed":202610281059,"semantic_template":"qf-mod28-commring-power-product/{\"a\":1,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,5,7,6,4,0],[2,3,5,6,0,4,7,1],[3,5,6,4,1,7,0,2],[4,7,0,1,5,2,3,6],[5,6,4,7,2,0,1,3],[6,4,7,0,3,1,2,5],[7,0,1,2,6,3,5,4]],\"b\":2,\"exponent\":2,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,4,5,0,2,4],[0,3,4,1,2,5,7,6],[0,4,5,2,5,0,4,2],[0,5,0,5,0,0,5,5],[0,6,2,7,4,5,1,3],[0,7,4,6,2,5,3,1]],\"n\":8}"},"fields":{"title":"A Power-of-Product Identity in a Qualified Commutative Ring","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 7, 6, 4, 0], [2, 3, 5, 6, 0, 4, 7, 1], [3, 5, 6, 4, 1, 7, 0, 2], [4, 7, 0, 1, 5, 2, 3, 6], [5, 6, 4, 7, 2, 0, 1, 3], [6, 4, 7, 0, 3, 1, 2, 5], [7, 0, 1, 2, 6, 3, 5, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 4, 5, 0, 2, 4], [0, 3, 4, 1, 2, 5, 7, 6], [0, 4, 5, 2, 5, 0, 4, 2], [0, 5, 0, 5, 0, 0, 5, 5], [0, 6, 2, 7, 4, 5, 1, 3], [0, 7, 4, 6, 2, 5, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=1, b=2, nonnegative exponent=2. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).","answer":{"kind":"expression","value":"(1, 2, 2, 2, 5, 5, Yes)","explanation":"Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (1, 2, 2, 2, 5, 5, Yes)."},"skills":["A Power-of-Product Identity in a Qualified Commutative Ring"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0060","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"204c2542f9efed9692b476425ead9bb859036696c1ff6fe2e1e74494336272e2","family_id":"qf-mod28-commring-power-product","seed":202610281060,"semantic_template":"qf-mod28-commring-power-product/{\"a\":2,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,5,0,6,7,4],[2,3,5,6,1,7,4,0],[3,5,6,7,2,4,0,1],[4,0,1,2,7,3,5,6],[5,6,7,4,3,0,1,2],[6,7,4,0,5,1,2,3],[7,4,0,1,6,2,3,5]],\"b\":6,\"exponent\":3,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,7,7,0,2,5],[0,3,7,1,6,5,4,2],[0,4,7,6,1,5,3,2],[0,5,0,5,5,0,5,0],[0,6,2,4,3,5,1,7],[0,7,5,2,2,0,7,5]],\"n\":8}"},"fields":{"title":"A Power-of-Product Identity in a Qualified Commutative Ring","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 0, 6, 7, 4], [2, 3, 5, 6, 1, 7, 4, 0], [3, 5, 6, 7, 2, 4, 0, 1], [4, 0, 1, 2, 7, 3, 5, 6], [5, 6, 7, 4, 3, 0, 1, 2], [6, 7, 4, 0, 5, 1, 2, 3], [7, 4, 0, 1, 6, 2, 3, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 7, 7, 0, 2, 5], [0, 3, 7, 1, 6, 5, 4, 2], [0, 4, 7, 6, 1, 5, 3, 2], [0, 5, 0, 5, 5, 0, 5, 0], [0, 6, 2, 4, 3, 5, 1, 7], [0, 7, 5, 2, 2, 0, 7, 5]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=2, b=6, nonnegative exponent=3. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).","answer":{"kind":"expression","value":"(2, 6, 3, 2, 0, 0, Yes)","explanation":"Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (2, 6, 3, 2, 0, 0, Yes)."},"skills":["A Power-of-Product Identity in a Qualified Commutative Ring"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0061","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2710d9a3f05691728b104c09101419e38d3beff0f60ff20eca8d1a7428aa00ab","family_id":"qf-mod28-commring-power-product","seed":202610281061,"semantic_template":"qf-mod28-commring-power-product/{\"a\":3,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,5,6,7,0,4],[2,3,5,7,0,4,1,6],[3,5,7,4,1,6,2,0],[4,6,0,1,5,2,7,3],[5,7,4,6,2,0,3,1],[6,0,1,2,7,3,4,5],[7,4,6,0,3,1,5,2]],\"b\":0,\"exponent\":4,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,4,5,0,4,2],[0,3,4,1,2,5,7,6],[0,4,5,2,5,0,2,4],[0,5,0,5,0,0,5,5],[0,6,4,7,2,5,1,3],[0,7,2,6,4,5,3,1]],\"n\":8}"},"fields":{"title":"A Power-of-Product Identity in a Qualified Commutative Ring","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 6, 7, 0, 4], [2, 3, 5, 7, 0, 4, 1, 6], [3, 5, 7, 4, 1, 6, 2, 0], [4, 6, 0, 1, 5, 2, 7, 3], [5, 7, 4, 6, 2, 0, 3, 1], [6, 0, 1, 2, 7, 3, 4, 5], [7, 4, 6, 0, 3, 1, 5, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 4, 5, 0, 4, 2], [0, 3, 4, 1, 2, 5, 7, 6], [0, 4, 5, 2, 5, 0, 2, 4], [0, 5, 0, 5, 0, 0, 5, 5], [0, 6, 4, 7, 2, 5, 1, 3], [0, 7, 2, 6, 4, 5, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=3, b=0, nonnegative exponent=4. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).","answer":{"kind":"expression","value":"(3, 0, 4, 0, 0, 0, Yes)","explanation":"Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (3, 0, 4, 0, 0, 0, Yes)."},"skills":["A Power-of-Product Identity in a Qualified Commutative Ring"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0062","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6122f6761bcd0f4ddd698c79c3bbf9a153b7b391c79fc14a569cdcd9497b0d02","family_id":"qf-mod28-commring-power-product","seed":202610281062,"semantic_template":"qf-mod28-commring-power-product/{\"a\":5,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,5,0,7,4,6],[2,3,5,7,1,6,0,4],[3,5,7,6,2,4,1,0],[4,0,1,2,6,3,7,5],[5,7,6,4,3,0,2,1],[6,4,0,1,7,2,5,3],[7,6,4,0,5,1,3,2]],\"b\":3,\"exponent\":5,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,6,6,0,5,2],[0,3,6,1,7,5,2,4],[0,4,6,7,1,5,2,3],[0,5,0,5,5,0,0,5],[0,6,5,2,2,0,5,6],[0,7,2,4,3,5,6,1]],\"n\":8}"},"fields":{"title":"A Power-of-Product Identity in a Qualified Commutative Ring","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 0, 7, 4, 6], [2, 3, 5, 7, 1, 6, 0, 4], [3, 5, 7, 6, 2, 4, 1, 0], [4, 0, 1, 2, 6, 3, 7, 5], [5, 7, 6, 4, 3, 0, 2, 1], [6, 4, 0, 1, 7, 2, 5, 3], [7, 6, 4, 0, 5, 1, 3, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 6, 6, 0, 5, 2], [0, 3, 6, 1, 7, 5, 2, 4], [0, 4, 6, 7, 1, 5, 2, 3], [0, 5, 0, 5, 5, 0, 0, 5], [0, 6, 5, 2, 2, 0, 5, 6], [0, 7, 2, 4, 3, 5, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=5, b=3, nonnegative exponent=5. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).","answer":{"kind":"expression","value":"(5, 3, 5, 5, 0, 0, Yes)","explanation":"Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (5, 3, 5, 5, 0, 0, Yes)."},"skills":["A Power-of-Product Identity in a Qualified Commutative Ring"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0063","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7ce324fb4c6015b686763e145c1787378d0cb59e6a775806020cfa817f870ecc","family_id":"qf-mod28-commring-power-product","seed":202610281063,"semantic_template":"qf-mod28-commring-power-product/{\"a\":4,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,6,5,7,4,0],[2,3,6,4,7,0,5,1],[3,6,4,5,0,1,7,2],[4,5,7,0,2,3,1,6],[5,7,0,1,3,6,2,4],[6,4,5,7,1,2,0,3],[7,0,1,2,6,4,3,5]],\"b\":5,\"exponent\":6,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,5,2,6,0,5],[0,3,5,1,7,2,6,4],[0,4,2,7,1,5,6,3],[0,5,6,2,5,6,0,2],[0,6,0,6,6,0,0,6],[0,7,5,4,3,2,6,1]],\"n\":8}"},"fields":{"title":"A Power-of-Product Identity in a Qualified Commutative Ring","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 5, 7, 4, 0], [2, 3, 6, 4, 7, 0, 5, 1], [3, 6, 4, 5, 0, 1, 7, 2], [4, 5, 7, 0, 2, 3, 1, 6], [5, 7, 0, 1, 3, 6, 2, 4], [6, 4, 5, 7, 1, 2, 0, 3], [7, 0, 1, 2, 6, 4, 3, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 5, 2, 6, 0, 5], [0, 3, 5, 1, 7, 2, 6, 4], [0, 4, 2, 7, 1, 5, 6, 3], [0, 5, 6, 2, 5, 6, 0, 2], [0, 6, 0, 6, 6, 0, 0, 6], [0, 7, 5, 4, 3, 2, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=4, b=5, nonnegative exponent=6. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).","answer":{"kind":"expression","value":"(4, 5, 6, 5, 0, 0, Yes)","explanation":"Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (4, 5, 6, 5, 0, 0, Yes)."},"skills":["A Power-of-Product Identity in a Qualified Commutative Ring"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0064","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6d30310da76bffbdd16ef18b5a5689979d854bbf14595e1328ae46396aa8835d","family_id":"qf-mod28-commring-power-product","seed":202610281064,"semantic_template":"qf-mod28-commring-power-product/{\"a\":7,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,6,7,0,4,5],[2,3,6,4,5,1,7,0],[3,6,4,7,0,2,5,1],[4,7,5,0,2,6,1,3],[5,0,1,2,6,7,3,4],[6,4,7,5,1,3,0,2],[7,5,0,1,3,4,2,6]],\"b\":1,\"exponent\":7,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,7,2,7,0,6],[0,3,7,1,5,4,6,2],[0,4,2,5,1,3,6,7],[0,5,7,4,3,1,6,2],[0,6,0,6,6,6,0,0],[0,7,6,2,7,2,0,6]],\"n\":8}"},"fields":{"title":"A Power-of-Product Identity in a Qualified Commutative Ring","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 7, 0, 4, 5], [2, 3, 6, 4, 5, 1, 7, 0], [3, 6, 4, 7, 0, 2, 5, 1], [4, 7, 5, 0, 2, 6, 1, 3], [5, 0, 1, 2, 6, 7, 3, 4], [6, 4, 7, 5, 1, 3, 0, 2], [7, 5, 0, 1, 3, 4, 2, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 7, 2, 7, 0, 6], [0, 3, 7, 1, 5, 4, 6, 2], [0, 4, 2, 5, 1, 3, 6, 7], [0, 5, 7, 4, 3, 1, 6, 2], [0, 6, 0, 6, 6, 6, 0, 0], [0, 7, 6, 2, 7, 2, 0, 6]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=7, b=1, nonnegative exponent=7. