# 有限交換環基礎 · 脈衝 R01

本批已通過 Schema、去重、來源快照與獨立答案重算；仍未標記為人工審定。

本題庫已有對應的中英逐步詳解。原題的 candidate 與英文 ai_candidate 狀態保留；AI 覆核、有限計算與 Lean 形式證明分開標示。

batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations · 2026-10-14 · 100 題



## 1. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-qualification；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,0,3,2,5,4,7,6],[2,3,0,1,6,7,4,5],[3,2,1,0,7,6,5,4],[4,5,6,7,0,1,2,3],[5,4,7,6,1,0,3,2],[6,7,4,5,2,3,0,1],[7,6,5,4,3,2,1,0]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,0,1,0,1,0,1],[0,2,0,2,0,2,0,2],[0,3,0,3,0,3,0,3],[0,0,2,2,4,4,6,6],[0,1,2,3,4,5,6,7],[0,2,2,0,4,6,6,4],[0,3,2,1,4,7,6,5]]}。請依 [n,zero,one,noncommuting_pairs,commutative,classification] 欄序回答。

查看答案與解釋(8, 0, 5, [[1, 2, 0, 2], [1, 3, 1, 3], [1, 6, 0, 2], [1, 7, 1, 3], [2, 3, 2, 0], [2, 4, 0, 2], [2, 6, 0, 2], [3, 4, 0, 2], [3, 7, 3, 1], [4, 6, 6, 4], [4, 7, 6, 4], [6, 7, 4, 6]], 否, unital_ring_not_commutative)

Wave4第二層核對=(8, 0, 5, [[1, 2, 0, 2], [1, 3, 1, 3], [1, 6, 0, 2], [1, 7, 1, 3], [2, 3, 2, 0], [2, 4, 0, 2], [2, 6, 0, 2], [3, 4, 0, 2], [3, 7, 3, 1], [4, 6, 6, 4], [4, 7, 6, 4], [6, 7, 4, 6]], 否, unital_ring_not_commutative)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0001/)

有限交換環 · qf-mod28-commring-qualification · candidate





## 2. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-qualification；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,4,5,7,0,6],[2,3,4,5,7,6,1,0],[3,4,5,7,6,0,2,1],[4,5,7,6,0,1,3,2],[5,7,6,0,1,2,4,3],[6,0,1,2,3,4,7,5],[7,6,0,1,2,3,5,4]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,7,0,2,7,4],[0,3,7,1,4,6,5,2],[0,4,0,4,0,4,4,0],[0,5,2,6,4,1,3,7],[0,6,7,5,4,3,1,2],[0,7,4,2,0,7,2,4]]}。請依 [n,zero,one,noncommuting_pairs,commutative,classification] 欄序回答。

查看答案與解釋(8, 0, 1, [], 是, commutative_ring)

Wave4第二層核對=(8, 0, 1, [], 是, commutative_ring)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0002/)

有限交換環 · qf-mod28-commring-qualification · candidate





## 3. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-qualification；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,4,6,7,5,0],[2,3,4,6,5,0,7,1],[3,4,6,5,7,1,0,2],[4,6,5,7,0,2,1,3],[5,7,0,1,2,4,3,6],[6,5,7,0,1,3,2,4],[7,0,1,2,3,6,4,5]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,5,0,4,2,5],[0,3,5,1,4,2,7,6],[0,4,0,4,0,0,4,4],[0,5,4,2,0,4,5,2],[0,6,2,7,4,5,1,3],[0,7,5,6,4,2,3,1]]}。請依 [n,zero,one,noncommuting_pairs,commutative,classification] 欄序回答。

查看答案與解釋(8, 0, 1, [], 是, commutative_ring)

Wave4第二層核對=(8, 0, 1, [], 是, commutative_ring)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0003/)

