{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261030-r01-finite-field-0001","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b7c86c9b632e7e5bcec77e0732469fbea4c875467544381e2a5a08b0b18107e1","family_id":"qf-mod30-field-qualification","seed":202610301001,"semantic_template":"qf-mod30-field-qualification/{\"addition\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,4,1,3],[0,3,1,4,2],[0,4,3,2,1]],\"n\":5}"},"fields":{"title":"Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 4, 1, 3], [0, 3, 1, 4, 2], [0, 4, 3, 2, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.","answer":{"kind":"expression","value":"(5, 0, 1, [], Yes)","explanation":"Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 1, [], Yes)."},"skills":["Audit Every Nonzero Inverse after Complete 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-20261030-r01-finite-field-0002","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5d1d5ca5a2f6cef176da1c6fea1cf0730ece05ee78f3a31b3a39284cb7e88729","family_id":"qf-mod30-field-qualification","seed":202610301002,"semantic_template":"qf-mod30-field-qualification/{\"addition\":[[0,1,2,3,4],[1,2,4,0,3],[2,4,3,1,0],[3,0,1,4,2],[4,3,0,2,1]],\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,3,4,1],[0,3,4,1,2],[0,4,1,2,3]],\"n\":5}"},"fields":{"title":"Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.","answer":{"kind":"expression","value":"(5, 0, 1, [], Yes)","explanation":"Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 1, [], Yes)."},"skills":["Audit Every Nonzero Inverse after Complete 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-20261030-r01-finite-field-0003","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"93b6e882e205001edfcc2bc78dd14b03c25833e468d95b767bd068ec9f655117","family_id":"qf-mod30-field-qualification","seed":202610301003,"semantic_template":"qf-mod30-field-qualification/{\"addition\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]],\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,4,1,3],[0,3,1,4,2],[0,4,3,2,1]],\"n\":5}"},"fields":{"title":"Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 4, 1, 3], [0, 3, 1, 4, 2], [0, 4, 3, 2, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.","answer":{"kind":"expression","value":"(5, 0, 1, [], Yes)","explanation":"Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 1, [], Yes)."},"skills":["Audit Every Nonzero Inverse after Complete 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-20261030-r01-finite-field-0004","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6daa03ba8369bedd70d656567be26727c3856ce23f38dd5b781205f9b161a15f","family_id":"qf-mod30-field-qualification","seed":202610301004,"semantic_template":"qf-mod30-field-qualification/{\"addition\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]],\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,1,4,3],[0,3,4,2,1],[0,4,3,1,2]],\"n\":5}"},"fields":{"title":"Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.","answer":{"kind":"expression","value":"(5, 0, 1, [], Yes)","explanation":"Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 1, [], Yes)."},"skills":["Audit Every Nonzero Inverse after Complete 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-20261030-r01-finite-field-0005","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"33c9ed540e0922d24cb2285280e63d521c36000f3966cd107a8657b93c4bc640","family_id":"qf-mod30-field-qualification","seed":202610301005,"semantic_template":"qf-mod30-field-qualification/{\"addition\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]],\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,3,4,1],[0,3,4,1,2],[0,4,1,2,3]],\"n\":5}"},"fields":{"title":"Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.","answer":{"kind":"expression","value":"(5, 0, 1, [], Yes)","explanation":"Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 1, [], Yes)."},"skills":["Audit Every Nonzero Inverse after Complete 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-20261030-r01-finite-field-0006","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1dbba169d2a7e7ada26e71ef798a487d2be52cd1eba5a14023f0cb958b32543d","family_id":"qf-mod30-field-qualification","seed":202610301006,"semantic_template":"qf-mod30-field-qualification/{\"addition\":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]],\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,1,4,3],[0,3,4,2,1],[0,4,3,1,2]],\"n\":5}"},"fields":{"title":"Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.","answer":{"kind":"expression","value":"(5, 0, 1, [], Yes)","explanation":"Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 1, [], Yes)."},"skills":["Audit Every Nonzero Inverse after Complete 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-20261030-r01-finite-field-0007","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"79704e418cbfbfe0505a7445333fc60c96f671bccb562fc8848213ae086fd967","family_id":"qf-mod30-field-qualification","seed":202610301007,"semantic_template":"qf-mod30-field-qualification/{\"addition\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]],\"multiplication\":[[0,0,0,0,0],[0,4,1,2,3],[0,1,2,3,4],[0,2,3,4,1],[0,3,4,1,2]],\"n\":5}"},"fields":{"title":"Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.","answer":{"kind":"expression","value":"(5, 0, 2, [], Yes)","explanation":"Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 2, [], Yes)."},"skills":["Audit Every Nonzero Inverse after Complete 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-20261030-r01-finite-field-0008","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0e39c8ea005d3d59d937ffbd16a7e22bb2cc74fa945f2000c88b248a7eb37731","family_id":"qf-mod30-field-qualification","seed":202610301008,"semantic_template":"qf-mod30-field-qualification/{\"addition\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]],\"multiplication\":[[0,0,0,0,0],[0,3,1,4,2],[0,1,2,3,4],[0,4,3,2,1],[0,2,4,1,3]],\"n\":5}"},"fields":{"title":"Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 1, 4, 2], [0, 1, 2, 3, 4], [0, 4, 3, 2, 1], [0, 2, 4, 1, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.","answer":{"kind":"expression","value":"(5, 0, 2, [], Yes)","explanation":"Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 2, [], Yes)."},"skills":["Audit Every Nonzero Inverse after Complete 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-20261030-r01-finite-field-0009","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"80e68e67619d15f5ca174df3b099d732220c928ab20a9bbefe03f95c33ccef2a","family_id":"qf-mod30-field-qualification","seed":202610301009,"semantic_template":"qf-mod30-field-qualification/{\"addition\":[[0,1,2,3,4],[1,2,4,0,3],[2,4,3,1,0],[3,0,1,4,2],[4,3,0,2,1]],\"multiplication\":[[0,0,0,0,0],[0,4,1,2,3],[0,1,2,3,4],[0,2,3,4,1],[0,3,4,1,2]],\"n\":5}"},"fields":{"title":"Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.","answer":{"kind":"expression","value":"(5, 0, 2, [], Yes)","explanation":"Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 2, [], Yes)."},"skills":["Audit Every Nonzero Inverse after Complete 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-20261030-r01-finite-field-0010","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f4beb7aea46ae831de95d27d0034044017b009287498cb058a5c5662f4a3ecdd","family_id":"qf-mod30-field-qualification","seed":202610301010,"semantic_template":"qf-mod30-field-qualification/{\"addition\":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]],\"multiplication\":[[0,0,0,0,0],[0,2,1,4,3],[0,1,2,3,4],[0,4,3,1,2],[0,3,4,2,1]],\"n\":5}"},"fields":{"title":"Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.","answer":{"kind":"expression","value":"(5, 0, 2, [], Yes)","explanation":"Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 2, [], Yes)."},"skills":["Audit Every Nonzero Inverse after Complete 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-20261030-r01-finite-field-0011","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a6caffff32b219d033b02f48fb5db2bcbab96b562e2c20b5db4ef59db54b2c3c","family_id":"qf-mod30-field-qualification","seed":202610301011,"semantic_template":"qf-mod30-field-qualification/{\"addition\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"multiplication\":[[0,0,0,0,0],[0,3,1,4,2],[0,1,2,3,4],[0,4,3,2,1],[0,2,4,1,3]],\"n\":5}"},"fields":{"title":"Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 1, 4, 2], [0, 1, 2, 3, 4], [0, 4, 3, 2, 1], [0, 2, 4, 1, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.","answer":{"kind":"expression","value":"(5, 0, 2, [], Yes)","explanation":"Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 2, [], Yes)."},"skills":["Audit Every Nonzero Inverse after Complete 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-20261030-r01-finite-field-0012","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bae09ec8aec06cc68d41f244149aa4656c0f9795a580c8f6fc2803010dd19acc","family_id":"qf-mod30-field-qualification","seed":202610301012,"semantic_template":"qf-mod30-field-qualification/{\"addition\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]],\"multiplication\":[[0,0,0,0,0],[0,2,1,4,3],[0,1,2,3,4],[0,4,3,1,2],[0,3,4,2,1]],\"n\":5}"},"fields":{"title":"Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.","answer":{"kind":"expression","value":"(5, 0, 2, [], Yes)","explanation":"Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 2, [], Yes)."},"skills":["Audit Every Nonzero Inverse after Complete 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-20261030-r01-finite-field-0013","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"68609a18a273edb48bdbaae8bc52c1b7402f2a3b71a4b35db6d4e434db0e25c9","family_id":"qf-mod30-field-qualification","seed":202610301013,"semantic_template":"qf-mod30-field-qualification/{\"addition\":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]],\"multiplication\":[[0,0,0,0,0],[0,4,3,1,2],[0,3,4,2,1],[0,1,2,3,4],[0,2,1,4,3]],\"n\":5}"},"fields":{"title":"Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.","answer":{"kind":"expression","value":"(5, 0, 3, [], Yes)","explanation":"Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 3, [], Yes)."},"skills":["Audit Every Nonzero Inverse after Complete 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-20261030-r01-finite-field-0014","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2f6696bdedddea04473d6bacad1fc06fda899f7940f6b2570696e91d5babbe0e","family_id":"qf-mod30-field-qualification","seed":202610301014,"semantic_template":"qf-mod30-field-qualification/{\"addition\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"multiplication\":[[0,0,0,0,0],[0,2,4,1,3],[0,4,3,2,1],[0,1,2,3,4],[0,3,1,4,2]],\"n\":5}"},"fields":{"title":"Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 4, 1, 3], [0, 4, 3, 2, 1], [0, 1, 2, 3, 4], [0, 3, 1, 4, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.","answer":{"kind":"expression","value":"(5, 0, 3, [], Yes)","explanation":"Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 3, [], Yes)."},"skills":["Audit Every Nonzero Inverse after Complete 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-20261030-r01-finite-field-0015","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2491da70925020998a58fca35ea2707601ce4ecf78a1967a700fdaea5f8678df","family_id":"qf-mod30-field-qualification","seed":202610301015,"semantic_template":"qf-mod30-field-qualification/{\"addition\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]],\"multiplication\":[[0,0,0,0,0],[0,4,3,1,2],[0,3,4,2,1],[0,1,2,3,4],[0,2,1,4,3]],\"n\":5}"},"fields":{"title":"Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.","answer":{"kind":"expression","value":"(5, 0, 3, [], Yes)","explanation":"Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 3, [], Yes)."