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).","answer":{"kind":"expression","value":"(7, 1, 7, 7, 0, 0, Yes)","explanation":"Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (7, 1, 7, 7, 0, 0, Yes)."},"skills":["A Power-of-Product Identity in a Qualified Commutative Ring"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0065","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"515389f1871312456bfa5f9bb17208ba38dcfb9ad6c19bb05a8199212e5d5b14","family_id":"qf-mod28-commring-power-product","seed":202610281065,"semantic_template":"qf-mod28-commring-power-product/{\"a\":7,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,6,7,4,5,0],[2,3,6,5,0,7,4,1],[3,6,5,4,1,0,7,2],[4,7,0,1,6,3,2,5],[5,4,7,0,3,2,1,6],[6,5,4,7,2,1,0,3],[7,0,1,2,5,6,3,4]],\"b\":6,\"exponent\":1,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,4,6,2,0,4],[0,3,4,1,2,7,6,5],[0,4,6,2,6,4,0,2],[0,5,2,7,4,1,6,3],[0,6,0,6,0,6,0,6],[0,7,4,5,2,3,6,1]],\"n\":8}"},"fields":{"title":"A Power-of-Product Identity in a Qualified Commutative Ring","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 7, 4, 5, 0], [2, 3, 6, 5, 0, 7, 4, 1], [3, 6, 5, 4, 1, 0, 7, 2], [4, 7, 0, 1, 6, 3, 2, 5], [5, 4, 7, 0, 3, 2, 1, 6], [6, 5, 4, 7, 2, 1, 0, 3], [7, 0, 1, 2, 5, 6, 3, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 4, 6, 2, 0, 4], [0, 3, 4, 1, 2, 7, 6, 5], [0, 4, 6, 2, 6, 4, 0, 2], [0, 5, 2, 7, 4, 1, 6, 3], [0, 6, 0, 6, 0, 6, 0, 6], [0, 7, 4, 5, 2, 3, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=7, b=6, nonnegative exponent=1. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).","answer":{"kind":"expression","value":"(7, 6, 1, 6, 6, 6, Yes)","explanation":"Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (7, 6, 1, 6, 6, 6, Yes)."},"skills":["A Power-of-Product Identity in a Qualified Commutative Ring"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0066","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"79be6eb0af4e4aba6948e3d8b3f041afa97c36bcdef0c59d52ead61bf42d27ce","family_id":"qf-mod28-commring-power-product","seed":202610281066,"semantic_template":"qf-mod28-commring-power-product/{\"a\":0,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,6,0,7,5,4],[2,3,6,5,1,4,7,0],[3,6,5,7,2,0,4,1],[4,0,1,2,7,6,3,5],[5,7,4,0,6,2,1,3],[6,5,7,4,3,1,0,2],[7,4,0,1,5,3,2,6]],\"b\":4,\"exponent\":2,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,7,7,2,0,6],[0,3,7,1,5,4,6,2],[0,4,7,5,1,3,6,2],[0,5,2,4,3,1,6,7],[0,6,0,6,6,6,0,0],[0,7,6,2,2,7,0,6]],\"n\":8}"},"fields":{"title":"A Power-of-Product Identity in a Qualified Commutative Ring","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 0, 7, 5, 4], [2, 3, 6, 5, 1, 4, 7, 0], [3, 6, 5, 7, 2, 0, 4, 1], [4, 0, 1, 2, 7, 6, 3, 5], [5, 7, 4, 0, 6, 2, 1, 3], [6, 5, 7, 4, 3, 1, 0, 2], [7, 4, 0, 1, 5, 3, 2, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 7, 7, 2, 0, 6], [0, 3, 7, 1, 5, 4, 6, 2], [0, 4, 7, 5, 1, 3, 6, 2], [0, 5, 2, 4, 3, 1, 6, 7], [0, 6, 0, 6, 6, 6, 0, 0], [0, 7, 6, 2, 2, 7, 0, 6]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=0, b=4, nonnegative exponent=2. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).","answer":{"kind":"expression","value":"(0, 4, 2, 0, 0, 0, Yes)","explanation":"Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (0, 4, 2, 0, 0, 0, Yes)."},"skills":["A Power-of-Product Identity in a Qualified Commutative Ring"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0067","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5241263f13c6c58dc46d7223a42f88f514be5c51b54d51cf0578a01eec0488c5","family_id":"qf-mod28-commring-power-product","seed":202610281067,"semantic_template":"qf-mod28-commring-power-product/{\"a\":1,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,6,5,0,7,4],[2,3,6,7,0,1,4,5],[3,6,7,4,1,2,5,0],[4,5,0,1,6,7,2,3],[5,0,1,2,7,4,3,6],[6,7,4,5,2,3,0,1],[7,4,5,0,3,6,1,2]],\"b\":2,\"exponent\":3,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,4,6,4,0,2],[0,3,4,1,2,7,6,5],[0,4,6,2,6,2,0,4],[0,5,4,7,2,1,6,3],[0,6,0,6,0,6,0,6],[0,7,2,5,4,3,6,1]],\"n\":8}"},"fields":{"title":"A Power-of-Product Identity in a Qualified Commutative Ring","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 5, 0, 7, 4], [2, 3, 6, 7, 0, 1, 4, 5], [3, 6, 7, 4, 1, 2, 5, 0], [4, 5, 0, 1, 6, 7, 2, 3], [5, 0, 1, 2, 7, 4, 3, 6], [6, 7, 4, 5, 2, 3, 0, 1], [7, 4, 5, 0, 3, 6, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 4, 6, 4, 0, 2], [0, 3, 4, 1, 2, 7, 6, 5], [0, 4, 6, 2, 6, 2, 0, 4], [0, 5, 4, 7, 2, 1, 6, 3], [0, 6, 0, 6, 0, 6, 0, 6], [0, 7, 2, 5, 4, 3, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=1, b=2, nonnegative exponent=3. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).","answer":{"kind":"expression","value":"(1, 2, 3, 2, 0, 0, Yes)","explanation":"Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (1, 2, 3, 2, 0, 0, Yes)."},"skills":["A Power-of-Product Identity in a Qualified Commutative Ring"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0068","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"67184ae3d1cae16c382920b544eec2cec9783be26b628d7c783a49ec07b28c4c","family_id":"qf-mod28-commring-power-product","seed":202610281068,"semantic_template":"qf-mod28-commring-power-product/{\"a\":2,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,6,0,4,7,5],[2,3,6,7,1,0,5,4],[3,6,7,5,2,1,4,0],[4,0,1,2,5,7,3,6],[5,4,0,1,7,6,2,3],[6,7,5,4,3,2,0,1],[7,5,4,0,6,3,1,2]],\"b\":7,\"exponent\":4,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,5,5,6,0,2],[0,3,5,1,7,2,6,4],[0,4,5,7,1,2,6,3],[0,5,6,2,2,6,0,5],[0,6,0,6,6,0,0,6],[0,7,2,4,3,5,6,1]],\"n\":8}"},"fields":{"title":"A Power-of-Product Identity in a Qualified Commutative Ring","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 0, 4, 7, 5], [2, 3, 6, 7, 1, 0, 5, 4], [3, 6, 7, 5, 2, 1, 4, 0], [4, 0, 1, 2, 5, 7, 3, 6], [5, 4, 0, 1, 7, 6, 2, 3], [6, 7, 5, 4, 3, 2, 0, 1], [7, 5, 4, 0, 6, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 5, 5, 6, 0, 2], [0, 3, 5, 1, 7, 2, 6, 4], [0, 4, 5, 7, 1, 2, 6, 3], [0, 5, 6, 2, 2, 6, 0, 5], [0, 6, 0, 6, 6, 0, 0, 6], [0, 7, 2, 4, 3, 5, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=2, b=7, nonnegative exponent=4. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).","answer":{"kind":"expression","value":"(2, 7, 4, 2, 0, 0, Yes)","explanation":"Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (2, 7, 4, 2, 0, 0, Yes)."},"skills":["A Power-of-Product Identity in a Qualified Commutative Ring"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0069","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d6f1fe990a4c4a578315e0fd6f3aa56c47cd0f8b0d21a93a2c431c6c7b700bb7","family_id":"qf-mod28-commring-power-product","seed":202610281069,"semantic_template":"qf-mod28-commring-power-product/{\"a\":3,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,7,5,6,0,4],[2,3,7,4,6,0,1,5],[3,7,4,5,0,1,2,6],[4,5,6,0,2,3,7,1],[5,6,0,1,3,7,4,2],[6,0,1,2,7,4,5,3],[7,4,5,6,1,2,3,0]],\"b\":0,\"exponent\":5,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,5,2,7,5,0],[0,3,5,1,6,2,4,7],[0,4,2,6,1,5,3,7],[0,5,7,2,5,7,2,0],[0,6,5,4,3,2,1,7],[0,7,0,7,7,0,7,0]],\"n\":8}"},"fields":{"title":"A Power-of-Product Identity in a Qualified Commutative Ring","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 5, 6, 0, 4], [2, 3, 7, 4, 6, 0, 1, 5], [3, 7, 4, 5, 0, 1, 2, 6], [4, 5, 6, 0, 2, 3, 7, 1], [5, 6, 0, 1, 3, 7, 4, 2], [6, 0, 1, 2, 7, 4, 5, 3], [7, 4, 5, 6, 1, 2, 3, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 5, 2, 7, 5, 0], [0, 3, 5, 1, 6, 2, 4, 7], [0, 4, 2, 6, 1, 5, 3, 7], [0, 5, 7, 2, 5, 7, 2, 0], [0, 6, 5, 4, 3, 2, 1, 7], [0, 7, 0, 7, 7, 0, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=3, b=0, nonnegative exponent=5. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).","answer":{"kind":"expression","value":"(3, 0, 5, 0, 0, 0, Yes)","explanation":"Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (3, 0, 5, 0, 0, 0, Yes)."},"skills":["A Power-of-Product Identity in a Qualified Commutative Ring"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0070","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fab3895ce7616124de95929a9337efc61aef45ad600ed489013b03c0fe12ffbe","family_id":"qf-mod28-commring-power-product","seed":202610281070,"semantic_template":"qf-mod28-commring-power-product/{\"a\":7,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,7,6,0,5,4],[2,3,7,4,5,1,0,6],[3,7,4,6,0,2,1,5],[4,6,5,0,2,7,3,1],[5,0,1,2,7,6,4,3],[6,5,0,1,3,4,7,2],[7,4,6,5,1,3,2,0]],\"b\":3,\"exponent\":6,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,6,2,6,7,0],[0,3,6,1,5,4,2,7],[0,4,2,5,1,3,6,7],[0,5,6,4,3,1,2,7],[0,6,7,2,6,2,7,0],[0,7,0,7,7,7,0,0]],\"n\":8}"},"fields":{"title":"A Power-of-Product Identity in a Qualified Commutative Ring","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 6, 0, 5, 4], [2, 3, 7, 4, 5, 1, 0, 6], [3, 7, 4, 6, 0, 2, 1, 5], [4, 6, 5, 0, 2, 7, 3, 1], [5, 0, 1, 2, 7, 6, 4, 3], [6, 5, 0, 1, 3, 4, 7, 2], [7, 4, 6, 5, 1, 3, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 6, 2, 6, 7, 0], [0, 3, 6, 1, 5, 4, 2, 7], [0, 4, 2, 5, 1, 3, 6, 7], [0, 5, 6, 4, 3, 1, 2, 7], [0, 6, 7, 2, 6, 2, 7, 0], [0, 7, 0, 7, 7, 7, 0, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=7, b=3, nonnegative exponent=6. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).","answer":{"kind":"expression","value":"(7, 3, 6, 7, 0, 0, Yes)","explanation":"Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (7, 3, 6, 7, 0, 0, Yes)."},"skills":["A Power-of-Product Identity in a Qualified Commutative Ring"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0071","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"846a2436eafc42f4285f3a8b8ba50dd90a0e537170edebcb2d7437b9c0210594","family_id":"qf-mod28-commring-power-product","seed":202610281071,"semantic_template":"qf-mod28-commring-power-product/{\"a\":5,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,7,6,4,0,5],[2,3,7,5,0,6,1,4],[3,7,5,4,1,0,2,6],[4,6,0,1,7,3,5,2],[5,4,6,0,3,2,7,1],[6,0,1,2,5,7,4,3],[7,5,4,6,2,1,3,0]],\"b\":4,\"exponent\":7,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,4,7,2,4,0],[0,3,4,1,2,6,5,7],[0,4,7,2,7,4,2,0],[0,5,2,6,4,1,3,7],[0,6,4,5,2,3,1,7],[0,7,0,7,0,7,7,0]],\"n\":8}"},"fields":{"title":"A Power-of-Product Identity in a Qualified Commutative Ring","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 6, 4, 0, 5], [2, 3, 7, 5, 0, 6, 1, 4], [3, 7, 5, 4, 1, 0, 2, 6], [4, 6, 0, 1, 7, 3, 5, 2], [5, 4, 6, 0, 3, 2, 7, 1], [6, 0, 1, 2, 5, 7, 4, 3], [7, 5, 4, 6, 2, 1, 3, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 4, 7, 2, 4, 0], [0, 3, 4, 1, 2, 6, 5, 7], [0, 4, 7, 2, 7, 4, 2, 0], [0, 5, 2, 6, 4, 1, 3, 7], [0, 6, 4, 5, 2, 3, 1, 7], [0, 7, 0, 7, 0, 7, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=5, b=4, nonnegative exponent=7. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).","answer":{"kind":"expression","value":"(5, 4, 7, 4, 0, 0, Yes)","explanation":"Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (5, 4, 7, 4, 0, 0, Yes)."},"skills":["A Power-of-Product Identity in a Qualified Commutative Ring"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0072","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"646dc18b0ca682756afef4cef8078b926fa253d2751cf1e51f4169c9a5f8e469","family_id":"qf-mod28-commring-power-product","seed":202610281072,"semantic_template":"qf-mod28-commring-power-product/{\"a\":6,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,7,0,6,4,5],[2,3,7,5,1,4,0,6],[3,7,5,6,2,0,1,4],[4,0,1,2,6,7,5,3],[5,6,4,0,7,2,3,1],[6,4,0,1,5,3,7,2],[7,5,6,4,3,1,2,0]],\"b\":1,\"exponent\":1,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,6,6,2,7,0],[0,3,6,1,5,4,2,7],[0,4,6,5,1,3,2,7],[0,5,2,4,3,1,6,7],[0,6,7,2,2,6,7,0],[0,7,0,7,7,7,0,0]],\"n\":8}"},"fields":{"title":"A Power-of-Product Identity in a Qualified Commutative Ring","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 0, 6, 4, 5], [2, 3, 7, 5, 1, 4, 0, 6], [3, 7, 5, 6, 2, 0, 1, 4], [4, 0, 1, 2, 6, 7, 5, 3], [5, 6, 4, 0, 7, 2, 3, 1], [6, 4, 0, 1, 5, 3, 7, 2], [7, 5, 6, 4, 3, 1, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 6, 6, 2, 7, 0], [0, 3, 6, 1, 5, 4, 2, 7], [0, 4, 6, 5, 1, 3, 2, 7], [0, 5, 2, 4, 3, 1, 6, 7], [0, 6, 7, 2, 2, 6, 7, 0], [0, 7, 0, 7, 7, 7, 0, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=6, b=1, nonnegative exponent=1. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).","answer":{"kind":"expression","value":"(6, 1, 1, 6, 6, 6, Yes)","explanation":"Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (6, 1, 1, 6, 6, 6, Yes)."},"skills":["A Power-of-Product Identity in a Qualified Commutative Ring"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0073","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a231f0fee1f34b53bc102c4059cada85f6c2bb9069049aebce45714a40236351","family_id":"qf-mod28-commring-power-product","seed":202610281073,"semantic_template":"qf-mod28-commring-power-product/{\"a\":5,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,7,5,0,4,6],[2,3,7,6,0,1,5,4],[3,7,6,4,1,2,0,5],[4,5,0,1,7,6,3,2],[5,0,1,2,6,4,7,3],[6,4,5,0,3,7,2,1],[7,6,4,5,2,3,1,0]],\"b\":7,\"exponent\":2,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,4,7,4,2,0],[0,3,4,1,2,6,5,7],[0,4,7,2,7,2,4,0],[0,5,4,6,2,1,3,7],[0,6,2,5,4,3,1,7],[0,7,0,7,0,7,7,0]],\"n\":8}"},"fields":{"title":"A Power-of-Product Identity in a Qualified Commutative Ring","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 5, 0, 4, 6], [2, 3, 7, 6, 0, 1, 5, 4], [3, 7, 6, 4, 1, 2, 0, 5], [4, 5, 0, 1, 7, 6, 3, 2], [5, 0, 1, 2, 6, 4, 7, 3], [6, 4, 5, 0, 3, 7, 2, 1], [7, 6, 4, 5, 2, 3, 1, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 4, 7, 4, 2, 0], [0, 3, 4, 1, 2, 6, 5, 7], [0, 4, 7, 2, 7, 2, 4, 0], [0, 5, 4, 6, 2, 1, 3, 7], [0, 6, 2, 5, 4, 3, 1, 7], [0, 7, 0, 7, 0, 7, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=5, b=7, nonnegative exponent=2. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).","answer":{"kind":"expression","value":"(5, 7, 2, 7, 0, 0, Yes)","explanation":"Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (5, 7, 2, 7, 0, 0, Yes)."},"skills":["A Power-of-Product Identity in a Qualified Commutative Ring"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0074","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0bf0cc1c4f39ae83569cfcc3f441cfb54d889f8aa1e70be00728b950ee11b4ec","family_id":"qf-mod28-commring-power-product","seed":202610281074,"semantic_template":"qf-mod28-commring-power-product/{\"a\":0,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,7,0,4,5,6],[2,3,7,6,1,0,4,5],[3,7,6,5,2,1,0,4],[4,0,1,2,5,6,7,3],[5,4,0,1,6,7,3,2],[6,5,4,0,7,3,2,1],[7,6,5,4,3,2,1,0]],\"b\":4,\"exponent\":3,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,5,5,7,2,0],[0,3,5,1,6,2,4,7],[0,4,5,6,1,2,3,7],[0,5,7,2,2,7,5,0],[0,6,2,4,3,5,1,7],[0,7,0,7,7,0,7,0]],\"n\":8}"},"fields":{"title":"A Power-of-Product Identity in a Qualified Commutative Ring","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 0, 4, 5, 6], [2, 3, 7, 6, 1, 0, 4, 5], [3, 7, 6, 5, 2, 1, 0, 4], [4, 0, 1, 2, 5, 6, 7, 3], [5, 4, 0, 1, 6, 7, 3, 2], [6, 5, 4, 0, 7, 3, 2, 1], [7, 6, 5, 4, 3, 2, 1, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 5, 5, 7, 2, 0], [0, 3, 5, 1, 6, 2, 4, 7], [0, 4, 5, 6, 1, 2, 3, 7], [0, 5, 7, 2, 2, 7, 5, 0], [0, 6, 2, 4, 3, 5, 1, 7], [0, 7, 0, 7, 7, 0, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=0, b=4, nonnegative exponent=3. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).","answer":{"kind":"expression","value":"(0, 4, 3, 0, 0, 0, Yes)","explanation":"Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (0, 4, 3, 0, 0, 0, Yes)."