有限交換環 · qf-mod28-commring-qualification · candidate





## 4. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-qualification；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,4,6,0,7,5],[2,3,4,6,7,1,5,0],[3,4,6,7,5,2,0,1],[4,6,7,5,0,3,1,2],[5,0,1,2,3,7,4,6],[6,7,5,0,1,4,2,3],[7,5,0,1,2,6,3,4]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,7,0,7,2,4],[0,3,7,1,4,6,5,2],[0,4,0,4,0,4,4,0],[0,5,7,6,4,1,3,2],[0,6,2,5,4,3,1,7],[0,7,4,2,0,2,7,4]]}。請依 [n,zero,one,noncommuting_pairs,commutative,classification] 欄序回答。

查看答案與解釋(8, 0, 1, [], 是, commutative_ring)

Wave4第二層核對=(8, 0, 1, [], 是, commutative_ring)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0004/)

有限交換環 · qf-mod28-commring-qualification · candidate





## 5. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-qualification；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,4,7,6,0,5],[2,3,4,7,5,0,1,6],[3,4,7,5,6,1,2,0],[4,7,5,6,0,2,3,1],[5,6,0,1,2,4,7,3],[6,0,1,2,3,7,5,4],[7,5,6,0,1,3,4,2]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,5,0,4,5,2],[0,3,5,1,4,2,7,6],[0,4,0,4,0,0,4,4],[0,5,4,2,0,4,2,5],[0,6,5,7,4,2,1,3],[0,7,2,6,4,5,3,1]]}。請依 [n,zero,one,noncommuting_pairs,commutative,classification] 欄序回答。

查看答案與解釋(8, 0, 1, [], 是, commutative_ring)

Wave4第二層核對=(8, 0, 1, [], 是, commutative_ring)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0005/)

有限交換環 · qf-mod28-commring-qualification · candidate





## 6. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-qualification；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,0,3,2,7,6,5,4],[2,3,0,1,6,7,4,5],[3,2,1,0,5,4,7,6],[4,7,6,5,0,3,2,1],[5,6,7,4,3,0,1,2],[6,5,4,7,2,1,0,3],[7,4,5,6,1,2,3,0]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,0,1,0,1,0,1],[0,2,0,2,0,2,0,2],[0,3,0,3,0,3,0,3],[0,0,2,2,4,6,6,4],[0,3,2,1,4,7,6,5],[0,2,2,0,4,4,6,6],[0,1,2,3,4,5,6,7]]}。請依 [n,zero,one,noncommuting_pairs,commutative,classification] 欄序回答。

查看答案與解釋(8, 0, 7, [[1, 2, 0, 2], [1, 3, 1, 3], [1, 5, 1, 3], [1, 6, 0, 2], [2, 3, 2, 0], [2, 4, 0, 2], [2, 6, 0, 2], [3, 4, 0, 2], [3, 5, 3, 1], [4, 5, 6, 4], [4, 6, 6, 4], [5, 6, 6, 4]], 否, unital_ring_not_commutative)

Wave4第二層核對=(8, 0, 7, [[1, 2, 0, 2], [1, 3, 1, 3], [1, 5, 1, 3], [1, 6, 0, 2], [2, 3, 2, 0], [2, 4, 0, 2], [2, 6, 0, 2], [3, 4, 0, 2], [3, 5, 3, 1], [4, 5, 6, 4], [4, 6, 6, 4], [5, 6, 6, 4]], 否, unital_ring_not_commutative)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0006/)

有限交換環 · qf-mod28-commring-qualification · candidate





## 7. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-qualification；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,5,6,4,7,0],[2,3,5,4,7,6,0,1],[3,5,4,6,0,7,1,2],[4,6,7,0,2,1,3,5],[5,4,6,7,1,0,2,3],[6,7,0,1,3,2,5,4],[7,0,1,2,5,3,4,6]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,6,2,0,5,6],[0,3,6,1,7,5,2,4],[0,4,2,7,1,5,6,3],[0,5,0,5,5,0,0,5],[0,6,5,2,6,0,5,2],[0,7,6,4,3,5,2,1]]}。請依 [n,zero,one,noncommuting_pairs,commutative,classification] 欄序回答。

查看答案與解釋(8, 0, 1, [], 是, commutative_ring)