},"skills":["Audit Every Nonzero Inverse after Complete 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-20261030-r01-finite-field-0016","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"567be79061d9c9924ca546c91fb845a270248206dc4561fa27d7d84aa69056a5","family_id":"qf-mod30-field-qualification","seed":202610301016,"semantic_template":"qf-mod30-field-qualification/{\"addition\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]],\"multiplication\":[[0,0,0,0,0],[0,3,4,1,2],[0,4,1,2,3],[0,1,2,3,4],[0,2,3,4,1]],\"n\":5}"},"fields":{"title":"Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.","answer":{"kind":"expression","value":"(5, 0, 3, [], Yes)","explanation":"Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 3, [], Yes)."},"skills":["Audit Every Nonzero Inverse after Complete 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-20261030-r01-finite-field-0017","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"22313f15d8ecb9d76b5e4a301e5bdcc0dd903df43f1a3ae99919c848ddd462e5","family_id":"qf-mod30-field-qualification","seed":202610301017,"semantic_template":"qf-mod30-field-qualification/{\"addition\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]],\"multiplication\":[[0,0,0,0,0],[0,2,4,1,3],[0,4,3,2,1],[0,1,2,3,4],[0,3,1,4,2]],\"n\":5}"},"fields":{"title":"Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 4, 1, 3], [0, 4, 3, 2, 1], [0, 1, 2, 3, 4], [0, 3, 1, 4, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.","answer":{"kind":"expression","value":"(5, 0, 3, [], Yes)","explanation":"Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 3, [], Yes)."},"skills":["Audit Every Nonzero Inverse after Complete 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-20261030-r01-finite-field-0018","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4ca08d7a7f3102d9fc56044fb3f630a5c804cdc31d27dabd557260a597033ebb","family_id":"qf-mod30-field-qualification","seed":202610301018,"semantic_template":"qf-mod30-field-qualification/{\"addition\":[[0,1,2,3,4],[1,2,4,0,3],[2,4,3,1,0],[3,0,1,4,2],[4,3,0,2,1]],\"multiplication\":[[0,0,0,0,0],[0,3,4,1,2],[0,4,1,2,3],[0,1,2,3,4],[0,2,3,4,1]],\"n\":5}"},"fields":{"title":"Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.","answer":{"kind":"expression","value":"(5, 0, 3, [], Yes)","explanation":"Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 3, [], Yes)."},"skills":["Audit Every Nonzero Inverse after Complete 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-20261030-r01-finite-field-0019","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d77c6f7bfe295b1e497d5a310c0b26e7ee75b9ee387e5b118771eb4caf52843d","family_id":"qf-mod30-field-qualification","seed":202610301019,"semantic_template":"qf-mod30-field-qualification/{\"addition\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]],\"multiplication\":[[0,0,0,0,0],[0,3,4,2,1],[0,4,3,1,2],[0,2,1,4,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.","answer":{"kind":"expression","value":"(5, 0, 4, [], Yes)","explanation":"Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 4, [], Yes)."},"skills":["Audit Every Nonzero Inverse after Complete 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-20261030-r01-finite-field-0020","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"38c3773d94b847b4d5e99c482dbfae51042aee76f328bdb5c8d33eb16ad043ac","family_id":"qf-mod30-field-qualification","seed":202610301020,"semantic_template":"qf-mod30-field-qualification/{\"addition\":[[0,1,2,3,4],[1,2,4,0,3],[2,4,3,1,0],[3,0,1,4,2],[4,3,0,2,1]],\"multiplication\":[[0,0,0,0,0],[0,2,3,4,1],[0,3,4,1,2],[0,4,1,2,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.","answer":{"kind":"expression","value":"(5, 0, 4, [], Yes)","explanation":"Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 4, [], Yes)."},"skills":["Audit Every Nonzero Inverse after Complete 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-20261030-r01-finite-field-0021","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d864b0918bdb7ea8b3f6fdff6a870b7ede0d2a0273a5ccd7251cbb3e6f22986d","family_id":"qf-mod30-field-qualification","seed":202610301021,"semantic_template":"qf-mod30-field-qualification/{\"addition\":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]],\"multiplication\":[[0,0,0,0,0],[0,3,4,2,1],[0,4,3,1,2],[0,2,1,4,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.","answer":{"kind":"expression","value":"(5, 0, 4, [], Yes)","explanation":"Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 4, [], Yes)."},"skills":["Audit Every Nonzero Inverse after Complete 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-20261030-r01-finite-field-0022","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b11200f0477ce06a05ad9e236df56d7518f44be3118da64f09d0b2b4eb85c976","family_id":"qf-mod30-field-qualification","seed":202610301022,"semantic_template":"qf-mod30-field-qualification/{\"addition\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]],\"multiplication\":[[0,0,0,0,0],[0,4,3,2,1],[0,3,1,4,2],[0,2,4,1,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 2, 1], [0, 3, 1, 4, 2], [0, 2, 4, 1, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.","answer":{"kind":"expression","value":"(5, 0, 4, [], Yes)","explanation":"Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 4, [], Yes)."},"skills":["Audit Every Nonzero Inverse after Complete 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-20261030-r01-finite-field-0023","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5d235df9ae8e6b20bea81b55fd93d5b14d0e880d502544b90b42e45dcde44a94","family_id":"qf-mod30-field-qualification","seed":202610301023,"semantic_template":"qf-mod30-field-qualification/{\"addition\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]],\"multiplication\":[[0,0,0,0,0],[0,2,3,4,1],[0,3,4,1,2],[0,4,1,2,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.","answer":{"kind":"expression","value":"(5, 0, 4, [], Yes)","explanation":"Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 4, [], Yes)."},"skills":["Audit Every Nonzero Inverse after Complete 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-20261030-r01-finite-field-0024","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4542e39b31c34f8848f73508178f3111e294f96d77bd5144f3691fc862c54970","family_id":"qf-mod30-field-qualification","seed":202610301024,"semantic_template":"qf-mod30-field-qualification/{\"addition\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"multiplication\":[[0,0,0,0,0],[0,4,3,2,1],[0,3,1,4,2],[0,2,4,1,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 2, 1], [0, 3, 1, 4, 2], [0, 2, 4, 1, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.","answer":{"kind":"expression","value":"(5, 0, 4, [], Yes)","explanation":"Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 0, 4, [], Yes)."},"skills":["Audit Every Nonzero Inverse after Complete 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-20261030-r01-finite-field-0025","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a89f9274601a525afa9f1092226c1b0e2a1a16d66a0f6b54d25e45aea8a0cd84","family_id":"qf-mod30-field-qualification","seed":202610301025,"semantic_template":"qf-mod30-field-qualification/{\"addition\":[[2,0,3,4,1],[0,1,2,3,4],[3,2,4,1,0],[4,3,1,0,2],[1,4,0,2,3]],\"multiplication\":[[0,1,2,3,4],[1,1,1,1,1],[2,1,4,0,3],[3,1,0,4,2],[4,1,3,2,0]],\"n\":5}"},"fields":{"title":"Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]] and multiplication=[[0, 1, 2, 3, 4], [1, 1, 1, 1, 1], [2, 1, 4, 0, 3], [3, 1, 0, 4, 2], [4, 1, 3, 2, 0]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,zero,one,missing_inverses,field?). List all nonzero labels without a two-sided inverse increasingly.","answer":{"kind":"expression","value":"(5, 1, 0, [], Yes)","explanation":"Find both actual identities and search all labels for each nonzero element's two-sided inverse. Missing inverses refute field qualification even when the parent is a commutative ring. The exact result, retaining the requested order, is (5, 1, 0, [], Yes)."},"skills":["Audit Every Nonzero Inverse after Complete 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-20261030-r01-finite-field-0026","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bdf126e11799598470ed8911d9c02108192fb95c2c02ed8556c7b5c4d9bea8af","family_id":"qf-mod30-field-inverse","seed":202610301026,"semantic_template":"qf-mod30-field-inverse/{\"addition\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"element\":1,\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,4,1,3],[0,3,1,4,2],[0,4,3,2,1]],\"n\":5}"},"fields":{"title":"Find a Two-Sided Inverse of the Specified Nonzero Element","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 4, 1, 3], [0, 3, 1, 4, 2], [0, 4, 3, 2, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=1 is nonzero. Give (element,one,inverse,found?).","answer":{"kind":"expression","value":"(1, 1, 1, Yes)","explanation":"Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (1, 1, 1, Yes)."},"skills":["Find a Two-Sided Inverse of the Specified Nonzero Element"],"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-20261030-r01-finite-field-0027","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fbe9ebd8230781c7bb53bf6f8c322f4be16c725ebacdd27d55dff546c9f88c6c","family_id":"qf-mod30-field-inverse","seed":202610301027,"semantic_template":"qf-mod30-field-inverse/{\"addition\":[[0,1,2,3,4],[1,2,4,0,3],[2,4,3,1,0],[3,0,1,4,2],[4,3,0,2,1]],\"element\":2,\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,3,4,1],[0,3,4,1,2],[0,4,1,2,3]],\"n\":5}"},"fields":{"title":"Find a Two-Sided Inverse of the Specified Nonzero Element","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=2 is nonzero. Give (element,one,inverse,found?).","answer":{"kind":"expression","value":"(2, 1, 4, Yes)","explanation":"Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (2, 1, 4, Yes)."},"skills":["Find a Two-Sided Inverse of the Specified Nonzero Element"],"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-20261030-r01-finite-field-0028","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c975fbe4d11ffd2739d965aa6447403ca716f8f6821488d86b0e35e3963f4111","family_id":"qf-mod30-field-inverse","seed":202610301028,"semantic_template":"qf-mod30-field-inverse/{\"addition\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]],\"element\":2,\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,4,1,3],[0,3,1,4,2],[0,4,3,2,1]],\"n\":5}"},"fields":{"title":"Find a Two-Sided Inverse of the Specified Nonzero Element","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 4, 1, 3], [0, 3, 1, 4, 2], [0, 4, 3, 2, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=2 is nonzero. Give (element,one,inverse,found?).","answer":{"kind":"expression","value":"(2, 1, 3, Yes)","explanation":"Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (2, 1, 3, Yes)."},"skills":["Find a Two-Sided Inverse of the Specified Nonzero Element"],"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-20261030-r01-finite-field-0029","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c952a757e11e71e400bcbb31844b9219a8752c449207453dd88bac288ff9e7b6","family_id":"qf-mod30-field-inverse","seed":202610301029,"semantic_template":"qf-mod30-field-inverse/{\"addition\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]],\"element\":2,\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,1,4,3],[0,3,4,2,1],[0,4,3,1,2]],\"n\":5}"},"fields":{"title":"Find a Two-Sided Inverse of the Specified Nonzero Element","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=2 is nonzero. Give (element,one,inverse,found?).","answer":{"kind":"expression","value":"(2, 1, 2, Yes)","explanation":"Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (2, 1, 2, Yes)."