},"skills":["A Power-of-Product Identity in a Qualified Commutative Ring"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0075","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"17ade8875b755b10c60350288671c570074409fb4bd56201349960803f93afd1","family_id":"qf-mod28-commring-power-product","seed":202610281075,"semantic_template":"qf-mod28-commring-power-product/{\"a\":1,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,4,5,3,6,7,0],[2,4,3,6,5,7,0,1],[3,5,6,0,7,1,2,4],[4,3,5,7,6,0,1,2],[5,6,7,1,0,2,4,3],[6,7,0,2,1,4,3,5],[7,0,1,4,2,3,5,6]],\"b\":2,\"exponent\":4,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,3,0,6,2,3,6],[0,3,0,0,3,3,0,3],[0,4,6,3,1,7,2,5],[0,5,2,3,7,1,6,4],[0,6,3,0,2,6,3,2],[0,7,6,3,5,4,2,1]],\"n\":8}"},"fields":{"title":"A Power-of-Product Identity in a Qualified Commutative Ring","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 4, 5, 3, 6, 7, 0], [2, 4, 3, 6, 5, 7, 0, 1], [3, 5, 6, 0, 7, 1, 2, 4], [4, 3, 5, 7, 6, 0, 1, 2], [5, 6, 7, 1, 0, 2, 4, 3], [6, 7, 0, 2, 1, 4, 3, 5], [7, 0, 1, 4, 2, 3, 5, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 3, 0, 6, 2, 3, 6], [0, 3, 0, 0, 3, 3, 0, 3], [0, 4, 6, 3, 1, 7, 2, 5], [0, 5, 2, 3, 7, 1, 6, 4], [0, 6, 3, 0, 2, 6, 3, 2], [0, 7, 6, 3, 5, 4, 2, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require multiplication to commute. a=1, b=2, nonnegative exponent=4. Give (a,b,exponent,ab,lhs=(ab)^exponent,rhs=a^exponent*b^exponent,equal?).","answer":{"kind":"expression","value":"(1, 2, 4, 2, 0, 0, Yes)","explanation":"Start each power at the actual multiplicative identity. At exponent zero all powers are the identity. This identity is used under the full commutative-ring premise, not to certify commutativity from one instance. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (1, 2, 4, 2, 0, 0, Yes)."},"skills":["A Power-of-Product Identity in a Qualified Commutative Ring"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0076","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ba5c307d1b043dbc2918d575c8cd03faca6fdec467ac9824bddcb6e4137356e1","family_id":"qf-mod28-commring-binomial-square","seed":202610281076,"semantic_template":"qf-mod28-commring-binomial-square/{\"a\":1,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2],[4,5,6,7,0,1,2,3],[5,6,7,0,1,2,3,4],[6,7,0,1,2,3,4,5],[7,0,1,2,3,4,5,6]],\"b\":2,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,6,0,2,4,6],[0,3,6,1,4,7,2,5],[0,4,0,4,0,4,0,4],[0,5,2,7,4,1,6,3],[0,6,4,2,0,6,4,2],[0,7,6,5,4,3,2,1]],\"n\":8}"},"fields":{"title":"Compute a Binomial Square Using Ring Operations, Not Label Arithmetic","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 0], [2, 3, 4, 5, 6, 7, 0, 1], [3, 4, 5, 6, 7, 0, 1, 2], [4, 5, 6, 7, 0, 1, 2, 3], [5, 6, 7, 0, 1, 2, 3, 4], [6, 7, 0, 1, 2, 3, 4, 5], [7, 0, 1, 2, 3, 4, 5, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 6, 0, 2, 4, 6], [0, 3, 6, 1, 4, 7, 2, 5], [0, 4, 0, 4, 0, 4, 0, 4], [0, 5, 2, 7, 4, 1, 6, 3], [0, 6, 4, 2, 0, 6, 4, 2], [0, 7, 6, 5, 4, 3, 2, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=1, b=2. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).","answer":{"kind":"expression","value":"(1, 2, 3, 1, 1, 2, 4, 1, Yes)","explanation":"Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (1, 2, 3, 1, 1, 2, 4, 1, Yes)."},"skills":["Compute a Binomial Square Using Ring Operations, Not Label Arithmetic"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0077","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2988d80dd1c8fe2044b0024edd06f86825705b2f11b2010cdc06142b2b0411f9","family_id":"qf-mod28-commring-binomial-square","seed":202610281077,"semantic_template":"qf-mod28-commring-binomial-square/{\"a\":2,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,4,5,7,0,6],[2,3,4,5,7,6,1,0],[3,4,5,7,6,0,2,1],[4,5,7,6,0,1,3,2],[5,7,6,0,1,2,4,3],[6,0,1,2,3,4,7,5],[7,6,0,1,2,3,5,4]],\"b\":5,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,7,0,2,7,4],[0,3,7,1,4,6,5,2],[0,4,0,4,0,4,4,0],[0,5,2,6,4,1,3,7],[0,6,7,5,4,3,1,2],[0,7,4,2,0,7,2,4]],\"n\":8}"},"fields":{"title":"Compute a Binomial Square Using Ring Operations, Not Label Arithmetic","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 7, 0, 6], [2, 3, 4, 5, 7, 6, 1, 0], [3, 4, 5, 7, 6, 0, 2, 1], [4, 5, 7, 6, 0, 1, 3, 2], [5, 7, 6, 0, 1, 2, 4, 3], [6, 0, 1, 2, 3, 4, 7, 5], [7, 6, 0, 1, 2, 3, 5, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 7, 0, 2, 7, 4], [0, 3, 7, 1, 4, 6, 5, 2], [0, 4, 0, 4, 0, 4, 4, 0], [0, 5, 2, 6, 4, 1, 3, 7], [0, 6, 7, 5, 4, 3, 1, 2], [0, 7, 4, 2, 0, 7, 2, 4]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=2, b=5. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).","answer":{"kind":"expression","value":"(2, 5, 6, 1, 4, 2, 1, 1, Yes)","explanation":"Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (2, 5, 6, 1, 4, 2, 1, 1, Yes)."},"skills":["Compute a Binomial Square Using Ring Operations, Not Label Arithmetic"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0078","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b74225221c085360b73d3879a060e7fcc3f9ae95ffd6824f10cf829050d901d8","family_id":"qf-mod28-commring-binomial-square","seed":202610281078,"semantic_template":"qf-mod28-commring-binomial-square/{\"a\":3,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,4,6,7,5,0],[2,3,4,6,5,0,7,1],[3,4,6,5,7,1,0,2],[4,6,5,7,0,2,1,3],[5,7,0,1,2,4,3,6],[6,5,7,0,1,3,2,4],[7,0,1,2,3,6,4,5]],\"b\":0,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,5,0,4,2,5],[0,3,5,1,4,2,7,6],[0,4,0,4,0,0,4,4],[0,5,4,2,0,4,5,2],[0,6,2,7,4,5,1,3],[0,7,5,6,4,2,3,1]],\"n\":8}"},"fields":{"title":"Compute a Binomial Square Using Ring Operations, Not Label Arithmetic","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 6, 7, 5, 0], [2, 3, 4, 6, 5, 0, 7, 1], [3, 4, 6, 5, 7, 1, 0, 2], [4, 6, 5, 7, 0, 2, 1, 3], [5, 7, 0, 1, 2, 4, 3, 6], [6, 5, 7, 0, 1, 3, 2, 4], [7, 0, 1, 2, 3, 6, 4, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 5, 0, 4, 2, 5], [0, 3, 5, 1, 4, 2, 7, 6], [0, 4, 0, 4, 0, 0, 4, 4], [0, 5, 4, 2, 0, 4, 5, 2], [0, 6, 2, 7, 4, 5, 1, 3], [0, 7, 5, 6, 4, 2, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=3, b=0. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).","answer":{"kind":"expression","value":"(3, 0, 3, 1, 1, 0, 0, 1, Yes)","explanation":"Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (3, 0, 3, 1, 1, 0, 0, 1, Yes)."},"skills":["Compute a Binomial Square Using Ring Operations, Not Label Arithmetic"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0079","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f22162b63fa440d013c26a59bf0d15af9308a3c8b1233b46c69dea9f23d24e2e","family_id":"qf-mod28-commring-binomial-square","seed":202610281079,"semantic_template":"qf-mod28-commring-binomial-square/{\"a\":4,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,4,6,0,7,5],[2,3,4,6,7,1,5,0],[3,4,6,7,5,2,0,1],[4,6,7,5,0,3,1,2],[5,0,1,2,3,7,4,6],[6,7,5,0,1,4,2,3],[7,5,0,1,2,6,3,4]],\"b\":3,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,7,0,7,2,4],[0,3,7,1,4,6,5,2],[0,4,0,4,0,4,4,0],[0,5,7,6,4,1,3,2],[0,6,2,5,4,3,1,7],[0,7,4,2,0,2,7,4]],\"n\":8}"},"fields":{"title":"Compute a Binomial Square Using Ring Operations, Not Label Arithmetic","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 6, 0, 7, 5], [2, 3, 4, 6, 7, 1, 5, 0], [3, 4, 6, 7, 5, 2, 0, 1], [4, 6, 7, 5, 0, 3, 1, 2], [5, 0, 1, 2, 3, 7, 4, 6], [6, 7, 5, 0, 1, 4, 2, 3], [7, 5, 0, 1, 2, 6, 3, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 7, 0, 7, 2, 4], [0, 3, 7, 1, 4, 6, 5, 2], [0, 4, 0, 4, 0, 4, 4, 0], [0, 5, 7, 6, 4, 1, 3, 2], [0, 6, 2, 5, 4, 3, 1, 7], [0, 7, 4, 2, 0, 2, 7, 4]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=4, b=3. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).","answer":{"kind":"expression","value":"(4, 3, 5, 1, 0, 4, 1, 1, Yes)","explanation":"Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (4, 3, 5, 1, 0, 4, 1, 1, Yes)."