Wave4第二層核對=(8, 0, 1, [], 是, commutative_ring)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0007/)

有限交換環 · qf-mod28-commring-qualification · candidate





## 8. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-qualification；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,5,7,4,0,6],[2,3,5,4,6,7,1,0],[3,5,4,7,0,6,2,1],[4,7,6,0,2,1,5,3],[5,4,7,6,1,0,3,2],[6,0,1,2,5,3,7,4],[7,6,0,1,3,2,4,5]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,7,2,0,7,5],[0,3,7,1,6,5,4,2],[0,4,2,6,1,5,3,7],[0,5,0,5,5,0,5,0],[0,6,7,4,3,5,1,2],[0,7,5,2,7,0,2,5]]}。請依 [n,zero,one,noncommuting_pairs,commutative,classification] 欄序回答。

查看答案與解釋(8, 0, 1, [], 是, commutative_ring)

Wave4第二層核對=(8, 0, 1, [], 是, commutative_ring)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0008/)

有限交換環 · qf-mod28-commring-qualification · candidate





## 9. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-qualification；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,5,7,6,4,0],[2,3,5,6,0,4,7,1],[3,5,6,4,1,7,0,2],[4,7,0,1,5,2,3,6],[5,6,4,7,2,0,1,3],[6,4,7,0,3,1,2,5],[7,0,1,2,6,3,5,4]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,4,5,0,2,4],[0,3,4,1,2,5,7,6],[0,4,5,2,5,0,4,2],[0,5,0,5,0,0,5,5],[0,6,2,7,4,5,1,3],[0,7,4,6,2,5,3,1]]}。請依 [n,zero,one,noncommuting_pairs,commutative,classification] 欄序回答。

查看答案與解釋(8, 0, 1, [], 是, commutative_ring)

Wave4第二層核對=(8, 0, 1, [], 是, commutative_ring)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0009/)

有限交換環 · qf-mod28-commring-qualification · candidate





## 10. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-qualification；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,5,0,6,7,4],[2,3,5,6,1,7,4,0],[3,5,6,7,2,4,0,1],[4,0,1,2,7,3,5,6],[5,6,7,4,3,0,1,2],[6,7,4,0,5,1,2,3],[7,4,0,1,6,2,3,5]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,7,7,0,2,5],[0,3,7,1,6,5,4,2],[0,4,7,6,1,5,3,2],[0,5,0,5,5,0,5,0],[0,6,2,4,3,5,1,7],[0,7,5,2,2,0,7,5]]}。請依 [n,zero,one,noncommuting_pairs,commutative,classification] 欄序回答。

查看答案與解釋(8, 0, 1, [], 是, commutative_ring)

Wave4第二層核對=(8, 0, 1, [], 是, commutative_ring)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0010/)

有限交換環 · qf-mod28-commring-qualification · candidate





## 11. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-qualification；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,0,3,2,6,7,4,5],[2,3,0,1,5,4,7,6],[3,2,1,0,7,6,5,4],[4,6,5,7,0,2,1,3],[5,7,4,6,2,0,3,1],[6,4,7,5,1,3,0,2],[7,5,6,4,3,1,2,0]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,0,1,0,0,1,1],[0,2,0,2,0,0,2,2],[0,3,0,3,0,0,3,3],[0,2,2,0,4,5,5,4],[0,0,2,2,4,5,4,5],[0,3,2,1,4,5,7,6],[0,1,2,3,4,5,6,7]]}。請依 [n,zero,one,noncommuting_pairs,commutative,classification] 欄序回答。

查看答案與解釋(8, 0, 7, [[1, 2, 0, 2], [1, 3, 1, 3], [1, 4, 0, 2], [1, 6, 1, 3], [2, 3, 2, 0], [2, 4, 0, 2], [2, 5, 0, 2], [3, 5, 0, 2], [3, 6, 3, 1], [4, 5, 5, 4], [4, 6, 5, 4], [5, 6, 4, 5]], 否, unital_ring_not_commutative)