},"skills":["Find a Two-Sided Inverse of the Specified Nonzero Element"],"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-20261030-r01-finite-field-0030","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8b751a4b3231bfaad995d56e93159f4fb3d4e06f5c3dde3b56a70064af34476c","family_id":"qf-mod30-field-inverse","seed":202610301030,"semantic_template":"qf-mod30-field-inverse/{\"addition\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]],\"element\":1,\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,3,4,1],[0,3,4,1,2],[0,4,1,2,3]],\"n\":5}"},"fields":{"title":"Find a Two-Sided Inverse of the Specified Nonzero Element","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=1 is nonzero. Give (element,one,inverse,found?).","answer":{"kind":"expression","value":"(1, 1, 1, Yes)","explanation":"Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (1, 1, 1, Yes)."},"skills":["Find a Two-Sided Inverse of the Specified Nonzero Element"],"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-20261030-r01-finite-field-0031","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1f2c7c765010a66d2cf5ad6c3e8b34ac7938cdae4b5a8711b7fe4b7f74ef7ce5","family_id":"qf-mod30-field-inverse","seed":202610301031,"semantic_template":"qf-mod30-field-inverse/{\"addition\":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]],\"element\":4,\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,1,4,3],[0,3,4,2,1],[0,4,3,1,2]],\"n\":5}"},"fields":{"title":"Find a Two-Sided Inverse of the Specified Nonzero Element","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=4 is nonzero. Give (element,one,inverse,found?).","answer":{"kind":"expression","value":"(4, 1, 3, Yes)","explanation":"Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (4, 1, 3, Yes)."},"skills":["Find a Two-Sided Inverse of the Specified Nonzero Element"],"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-20261030-r01-finite-field-0032","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5b57772fac2dca6e73d1c48ad57fc61e333536ec2973467caa94079189179436","family_id":"qf-mod30-field-inverse","seed":202610301032,"semantic_template":"qf-mod30-field-inverse/{\"addition\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]],\"element\":3,\"multiplication\":[[0,0,0,0,0],[0,4,1,2,3],[0,1,2,3,4],[0,2,3,4,1],[0,3,4,1,2]],\"n\":5}"},"fields":{"title":"Find a Two-Sided Inverse of the Specified Nonzero Element","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=3 is nonzero. Give (element,one,inverse,found?).","answer":{"kind":"expression","value":"(3, 2, 1, Yes)","explanation":"Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (3, 2, 1, Yes)."},"skills":["Find a Two-Sided Inverse of the Specified Nonzero Element"],"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-20261030-r01-finite-field-0033","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3c2c392a0085de8e5e8744ad7d507f6fd2b29c0717bcdb49d555b7c64caaa9e3","family_id":"qf-mod30-field-inverse","seed":202610301033,"semantic_template":"qf-mod30-field-inverse/{\"addition\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]],\"element\":3,\"multiplication\":[[0,0,0,0,0],[0,3,1,4,2],[0,1,2,3,4],[0,4,3,2,1],[0,2,4,1,3]],\"n\":5}"},"fields":{"title":"Find a Two-Sided Inverse of the Specified Nonzero Element","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 1, 4, 2], [0, 1, 2, 3, 4], [0, 4, 3, 2, 1], [0, 2, 4, 1, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=3 is nonzero. Give (element,one,inverse,found?).","answer":{"kind":"expression","value":"(3, 2, 3, Yes)","explanation":"Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (3, 2, 3, Yes)."},"skills":["Find a Two-Sided Inverse of the Specified Nonzero Element"],"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-20261030-r01-finite-field-0034","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5570dbefbbedfdb8c1b7fee748d090629ea3336f178d9feac6462013c9cf75f0","family_id":"qf-mod30-field-inverse","seed":202610301034,"semantic_template":"qf-mod30-field-inverse/{\"addition\":[[0,1,2,3,4],[1,2,4,0,3],[2,4,3,1,0],[3,0,1,4,2],[4,3,0,2,1]],\"element\":2,\"multiplication\":[[0,0,0,0,0],[0,4,1,2,3],[0,1,2,3,4],[0,2,3,4,1],[0,3,4,1,2]],\"n\":5}"},"fields":{"title":"Find a Two-Sided Inverse of the Specified Nonzero Element","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=2 is nonzero. Give (element,one,inverse,found?).","answer":{"kind":"expression","value":"(2, 2, 2, Yes)","explanation":"Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (2, 2, 2, Yes)."},"skills":["Find a Two-Sided Inverse of the Specified Nonzero Element"],"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-20261030-r01-finite-field-0035","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ff2cc665fc6f427b7040267b0266722facfa56070532354a75bd86330a805f39","family_id":"qf-mod30-field-inverse","seed":202610301035,"semantic_template":"qf-mod30-field-inverse/{\"addition\":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]],\"element\":3,\"multiplication\":[[0,0,0,0,0],[0,2,1,4,3],[0,1,2,3,4],[0,4,3,1,2],[0,3,4,2,1]],\"n\":5}"},"fields":{"title":"Find a Two-Sided Inverse of the Specified Nonzero Element","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=3 is nonzero. Give (element,one,inverse,found?).","answer":{"kind":"expression","value":"(3, 2, 4, Yes)","explanation":"Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (3, 2, 4, Yes)."},"skills":["Find a Two-Sided Inverse of the Specified Nonzero Element"],"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-20261030-r01-finite-field-0036","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"da01cb64062c5ba4afefe44c9ceff26329183b2c2845d9b48aae880ba7bb62b7","family_id":"qf-mod30-field-inverse","seed":202610301036,"semantic_template":"qf-mod30-field-inverse/{\"addition\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"element\":1,\"multiplication\":[[0,0,0,0,0],[0,3,1,4,2],[0,1,2,3,4],[0,4,3,2,1],[0,2,4,1,3]],\"n\":5}"},"fields":{"title":"Find a Two-Sided Inverse of the Specified Nonzero Element","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 1, 4, 2], [0, 1, 2, 3, 4], [0, 4, 3, 2, 1], [0, 2, 4, 1, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=1 is nonzero. Give (element,one,inverse,found?).","answer":{"kind":"expression","value":"(1, 2, 4, Yes)","explanation":"Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (1, 2, 4, Yes)."},"skills":["Find a Two-Sided Inverse of the Specified Nonzero Element"],"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-20261030-r01-finite-field-0037","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a52cf0cf5c9cdf4fe211bb69e7f78bee5caf05a9d725e52d6aa00382477d623b","family_id":"qf-mod30-field-inverse","seed":202610301037,"semantic_template":"qf-mod30-field-inverse/{\"addition\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]],\"element\":1,\"multiplication\":[[0,0,0,0,0],[0,2,1,4,3],[0,1,2,3,4],[0,4,3,1,2],[0,3,4,2,1]],\"n\":5}"},"fields":{"title":"Find a Two-Sided Inverse of the Specified Nonzero Element","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=1 is nonzero. Give (element,one,inverse,found?).","answer":{"kind":"expression","value":"(1, 2, 1, Yes)","explanation":"Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (1, 2, 1, Yes)."},"skills":["Find a Two-Sided Inverse of the Specified Nonzero Element"],"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-20261030-r01-finite-field-0038","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4456d8f601e3368010519f8c61a6d78e0ce64840751023521eafb5e7104bd2d3","family_id":"qf-mod30-field-inverse","seed":202610301038,"semantic_template":"qf-mod30-field-inverse/{\"addition\":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]],\"element\":3,\"multiplication\":[[0,0,0,0,0],[0,4,3,1,2],[0,3,4,2,1],[0,1,2,3,4],[0,2,1,4,3]],\"n\":5}"},"fields":{"title":"Find a Two-Sided Inverse of the Specified Nonzero Element","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=3 is nonzero. Give (element,one,inverse,found?).","answer":{"kind":"expression","value":"(3, 3, 3, Yes)","explanation":"Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (3, 3, 3, Yes)."},"skills":["Find a Two-Sided Inverse of the Specified Nonzero Element"],"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-20261030-r01-finite-field-0039","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3d22bc21511353b34041acf5388df224a85f7019e66a4d244200419cfee21df3","family_id":"qf-mod30-field-inverse","seed":202610301039,"semantic_template":"qf-mod30-field-inverse/{\"addition\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"element\":1,\"multiplication\":[[0,0,0,0,0],[0,2,4,1,3],[0,4,3,2,1],[0,1,2,3,4],[0,3,1,4,2]],\"n\":5}"},"fields":{"title":"Find a Two-Sided Inverse of the Specified Nonzero Element","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 4, 1, 3], [0, 4, 3, 2, 1], [0, 1, 2, 3, 4], [0, 3, 1, 4, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=1 is nonzero. Give (element,one,inverse,found?).","answer":{"kind":"expression","value":"(1, 3, 4, Yes)","explanation":"Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (1, 3, 4, Yes)."},"skills":["Find a Two-Sided Inverse of the Specified Nonzero Element"],"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-20261030-r01-finite-field-0040","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1061730b44afe3c5ed763d6a7ed711ca664cda391e612ac7166ebc8b0334737d","family_id":"qf-mod30-field-inverse","seed":202610301040,"semantic_template":"qf-mod30-field-inverse/{\"addition\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]],\"element\":1,\"multiplication\":[[0,0,0,0,0],[0,4,3,1,2],[0,3,4,2,1],[0,1,2,3,4],[0,2,1,4,3]],\"n\":5}"},"fields":{"title":"Find a Two-Sided Inverse of the Specified Nonzero Element","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=1 is nonzero. Give (element,one,inverse,found?).","answer":{"kind":"expression","value":"(1, 3, 2, Yes)","explanation":"Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (1, 3, 2, Yes)."},"skills":["Find a Two-Sided Inverse of the Specified Nonzero Element"],"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-20261030-r01-finite-field-0041","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2258a6aa679464a03b7afcc86c15631232658ff48d43d4031d0f4ebca099af35","family_id":"qf-mod30-field-inverse","seed":202610301041,"semantic_template":"qf-mod30-field-inverse/{\"addition\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]],\"element\":1,\"multiplication\":[[0,0,0,0,0],[0,3,4,1,2],[0,4,1,2,3],[0,1,2,3,4],[0,2,3,4,1]],\"n\":5}"},"fields":{"title":"Find a Two-Sided Inverse of the Specified Nonzero Element","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=1 is nonzero. Give (element,one,inverse,found?).","answer":{"kind":"expression","value":"(1, 3, 1, Yes)","explanation":"Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (1, 3, 1, Yes)."