},"skills":["Compute a Binomial Square Using Ring Operations, Not Label Arithmetic"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0080","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3c40769cbd18f42c7ebc6aee5cb52b5a1a972e227d7cb6c09a0c6ee2905bc04f","family_id":"qf-mod28-commring-binomial-square","seed":202610281080,"semantic_template":"qf-mod28-commring-binomial-square/{\"a\":7,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,4,7,6,0,5],[2,3,4,7,5,0,1,6],[3,4,7,5,6,1,2,0],[4,7,5,6,0,2,3,1],[5,6,0,1,2,4,7,3],[6,0,1,2,3,7,5,4],[7,5,6,0,1,3,4,2]],\"b\":5,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,5,0,4,5,2],[0,3,5,1,4,2,7,6],[0,4,0,4,0,0,4,4],[0,5,4,2,0,4,2,5],[0,6,5,7,4,2,1,3],[0,7,2,6,4,5,3,1]],\"n\":8}"},"fields":{"title":"Compute a Binomial Square Using Ring Operations, Not Label Arithmetic","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 7, 6, 0, 5], [2, 3, 4, 7, 5, 0, 1, 6], [3, 4, 7, 5, 6, 1, 2, 0], [4, 7, 5, 6, 0, 2, 3, 1], [5, 6, 0, 1, 2, 4, 7, 3], [6, 0, 1, 2, 3, 7, 5, 4], [7, 5, 6, 0, 1, 3, 4, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 5, 0, 4, 5, 2], [0, 3, 5, 1, 4, 2, 7, 6], [0, 4, 0, 4, 0, 0, 4, 4], [0, 5, 4, 2, 0, 4, 2, 5], [0, 6, 5, 7, 4, 2, 1, 3], [0, 7, 2, 6, 4, 5, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=7, b=5. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).","answer":{"kind":"expression","value":"(7, 5, 3, 1, 1, 5, 4, 1, Yes)","explanation":"Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (7, 5, 3, 1, 1, 5, 4, 1, Yes)."},"skills":["Compute a Binomial Square Using Ring Operations, Not Label Arithmetic"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0081","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"faf7cc6e5408f4f09f59d46014a168649b95e48f10388cdb82733cdd8f73af21","family_id":"qf-mod28-commring-binomial-square","seed":202610281081,"semantic_template":"qf-mod28-commring-binomial-square/{\"a\":6,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,4,7,0,5,6],[2,3,4,7,6,1,0,5],[3,4,7,6,5,2,1,0],[4,7,6,5,0,3,2,1],[5,0,1,2,3,6,7,4],[6,5,0,1,2,7,4,3],[7,6,5,0,1,4,3,2]],\"b\":1,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,6,0,6,4,2],[0,3,6,1,4,7,2,5],[0,4,0,4,0,4,0,4],[0,5,6,7,4,1,2,3],[0,6,4,2,0,2,4,6],[0,7,2,5,4,3,6,1]],\"n\":8}"},"fields":{"title":"Compute a Binomial Square Using Ring Operations, Not Label Arithmetic","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 7, 0, 5, 6], [2, 3, 4, 7, 6, 1, 0, 5], [3, 4, 7, 6, 5, 2, 1, 0], [4, 7, 6, 5, 0, 3, 2, 1], [5, 0, 1, 2, 3, 6, 7, 4], [6, 5, 0, 1, 2, 7, 4, 3], [7, 6, 5, 0, 1, 4, 3, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 4, 6, 0, 6, 4, 2], [0, 3, 6, 1, 4, 7, 2, 5], [0, 4, 0, 4, 0, 4, 0, 4], [0, 5, 6, 7, 4, 1, 2, 3], [0, 6, 4, 2, 0, 2, 4, 6], [0, 7, 2, 5, 4, 3, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=6, b=1. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).","answer":{"kind":"expression","value":"(6, 1, 5, 1, 4, 6, 1, 1, Yes)","explanation":"Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (6, 1, 5, 1, 4, 6, 1, 1, Yes)."},"skills":["Compute a Binomial Square Using Ring Operations, Not Label Arithmetic"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0082","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"178e22b82815db3ebf8cc37eea8a1d79f1339269eca0be9942f7892c01c2b35f","family_id":"qf-mod28-commring-binomial-square","seed":202610281082,"semantic_template":"qf-mod28-commring-binomial-square/{\"a\":7,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,5,6,4,7,0],[2,3,5,4,7,6,0,1],[3,5,4,6,0,7,1,2],[4,6,7,0,2,1,3,5],[5,4,6,7,1,0,2,3],[6,7,0,1,3,2,5,4],[7,0,1,2,5,3,4,6]],\"b\":5,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,6,2,0,5,6],[0,3,6,1,7,5,2,4],[0,4,2,7,1,5,6,3],[0,5,0,5,5,0,0,5],[0,6,5,2,6,0,5,2],[0,7,6,4,3,5,2,1]],\"n\":8}"},"fields":{"title":"Compute a Binomial Square Using Ring Operations, Not Label Arithmetic","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 6, 4, 7, 0], [2, 3, 5, 4, 7, 6, 0, 1], [3, 5, 4, 6, 0, 7, 1, 2], [4, 6, 7, 0, 2, 1, 3, 5], [5, 4, 6, 7, 1, 0, 2, 3], [6, 7, 0, 1, 3, 2, 5, 4], [7, 0, 1, 2, 5, 3, 4, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 6, 2, 0, 5, 6], [0, 3, 6, 1, 7, 5, 2, 4], [0, 4, 2, 7, 1, 5, 6, 3], [0, 5, 0, 5, 5, 0, 0, 5], [0, 6, 5, 2, 6, 0, 5, 2], [0, 7, 6, 4, 3, 5, 2, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=7, b=5. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).","answer":{"kind":"expression","value":"(7, 5, 3, 1, 1, 5, 0, 1, Yes)","explanation":"Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (7, 5, 3, 1, 1, 5, 0, 1, Yes)."},"skills":["Compute a Binomial Square Using Ring Operations, Not Label Arithmetic"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0083","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6fc9a652bc81de37d9a1b497fc70e1bf9223568f3e7d9cf419abdeb19742b076","family_id":"qf-mod28-commring-binomial-square","seed":202610281083,"semantic_template":"qf-mod28-commring-binomial-square/{\"a\":0,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,5,7,4,0,6],[2,3,5,4,6,7,1,0],[3,5,4,7,0,6,2,1],[4,7,6,0,2,1,5,3],[5,4,7,6,1,0,3,2],[6,0,1,2,5,3,7,4],[7,6,0,1,3,2,4,5]],\"b\":6,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,7,2,0,7,5],[0,3,7,1,6,5,4,2],[0,4,2,6,1,5,3,7],[0,5,0,5,5,0,5,0],[0,6,7,4,3,5,1,2],[0,7,5,2,7,0,2,5]],\"n\":8}"},"fields":{"title":"Compute a Binomial Square Using Ring Operations, Not Label Arithmetic","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 7, 4, 0, 6], [2, 3, 5, 4, 6, 7, 1, 0], [3, 5, 4, 7, 0, 6, 2, 1], [4, 7, 6, 0, 2, 1, 5, 3], [5, 4, 7, 6, 1, 0, 3, 2], [6, 0, 1, 2, 5, 3, 7, 4], [7, 6, 0, 1, 3, 2, 4, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 7, 2, 0, 7, 5], [0, 3, 7, 1, 6, 5, 4, 2], [0, 4, 2, 6, 1, 5, 3, 7], [0, 5, 0, 5, 5, 0, 5, 0], [0, 6, 7, 4, 3, 5, 1, 2], [0, 7, 5, 2, 7, 0, 2, 5]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=0, b=6. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).","answer":{"kind":"expression","value":"(0, 6, 6, 1, 0, 0, 1, 1, Yes)","explanation":"Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (0, 6, 6, 1, 0, 0, 1, 1, Yes)."},"skills":["Compute a Binomial Square Using Ring Operations, Not Label Arithmetic"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0084","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"81372a6a6d909ea67df106ab61c0ddf074a6de4c5063ad6564a0af085313fc23","family_id":"qf-mod28-commring-binomial-square","seed":202610281084,"semantic_template":"qf-mod28-commring-binomial-square/{\"a\":1,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,5,7,6,4,0],[2,3,5,6,0,4,7,1],[3,5,6,4,1,7,0,2],[4,7,0,1,5,2,3,6],[5,6,4,7,2,0,1,3],[6,4,7,0,3,1,2,5],[7,0,1,2,6,3,5,4]],\"b\":2,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,4,5,0,2,4],[0,3,4,1,2,5,7,6],[0,4,5,2,5,0,4,2],[0,5,0,5,0,0,5,5],[0,6,2,7,4,5,1,3],[0,7,4,6,2,5,3,1]],\"n\":8}"},"fields":{"title":"Compute a Binomial Square Using Ring Operations, Not Label Arithmetic","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 7, 6, 4, 0], [2, 3, 5, 6, 0, 4, 7, 1], [3, 5, 6, 4, 1, 7, 0, 2], [4, 7, 0, 1, 5, 2, 3, 6], [5, 6, 4, 7, 2, 0, 1, 3], [6, 4, 7, 0, 3, 1, 2, 5], [7, 0, 1, 2, 6, 3, 5, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 4, 5, 0, 2, 4], [0, 3, 4, 1, 2, 5, 7, 6], [0, 4, 5, 2, 5, 0, 4, 2], [0, 5, 0, 5, 0, 0, 5, 5], [0, 6, 2, 7, 4, 5, 1, 3], [0, 7, 4, 6, 2, 5, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=1, b=2. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).","answer":{"kind":"expression","value":"(1, 2, 3, 1, 1, 2, 5, 1, Yes)","explanation":"Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (1, 2, 3, 1, 1, 2, 5, 1, Yes)."},"skills":["Compute a Binomial Square Using Ring Operations, Not Label Arithmetic"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0085","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"db9959cd9c789c0be38ccd198965d1c2b150add28c5d11d3059604b764d9986d","family_id":"qf-mod28-commring-binomial-square","seed":202610281085,"semantic_template":"qf-mod28-commring-binomial-square/{\"a\":2,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,5,0,6,7,4],[2,3,5,6,1,7,4,0],[3,5,6,7,2,4,0,1],[4,0,1,2,7,3,5,6],[5,6,7,4,3,0,1,2],[6,7,4,0,5,1,2,3],[7,4,0,1,6,2,3,5]],\"b\":6,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,7,7,0,2,5],[0,3,7,1,6,5,4,2],[0,4,7,6,1,5,3,2],[0,5,0,5,5,0,5,0],[0,6,2,4,3,5,1,7],[0,7,5,2,2,0,7,5]],\"n\":8}"},"fields":{"title":"Compute a Binomial Square Using Ring Operations, Not Label Arithmetic","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 0, 6, 7, 4], [2, 3, 5, 6, 1, 7, 4, 0], [3, 5, 6, 7, 2, 4, 0, 1], [4, 0, 1, 2, 7, 3, 5, 6], [5, 6, 7, 4, 3, 0, 1, 2], [6, 7, 4, 0, 5, 1, 2, 3], [7, 4, 0, 1, 6, 2, 3, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 7, 7, 0, 2, 5], [0, 3, 7, 1, 6, 5, 4, 2], [0, 4, 7, 6, 1, 5, 3, 2], [0, 5, 0, 5, 5, 0, 5, 0], [0, 6, 2, 4, 3, 5, 1, 7], [0, 7, 5, 2, 2, 0, 7, 5]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=2, b=6. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).","answer":{"kind":"expression","value":"(2, 6, 4, 1, 5, 2, 1, 1, Yes)","explanation":"Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (2, 6, 4, 1, 5, 2, 1, 1, Yes)."},"skills":["Compute a Binomial Square Using Ring Operations, Not Label Arithmetic"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0086","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3946ce1ce1a87b8986f1220e51427429dbe553cefc274db3f86ebb20869c375e","family_id":"qf-mod28-commring-binomial-square","seed":202610281086,"semantic_template":"qf-mod28-commring-binomial-square/{\"a\":3,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,5,6,7,0,4],[2,3,5,7,0,4,1,6],[3,5,7,4,1,6,2,0],[4,6,0,1,5,2,7,3],[5,7,4,6,2,0,3,1],[6,0,1,2,7,3,4,5],[7,4,6,0,3,1,5,2]],\"b\":0,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,4,5,0,4,2],[0,3,4,1,2,5,7,6],[0,4,5,2,5,0,2,4],[0,5,0,5,0,0,5,5],[0,6,4,7,2,5,1,3],[0,7,2,6,4,5,3,1]],\"n\":8}"},"fields":{"title":"Compute a Binomial Square Using Ring Operations, Not Label Arithmetic","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 6, 7, 0, 4], [2, 3, 5, 7, 0, 4, 1, 6], [3, 5, 7, 4, 1, 6, 2, 0], [4, 6, 0, 1, 5, 2, 7, 3], [5, 7, 4, 6, 2, 0, 3, 1], [6, 0, 1, 2, 7, 3, 4, 5], [7, 4, 6, 0, 3, 1, 5, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 4, 5, 0, 4, 2], [0, 3, 4, 1, 2, 5, 7, 6], [0, 4, 5, 2, 5, 0, 2, 4], [0, 5, 0, 5, 0, 0, 5, 5], [0, 6, 4, 7, 2, 5, 1, 3], [0, 7, 2, 6, 4, 5, 3, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=3, b=0. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).","answer":{"kind":"expression","value":"(3, 0, 3, 1, 1, 0, 0, 1, Yes)","explanation":"Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (3, 0, 3, 1, 1, 0, 0, 1, Yes)."},"skills":["Compute a Binomial Square Using Ring Operations, Not Label Arithmetic"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0087","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7b9afb35d7a93bbc4fc22f6ff7f65924b78e5dcd85bccc9e3069fb901436cd00","family_id":"qf-mod28-commring-binomial-square","seed":202610281087,"semantic_template":"qf-mod28-commring-binomial-square/{\"a\":5,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,5,0,7,4,6],[2,3,5,7,1,6,0,4],[3,5,7,6,2,4,1,0],[4,0,1,2,6,3,7,5],[5,7,6,4,3,0,2,1],[6,4,0,1,7,2,5,3],[7,6,4,0,5,1,3,2]],\"b\":3,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,6,6,0,5,2],[0,3,6,1,7,5,2,4],[0,4,6,7,1,5,2,3],[0,5,0,5,5,0,0,5],[0,6,5,2,2,0,5,6],[0,7,2,4,3,5,6,1]],\"n\":8}"},"fields":{"title":"Compute a Binomial Square Using Ring Operations, Not Label Arithmetic","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 5, 0, 7, 4, 6], [2, 3, 5, 7, 1, 6, 0, 4], [3, 5, 7, 6, 2, 4, 1, 0], [4, 0, 1, 2, 6, 3, 7, 5], [5, 7, 6, 4, 3, 0, 2, 1], [6, 4, 0, 1, 7, 2, 5, 3], [7, 6, 4, 0, 5, 1, 3, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 5, 6, 6, 0, 5, 2], [0, 3, 6, 1, 7, 5, 2, 4], [0, 4, 6, 7, 1, 5, 2, 3], [0, 5, 0, 5, 5, 0, 0, 5], [0, 6, 5, 2, 2, 0, 5, 6], [0, 7, 2, 4, 3, 5, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=5, b=3. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).","answer":{"kind":"expression","value":"(5, 3, 4, 1, 0, 5, 1, 1, Yes)","explanation":"Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (5, 3, 4, 1, 0, 5, 1, 1, Yes)."},"skills":["Compute a Binomial Square Using Ring Operations, Not Label Arithmetic"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0088","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1e2765502e532f2bd2c044948b56507b6c836344593950cf9fd0ab9bc616f438","family_id":"qf-mod28-commring-binomial-square","seed":202610281088,"semantic_template":"qf-mod28-commring-binomial-square/{\"a\":4,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,6,5,7,4,0],[2,3,6,4,7,0,5,1],[3,6,4,5,0,1,7,2],[4,5,7,0,2,3,1,6],[5,7,0,1,3,6,2,4],[6,4,5,7,1,2,0,3],[7,0,1,2,6,4,3,5]],\"b\":5,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,5,2,6,0,5],[0,3,5,1,7,2,6,4],[0,4,2,7,1,5,6,3],[0,5,6,2,5,6,0,2],[0,6,0,6,6,0,0,6],[0,7,5,4,3,2,6,1]],\"n\":8}"},"fields":{"title":"Compute a Binomial Square Using Ring Operations, Not Label Arithmetic","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 5, 7, 4, 0], [2, 3, 6, 4, 7, 0, 5, 1], [3, 6, 4, 5, 0, 1, 7, 2], [4, 5, 7, 0, 2, 3, 1, 6], [5, 7, 0, 1, 3, 6, 2, 4], [6, 4, 5, 7, 1, 2, 0, 3], [7, 0, 1, 2, 6, 4, 3, 5]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 5, 2, 6, 0, 5], [0, 3, 5, 1, 7, 2, 6, 4], [0, 4, 2, 7, 1, 5, 6, 3], [0, 5, 6, 2, 5, 6, 0, 2], [0, 6, 0, 6, 6, 0, 0, 6], [0, 7, 5, 4, 3, 2, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=4, b=5. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).","answer":{"kind":"expression","value":"(4, 5, 3, 1, 1, 5, 6, 1, Yes)","explanation":"Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (4, 5, 3, 1, 1, 5, 6, 1, Yes)."},"skills":["Compute a Binomial Square Using Ring Operations, Not Label Arithmetic"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0089","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"210dd54a14830349d432f520a94d17197f67442b02594f15a8395079687f3ce9","family_id":"qf-mod28-commring-binomial-square","seed":202610281089,"semantic_template":"qf-mod28-commring-binomial-square/{\"a\":7,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,6,7,0,4,5],[2,3,6,4,5,1,7,0],[3,6,4,7,0,2,5,1],[4,7,5,0,2,6,1,3],[5,0,1,2,6,7,3,4],[6,4,7,5,1,3,0,2],[7,5,0,1,3,4,2,6]],\"b\":1,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,7,2,7,0,6],[0,3,7,1,5,4,6,2],[0,4,2,5,1,3,6,7],[0,5,7,4,3,1,6,2],[0,6,0,6,6,6,0,0],[0,7,6,2,7,2,0,6]],\"n\":8}"},"fields":{"title":"Compute a Binomial Square Using Ring Operations, Not Label Arithmetic","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 7, 0, 4, 5], [2, 3, 6, 4, 5, 1, 7, 0], [3, 6, 4, 7, 0, 2, 5, 1], [4, 7, 5, 0, 2, 6, 1, 3], [5, 0, 1, 2, 6, 7, 3, 4], [6, 4, 7, 5, 1, 3, 0, 2], [7, 5, 0, 1, 3, 4, 2, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 7, 2, 7, 0, 6], [0, 3, 7, 1, 5, 4, 6, 2], [0, 4, 2, 5, 1, 3, 6, 7], [0, 5, 7, 4, 3, 1, 6, 2], [0, 6, 0, 6, 6, 6, 0, 0], [0, 7, 6, 2, 7, 2, 0, 6]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=7, b=1. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).","answer":{"kind":"expression","value":"(7, 1, 5, 1, 6, 7, 1, 1, Yes)","explanation":"Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (7, 1, 5, 1, 6, 7, 1, 1, Yes)."},"skills":["Compute a Binomial Square Using Ring Operations, Not Label Arithmetic"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0090","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"384fd51296c2f03eea06b68929e3dc211871384ab7e3f858e9ddd64b4aa29e0b","family_id":"qf-mod28-commring-binomial-square","seed":202610281090,"semantic_template":"qf-mod28-commring-binomial-square/{\"a\":7,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,6,7,4,5,0],[2,3,6,5,0,7,4,1],[3,6,5,4,1,0,7,2],[4,7,0,1,6,3,2,5],[5,4,7,0,3,2,1,6],[6,5,4,7,2,1,0,3],[7,0,1,2,5,6,3,4]],\"b\":6,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,4,6,2,0,4],[0,3,4,1,2,7,6,5],[0,4,6,2,6,4,0,2],[0,5,2,7,4,1,6,3],[0,6,0,6,0,6,0,6],[0,7,4,5,2,3,6,1]],\"n\":8}"},"fields":{"title":"Compute a Binomial Square Using Ring Operations, Not Label Arithmetic","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 7, 4, 5, 0], [2, 3, 6, 5, 0, 7, 4, 1], [3, 6, 5, 4, 1, 0, 7, 2], [4, 7, 0, 1, 6, 3, 2, 5], [5, 4, 7, 0, 3, 2, 1, 6], [6, 5, 4, 7, 2, 1, 0, 3], [7, 0, 1, 2, 5, 6, 3, 4]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 4, 6, 2, 0, 4], [0, 3, 4, 1, 2, 7, 6, 5], [0, 4, 6, 2, 6, 4, 0, 2], [0, 5, 2, 7, 4, 1, 6, 3], [0, 6, 0, 6, 0, 6, 0, 6], [0, 7, 4, 5, 2, 3, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=7, b=6. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).","answer":{"kind":"expression","value":"(7, 6, 3, 1, 1, 6, 0, 1, Yes)","explanation":"Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (7, 6, 3, 1, 1, 6, 0, 1, Yes)."