Wave4第二層核對=(8, 0, 7, [[1, 2, 0, 2], [1, 3, 1, 3], [1, 4, 0, 2], [1, 6, 1, 3], [2, 3, 2, 0], [2, 4, 0, 2], [2, 5, 0, 2], [3, 5, 0, 2], [3, 6, 3, 1], [4, 5, 5, 4], [4, 6, 5, 4], [5, 6, 4, 5]], 否, unital_ring_not_commutative)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0011/)

有限交換環 · qf-mod28-commring-qualification · candidate





## 12. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-qualification；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,5,0,7,4,6],[2,3,5,7,1,6,0,4],[3,5,7,6,2,4,1,0],[4,0,1,2,6,3,7,5],[5,7,6,4,3,0,2,1],[6,4,0,1,7,2,5,3],[7,6,4,0,5,1,3,2]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,6,6,0,5,2],[0,3,6,1,7,5,2,4],[0,4,6,7,1,5,2,3],[0,5,0,5,5,0,0,5],[0,6,5,2,2,0,5,6],[0,7,2,4,3,5,6,1]]}。請依 [n,zero,one,noncommuting_pairs,commutative,classification] 欄序回答。

查看答案與解釋(8, 0, 1, [], 是, commutative_ring)

Wave4第二層核對=(8, 0, 1, [], 是, commutative_ring)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0012/)

有限交換環 · qf-mod28-commring-qualification · candidate





## 13. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-qualification；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,6,5,7,4,0],[2,3,6,4,7,0,5,1],[3,6,4,5,0,1,7,2],[4,5,7,0,2,3,1,6],[5,7,0,1,3,6,2,4],[6,4,5,7,1,2,0,3],[7,0,1,2,6,4,3,5]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,5,2,6,0,5],[0,3,5,1,7,2,6,4],[0,4,2,7,1,5,6,3],[0,5,6,2,5,6,0,2],[0,6,0,6,6,0,0,6],[0,7,5,4,3,2,6,1]]}。請依 [n,zero,one,noncommuting_pairs,commutative,classification] 欄序回答。

查看答案與解釋(8, 0, 1, [], 是, commutative_ring)

Wave4第二層核對=(8, 0, 1, [], 是, commutative_ring)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0013/)

有限交換環 · qf-mod28-commring-qualification · candidate





## 14. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-qualification；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,6,7,0,4,5],[2,3,6,4,5,1,7,0],[3,6,4,7,0,2,5,1],[4,7,5,0,2,6,1,3],[5,0,1,2,6,7,3,4],[6,4,7,5,1,3,0,2],[7,5,0,1,3,4,2,6]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,7,2,7,0,6],[0,3,7,1,5,4,6,2],[0,4,2,5,1,3,6,7],[0,5,7,4,3,1,6,2],[0,6,0,6,6,6,0,0],[0,7,6,2,7,2,0,6]]}。請依 [n,zero,one,noncommuting_pairs,commutative,classification] 欄序回答。

查看答案與解釋(8, 0, 1, [], 是, commutative_ring)

Wave4第二層核對=(8, 0, 1, [], 是, commutative_ring)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0014/)

有限交換環 · qf-mod28-commring-qualification · candidate





## 15. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-qualification；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,6,7,4,5,0],[2,3,6,5,0,7,4,1],[3,6,5,4,1,0,7,2],[4,7,0,1,6,3,2,5],[5,4,7,0,3,2,1,6],[6,5,4,7,2,1,0,3],[7,0,1,2,5,6,3,4]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,4,6,2,0,4],[0,3,4,1,2,7,6,5],[0,4,6,2,6,4,0,2],[0,5,2,7,4,1,6,3],[0,6,0,6,0,6,0,6],[0,7,4,5,2,3,6,1]]}。請依 [n,zero,one,noncommuting_pairs,commutative,classification] 欄序回答。

查看答案與解釋(8, 0, 1, [], 是, commutative_ring)

Wave4第二層核對=(8, 0, 1, [], 是, commutative_ring)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0015/)