},"skills":["Find a Two-Sided Inverse of the Specified Nonzero Element"],"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-20261030-r01-finite-field-0042","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4ae640462d300f92b3dbfc9a42c88b8eb89f6eb7a3fa3f59055110a7218efd02","family_id":"qf-mod30-field-inverse","seed":202610301042,"semantic_template":"qf-mod30-field-inverse/{\"addition\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]],\"element\":3,\"multiplication\":[[0,0,0,0,0],[0,2,4,1,3],[0,4,3,2,1],[0,1,2,3,4],[0,3,1,4,2]],\"n\":5}"},"fields":{"title":"Find a Two-Sided Inverse of the Specified Nonzero Element","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 4, 1, 3], [0, 4, 3, 2, 1], [0, 1, 2, 3, 4], [0, 3, 1, 4, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=3 is nonzero. Give (element,one,inverse,found?).","answer":{"kind":"expression","value":"(3, 3, 3, Yes)","explanation":"Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (3, 3, 3, Yes)."},"skills":["Find a Two-Sided Inverse of the Specified Nonzero Element"],"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-20261030-r01-finite-field-0043","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9a24c23d932a374d2ab0b5f0271bfe68ead4eef90ab62b8e91b62f9d84813ca4","family_id":"qf-mod30-field-inverse","seed":202610301043,"semantic_template":"qf-mod30-field-inverse/{\"addition\":[[0,1,2,3,4],[1,2,4,0,3],[2,4,3,1,0],[3,0,1,4,2],[4,3,0,2,1]],\"element\":4,\"multiplication\":[[0,0,0,0,0],[0,3,4,1,2],[0,4,1,2,3],[0,1,2,3,4],[0,2,3,4,1]],\"n\":5}"},"fields":{"title":"Find a Two-Sided Inverse of the Specified Nonzero Element","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=4 is nonzero. Give (element,one,inverse,found?).","answer":{"kind":"expression","value":"(4, 3, 2, Yes)","explanation":"Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (4, 3, 2, Yes)."},"skills":["Find a Two-Sided Inverse of the Specified Nonzero Element"],"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-20261030-r01-finite-field-0044","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a2389db763eda376ea68d50bcf739cedfa62bc42c7708170b7b1cddb2e034732","family_id":"qf-mod30-field-inverse","seed":202610301044,"semantic_template":"qf-mod30-field-inverse/{\"addition\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]],\"element\":2,\"multiplication\":[[0,0,0,0,0],[0,3,4,2,1],[0,4,3,1,2],[0,2,1,4,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Find a Two-Sided Inverse of the Specified Nonzero Element","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=2 is nonzero. Give (element,one,inverse,found?).","answer":{"kind":"expression","value":"(2, 4, 1, Yes)","explanation":"Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (2, 4, 1, Yes)."},"skills":["Find a Two-Sided Inverse of the Specified Nonzero Element"],"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-20261030-r01-finite-field-0045","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"30c54a98c4e76cc7b1212e97775c71d1362405ecadc2f192cf5439d55ec1e0ca","family_id":"qf-mod30-field-inverse","seed":202610301045,"semantic_template":"qf-mod30-field-inverse/{\"addition\":[[0,1,2,3,4],[1,2,4,0,3],[2,4,3,1,0],[3,0,1,4,2],[4,3,0,2,1]],\"element\":2,\"multiplication\":[[0,0,0,0,0],[0,2,3,4,1],[0,3,4,1,2],[0,4,1,2,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Find a Two-Sided Inverse of the Specified Nonzero Element","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=2 is nonzero. Give (element,one,inverse,found?).","answer":{"kind":"expression","value":"(2, 4, 2, Yes)","explanation":"Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (2, 4, 2, Yes)."},"skills":["Find a Two-Sided Inverse of the Specified Nonzero Element"],"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-20261030-r01-finite-field-0046","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d4d12563d265c0eb994ae428dccc5fa1ea8614b10f545cdaf8caa834ee0d3a60","family_id":"qf-mod30-field-inverse","seed":202610301046,"semantic_template":"qf-mod30-field-inverse/{\"addition\":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]],\"element\":4,\"multiplication\":[[0,0,0,0,0],[0,3,4,2,1],[0,4,3,1,2],[0,2,1,4,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Find a Two-Sided Inverse of the Specified Nonzero Element","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=4 is nonzero. Give (element,one,inverse,found?).","answer":{"kind":"expression","value":"(4, 4, 4, Yes)","explanation":"Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (4, 4, 4, Yes)."},"skills":["Find a Two-Sided Inverse of the Specified Nonzero Element"],"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-20261030-r01-finite-field-0047","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2f63b5f687b8cfd3b0323fe198fa0dcb0d73fd79a7a48b9dfd993bdc3d7292da","family_id":"qf-mod30-field-inverse","seed":202610301047,"semantic_template":"qf-mod30-field-inverse/{\"addition\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]],\"element\":2,\"multiplication\":[[0,0,0,0,0],[0,4,3,2,1],[0,3,1,4,2],[0,2,4,1,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Find a Two-Sided Inverse of the Specified Nonzero Element","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 2, 1], [0, 3, 1, 4, 2], [0, 2, 4, 1, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=2 is nonzero. Give (element,one,inverse,found?).","answer":{"kind":"expression","value":"(2, 4, 3, Yes)","explanation":"Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (2, 4, 3, Yes)."},"skills":["Find a Two-Sided Inverse of the Specified Nonzero Element"],"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-20261030-r01-finite-field-0048","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dd29c7e5d9f459fcce6d278347c6d2bfbcd7e181f87e5217512555f78d9d185b","family_id":"qf-mod30-field-inverse","seed":202610301048,"semantic_template":"qf-mod30-field-inverse/{\"addition\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]],\"element\":1,\"multiplication\":[[0,0,0,0,0],[0,2,3,4,1],[0,3,4,1,2],[0,4,1,2,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Find a Two-Sided Inverse of the Specified Nonzero Element","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=1 is nonzero. Give (element,one,inverse,found?).","answer":{"kind":"expression","value":"(1, 4, 3, Yes)","explanation":"Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (1, 4, 3, Yes)."},"skills":["Find a Two-Sided Inverse of the Specified Nonzero Element"],"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-20261030-r01-finite-field-0049","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"39dcec66d6db991a352925fa8bb83f77758381a54bca6d95f83519194eac50be","family_id":"qf-mod30-field-inverse","seed":202610301049,"semantic_template":"qf-mod30-field-inverse/{\"addition\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"element\":1,\"multiplication\":[[0,0,0,0,0],[0,4,3,2,1],[0,3,1,4,2],[0,2,4,1,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Find a Two-Sided Inverse of the Specified Nonzero Element","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 2, 1], [0, 3, 1, 4, 2], [0, 2, 4, 1, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=1 is nonzero. Give (element,one,inverse,found?).","answer":{"kind":"expression","value":"(1, 4, 1, Yes)","explanation":"Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (1, 4, 1, Yes)."},"skills":["Find a Two-Sided Inverse of the Specified Nonzero Element"],"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-20261030-r01-finite-field-0050","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7be556561117a004e97d3a69b34a64b62793d7e2be50b1839a16e92267e7ded3","family_id":"qf-mod30-field-inverse","seed":202610301050,"semantic_template":"qf-mod30-field-inverse/{\"addition\":[[2,0,3,4,1],[0,1,2,3,4],[3,2,4,1,0],[4,3,1,0,2],[1,4,0,2,3]],\"element\":0,\"multiplication\":[[0,1,2,3,4],[1,1,1,1,1],[2,1,4,0,3],[3,1,0,4,2],[4,1,3,2,0]],\"n\":5}"},"fields":{"title":"Find a Two-Sided Inverse of the Specified Nonzero Element","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]] and multiplication=[[0, 1, 2, 3, 4], [1, 1, 1, 1, 1], [2, 1, 4, 0, 3], [3, 1, 0, 4, 2], [4, 1, 3, 2, 0]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. element=0 is nonzero. Give (element,one,inverse,found?).","answer":{"kind":"expression","value":"(0, 0, 0, Yes)","explanation":"Search carrier labels increasingly and return the first element whose left and right products are the actual multiplicative identity. The field premise provides existence; zero is rejected rather than assigned an inverse. The exact result, retaining the requested order, is (0, 0, 0, Yes)."},"skills":["Find a Two-Sided Inverse of the Specified Nonzero Element"],"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-20261030-r01-finite-field-0051","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b0c3de643b59f9c641f1af67cf96d670f0d36911254e3cf8c2e2735a86d26e3a","family_id":"qf-mod30-field-division","seed":202610301051,"semantic_template":"qf-mod30-field-division/{\"addition\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"denominator\":1,\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,4,1,3],[0,3,1,4,2],[0,4,3,2,1]],\"n\":5,\"numerator\":1}"},"fields":{"title":"Field Division Uses a Nonzero Denominator and the Same Table","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 4, 1, 3], [0, 3, 1, 4, 2], [0, 4, 3, 2, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=1, denominator=1 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).","answer":{"kind":"expression","value":"(1, 1, 1, 1, 1, Yes)","explanation":"Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (1, 1, 1, 1, 1, Yes)."},"skills":["Field Division Uses a Nonzero Denominator and the Same Table"],"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-20261030-r01-finite-field-0052","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cd28daf312d20b3e232c0b7a2e26ec40dc0e906ac8e458905a8a7e95f08c2762","family_id":"qf-mod30-field-division","seed":202610301052,"semantic_template":"qf-mod30-field-division/{\"addition\":[[0,1,2,3,4],[1,2,4,0,3],[2,4,3,1,0],[3,0,1,4,2],[4,3,0,2,1]],\"denominator\":2,\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,3,4,1],[0,3,4,1,2],[0,4,1,2,3]],\"n\":5,\"numerator\":2}"},"fields":{"title":"Field Division Uses a Nonzero Denominator and the Same Table","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=2, denominator=2 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).","answer":{"kind":"expression","value":"(2, 2, 4, 1, 2, Yes)","explanation":"Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (2, 2, 4, 1, 2, Yes)."},"skills":["Field Division Uses a Nonzero Denominator and the Same Table"],"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-20261030-r01-finite-field-0053","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a3214b75322d09eba68c624b47509b781d252bdd1557fa95dff9915e1da387ad","family_id":"qf-mod30-field-division","seed":202610301053,"semantic_template":"qf-mod30-field-division/{\"addition\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]],\"denominator\":2,\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,4,1,3],[0,3,1,4,2],[0,4,3,2,1]],\"n\":5,\"numerator\":2}"},"fields":{"title":"Field Division Uses a Nonzero Denominator and the Same Table","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 4, 1, 3], [0, 3, 1, 4, 2], [0, 4, 3, 2, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=2, denominator=2 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).","answer":{"kind":"expression","value":"(2, 2, 3, 1, 2, Yes)","explanation":"Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (2, 2, 3, 1, 2, Yes)."