# Complete Nonzero Inverses, Safe Division and the Finite-Domain Theorem

This batch has passed Schema, deduplication, source-snapshot, and independent-answer recomputation checks; it is still not marked as human-reviewed.

This question bank has corresponding bilingual step-by-step derivations. Source candidate and English ai_candidate states are retained. AI review, finite computation and Lean formal proof are labelled separately.

batch-20261014-advance-20261030-r01-finite-field-foundations · 2026-10-14 · 100 questions

This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt.



## 1. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

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.

View answer and explanation(5, 0, 1, [], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0001/)

Finite field · qf-mod30-field-qualification · candidate





## 2. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

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.

View answer and explanation(5, 0, 1, [], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0002/)

Finite field · qf-mod30-field-qualification · candidate





## 3. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

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.

View answer and explanation(5, 0, 1, [], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0003/)

Finite field · qf-mod30-field-qualification · candidate





## 4. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

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.

View answer and explanation(5, 0, 1, [], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0004/)

Finite field · qf-mod30-field-qualification · candidate





## 5. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

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.

View answer and explanation(5, 0, 1, [], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0005/)

Finite field · qf-mod30-field-qualification · candidate





## 6. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

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.

View answer and explanation(5, 0, 1, [], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0006/)

Finite field · qf-mod30-field-qualification · candidate





## 7. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

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.

View answer and explanation(5, 0, 2, [], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0007/)

Finite field · qf-mod30-field-qualification · candidate





## 8. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

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.

View answer and explanation(5, 0, 2, [], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0008/)

Finite field · qf-mod30-field-qualification · candidate





## 9. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

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.

View answer and explanation(5, 0, 2, [], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0009/)

Finite field · qf-mod30-field-qualification · candidate





## 10. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

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.

View answer and explanation(5, 0, 2, [], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0010/)

Finite field · qf-mod30-field-qualification · candidate





## 11. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

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.

View answer and explanation(5, 0, 2, [], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0011/)

Finite field · qf-mod30-field-qualification · candidate





## 12. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

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.

View answer and explanation(5, 0, 2, [], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0012/)

Finite field · qf-mod30-field-qualification · candidate





## 13. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

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.

View answer and explanation(5, 0, 3, [], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0013/)

Finite field · qf-mod30-field-qualification · candidate





## 14. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

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.

View answer and explanation(5, 0, 3, [], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0014/)

Finite field · qf-mod30-field-qualification · candidate





## 15. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

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.

View answer and explanation(5, 0, 3, [], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0015/)

Finite field · qf-mod30-field-qualification · candidate





## 16. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

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.

View answer and explanation(5, 0, 3, [], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0016/)

Finite field · qf-mod30-field-qualification · candidate





## 17. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

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.

View answer and explanation(5, 0, 3, [], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0017/)

Finite field · qf-mod30-field-qualification · candidate





## 18. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

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.

View answer and explanation(5, 0, 3, [], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0018/)

Finite field · qf-mod30-field-qualification · candidate





## 19. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

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.

View answer and explanation(5, 0, 4, [], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0019/)

Finite field · qf-mod30-field-qualification · candidate





## 20. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

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.

View answer and explanation(5, 0, 4, [], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0020/)

Finite field · qf-mod30-field-qualification · candidate





## 21. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

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.

View answer and explanation(5, 0, 4, [], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0021/)

Finite field · qf-mod30-field-qualification · candidate





## 22. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

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.

View answer and explanation(5, 0, 4, [], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0022/)

Finite field · qf-mod30-field-qualification · candidate





## 23. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

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.

View answer and explanation(5, 0, 4, [], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0023/)

Finite field · qf-mod30-field-qualification · candidate





## 24. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

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.

View answer and explanation(5, 0, 4, [], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0024/)

Finite field · qf-mod30-field-qualification · candidate





## 25. Audit Every Nonzero Inverse after Complete Commutative-Ring Qualification

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.

View answer and explanation(5, 1, 0, [], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0025/)

Finite field · qf-mod30-field-qualification · candidate





## 26. Find a Two-Sided Inverse of the Specified Nonzero Element

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?).

View answer and explanation(1, 1, 1, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0026/)

Finite field · qf-mod30-field-inverse · candidate





## 27. Find a Two-Sided Inverse of the Specified Nonzero Element

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?).

View answer and explanation(2, 1, 4, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0027/)

Finite field · qf-mod30-field-inverse · candidate





## 28. Find a Two-Sided Inverse of the Specified Nonzero Element

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?).

View answer and explanation(2, 1, 3, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0028/)

Finite field · qf-mod30-field-inverse · candidate





## 29. Find a Two-Sided Inverse of the Specified Nonzero Element

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?).