有限交換環 · qf-mod28-commring-qualification · candidate





## 16. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-qualification；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,0,3,2,7,6,5,4],[2,3,0,1,5,4,7,6],[3,2,1,0,6,7,4,5],[4,7,5,6,0,2,3,1],[5,6,4,7,2,0,1,3],[6,5,7,4,3,1,0,2],[7,4,6,5,1,3,2,0]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,0,1,1,1,0,0],[0,2,0,2,2,2,0,0],[0,3,0,3,3,3,0,0],[0,3,2,1,5,4,6,7],[0,1,2,3,4,5,6,7],[0,0,2,2,7,6,6,7],[0,2,2,0,6,7,6,7]]}。請依 [n,zero,one,noncommuting_pairs,commutative,classification] 欄序回答。

查看答案與解釋(8, 0, 5, [[1, 2, 0, 2], [1, 3, 1, 3], [1, 4, 1, 3], [1, 7, 0, 2], [2, 3, 2, 0], [2, 6, 0, 2], [2, 7, 0, 2], [3, 4, 3, 1], [3, 6, 0, 2], [4, 6, 6, 7], [4, 7, 7, 6], [6, 7, 7, 6]], 否, unital_ring_not_commutative)

Wave4第二層核對=(8, 0, 5, [[1, 2, 0, 2], [1, 3, 1, 3], [1, 4, 1, 3], [1, 7, 0, 2], [2, 3, 2, 0], [2, 6, 0, 2], [2, 7, 0, 2], [3, 4, 3, 1], [3, 6, 0, 2], [4, 6, 6, 7], [4, 7, 7, 6], [6, 7, 7, 6]], 否, unital_ring_not_commutative)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0016/)

有限交換環 · qf-mod28-commring-qualification · candidate





## 17. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-qualification；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,6,5,0,7,4],[2,3,6,7,0,1,4,5],[3,6,7,4,1,2,5,0],[4,5,0,1,6,7,2,3],[5,0,1,2,7,4,3,6],[6,7,4,5,2,3,0,1],[7,4,5,0,3,6,1,2]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,4,6,4,0,2],[0,3,4,1,2,7,6,5],[0,4,6,2,6,2,0,4],[0,5,4,7,2,1,6,3],[0,6,0,6,0,6,0,6],[0,7,2,5,4,3,6,1]]}。請依 [n,zero,one,noncommuting_pairs,commutative,classification] 欄序回答。

查看答案與解釋(8, 0, 1, [], 是, commutative_ring)

Wave4第二層核對=(8, 0, 1, [], 是, commutative_ring)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0017/)

有限交換環 · qf-mod28-commring-qualification · candidate





## 18. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-qualification；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,6,0,4,7,5],[2,3,6,7,1,0,5,4],[3,6,7,5,2,1,4,0],[4,0,1,2,5,7,3,6],[5,4,0,1,7,6,2,3],[6,7,5,4,3,2,0,1],[7,5,4,0,6,3,1,2]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,5,5,6,0,2],[0,3,5,1,7,2,6,4],[0,4,5,7,1,2,6,3],[0,5,6,2,2,6,0,5],[0,6,0,6,6,0,0,6],[0,7,2,4,3,5,6,1]]}。請依 [n,zero,one,noncommuting_pairs,commutative,classification] 欄序回答。

查看答案與解釋(8, 0, 1, [], 是, commutative_ring)

Wave4第二層核對=(8, 0, 1, [], 是, commutative_ring)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0018/)

有限交換環 · qf-mod28-commring-qualification · candidate





## 19. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-qualification；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,7,5,6,0,4],[2,3,7,4,6,0,1,5],[3,7,4,5,0,1,2,6],[4,5,6,0,2,3,7,1],[5,6,0,1,3,7,4,2],[6,0,1,2,7,4,5,3],[7,4,5,6,1,2,3,0]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,5,2,7,5,0],[0,3,5,1,6,2,4,7],[0,4,2,6,1,5,3,7],[0,5,7,2,5,7,2,0],[0,6,5,4,3,2,1,7],[0,7,0,7,7,0,7,0]]}。請依 [n,zero,one,noncommuting_pairs,commutative,classification] 欄序回答。