},"skills":["Field Division Uses a Nonzero Denominator and the Same Table"],"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-20261030-r01-finite-field-0054","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2aa96bb57babd9a090f6fb25007f1b9db30f3bde39a9efc3cfdc72b94e6915b3","family_id":"qf-mod30-field-division","seed":202610301054,"semantic_template":"qf-mod30-field-division/{\"addition\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]],\"denominator\":2,\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,1,4,3],[0,3,4,2,1],[0,4,3,1,2]],\"n\":5,\"numerator\":2}"},"fields":{"title":"Field Division Uses a Nonzero Denominator and the Same Table","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=2, denominator=2 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).","answer":{"kind":"expression","value":"(2, 2, 2, 1, 2, Yes)","explanation":"Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (2, 2, 2, 1, 2, Yes)."},"skills":["Field Division Uses a Nonzero Denominator and the Same Table"],"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-20261030-r01-finite-field-0055","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3de371c01eba31515f28c7b2de743a2a2e9b78cdfe49ff0ce0ce4370b2dfc422","family_id":"qf-mod30-field-division","seed":202610301055,"semantic_template":"qf-mod30-field-division/{\"addition\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]],\"denominator\":1,\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,3,4,1],[0,3,4,1,2],[0,4,1,2,3]],\"n\":5,\"numerator\":0}"},"fields":{"title":"Field Division Uses a Nonzero Denominator and the Same Table","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=0, denominator=1 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).","answer":{"kind":"expression","value":"(0, 1, 1, 0, 0, Yes)","explanation":"Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (0, 1, 1, 0, 0, Yes)."},"skills":["Field Division Uses a Nonzero Denominator and the Same Table"],"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-20261030-r01-finite-field-0056","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"28eddfb47d1e66fb9e3778f53151e41c8fbbe86f9890763d06086fd1d9b0f129","family_id":"qf-mod30-field-division","seed":202610301056,"semantic_template":"qf-mod30-field-division/{\"addition\":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]],\"denominator\":4,\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,1,4,3],[0,3,4,2,1],[0,4,3,1,2]],\"n\":5,\"numerator\":1}"},"fields":{"title":"Field Division Uses a Nonzero Denominator and the Same Table","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=1, denominator=4 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).","answer":{"kind":"expression","value":"(1, 4, 3, 3, 1, Yes)","explanation":"Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (1, 4, 3, 3, 1, Yes)."},"skills":["Field Division Uses a Nonzero Denominator and the Same Table"],"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-20261030-r01-finite-field-0057","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d0424188f95ad7e4c95bd4b37be51df2a6199fae24dc6d9054ce6f75aba60167","family_id":"qf-mod30-field-division","seed":202610301057,"semantic_template":"qf-mod30-field-division/{\"addition\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]],\"denominator\":3,\"multiplication\":[[0,0,0,0,0],[0,4,1,2,3],[0,1,2,3,4],[0,2,3,4,1],[0,3,4,1,2]],\"n\":5,\"numerator\":1}"},"fields":{"title":"Field Division Uses a Nonzero Denominator and the Same Table","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=1, denominator=3 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).","answer":{"kind":"expression","value":"(1, 3, 1, 4, 1, Yes)","explanation":"Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (1, 3, 1, 4, 1, Yes)."},"skills":["Field Division Uses a Nonzero Denominator and the Same Table"],"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-20261030-r01-finite-field-0058","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5456381a36c6ca827324c913d7ffe260efaa9ddd933360f54bef0d2bf27ca7a5","family_id":"qf-mod30-field-division","seed":202610301058,"semantic_template":"qf-mod30-field-division/{\"addition\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]],\"denominator\":3,\"multiplication\":[[0,0,0,0,0],[0,3,1,4,2],[0,1,2,3,4],[0,4,3,2,1],[0,2,4,1,3]],\"n\":5,\"numerator\":4}"},"fields":{"title":"Field Division Uses a Nonzero Denominator and the Same Table","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 1, 4, 2], [0, 1, 2, 3, 4], [0, 4, 3, 2, 1], [0, 2, 4, 1, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=4, denominator=3 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).","answer":{"kind":"expression","value":"(4, 3, 3, 1, 4, Yes)","explanation":"Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (4, 3, 3, 1, 4, Yes)."},"skills":["Field Division Uses a Nonzero Denominator and the Same Table"],"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-20261030-r01-finite-field-0059","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e8f575f6c899a62059e34c432463d0cb7d066c493cee6ca364f6aab5a4893629","family_id":"qf-mod30-field-division","seed":202610301059,"semantic_template":"qf-mod30-field-division/{\"addition\":[[0,1,2,3,4],[1,2,4,0,3],[2,4,3,1,0],[3,0,1,4,2],[4,3,0,2,1]],\"denominator\":2,\"multiplication\":[[0,0,0,0,0],[0,4,1,2,3],[0,1,2,3,4],[0,2,3,4,1],[0,3,4,1,2]],\"n\":5,\"numerator\":4}"},"fields":{"title":"Field Division Uses a Nonzero Denominator and the Same Table","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=4, denominator=2 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).","answer":{"kind":"expression","value":"(4, 2, 2, 4, 4, Yes)","explanation":"Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (4, 2, 2, 4, 4, Yes)."},"skills":["Field Division Uses a Nonzero Denominator and the Same Table"],"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-20261030-r01-finite-field-0060","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7ca383203c956b13e90c57221f9ced05ebe7f0bf3cb1807c495b8950443aaa55","family_id":"qf-mod30-field-division","seed":202610301060,"semantic_template":"qf-mod30-field-division/{\"addition\":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]],\"denominator\":3,\"multiplication\":[[0,0,0,0,0],[0,2,1,4,3],[0,1,2,3,4],[0,4,3,1,2],[0,3,4,2,1]],\"n\":5,\"numerator\":0}"},"fields":{"title":"Field Division Uses a Nonzero Denominator and the Same Table","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=0, denominator=3 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).","answer":{"kind":"expression","value":"(0, 3, 4, 0, 0, Yes)","explanation":"Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (0, 3, 4, 0, 0, Yes)."},"skills":["Field Division Uses a Nonzero Denominator and the Same Table"],"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-20261030-r01-finite-field-0061","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9044183f2c469032b440877c77f7a12f63b6c13146534aef468d1d8925c7e00b","family_id":"qf-mod30-field-division","seed":202610301061,"semantic_template":"qf-mod30-field-division/{\"addition\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"denominator\":1,\"multiplication\":[[0,0,0,0,0],[0,3,1,4,2],[0,1,2,3,4],[0,4,3,2,1],[0,2,4,1,3]],\"n\":5,\"numerator\":2}"},"fields":{"title":"Field Division Uses a Nonzero Denominator and the Same Table","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 1, 4, 2], [0, 1, 2, 3, 4], [0, 4, 3, 2, 1], [0, 2, 4, 1, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=2, denominator=1 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).","answer":{"kind":"expression","value":"(2, 1, 4, 4, 2, Yes)","explanation":"Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (2, 1, 4, 4, 2, Yes)."},"skills":["Field Division Uses a Nonzero Denominator and the Same Table"],"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-20261030-r01-finite-field-0062","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a5fde92e125a480463b1aa092611742d5e99a258398161e5acd39d61f515c62e","family_id":"qf-mod30-field-division","seed":202610301062,"semantic_template":"qf-mod30-field-division/{\"addition\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]],\"denominator\":1,\"multiplication\":[[0,0,0,0,0],[0,2,1,4,3],[0,1,2,3,4],[0,4,3,1,2],[0,3,4,2,1]],\"n\":5,\"numerator\":4}"},"fields":{"title":"Field Division Uses a Nonzero Denominator and the Same Table","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=4, denominator=1 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).","answer":{"kind":"expression","value":"(4, 1, 1, 3, 4, Yes)","explanation":"Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (4, 1, 1, 3, 4, Yes)."},"skills":["Field Division Uses a Nonzero Denominator and the Same Table"],"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-20261030-r01-finite-field-0063","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b900521df2cfaa761b7b1797406a8b7faf6d227fd035ed20184c3fcb87a31998","family_id":"qf-mod30-field-division","seed":202610301063,"semantic_template":"qf-mod30-field-division/{\"addition\":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]],\"denominator\":3,\"multiplication\":[[0,0,0,0,0],[0,4,3,1,2],[0,3,4,2,1],[0,1,2,3,4],[0,2,1,4,3]],\"n\":5,\"numerator\":2}"},"fields":{"title":"Field Division Uses a Nonzero Denominator and the Same Table","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=2, denominator=3 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).","answer":{"kind":"expression","value":"(2, 3, 3, 2, 2, Yes)","explanation":"Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (2, 3, 3, 2, 2, Yes)."},"skills":["Field Division Uses a Nonzero Denominator and the Same Table"],"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-20261030-r01-finite-field-0064","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d9b01ba3fd81b2d202b722a2599b226a647716c779af39144d9f00a446a21357","family_id":"qf-mod30-field-division","seed":202610301064,"semantic_template":"qf-mod30-field-division/{\"addition\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"denominator\":1,\"multiplication\":[[0,0,0,0,0],[0,2,4,1,3],[0,4,3,2,1],[0,1,2,3,4],[0,3,1,4,2]],\"n\":5,\"numerator\":2}"},"fields":{"title":"Field Division Uses a Nonzero Denominator and the Same Table","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 4, 1, 3], [0, 4, 3, 2, 1], [0, 1, 2, 3, 4], [0, 3, 1, 4, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=2, denominator=1 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).","answer":{"kind":"expression","value":"(2, 1, 4, 1, 2, Yes)","explanation":"Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (2, 1, 4, 1, 2, Yes)."},"skills":["Field Division Uses a Nonzero Denominator and the Same Table"],"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-20261030-r01-finite-field-0065","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"21221b5c0ae82161dba75710ffc689e2ef5195a5f734611c671bc74cca221430","family_id":"qf-mod30-field-division","seed":202610301065,"semantic_template":"qf-mod30-field-division/{\"addition\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]],\"denominator\":1,\"multiplication\":[[0,0,0,0,0],[0,4,3,1,2],[0,3,4,2,1],[0,1,2,3,4],[0,2,1,4,3]],\"n\":5,\"numerator\":0}"},"fields":{"title":"Field Division Uses a Nonzero Denominator and the Same Table","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=0, denominator=1 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).","answer":{"kind":"expression","value":"(0, 1, 2, 0, 0, Yes)","explanation":"Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (0, 1, 2, 0, 0, Yes)."