View answer and explanation(2, 1, 2, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0029/)

Finite field · qf-mod30-field-inverse · candidate





## 30. Find a Two-Sided Inverse of the Specified Nonzero Element

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?).

View answer and explanation(1, 1, 1, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0030/)

Finite field · qf-mod30-field-inverse · candidate





## 31. Find a Two-Sided Inverse of the Specified Nonzero Element

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?).

View answer and explanation(4, 1, 3, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0031/)

Finite field · qf-mod30-field-inverse · candidate





## 32. Find a Two-Sided Inverse of the Specified Nonzero Element

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?).

View answer and explanation(3, 2, 1, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0032/)

Finite field · qf-mod30-field-inverse · candidate





## 33. Find a Two-Sided Inverse of the Specified Nonzero Element

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?).

View answer and explanation(3, 2, 3, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0033/)

Finite field · qf-mod30-field-inverse · candidate





## 34. Find a Two-Sided Inverse of the Specified Nonzero Element

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?).

View answer and explanation(2, 2, 2, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0034/)

Finite field · qf-mod30-field-inverse · candidate





## 35. Find a Two-Sided Inverse of the Specified Nonzero Element

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?).

View answer and explanation(3, 2, 4, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0035/)

Finite field · qf-mod30-field-inverse · candidate





## 36. Find a Two-Sided Inverse of the Specified Nonzero Element

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?).

View answer and explanation(1, 2, 4, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0036/)

Finite field · qf-mod30-field-inverse · candidate





## 37. Find a Two-Sided Inverse of the Specified Nonzero Element

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?).

View answer and explanation(1, 2, 1, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0037/)

Finite field · qf-mod30-field-inverse · candidate





## 38. Find a Two-Sided Inverse of the Specified Nonzero Element

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?).

View answer and explanation(3, 3, 3, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0038/)

Finite field · qf-mod30-field-inverse · candidate





## 39. Find a Two-Sided Inverse of the Specified Nonzero Element

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?).

View answer and explanation(1, 3, 4, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0039/)

Finite field · qf-mod30-field-inverse · candidate





## 40. Find a Two-Sided Inverse of the Specified Nonzero Element

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?).

View answer and explanation(1, 3, 2, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0040/)

Finite field · qf-mod30-field-inverse · candidate





## 41. Find a Two-Sided Inverse of the Specified Nonzero Element

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?).

View answer and explanation(1, 3, 1, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0041/)

Finite field · qf-mod30-field-inverse · candidate





## 42. Find a Two-Sided Inverse of the Specified Nonzero Element

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?).

View answer and explanation(3, 3, 3, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0042/)

Finite field · qf-mod30-field-inverse · candidate





## 43. Find a Two-Sided Inverse of the Specified Nonzero Element

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?).

View answer and explanation(4, 3, 2, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0043/)

Finite field · qf-mod30-field-inverse · candidate





## 44. Find a Two-Sided Inverse of the Specified Nonzero Element

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?).

View answer and explanation(2, 4, 1, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0044/)

Finite field · qf-mod30-field-inverse · candidate





## 45. Find a Two-Sided Inverse of the Specified Nonzero Element

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?).

View answer and explanation(2, 4, 2, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0045/)

Finite field · qf-mod30-field-inverse · candidate





## 46. Find a Two-Sided Inverse of the Specified Nonzero Element

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?).

View answer and explanation(4, 4, 4, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0046/)

Finite field · qf-mod30-field-inverse · candidate





## 47. Find a Two-Sided Inverse of the Specified Nonzero Element

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?).

View answer and explanation(2, 4, 3, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0047/)

Finite field · qf-mod30-field-inverse · candidate





## 48. Find a Two-Sided Inverse of the Specified Nonzero Element

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?).

View answer and explanation(1, 4, 3, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0048/)

Finite field · qf-mod30-field-inverse · candidate





## 49. Find a Two-Sided Inverse of the Specified Nonzero Element

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?).

View answer and explanation(1, 4, 1, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0049/)

Finite field · qf-mod30-field-inverse · candidate





## 50. Find a Two-Sided Inverse of the Specified Nonzero Element

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?).

View answer and explanation(0, 0, 0, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0050/)

Finite field · qf-mod30-field-inverse · candidate





## 51. Field Division Uses a Nonzero Denominator and the Same Table

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?).

View answer and explanation(1, 1, 1, 1, 1, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0051/)

Finite field · qf-mod30-field-division · candidate





## 52. Field Division Uses a Nonzero Denominator and the Same Table

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?).

View answer and explanation(2, 2, 4, 1, 2, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0052/)

Finite field · qf-mod30-field-division · candidate





## 53. Field Division Uses a Nonzero Denominator and the Same Table

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?).

View answer and explanation(2, 2, 3, 1, 2, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0053/)

Finite field · qf-mod30-field-division · candidate





## 54. Field Division Uses a Nonzero Denominator and the Same Table

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?).