查看答案與解釋(8, 0, 1, [], 是, commutative_ring)

Wave4第二層核對=(8, 0, 1, [], 是, commutative_ring)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0019/)

有限交換環 · qf-mod28-commring-qualification · candidate





## 20. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-qualification；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,7,6,0,5,4],[2,3,7,4,5,1,0,6],[3,7,4,6,0,2,1,5],[4,6,5,0,2,7,3,1],[5,0,1,2,7,6,4,3],[6,5,0,1,3,4,7,2],[7,4,6,5,1,3,2,0]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,6,2,6,7,0],[0,3,6,1,5,4,2,7],[0,4,2,5,1,3,6,7],[0,5,6,4,3,1,2,7],[0,6,7,2,6,2,7,0],[0,7,0,7,7,7,0,0]]}。請依 [n,zero,one,noncommuting_pairs,commutative,classification] 欄序回答。

查看答案與解釋(8, 0, 1, [], 是, commutative_ring)

Wave4第二層核對=(8, 0, 1, [], 是, commutative_ring)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0020/)

有限交換環 · qf-mod28-commring-qualification · candidate





## 21. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-qualification；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,0,3,2,6,7,4,5],[2,3,0,1,7,6,5,4],[3,2,1,0,5,4,7,6],[4,6,7,5,0,3,1,2],[5,7,6,4,3,0,2,1],[6,4,5,7,1,2,0,3],[7,5,4,6,2,1,3,0]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,0,1,0,1,1,0],[0,2,0,2,0,2,2,0],[0,3,0,3,0,3,3,0],[0,2,2,0,4,4,7,7],[0,1,2,3,4,5,6,7],[0,3,2,1,4,6,5,7],[0,0,2,2,4,7,4,7]]}。請依 [n,zero,one,noncommuting_pairs,commutative,classification] 欄序回答。

查看答案與解釋(8, 0, 5, [[1, 2, 0, 2], [1, 3, 1, 3], [1, 4, 0, 2], [1, 6, 1, 3], [2, 3, 2, 0], [2, 4, 0, 2], [2, 7, 0, 2], [3, 6, 3, 1], [3, 7, 0, 2], [4, 6, 7, 4], [4, 7, 7, 4], [6, 7, 7, 4]], 否, unital_ring_not_commutative)

Wave4第二層核對=(8, 0, 5, [[1, 2, 0, 2], [1, 3, 1, 3], [1, 4, 0, 2], [1, 6, 1, 3], [2, 3, 2, 0], [2, 4, 0, 2], [2, 7, 0, 2], [3, 6, 3, 1], [3, 7, 0, 2], [4, 6, 7, 4], [4, 7, 7, 4], [6, 7, 7, 4]], 否, unital_ring_not_commutative)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0021/)

有限交換環 · qf-mod28-commring-qualification · candidate





## 22. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-qualification；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,7,0,6,4,5],[2,3,7,5,1,4,0,6],[3,7,5,6,2,0,1,4],[4,0,1,2,6,7,5,3],[5,6,4,0,7,2,3,1],[6,4,0,1,5,3,7,2],[7,5,6,4,3,1,2,0]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,6,6,2,7,0],[0,3,6,1,5,4,2,7],[0,4,6,5,1,3,2,7],[0,5,2,4,3,1,6,7],[0,6,7,2,2,6,7,0],[0,7,0,7,7,7,0,0]]}。請依 [n,zero,one,noncommuting_pairs,commutative,classification] 欄序回答。

查看答案與解釋(8, 0, 1, [], 是, commutative_ring)

Wave4第二層核對=(8, 0, 1, [], 是, commutative_ring)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0022/)