},"skills":["Field Division Uses a Nonzero Denominator and the Same Table"],"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-20261030-r01-finite-field-0066","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c5c2de447438a8b52058376b7e730d3324a7d958416313962a817a86c6ebf284","family_id":"qf-mod30-field-division","seed":202610301066,"semantic_template":"qf-mod30-field-division/{\"addition\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]],\"denominator\":1,\"multiplication\":[[0,0,0,0,0],[0,3,4,1,2],[0,4,1,2,3],[0,1,2,3,4],[0,2,3,4,1]],\"n\":5,\"numerator\":3}"},"fields":{"title":"Field Division Uses a Nonzero Denominator and the Same Table","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=3, denominator=1 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).","answer":{"kind":"expression","value":"(3, 1, 1, 1, 3, Yes)","explanation":"Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (3, 1, 1, 1, 3, Yes)."},"skills":["Field Division Uses a Nonzero Denominator and the Same Table"],"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-20261030-r01-finite-field-0067","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a38b84b8ea92e8404b9ef72450d5bd8dcd6d802f5a969ed1cb522b68e900ba19","family_id":"qf-mod30-field-division","seed":202610301067,"semantic_template":"qf-mod30-field-division/{\"addition\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]],\"denominator\":3,\"multiplication\":[[0,0,0,0,0],[0,2,4,1,3],[0,4,3,2,1],[0,1,2,3,4],[0,3,1,4,2]],\"n\":5,\"numerator\":4}"},"fields":{"title":"Field Division Uses a Nonzero Denominator and the Same Table","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 4, 1, 3], [0, 4, 3, 2, 1], [0, 1, 2, 3, 4], [0, 3, 1, 4, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=4, denominator=3 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).","answer":{"kind":"expression","value":"(4, 3, 3, 4, 4, Yes)","explanation":"Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (4, 3, 3, 4, 4, Yes)."},"skills":["Field Division Uses a Nonzero Denominator and the Same Table"],"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-20261030-r01-finite-field-0068","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"de5c9d850492cbcb3f5489814a51065366a81894fb599e4946050d4c2e2e2335","family_id":"qf-mod30-field-division","seed":202610301068,"semantic_template":"qf-mod30-field-division/{\"addition\":[[0,1,2,3,4],[1,2,4,0,3],[2,4,3,1,0],[3,0,1,4,2],[4,3,0,2,1]],\"denominator\":4,\"multiplication\":[[0,0,0,0,0],[0,3,4,1,2],[0,4,1,2,3],[0,1,2,3,4],[0,2,3,4,1]],\"n\":5,\"numerator\":2}"},"fields":{"title":"Field Division Uses a Nonzero Denominator and the Same Table","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=2, denominator=4 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).","answer":{"kind":"expression","value":"(2, 4, 2, 1, 2, Yes)","explanation":"Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (2, 4, 2, 1, 2, Yes)."},"skills":["Field Division Uses a Nonzero Denominator and the Same Table"],"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-20261030-r01-finite-field-0069","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5db3e6869144de50015ea28e7fa94b55633c8fc89eb6e71cc7ee66704252a338","family_id":"qf-mod30-field-division","seed":202610301069,"semantic_template":"qf-mod30-field-division/{\"addition\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]],\"denominator\":2,\"multiplication\":[[0,0,0,0,0],[0,3,4,2,1],[0,4,3,1,2],[0,2,1,4,3],[0,1,2,3,4]],\"n\":5,\"numerator\":3}"},"fields":{"title":"Field Division Uses a Nonzero Denominator and the Same Table","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=3, denominator=2 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).","answer":{"kind":"expression","value":"(3, 2, 1, 2, 3, Yes)","explanation":"Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (3, 2, 1, 2, 3, Yes)."},"skills":["Field Division Uses a Nonzero Denominator and the Same Table"],"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-20261030-r01-finite-field-0070","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9e7bae4eacefa0ac0f878fd530b02efd3fbe85141c370e9229eaa3269d3ba31d","family_id":"qf-mod30-field-division","seed":202610301070,"semantic_template":"qf-mod30-field-division/{\"addition\":[[0,1,2,3,4],[1,2,4,0,3],[2,4,3,1,0],[3,0,1,4,2],[4,3,0,2,1]],\"denominator\":2,\"multiplication\":[[0,0,0,0,0],[0,2,3,4,1],[0,3,4,1,2],[0,4,1,2,3],[0,1,2,3,4]],\"n\":5,\"numerator\":0}"},"fields":{"title":"Field Division Uses a Nonzero Denominator and the Same Table","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=0, denominator=2 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).","answer":{"kind":"expression","value":"(0, 2, 2, 0, 0, Yes)","explanation":"Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (0, 2, 2, 0, 0, Yes)."},"skills":["Field Division Uses a Nonzero Denominator and the Same Table"],"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-20261030-r01-finite-field-0071","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0b97bee8badabc0964aa3a43026443c79abdba77c2ea2d63e6e219903ee0c521","family_id":"qf-mod30-field-division","seed":202610301071,"semantic_template":"qf-mod30-field-division/{\"addition\":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]],\"denominator\":4,\"multiplication\":[[0,0,0,0,0],[0,3,4,2,1],[0,4,3,1,2],[0,2,1,4,3],[0,1,2,3,4]],\"n\":5,\"numerator\":4}"},"fields":{"title":"Field Division Uses a Nonzero Denominator and the Same Table","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=4, denominator=4 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).","answer":{"kind":"expression","value":"(4, 4, 4, 4, 4, Yes)","explanation":"Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (4, 4, 4, 4, 4, Yes)."},"skills":["Field Division Uses a Nonzero Denominator and the Same Table"],"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-20261030-r01-finite-field-0072","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d40b5d692a6f38c5e2da2275c0f5b7ac799099f1b5c43fc1b296cc5807fea77a","family_id":"qf-mod30-field-division","seed":202610301072,"semantic_template":"qf-mod30-field-division/{\"addition\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]],\"denominator\":2,\"multiplication\":[[0,0,0,0,0],[0,4,3,2,1],[0,3,1,4,2],[0,2,4,1,3],[0,1,2,3,4]],\"n\":5,\"numerator\":2}"},"fields":{"title":"Field Division Uses a Nonzero Denominator and the Same Table","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 2, 1], [0, 3, 1, 4, 2], [0, 2, 4, 1, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=2, denominator=2 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).","answer":{"kind":"expression","value":"(2, 2, 3, 4, 2, Yes)","explanation":"Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (2, 2, 3, 4, 2, Yes)."},"skills":["Field Division Uses a Nonzero Denominator and the Same Table"],"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-20261030-r01-finite-field-0073","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1ffcf03b0374ea82df8983ac66cab73fd17926226d9c3d89d2caa78501266abd","family_id":"qf-mod30-field-division","seed":202610301073,"semantic_template":"qf-mod30-field-division/{\"addition\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]],\"denominator\":1,\"multiplication\":[[0,0,0,0,0],[0,2,3,4,1],[0,3,4,1,2],[0,4,1,2,3],[0,1,2,3,4]],\"n\":5,\"numerator\":1}"},"fields":{"title":"Field Division Uses a Nonzero Denominator and the Same Table","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=1, denominator=1 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).","answer":{"kind":"expression","value":"(1, 1, 3, 4, 1, Yes)","explanation":"Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (1, 1, 3, 4, 1, Yes)."},"skills":["Field Division Uses a Nonzero Denominator and the Same Table"],"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-20261030-r01-finite-field-0074","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"791c66ca503614419c794260e99276204544e167fc4c557e23e22c3015b34238","family_id":"qf-mod30-field-division","seed":202610301074,"semantic_template":"qf-mod30-field-division/{\"addition\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"denominator\":1,\"multiplication\":[[0,0,0,0,0],[0,4,3,2,1],[0,3,1,4,2],[0,2,4,1,3],[0,1,2,3,4]],\"n\":5,\"numerator\":1}"},"fields":{"title":"Field Division Uses a Nonzero Denominator and the Same Table","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 2, 1], [0, 3, 1, 4, 2], [0, 2, 4, 1, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=1, denominator=1 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).","answer":{"kind":"expression","value":"(1, 1, 1, 4, 1, Yes)","explanation":"Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (1, 1, 1, 4, 1, Yes)."},"skills":["Field Division Uses a Nonzero Denominator and the Same Table"],"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-20261030-r01-finite-field-0075","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"14c3bb2c24dc840c41f13eada3af83f3885da3abd43508d8508411a15a3603eb","family_id":"qf-mod30-field-division","seed":202610301075,"semantic_template":"qf-mod30-field-division/{\"addition\":[[2,0,3,4,1],[0,1,2,3,4],[3,2,4,1,0],[4,3,1,0,2],[1,4,0,2,3]],\"denominator\":0,\"multiplication\":[[0,1,2,3,4],[1,1,1,1,1],[2,1,4,0,3],[3,1,0,4,2],[4,1,3,2,0]],\"n\":5,\"numerator\":1}"},"fields":{"title":"Field Division Uses a Nonzero Denominator and the Same Table","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]] and multiplication=[[0, 1, 2, 3, 4], [1, 1, 1, 1, 1], [2, 1, 4, 0, 3], [3, 1, 0, 4, 2], [4, 1, 3, 2, 0]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Under field qualification, numerator=1, denominator=0 is nonzero. Give (numerator,denominator,inverse,quotient,check,pass?).","answer":{"kind":"expression","value":"(1, 0, 0, 1, 1, Yes)","explanation":"Multiply the numerator by the denominator's inverse, then multiply the quotient by the denominator in the same table to recover the numerator. Ordinary arithmetic on the carrier labels is not this operation. The exact result, retaining the requested order, is (1, 0, 0, 1, 1, Yes)."},"skills":["Field Division Uses a Nonzero Denominator and the Same Table"],"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-20261030-r01-finite-field-0076","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dc35682a549b0ba3c02c1397b329c7b1cd935ebb7f20c518bd7423e7f030d4a0","family_id":"qf-mod30-field-finite-domain-theorem","seed":202610301076,"semantic_template":"qf-mod30-field-finite-domain-theorem/{\"addition\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,4,1,3],[0,3,1,4,2],[0,4,3,2,1]],\"n\":5}"},"fields":{"title":"Do Not Skip the Finite Injective-to-Surjective Step","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 4, 1, 3], [0, 3, 1, 4, 2], [0, 4, 3, 2, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.","answer":{"kind":"expression","value":"(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)","explanation":"Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)."},"skills":["Do Not Skip the Finite Injective-to-Surjective Step"],"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-20261030-r01-finite-field-0077","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ba04bc6365524cfac69573d2965c4654a5276c72702a892991de386eb0f68c12","family_id":"qf-mod30-field-finite-domain-theorem","seed":202610301077,"semantic_template":"qf-mod30-field-finite-domain-theorem/{\"addition\":[[0,1,2,3,4],[1,2,4,0,3],[2,4,3,1,0],[3,0,1,4,2],[4,3,0,2,1]],\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,3,4,1],[0,3,4,1,2],[0,4,1,2,3]],\"n\":5}"},"fields":{"title":"Do Not Skip the Finite Injective-to-Surjective Step","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.","answer":{"kind":"expression","value":"(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)","explanation":"Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)."