},"skills":["Compute a Binomial Square Using Ring Operations, Not Label Arithmetic"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0091","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6d06cc6604de26bb439cb6b2279525ec3f74615fcea5b8ecc37e262093c855e6","family_id":"qf-mod28-commring-binomial-square","seed":202610281091,"semantic_template":"qf-mod28-commring-binomial-square/{\"a\":0,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,6,0,7,5,4],[2,3,6,5,1,4,7,0],[3,6,5,7,2,0,4,1],[4,0,1,2,7,6,3,5],[5,7,4,0,6,2,1,3],[6,5,7,4,3,1,0,2],[7,4,0,1,5,3,2,6]],\"b\":4,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,7,7,2,0,6],[0,3,7,1,5,4,6,2],[0,4,7,5,1,3,6,2],[0,5,2,4,3,1,6,7],[0,6,0,6,6,6,0,0],[0,7,6,2,2,7,0,6]],\"n\":8}"},"fields":{"title":"Compute a Binomial Square Using Ring Operations, Not Label Arithmetic","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 0, 7, 5, 4], [2, 3, 6, 5, 1, 4, 7, 0], [3, 6, 5, 7, 2, 0, 4, 1], [4, 0, 1, 2, 7, 6, 3, 5], [5, 7, 4, 0, 6, 2, 1, 3], [6, 5, 7, 4, 3, 1, 0, 2], [7, 4, 0, 1, 5, 3, 2, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 7, 7, 2, 0, 6], [0, 3, 7, 1, 5, 4, 6, 2], [0, 4, 7, 5, 1, 3, 6, 2], [0, 5, 2, 4, 3, 1, 6, 7], [0, 6, 0, 6, 6, 6, 0, 0], [0, 7, 6, 2, 2, 7, 0, 6]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=0, b=4. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).","answer":{"kind":"expression","value":"(0, 4, 4, 1, 0, 0, 1, 1, Yes)","explanation":"Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (0, 4, 4, 1, 0, 0, 1, 1, Yes)."},"skills":["Compute a Binomial Square Using Ring Operations, Not Label Arithmetic"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0092","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2414cb0336e9fbd0d0a966c1f073521a81f9563decf54df8bbe34f97e2fb62b4","family_id":"qf-mod28-commring-binomial-square","seed":202610281092,"semantic_template":"qf-mod28-commring-binomial-square/{\"a\":1,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,6,5,0,7,4],[2,3,6,7,0,1,4,5],[3,6,7,4,1,2,5,0],[4,5,0,1,6,7,2,3],[5,0,1,2,7,4,3,6],[6,7,4,5,2,3,0,1],[7,4,5,0,3,6,1,2]],\"b\":2,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,4,6,4,0,2],[0,3,4,1,2,7,6,5],[0,4,6,2,6,2,0,4],[0,5,4,7,2,1,6,3],[0,6,0,6,0,6,0,6],[0,7,2,5,4,3,6,1]],\"n\":8}"},"fields":{"title":"Compute a Binomial Square Using Ring Operations, Not Label Arithmetic","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 5, 0, 7, 4], [2, 3, 6, 7, 0, 1, 4, 5], [3, 6, 7, 4, 1, 2, 5, 0], [4, 5, 0, 1, 6, 7, 2, 3], [5, 0, 1, 2, 7, 4, 3, 6], [6, 7, 4, 5, 2, 3, 0, 1], [7, 4, 5, 0, 3, 6, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 4, 6, 4, 0, 2], [0, 3, 4, 1, 2, 7, 6, 5], [0, 4, 6, 2, 6, 2, 0, 4], [0, 5, 4, 7, 2, 1, 6, 3], [0, 6, 0, 6, 0, 6, 0, 6], [0, 7, 2, 5, 4, 3, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=1, b=2. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).","answer":{"kind":"expression","value":"(1, 2, 3, 1, 1, 2, 6, 1, Yes)","explanation":"Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (1, 2, 3, 1, 1, 2, 6, 1, Yes)."},"skills":["Compute a Binomial Square Using Ring Operations, Not Label Arithmetic"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0093","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b152d89a5fba3832ff53db19507ca01ac51185f4a8fae03a83e11c755aafb766","family_id":"qf-mod28-commring-binomial-square","seed":202610281093,"semantic_template":"qf-mod28-commring-binomial-square/{\"a\":2,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,6,0,4,7,5],[2,3,6,7,1,0,5,4],[3,6,7,5,2,1,4,0],[4,0,1,2,5,7,3,6],[5,4,0,1,7,6,2,3],[6,7,5,4,3,2,0,1],[7,5,4,0,6,3,1,2]],\"b\":7,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,5,5,6,0,2],[0,3,5,1,7,2,6,4],[0,4,5,7,1,2,6,3],[0,5,6,2,2,6,0,5],[0,6,0,6,6,0,0,6],[0,7,2,4,3,5,6,1]],\"n\":8}"},"fields":{"title":"Compute a Binomial Square Using Ring Operations, Not Label Arithmetic","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 6, 0, 4, 7, 5], [2, 3, 6, 7, 1, 0, 5, 4], [3, 6, 7, 5, 2, 1, 4, 0], [4, 0, 1, 2, 5, 7, 3, 6], [5, 4, 0, 1, 7, 6, 2, 3], [6, 7, 5, 4, 3, 2, 0, 1], [7, 5, 4, 0, 6, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 6, 5, 5, 6, 0, 2], [0, 3, 5, 1, 7, 2, 6, 4], [0, 4, 5, 7, 1, 2, 6, 3], [0, 5, 6, 2, 2, 6, 0, 5], [0, 6, 0, 6, 6, 0, 0, 6], [0, 7, 2, 4, 3, 5, 6, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=2, b=7. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).","answer":{"kind":"expression","value":"(2, 7, 4, 1, 6, 2, 1, 1, Yes)","explanation":"Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (2, 7, 4, 1, 6, 2, 1, 1, Yes)."},"skills":["Compute a Binomial Square Using Ring Operations, Not Label Arithmetic"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0094","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0770bc0acf88c5fd2a131190f8555e82a723eb57b74a772b5d0a628137710c69","family_id":"qf-mod28-commring-binomial-square","seed":202610281094,"semantic_template":"qf-mod28-commring-binomial-square/{\"a\":3,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,7,5,6,0,4],[2,3,7,4,6,0,1,5],[3,7,4,5,0,1,2,6],[4,5,6,0,2,3,7,1],[5,6,0,1,3,7,4,2],[6,0,1,2,7,4,5,3],[7,4,5,6,1,2,3,0]],\"b\":0,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,5,2,7,5,0],[0,3,5,1,6,2,4,7],[0,4,2,6,1,5,3,7],[0,5,7,2,5,7,2,0],[0,6,5,4,3,2,1,7],[0,7,0,7,7,0,7,0]],\"n\":8}"},"fields":{"title":"Compute a Binomial Square Using Ring Operations, Not Label Arithmetic","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 5, 6, 0, 4], [2, 3, 7, 4, 6, 0, 1, 5], [3, 7, 4, 5, 0, 1, 2, 6], [4, 5, 6, 0, 2, 3, 7, 1], [5, 6, 0, 1, 3, 7, 4, 2], [6, 0, 1, 2, 7, 4, 5, 3], [7, 4, 5, 6, 1, 2, 3, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 5, 2, 7, 5, 0], [0, 3, 5, 1, 6, 2, 4, 7], [0, 4, 2, 6, 1, 5, 3, 7], [0, 5, 7, 2, 5, 7, 2, 0], [0, 6, 5, 4, 3, 2, 1, 7], [0, 7, 0, 7, 7, 0, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=3, b=0. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).","answer":{"kind":"expression","value":"(3, 0, 3, 1, 1, 0, 0, 1, Yes)","explanation":"Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (3, 0, 3, 1, 1, 0, 0, 1, Yes)."},"skills":["Compute a Binomial Square Using Ring Operations, Not Label Arithmetic"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0095","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ed8506d5fd1a3b337f35a6652727d16c22a38f19501bff594201146b7571996a","family_id":"qf-mod28-commring-binomial-square","seed":202610281095,"semantic_template":"qf-mod28-commring-binomial-square/{\"a\":7,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,7,6,0,5,4],[2,3,7,4,5,1,0,6],[3,7,4,6,0,2,1,5],[4,6,5,0,2,7,3,1],[5,0,1,2,7,6,4,3],[6,5,0,1,3,4,7,2],[7,4,6,5,1,3,2,0]],\"b\":3,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,6,2,6,7,0],[0,3,6,1,5,4,2,7],[0,4,2,5,1,3,6,7],[0,5,6,4,3,1,2,7],[0,6,7,2,6,2,7,0],[0,7,0,7,7,7,0,0]],\"n\":8}"},"fields":{"title":"Compute a Binomial Square Using Ring Operations, Not Label Arithmetic","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 6, 0, 5, 4], [2, 3, 7, 4, 5, 1, 0, 6], [3, 7, 4, 6, 0, 2, 1, 5], [4, 6, 5, 0, 2, 7, 3, 1], [5, 0, 1, 2, 7, 6, 4, 3], [6, 5, 0, 1, 3, 4, 7, 2], [7, 4, 6, 5, 1, 3, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 6, 2, 6, 7, 0], [0, 3, 6, 1, 5, 4, 2, 7], [0, 4, 2, 5, 1, 3, 6, 7], [0, 5, 6, 4, 3, 1, 2, 7], [0, 6, 7, 2, 6, 2, 7, 0], [0, 7, 0, 7, 7, 7, 0, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=7, b=3. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).","answer":{"kind":"expression","value":"(7, 3, 5, 1, 0, 7, 1, 1, Yes)","explanation":"Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (7, 3, 5, 1, 0, 7, 1, 1, Yes)."