有限交換環 · qf-mod28-commring-qualification · candidate





## 23. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-qualification；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,7,5,0,4,6],[2,3,7,6,0,1,5,4],[3,7,6,4,1,2,0,5],[4,5,0,1,7,6,3,2],[5,0,1,2,6,4,7,3],[6,4,5,0,3,7,2,1],[7,6,4,5,2,3,1,0]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,4,7,4,2,0],[0,3,4,1,2,6,5,7],[0,4,7,2,7,2,4,0],[0,5,4,6,2,1,3,7],[0,6,2,5,4,3,1,7],[0,7,0,7,0,7,7,0]]}。請依 [n,zero,one,noncommuting_pairs,commutative,classification] 欄序回答。

查看答案與解釋(8, 0, 1, [], 是, commutative_ring)

Wave4第二層核對=(8, 0, 1, [], 是, commutative_ring)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0023/)

有限交換環 · qf-mod28-commring-qualification · candidate





## 24. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-qualification；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,7,0,4,5,6],[2,3,7,6,1,0,4,5],[3,7,6,5,2,1,0,4],[4,0,1,2,5,6,7,3],[5,4,0,1,6,7,3,2],[6,5,4,0,7,3,2,1],[7,6,5,4,3,2,1,0]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,5,5,7,2,0],[0,3,5,1,6,2,4,7],[0,4,5,6,1,2,3,7],[0,5,7,2,2,7,5,0],[0,6,2,4,3,5,1,7],[0,7,0,7,7,0,7,0]]}。請依 [n,zero,one,noncommuting_pairs,commutative,classification] 欄序回答。

查看答案與解釋(8, 0, 1, [], 是, commutative_ring)

Wave4第二層核對=(8, 0, 1, [], 是, commutative_ring)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0024/)

有限交換環 · qf-mod28-commring-qualification · candidate





## 25. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-qualification；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,4,5,3,6,7,0],[2,4,3,6,5,7,0,1],[3,5,6,0,7,1,2,4],[4,3,5,7,6,0,1,2],[5,6,7,1,0,2,4,3],[6,7,0,2,1,4,3,5],[7,0,1,4,2,3,5,6]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,3,0,6,2,3,6],[0,3,0,0,3,3,0,3],[0,4,6,3,1,7,2,5],[0,5,2,3,7,1,6,4],[0,6,3,0,2,6,3,2],[0,7,6,3,5,4,2,1]]}。請依 [n,zero,one,noncommuting_pairs,commutative,classification] 欄序回答。

查看答案與解釋(8, 0, 1, [], 是, commutative_ring)

Wave4第二層核對=(8, 0, 1, [], 是, commutative_ring)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0025/)

有限交換環 · qf-mod28-commring-qualification · candidate





## 26. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-product-swap；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2],[4,5,6,7,0,1,2,3],[5,6,7,0,1,2,3,4],[6,7,0,1,2,3,4,5],[7,0,1,2,3,4,5,6]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,6,0,2,4,6],[0,3,6,1,4,7,2,5],[0,4,0,4,0,4,0,4],[0,5,2,7,4,1,6,3],[0,6,4,2,0,6,4,2],[0,7,6,5,4,3,2,1]],"a":1,"b":2}。請依 [a,b,ab,ba,equal] 欄序回答。

查看答案與解釋(1, 2, 2, 2, 是)

Wave4第二層核對=(1, 2, 2, 2, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0026/)

有限交換環 · qf-mod28-commring-product-swap · candidate





## 27. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-product-swap；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,4,5,7,0,6],[2,3,4,5,7,6,1,0],[3,4,5,7,6,0,2,1],[4,5,7,6,0,1,3,2],[5,7,6,0,1,2,4,3],[6,0,1,2,3,4,7,5],[7,6,0,1,2,3,5,4]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,7,0,2,7,4],[0,3,7,1,4,6,5,2],[0,4,0,4,0,4,4,0],[0,5,2,6,4,1,3,7],[0,6,7,5,4,3,1,2],[0,7,4,2,0,7,2,4]],"a":2,"b":5}。請依 [a,b,ab,ba,equal] 欄序回答。

查看答案與解釋(2, 5, 2, 2, 是)

Wave4第二層核對=(2, 5, 2, 2, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0027/)