},"skills":["Do Not Skip the Finite Injective-to-Surjective Step"],"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-20261030-r01-finite-field-0078","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1875af464f8406a91e9e8240a432ba2e7ba11d68b690d1e136107ff0580f7b39","family_id":"qf-mod30-field-finite-domain-theorem","seed":202610301078,"semantic_template":"qf-mod30-field-finite-domain-theorem/{\"addition\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]],\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,4,1,3],[0,3,1,4,2],[0,4,3,2,1]],\"n\":5}"},"fields":{"title":"Do Not Skip the Finite Injective-to-Surjective Step","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 4, 1, 3], [0, 3, 1, 4, 2], [0, 4, 3, 2, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.","answer":{"kind":"expression","value":"(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)","explanation":"Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)."},"skills":["Do Not Skip the Finite Injective-to-Surjective Step"],"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-20261030-r01-finite-field-0079","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b41619fdac1053d4aab9229d750c2e4b5eaee8be59fafa365b297b2d651bf12e","family_id":"qf-mod30-field-finite-domain-theorem","seed":202610301079,"semantic_template":"qf-mod30-field-finite-domain-theorem/{\"addition\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]],\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,1,4,3],[0,3,4,2,1],[0,4,3,1,2]],\"n\":5}"},"fields":{"title":"Do Not Skip the Finite Injective-to-Surjective Step","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.","answer":{"kind":"expression","value":"(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)","explanation":"Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)."},"skills":["Do Not Skip the Finite Injective-to-Surjective Step"],"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-20261030-r01-finite-field-0080","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b304eaa22cd320f20b6de11bbeb29ccdf5871ba709bd0a3d2dfd509f6d308f01","family_id":"qf-mod30-field-finite-domain-theorem","seed":202610301080,"semantic_template":"qf-mod30-field-finite-domain-theorem/{\"addition\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]],\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,3,4,1],[0,3,4,1,2],[0,4,1,2,3]],\"n\":5}"},"fields":{"title":"Do Not Skip the Finite Injective-to-Surjective Step","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.","answer":{"kind":"expression","value":"(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)","explanation":"Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)."},"skills":["Do Not Skip the Finite Injective-to-Surjective Step"],"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-20261030-r01-finite-field-0081","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"78a9ab6ee25c5f45fb1ee7ee7b135487b4971c5fd41b4a7bfa79a2543f2a170e","family_id":"qf-mod30-field-finite-domain-theorem","seed":202610301081,"semantic_template":"qf-mod30-field-finite-domain-theorem/{\"addition\":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]],\"multiplication\":[[0,0,0,0,0],[0,1,2,3,4],[0,2,1,4,3],[0,3,4,2,1],[0,4,3,1,2]],\"n\":5}"},"fields":{"title":"Do Not Skip the Finite Injective-to-Surjective Step","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.","answer":{"kind":"expression","value":"(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)","explanation":"Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)."},"skills":["Do Not Skip the Finite Injective-to-Surjective Step"],"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-20261030-r01-finite-field-0082","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"746ef14536c7fdb9e28c19dce28127e46609f9a9f469127224cb9f2120c75fd8","family_id":"qf-mod30-field-finite-domain-theorem","seed":202610301082,"semantic_template":"qf-mod30-field-finite-domain-theorem/{\"addition\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]],\"multiplication\":[[0,0,0,0,0],[0,4,1,2,3],[0,1,2,3,4],[0,2,3,4,1],[0,3,4,1,2]],\"n\":5}"},"fields":{"title":"Do Not Skip the Finite Injective-to-Surjective Step","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.","answer":{"kind":"expression","value":"(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)","explanation":"Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)."},"skills":["Do Not Skip the Finite Injective-to-Surjective Step"],"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-20261030-r01-finite-field-0083","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a4182a524e1f3e15457670bc1c4e60de7abec9d88b9e4ea3305ced36c8daa8af","family_id":"qf-mod30-field-finite-domain-theorem","seed":202610301083,"semantic_template":"qf-mod30-field-finite-domain-theorem/{\"addition\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]],\"multiplication\":[[0,0,0,0,0],[0,3,1,4,2],[0,1,2,3,4],[0,4,3,2,1],[0,2,4,1,3]],\"n\":5}"},"fields":{"title":"Do Not Skip the Finite Injective-to-Surjective Step","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 1, 4, 2], [0, 1, 2, 3, 4], [0, 4, 3, 2, 1], [0, 2, 4, 1, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.","answer":{"kind":"expression","value":"(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)","explanation":"Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)."},"skills":["Do Not Skip the Finite Injective-to-Surjective Step"],"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-20261030-r01-finite-field-0084","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ebde5b3dcc21f7bad3ebdea85238f61702915819433526c9a2ca2a4fd8fd99e6","family_id":"qf-mod30-field-finite-domain-theorem","seed":202610301084,"semantic_template":"qf-mod30-field-finite-domain-theorem/{\"addition\":[[0,1,2,3,4],[1,2,4,0,3],[2,4,3,1,0],[3,0,1,4,2],[4,3,0,2,1]],\"multiplication\":[[0,0,0,0,0],[0,4,1,2,3],[0,1,2,3,4],[0,2,3,4,1],[0,3,4,1,2]],\"n\":5}"},"fields":{"title":"Do Not Skip the Finite Injective-to-Surjective Step","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.","answer":{"kind":"expression","value":"(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)","explanation":"Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)."},"skills":["Do Not Skip the Finite Injective-to-Surjective Step"],"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-20261030-r01-finite-field-0085","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9376a648a2c71ecf470a97d9adc0a7cea7aa3660fa4ce5e9ef989144fd593f82","family_id":"qf-mod30-field-finite-domain-theorem","seed":202610301085,"semantic_template":"qf-mod30-field-finite-domain-theorem/{\"addition\":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]],\"multiplication\":[[0,0,0,0,0],[0,2,1,4,3],[0,1,2,3,4],[0,4,3,1,2],[0,3,4,2,1]],\"n\":5}"},"fields":{"title":"Do Not Skip the Finite Injective-to-Surjective Step","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.","answer":{"kind":"expression","value":"(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)","explanation":"Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)."},"skills":["Do Not Skip the Finite Injective-to-Surjective Step"],"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-20261030-r01-finite-field-0086","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4015811f7bd7eeff7c90f12c1cbedd5f50a4aa261e4487f0ff4b14dfbbc9beb8","family_id":"qf-mod30-field-finite-domain-theorem","seed":202610301086,"semantic_template":"qf-mod30-field-finite-domain-theorem/{\"addition\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"multiplication\":[[0,0,0,0,0],[0,3,1,4,2],[0,1,2,3,4],[0,4,3,2,1],[0,2,4,1,3]],\"n\":5}"},"fields":{"title":"Do Not Skip the Finite Injective-to-Surjective Step","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 1, 4, 2], [0, 1, 2, 3, 4], [0, 4, 3, 2, 1], [0, 2, 4, 1, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.","answer":{"kind":"expression","value":"(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)","explanation":"Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)."},"skills":["Do Not Skip the Finite Injective-to-Surjective Step"],"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-20261030-r01-finite-field-0087","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"81309a4d84f3470f81dacd8a1e677a09495bf80e577d0d86cf16b95f6e7ba8cf","family_id":"qf-mod30-field-finite-domain-theorem","seed":202610301087,"semantic_template":"qf-mod30-field-finite-domain-theorem/{\"addition\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]],\"multiplication\":[[0,0,0,0,0],[0,2,1,4,3],[0,1,2,3,4],[0,4,3,1,2],[0,3,4,2,1]],\"n\":5}"},"fields":{"title":"Do Not Skip the Finite Injective-to-Surjective Step","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.","answer":{"kind":"expression","value":"(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)","explanation":"Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)."},"skills":["Do Not Skip the Finite Injective-to-Surjective Step"],"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-20261030-r01-finite-field-0088","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"700f0cf77e7038c8324512d3025f2b4f7f2d1c0bd3082e02058c617cdd8f9e0c","family_id":"qf-mod30-field-finite-domain-theorem","seed":202610301088,"semantic_template":"qf-mod30-field-finite-domain-theorem/{\"addition\":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]],\"multiplication\":[[0,0,0,0,0],[0,4,3,1,2],[0,3,4,2,1],[0,1,2,3,4],[0,2,1,4,3]],\"n\":5}"},"fields":{"title":"Do Not Skip the Finite Injective-to-Surjective Step","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.","answer":{"kind":"expression","value":"(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)","explanation":"Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)."},"skills":["Do Not Skip the Finite Injective-to-Surjective Step"],"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-20261030-r01-finite-field-0089","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3ec2a1a33be54cc1ddd75798b6f43f857ed9679185801e6155efba2f460de252","family_id":"qf-mod30-field-finite-domain-theorem","seed":202610301089,"semantic_template":"qf-mod30-field-finite-domain-theorem/{\"addition\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"multiplication\":[[0,0,0,0,0],[0,2,4,1,3],[0,4,3,2,1],[0,1,2,3,4],[0,3,1,4,2]],\"n\":5}"},"fields":{"title":"Do Not Skip the Finite Injective-to-Surjective Step","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 4, 1, 3], [0, 4, 3, 2, 1], [0, 1, 2, 3, 4], [0, 3, 1, 4, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.","answer":{"kind":"expression","value":"(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)","explanation":"Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)."},"skills":["Do Not Skip the Finite Injective-to-Surjective Step"],"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-20261030-r01-finite-field-0090","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f8bc19f6d8fd9677a44767181c7a7e1e09ac5c811268a12d3d23a4a819363525","family_id":"qf-mod30-field-finite-domain-theorem","seed":202610301090,"semantic_template":"qf-mod30-field-finite-domain-theorem/{\"addition\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]],\"multiplication\":[[0,0,0,0,0],[0,4,3,1,2],[0,3,4,2,1],[0,1,2,3,4],[0,2,1,4,3]],\"n\":5}"},"fields":{"title":"Do Not Skip the Finite Injective-to-Surjective Step","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 1, 2], [0, 3, 4, 2, 1], [0, 1, 2, 3, 4], [0, 2, 1, 4, 3]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.","answer":{"kind":"expression","value":"(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)","explanation":"Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)."