View answer and explanation(2, 2, 2, 1, 2, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0054/)

Finite field · qf-mod30-field-division · candidate





## 55. Field Division Uses a Nonzero Denominator and the Same Table

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?).

View answer and explanation(0, 1, 1, 0, 0, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0055/)

Finite field · qf-mod30-field-division · candidate





## 56. Field Division Uses a Nonzero Denominator and the Same Table

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?).

View answer and explanation(1, 4, 3, 3, 1, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0056/)

Finite field · qf-mod30-field-division · candidate





## 57. Field Division Uses a Nonzero Denominator and the Same Table

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?).

View answer and explanation(1, 3, 1, 4, 1, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0057/)

Finite field · qf-mod30-field-division · candidate





## 58. Field Division Uses a Nonzero Denominator and the Same Table

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?).

View answer and explanation(4, 3, 3, 1, 4, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0058/)

Finite field · qf-mod30-field-division · candidate





## 59. Field Division Uses a Nonzero Denominator and the Same Table

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?).

View answer and explanation(4, 2, 2, 4, 4, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0059/)

Finite field · qf-mod30-field-division · candidate





## 60. Field Division Uses a Nonzero Denominator and the Same Table

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?).

View answer and explanation(0, 3, 4, 0, 0, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0060/)

Finite field · qf-mod30-field-division · candidate





## 61. Field Division Uses a Nonzero Denominator and the Same Table

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?).

View answer and explanation(2, 1, 4, 4, 2, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0061/)

Finite field · qf-mod30-field-division · candidate





## 62. Field Division Uses a Nonzero Denominator and the Same Table

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?).

View answer and explanation(4, 1, 1, 3, 4, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0062/)

Finite field · qf-mod30-field-division · candidate





## 63. Field Division Uses a Nonzero Denominator and the Same Table

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?).

View answer and explanation(2, 3, 3, 2, 2, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0063/)

Finite field · qf-mod30-field-division · candidate





## 64. Field Division Uses a Nonzero Denominator and the Same Table

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?).

View answer and explanation(2, 1, 4, 1, 2, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0064/)

Finite field · qf-mod30-field-division · candidate





## 65. Field Division Uses a Nonzero Denominator and the Same Table

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?).

View answer and explanation(0, 1, 2, 0, 0, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0065/)

Finite field · qf-mod30-field-division · candidate





## 66. Field Division Uses a Nonzero Denominator and the Same Table

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?).

View answer and explanation(3, 1, 1, 1, 3, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0066/)

Finite field · qf-mod30-field-division · candidate





## 67. Field Division Uses a Nonzero Denominator and the Same Table

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?).

View answer and explanation(4, 3, 3, 4, 4, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0067/)

Finite field · qf-mod30-field-division · candidate





## 68. Field Division Uses a Nonzero Denominator and the Same Table

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?).

View answer and explanation(2, 4, 2, 1, 2, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0068/)

Finite field · qf-mod30-field-division · candidate





## 69. Field Division Uses a Nonzero Denominator and the Same Table

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?).

View answer and explanation(3, 2, 1, 2, 3, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0069/)

Finite field · qf-mod30-field-division · candidate





## 70. Field Division Uses a Nonzero Denominator and the Same Table

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?).

View answer and explanation(0, 2, 2, 0, 0, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0070/)

Finite field · qf-mod30-field-division · candidate





## 71. Field Division Uses a Nonzero Denominator and the Same Table

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?).

View answer and explanation(4, 4, 4, 4, 4, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0071/)

Finite field · qf-mod30-field-division · candidate





## 72. Field Division Uses a Nonzero Denominator and the Same Table

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?).

View answer and explanation(2, 2, 3, 4, 2, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0072/)

Finite field · qf-mod30-field-division · candidate





## 73. Field Division Uses a Nonzero Denominator and the Same Table

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?).

View answer and explanation(1, 1, 3, 4, 1, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0073/)

Finite field · qf-mod30-field-division · candidate





## 74. Field Division Uses a Nonzero Denominator and the Same Table

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?).

View answer and explanation(1, 1, 1, 4, 1, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0074/)

Finite field · qf-mod30-field-division · candidate





## 75. Field Division Uses a Nonzero Denominator and the Same Table

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?).

View answer and explanation(1, 0, 0, 1, 1, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0075/)

Finite field · qf-mod30-field-division · candidate





## 76. Do Not Skip the Finite Injective-to-Surjective Step

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.

View answer and explanation(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0076/)

Finite field · qf-mod30-field-finite-domain-theorem · candidate





## 77. Do Not Skip the Finite Injective-to-Surjective Step

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.