},"skills":["Compute a Binomial Square Using Ring Operations, Not Label Arithmetic"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0096","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"89c366ca3dcfc8ba4d65227ed2d9472c7165632eb08fecb47b9b84fc5adc731c","family_id":"qf-mod28-commring-binomial-square","seed":202610281096,"semantic_template":"qf-mod28-commring-binomial-square/{\"a\":5,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,7,6,4,0,5],[2,3,7,5,0,6,1,4],[3,7,5,4,1,0,2,6],[4,6,0,1,7,3,5,2],[5,4,6,0,3,2,7,1],[6,0,1,2,5,7,4,3],[7,5,4,6,2,1,3,0]],\"b\":4,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,4,7,2,4,0],[0,3,4,1,2,6,5,7],[0,4,7,2,7,4,2,0],[0,5,2,6,4,1,3,7],[0,6,4,5,2,3,1,7],[0,7,0,7,0,7,7,0]],\"n\":8}"},"fields":{"title":"Compute a Binomial Square Using Ring Operations, Not Label Arithmetic","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 6, 4, 0, 5], [2, 3, 7, 5, 0, 6, 1, 4], [3, 7, 5, 4, 1, 0, 2, 6], [4, 6, 0, 1, 7, 3, 5, 2], [5, 4, 6, 0, 3, 2, 7, 1], [6, 0, 1, 2, 5, 7, 4, 3], [7, 5, 4, 6, 2, 1, 3, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 4, 7, 2, 4, 0], [0, 3, 4, 1, 2, 6, 5, 7], [0, 4, 7, 2, 7, 4, 2, 0], [0, 5, 2, 6, 4, 1, 3, 7], [0, 6, 4, 5, 2, 3, 1, 7], [0, 7, 0, 7, 0, 7, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=5, b=4. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).","answer":{"kind":"expression","value":"(5, 4, 3, 1, 1, 4, 7, 1, Yes)","explanation":"Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (5, 4, 3, 1, 1, 4, 7, 1, Yes)."},"skills":["Compute a Binomial Square Using Ring Operations, Not Label Arithmetic"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0097","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d9b000d9fe82bd7b86e8a9a298afa8bec36bd63d598c980ff769f64954270589","family_id":"qf-mod28-commring-binomial-square","seed":202610281097,"semantic_template":"qf-mod28-commring-binomial-square/{\"a\":6,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,7,0,6,4,5],[2,3,7,5,1,4,0,6],[3,7,5,6,2,0,1,4],[4,0,1,2,6,7,5,3],[5,6,4,0,7,2,3,1],[6,4,0,1,5,3,7,2],[7,5,6,4,3,1,2,0]],\"b\":1,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,6,6,2,7,0],[0,3,6,1,5,4,2,7],[0,4,6,5,1,3,2,7],[0,5,2,4,3,1,6,7],[0,6,7,2,2,6,7,0],[0,7,0,7,7,7,0,0]],\"n\":8}"},"fields":{"title":"Compute a Binomial Square Using Ring Operations, Not Label Arithmetic","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 0, 6, 4, 5], [2, 3, 7, 5, 1, 4, 0, 6], [3, 7, 5, 6, 2, 0, 1, 4], [4, 0, 1, 2, 6, 7, 5, 3], [5, 6, 4, 0, 7, 2, 3, 1], [6, 4, 0, 1, 5, 3, 7, 2], [7, 5, 6, 4, 3, 1, 2, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 6, 6, 2, 7, 0], [0, 3, 6, 1, 5, 4, 2, 7], [0, 4, 6, 5, 1, 3, 2, 7], [0, 5, 2, 4, 3, 1, 6, 7], [0, 6, 7, 2, 2, 6, 7, 0], [0, 7, 0, 7, 7, 7, 0, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=6, b=1. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).","answer":{"kind":"expression","value":"(6, 1, 4, 1, 7, 6, 1, 1, Yes)","explanation":"Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (6, 1, 4, 1, 7, 6, 1, 1, Yes)."},"skills":["Compute a Binomial Square Using Ring Operations, Not Label Arithmetic"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0098","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d804d47ddbcc058f38242d634308977a9558e1335ba70e54011c90379c5f7970","family_id":"qf-mod28-commring-binomial-square","seed":202610281098,"semantic_template":"qf-mod28-commring-binomial-square/{\"a\":5,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,7,5,0,4,6],[2,3,7,6,0,1,5,4],[3,7,6,4,1,2,0,5],[4,5,0,1,7,6,3,2],[5,0,1,2,6,4,7,3],[6,4,5,0,3,7,2,1],[7,6,4,5,2,3,1,0]],\"b\":7,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,4,7,4,2,0],[0,3,4,1,2,6,5,7],[0,4,7,2,7,2,4,0],[0,5,4,6,2,1,3,7],[0,6,2,5,4,3,1,7],[0,7,0,7,0,7,7,0]],\"n\":8}"},"fields":{"title":"Compute a Binomial Square Using Ring Operations, Not Label Arithmetic","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 5, 0, 4, 6], [2, 3, 7, 6, 0, 1, 5, 4], [3, 7, 6, 4, 1, 2, 0, 5], [4, 5, 0, 1, 7, 6, 3, 2], [5, 0, 1, 2, 6, 4, 7, 3], [6, 4, 5, 0, 3, 7, 2, 1], [7, 6, 4, 5, 2, 3, 1, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 4, 7, 4, 2, 0], [0, 3, 4, 1, 2, 6, 5, 7], [0, 4, 7, 2, 7, 2, 4, 0], [0, 5, 4, 6, 2, 1, 3, 7], [0, 6, 2, 5, 4, 3, 1, 7], [0, 7, 0, 7, 0, 7, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=5, b=7. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).","answer":{"kind":"expression","value":"(5, 7, 3, 1, 1, 7, 0, 1, Yes)","explanation":"Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (5, 7, 3, 1, 1, 7, 0, 1, Yes)."},"skills":["Compute a Binomial Square Using Ring Operations, Not Label Arithmetic"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0099","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d77eab225bd5557e8385c535a57c61970d9d4ac5548484f4504105d9100c8190","family_id":"qf-mod28-commring-binomial-square","seed":202610281099,"semantic_template":"qf-mod28-commring-binomial-square/{\"a\":0,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,3,7,0,4,5,6],[2,3,7,6,1,0,4,5],[3,7,6,5,2,1,0,4],[4,0,1,2,5,6,7,3],[5,4,0,1,6,7,3,2],[6,5,4,0,7,3,2,1],[7,6,5,4,3,2,1,0]],\"b\":4,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,5,5,7,2,0],[0,3,5,1,6,2,4,7],[0,4,5,6,1,2,3,7],[0,5,7,2,2,7,5,0],[0,6,2,4,3,5,1,7],[0,7,0,7,7,0,7,0]],\"n\":8}"},"fields":{"title":"Compute a Binomial Square Using Ring Operations, Not Label Arithmetic","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 7, 0, 4, 5, 6], [2, 3, 7, 6, 1, 0, 4, 5], [3, 7, 6, 5, 2, 1, 0, 4], [4, 0, 1, 2, 5, 6, 7, 3], [5, 4, 0, 1, 6, 7, 3, 2], [6, 5, 4, 0, 7, 3, 2, 1], [7, 6, 5, 4, 3, 2, 1, 0]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 7, 5, 5, 7, 2, 0], [0, 3, 5, 1, 6, 2, 4, 7], [0, 4, 5, 6, 1, 2, 3, 7], [0, 5, 7, 2, 2, 7, 5, 0], [0, 6, 2, 4, 3, 5, 1, 7], [0, 7, 0, 7, 7, 0, 7, 0]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=0, b=4. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).","answer":{"kind":"expression","value":"(0, 4, 4, 1, 0, 0, 1, 1, Yes)","explanation":"Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (0, 4, 4, 1, 0, 0, 1, 1, Yes)."},"skills":["Compute a Binomial Square Using Ring Operations, Not Label Arithmetic"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261028-r01-finite-commutative-ring-0100","batch_id":"batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bd6c27d28efec1accaf87e8a3660a4f7f28ad5fb18c57328340575650ab16473","family_id":"qf-mod28-commring-binomial-square","seed":202610281100,"semantic_template":"qf-mod28-commring-binomial-square/{\"a\":1,\"addition\":[[0,1,2,3,4,5,6,7],[1,2,4,5,3,6,7,0],[2,4,3,6,5,7,0,1],[3,5,6,0,7,1,2,4],[4,3,5,7,6,0,1,2],[5,6,7,1,0,2,4,3],[6,7,0,2,1,4,3,5],[7,0,1,4,2,3,5,6]],\"b\":2,\"multiplication\":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,3,0,6,2,3,6],[0,3,0,0,3,3,0,3],[0,4,6,3,1,7,2,5],[0,5,2,3,7,1,6,4],[0,6,3,0,2,6,3,2],[0,7,6,3,5,4,2,1]],\"n\":8}"},"fields":{"title":"Compute a Binomial Square Using Ring Operations, Not Label Arithmetic","stem":"Use the same qualified complete finite unital ring on Fin(8)={0,…,7}, with addition=[[0, 1, 2, 3, 4, 5, 6, 7], [1, 2, 4, 5, 3, 6, 7, 0], [2, 4, 3, 6, 5, 7, 0, 1], [3, 5, 6, 0, 7, 1, 2, 4], [4, 3, 5, 7, 6, 0, 1, 2], [5, 6, 7, 1, 0, 2, 4, 3], [6, 7, 0, 2, 1, 4, 3, 5], [7, 0, 1, 4, 2, 3, 5, 6]] and multiplication=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7], [0, 2, 3, 0, 6, 2, 3, 6], [0, 3, 0, 0, 3, 3, 0, 3], [0, 4, 6, 3, 1, 7, 2, 5], [0, 5, 2, 3, 7, 1, 6, 4], [0, 6, 3, 0, 2, 6, 3, 2], [0, 7, 6, 3, 5, 4, 2, 1]] in label order. All unital-ring axioms, including distinct two-sided identities, must already hold. The identity labels need not be 0 and 1. Commutativity is a separate multiplication obligation; even a commutative ring may have nonzero zero divisors and need not be a field. Require commutativity. a=1, b=2. Give (a,b,sum=a+b,lhs=sum*sum,a2,ab,b2,rhs=a2+ab+ab+b2,equal?).","answer":{"kind":"expression","value":"(1, 2, 4, 1, 1, 2, 3, 1, Yes)","explanation":"Use the addition and multiplication tables throughout. The two copies of ab are added in the ring; do not substitute a carrier label called 2 for repeated addition. Zero divisors do not invalidate this identity. An exact finite or named-formula slice does not automatically establish an arbitrary infinite theorem. The exact result, retaining the requested order, is (1, 2, 4, 1, 1, 2, 3, 1, Yes)."},"skills":["Compute a Binomial Square Using Ring Operations, Not Label Arithmetic"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}