},"skills":["Do Not Skip the Finite Injective-to-Surjective Step"],"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-20261030-r01-finite-field-0091","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"21eefa28a72dc67527507c7f01091af8bca2163df5c2c666445f26cfe07abe72","family_id":"qf-mod30-field-finite-domain-theorem","seed":202610301091,"semantic_template":"qf-mod30-field-finite-domain-theorem/{\"addition\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]],\"multiplication\":[[0,0,0,0,0],[0,3,4,1,2],[0,4,1,2,3],[0,1,2,3,4],[0,2,3,4,1]],\"n\":5}"},"fields":{"title":"Do Not Skip the Finite Injective-to-Surjective Step","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.","answer":{"kind":"expression","value":"(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)","explanation":"Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)."},"skills":["Do Not Skip the Finite Injective-to-Surjective Step"],"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-20261030-r01-finite-field-0092","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c83be65bd258ba0c02bc17ae8c7ffbc70caae7ef1f5b08754f42df4043cba201","family_id":"qf-mod30-field-finite-domain-theorem","seed":202610301092,"semantic_template":"qf-mod30-field-finite-domain-theorem/{\"addition\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]],\"multiplication\":[[0,0,0,0,0],[0,2,4,1,3],[0,4,3,2,1],[0,1,2,3,4],[0,3,1,4,2]],\"n\":5}"},"fields":{"title":"Do Not Skip the Finite Injective-to-Surjective Step","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 4, 1, 3], [0, 4, 3, 2, 1], [0, 1, 2, 3, 4], [0, 3, 1, 4, 2]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.","answer":{"kind":"expression","value":"(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)","explanation":"Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)."},"skills":["Do Not Skip the Finite Injective-to-Surjective Step"],"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-20261030-r01-finite-field-0093","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c7119c7705a6e0d851742b803acb9cd2a90f551235134e1fdb5da3549cad7267","family_id":"qf-mod30-field-finite-domain-theorem","seed":202610301093,"semantic_template":"qf-mod30-field-finite-domain-theorem/{\"addition\":[[0,1,2,3,4],[1,2,4,0,3],[2,4,3,1,0],[3,0,1,4,2],[4,3,0,2,1]],\"multiplication\":[[0,0,0,0,0],[0,3,4,1,2],[0,4,1,2,3],[0,1,2,3,4],[0,2,3,4,1]],\"n\":5}"},"fields":{"title":"Do Not Skip the Finite Injective-to-Surjective Step","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4], [0, 2, 3, 4, 1]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.","answer":{"kind":"expression","value":"(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)","explanation":"Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)."},"skills":["Do Not Skip the Finite Injective-to-Surjective Step"],"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-20261030-r01-finite-field-0094","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"189dd8707cec495e8c7c14ecc971cbac4c70fe192a857d2eb47d5ef0c63dbbfe","family_id":"qf-mod30-field-finite-domain-theorem","seed":202610301094,"semantic_template":"qf-mod30-field-finite-domain-theorem/{\"addition\":[[0,1,2,3,4],[1,3,0,4,2],[2,0,4,1,3],[3,4,1,2,0],[4,2,3,0,1]],\"multiplication\":[[0,0,0,0,0],[0,3,4,2,1],[0,4,3,1,2],[0,2,1,4,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Do Not Skip the Finite Injective-to-Surjective Step","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 0, 4, 2], [2, 0, 4, 1, 3], [3, 4, 1, 2, 0], [4, 2, 3, 0, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.","answer":{"kind":"expression","value":"(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)","explanation":"Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)."},"skills":["Do Not Skip the Finite Injective-to-Surjective Step"],"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-20261030-r01-finite-field-0095","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4f49101ccd6a8c401e0974d9d06e122f4c861d781f57e7d1a974143a28eafb31","family_id":"qf-mod30-field-finite-domain-theorem","seed":202610301095,"semantic_template":"qf-mod30-field-finite-domain-theorem/{\"addition\":[[0,1,2,3,4],[1,2,4,0,3],[2,4,3,1,0],[3,0,1,4,2],[4,3,0,2,1]],\"multiplication\":[[0,0,0,0,0],[0,2,3,4,1],[0,3,4,1,2],[0,4,1,2,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Do Not Skip the Finite Injective-to-Surjective Step","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 4, 0, 3], [2, 4, 3, 1, 0], [3, 0, 1, 4, 2], [4, 3, 0, 2, 1]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.","answer":{"kind":"expression","value":"(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)","explanation":"Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)."},"skills":["Do Not Skip the Finite Injective-to-Surjective Step"],"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-20261030-r01-finite-field-0096","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"99a490cbbee175c34335d5465a58eb9d9b781a929576588a5636a62cc1007dbe","family_id":"qf-mod30-field-finite-domain-theorem","seed":202610301096,"semantic_template":"qf-mod30-field-finite-domain-theorem/{\"addition\":[[0,1,2,3,4],[1,4,0,2,3],[2,0,3,4,1],[3,2,4,1,0],[4,3,1,0,2]],\"multiplication\":[[0,0,0,0,0],[0,3,4,2,1],[0,4,3,1,2],[0,2,1,4,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Do Not Skip the Finite Injective-to-Surjective Step","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 0, 2, 3], [2, 0, 3, 4, 1], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 3, 4, 2, 1], [0, 4, 3, 1, 2], [0, 2, 1, 4, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.","answer":{"kind":"expression","value":"(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)","explanation":"Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)."},"skills":["Do Not Skip the Finite Injective-to-Surjective Step"],"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-20261030-r01-finite-field-0097","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6ce309f09e3cd7064d8c6cb40c4bad0ce502ef54b672414d4ef7648decfdc095","family_id":"qf-mod30-field-finite-domain-theorem","seed":202610301097,"semantic_template":"qf-mod30-field-finite-domain-theorem/{\"addition\":[[0,1,2,3,4],[1,3,4,2,0],[2,4,1,0,3],[3,2,0,4,1],[4,0,3,1,2]],\"multiplication\":[[0,0,0,0,0],[0,4,3,2,1],[0,3,1,4,2],[0,2,4,1,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Do Not Skip the Finite Injective-to-Surjective Step","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 3, 4, 2, 0], [2, 4, 1, 0, 3], [3, 2, 0, 4, 1], [4, 0, 3, 1, 2]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 2, 1], [0, 3, 1, 4, 2], [0, 2, 4, 1, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.","answer":{"kind":"expression","value":"(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)","explanation":"Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)."},"skills":["Do Not Skip the Finite Injective-to-Surjective Step"],"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-20261030-r01-finite-field-0098","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5e8f686ac33c96737c75e22ec50e270a01c57adf0799d210a2435c333f0aee0a","family_id":"qf-mod30-field-finite-domain-theorem","seed":202610301098,"semantic_template":"qf-mod30-field-finite-domain-theorem/{\"addition\":[[0,1,2,3,4],[1,4,3,0,2],[2,3,1,4,0],[3,0,4,2,1],[4,2,0,1,3]],\"multiplication\":[[0,0,0,0,0],[0,2,3,4,1],[0,3,4,1,2],[0,4,1,2,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Do Not Skip the Finite Injective-to-Surjective Step","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 4, 3, 0, 2], [2, 3, 1, 4, 0], [3, 0, 4, 2, 1], [4, 2, 0, 1, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 2, 3, 4, 1], [0, 3, 4, 1, 2], [0, 4, 1, 2, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.","answer":{"kind":"expression","value":"(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)","explanation":"Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)."},"skills":["Do Not Skip the Finite Injective-to-Surjective Step"],"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-20261030-r01-finite-field-0099","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4d1f7485f5cb8742233de3d2492cc74442eaa2984fc0218403264b40919c73c8","family_id":"qf-mod30-field-finite-domain-theorem","seed":202610301099,"semantic_template":"qf-mod30-field-finite-domain-theorem/{\"addition\":[[0,1,2,3,4],[1,2,3,4,0],[2,3,4,0,1],[3,4,0,1,2],[4,0,1,2,3]],\"multiplication\":[[0,0,0,0,0],[0,4,3,2,1],[0,3,1,4,2],[0,2,4,1,3],[0,1,2,3,4]],\"n\":5}"},"fields":{"title":"Do Not Skip the Finite Injective-to-Surjective Step","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[0, 1, 2, 3, 4], [1, 2, 3, 4, 0], [2, 3, 4, 0, 1], [3, 4, 0, 1, 2], [4, 0, 1, 2, 3]] and multiplication=[[0, 0, 0, 0, 0], [0, 4, 3, 2, 1], [0, 3, 1, 4, 2], [0, 2, 4, 1, 3], [0, 1, 2, 3, 4]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.","answer":{"kind":"expression","value":"(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)","explanation":"Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)."},"skills":["Do Not Skip the Finite Injective-to-Surjective Step"],"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-20261030-r01-finite-field-0100","batch_id":"batch-20261014-advance-20261030-r01-finite-field-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c2173ad6129fe98de4a8beabeb50ca9d434d2672976f47a5e95dd02177308249","family_id":"qf-mod30-field-finite-domain-theorem","seed":202610301100,"semantic_template":"qf-mod30-field-finite-domain-theorem/{\"addition\":[[2,0,3,4,1],[0,1,2,3,4],[3,2,4,1,0],[4,3,1,0,2],[1,4,0,2,3]],\"multiplication\":[[0,1,2,3,4],[1,1,1,1,1],[2,1,4,0,3],[3,1,0,4,2],[4,1,3,2,0]],\"n\":5}"},"fields":{"title":"Do Not Skip the Finite Injective-to-Surjective Step","stem":"Use the same already qualified complete finite commutative unital-ring tables on Fin(5)={0,…,4}, addition=[[2, 0, 3, 4, 1], [0, 1, 2, 3, 4], [3, 2, 4, 1, 0], [4, 3, 1, 0, 2], [1, 4, 0, 2, 3]] and multiplication=[[0, 1, 2, 3, 4], [1, 1, 1, 1, 1], [2, 1, 4, 0, 3], [3, 1, 0, 4, 2], [4, 1, 3, 2, 0]] in carrier-label order. The actual zero and one are distinct and need not be labels 0 and 1. The inverse audit is additional to, not a replacement for, those ring axioms. Division by zero is never defined. A finite field has prime-power order, not necessarily prime order. Give (n,missing,field?,theorem_step,field_order_prime_power_not_necessarily_prime). theorem_step is finite_domain_injective_to_surjective_theorem or missing_nonzero_inverse.","answer":{"kind":"expression","value":"(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)","explanation":"Record every missing nonzero inverse. In a finite integral domain, nonzero multiplication is injective and therefore surjective, yielding an inverse. A table with missing inverses does not qualify; this does not invalidate that finite-domain theorem. The exact result, retaining the requested order, is (5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)."},"skills":["Do Not Skip the Finite Injective-to-Surjective Step"],"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."}}