View answer and explanation(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0077/)

Finite field · qf-mod30-field-finite-domain-theorem · candidate





## 78. Do Not Skip the Finite Injective-to-Surjective Step

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.

View answer and explanation(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0078/)

Finite field · qf-mod30-field-finite-domain-theorem · candidate





## 79. Do Not Skip the Finite Injective-to-Surjective Step

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.

View answer and explanation(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0079/)

Finite field · qf-mod30-field-finite-domain-theorem · candidate





## 80. Do Not Skip the Finite Injective-to-Surjective Step

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.

View answer and explanation(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0080/)

Finite field · qf-mod30-field-finite-domain-theorem · candidate





## 81. Do Not Skip the Finite Injective-to-Surjective Step

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.

View answer and explanation(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0081/)

Finite field · qf-mod30-field-finite-domain-theorem · candidate





## 82. Do Not Skip the Finite Injective-to-Surjective Step

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.

View answer and explanation(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0082/)

Finite field · qf-mod30-field-finite-domain-theorem · candidate





## 83. Do Not Skip the Finite Injective-to-Surjective Step

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.

View answer and explanation(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0083/)

Finite field · qf-mod30-field-finite-domain-theorem · candidate





## 84. Do Not Skip the Finite Injective-to-Surjective Step

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.

View answer and explanation(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0084/)

Finite field · qf-mod30-field-finite-domain-theorem · candidate





## 85. Do Not Skip the Finite Injective-to-Surjective Step

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.

View answer and explanation(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0085/)

Finite field · qf-mod30-field-finite-domain-theorem · candidate





## 86. Do Not Skip the Finite Injective-to-Surjective Step

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.

View answer and explanation(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0086/)

Finite field · qf-mod30-field-finite-domain-theorem · candidate





## 87. Do Not Skip the Finite Injective-to-Surjective Step

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.

View answer and explanation(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0087/)

Finite field · qf-mod30-field-finite-domain-theorem · candidate





## 88. Do Not Skip the Finite Injective-to-Surjective Step

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.

View answer and explanation(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0088/)

Finite field · qf-mod30-field-finite-domain-theorem · candidate





## 89. Do Not Skip the Finite Injective-to-Surjective Step

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.

View answer and explanation(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0089/)

Finite field · qf-mod30-field-finite-domain-theorem · candidate





## 90. Do Not Skip the Finite Injective-to-Surjective Step

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.

View answer and explanation(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0090/)

Finite field · qf-mod30-field-finite-domain-theorem · candidate





## 91. Do Not Skip the Finite Injective-to-Surjective Step

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.

View answer and explanation(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0091/)

Finite field · qf-mod30-field-finite-domain-theorem · candidate





## 92. Do Not Skip the Finite Injective-to-Surjective Step

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.

View answer and explanation(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0092/)

Finite field · qf-mod30-field-finite-domain-theorem · candidate





## 93. Do Not Skip the Finite Injective-to-Surjective Step

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.

View answer and explanation(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0093/)

Finite field · qf-mod30-field-finite-domain-theorem · candidate





## 94. Do Not Skip the Finite Injective-to-Surjective Step

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.

View answer and explanation(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0094/)

Finite field · qf-mod30-field-finite-domain-theorem · candidate





## 95. Do Not Skip the Finite Injective-to-Surjective Step

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.

View answer and explanation(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0095/)

Finite field · qf-mod30-field-finite-domain-theorem · candidate





## 96. Do Not Skip the Finite Injective-to-Surjective Step

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.

View answer and explanation(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0096/)

Finite field · qf-mod30-field-finite-domain-theorem · candidate





## 97. Do Not Skip the Finite Injective-to-Surjective Step

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.

View answer and explanation(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0097/)

Finite field · qf-mod30-field-finite-domain-theorem · candidate





## 98. Do Not Skip the Finite Injective-to-Surjective Step

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.

View answer and explanation(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0098/)

Finite field · qf-mod30-field-finite-domain-theorem · candidate





## 99. Do Not Skip the Finite Injective-to-Surjective Step

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.

View answer and explanation(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0099/)

Finite field · qf-mod30-field-finite-domain-theorem · candidate





## 100. Do Not Skip the Finite Injective-to-Surjective Step

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.

View answer and explanation(5, [], Yes, finite_domain_injective_to_surjective_theorem, field_order_prime_power_not_necessarily_prime)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261014-advance-20261030-r01-finite-field-0100/)

Finite field · qf-mod30-field-finite-domain-theorem · candidate



[Canonical question JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261030-r01-finite-field-foundations/questions.jsonl) · [English JSONL](https://ocme.evemisslab.com/data/i18n/questions/batch-20261014-advance-20261030-r01-finite-field-foundations.en.jsonl) · [Validation report](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261030-r01-finite-field-foundations/validation-report.json)