有限交換環 · qf-mod28-commring-product-swap · candidate





## 28. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-product-swap；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,4,6,7,5,0],[2,3,4,6,5,0,7,1],[3,4,6,5,7,1,0,2],[4,6,5,7,0,2,1,3],[5,7,0,1,2,4,3,6],[6,5,7,0,1,3,2,4],[7,0,1,2,3,6,4,5]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,5,0,4,2,5],[0,3,5,1,4,2,7,6],[0,4,0,4,0,0,4,4],[0,5,4,2,0,4,5,2],[0,6,2,7,4,5,1,3],[0,7,5,6,4,2,3,1]],"a":3,"b":0}。請依 [a,b,ab,ba,equal] 欄序回答。

查看答案與解釋(3, 0, 0, 0, 是)

Wave4第二層核對=(3, 0, 0, 0, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0028/)

有限交換環 · qf-mod28-commring-product-swap · candidate





## 29. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-product-swap；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,4,6,0,7,5],[2,3,4,6,7,1,5,0],[3,4,6,7,5,2,0,1],[4,6,7,5,0,3,1,2],[5,0,1,2,3,7,4,6],[6,7,5,0,1,4,2,3],[7,5,0,1,2,6,3,4]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,7,0,7,2,4],[0,3,7,1,4,6,5,2],[0,4,0,4,0,4,4,0],[0,5,7,6,4,1,3,2],[0,6,2,5,4,3,1,7],[0,7,4,2,0,2,7,4]],"a":4,"b":3}。請依 [a,b,ab,ba,equal] 欄序回答。

查看答案與解釋(4, 3, 4, 4, 是)

Wave4第二層核對=(4, 3, 4, 4, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0029/)

有限交換環 · qf-mod28-commring-product-swap · candidate





## 30. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-product-swap；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,4,7,6,0,5],[2,3,4,7,5,0,1,6],[3,4,7,5,6,1,2,0],[4,7,5,6,0,2,3,1],[5,6,0,1,2,4,7,3],[6,0,1,2,3,7,5,4],[7,5,6,0,1,3,4,2]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,5,0,4,5,2],[0,3,5,1,4,2,7,6],[0,4,0,4,0,0,4,4],[0,5,4,2,0,4,2,5],[0,6,5,7,4,2,1,3],[0,7,2,6,4,5,3,1]],"a":7,"b":5}。請依 [a,b,ab,ba,equal] 欄序回答。

查看答案與解釋(7, 5, 5, 5, 是)

Wave4第二層核對=(7, 5, 5, 5, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0030/)

有限交換環 · qf-mod28-commring-product-swap · candidate





## 31. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-product-swap；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,4,7,0,5,6],[2,3,4,7,6,1,0,5],[3,4,7,6,5,2,1,0],[4,7,6,5,0,3,2,1],[5,0,1,2,3,6,7,4],[6,5,0,1,2,7,4,3],[7,6,5,0,1,4,3,2]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,6,0,6,4,2],[0,3,6,1,4,7,2,5],[0,4,0,4,0,4,0,4],[0,5,6,7,4,1,2,3],[0,6,4,2,0,2,4,6],[0,7,2,5,4,3,6,1]],"a":6,"b":1}。請依 [a,b,ab,ba,equal] 欄序回答。

查看答案與解釋(6, 1, 6, 6, 是)

Wave4第二層核對=(6, 1, 6, 6, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0031/)

有限交換環 · qf-mod28-commring-product-swap · candidate





## 32. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-product-swap；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,5,6,4,7,0],[2,3,5,4,7,6,0,1],[3,5,4,6,0,7,1,2],[4,6,7,0,2,1,3,5],[5,4,6,7,1,0,2,3],[6,7,0,1,3,2,5,4],[7,0,1,2,5,3,4,6]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,6,2,0,5,6],[0,3,6,1,7,5,2,4],[0,4,2,7,1,5,6,3],[0,5,0,5,5,0,0,5],[0,6,5,2,6,0,5,2],[0,7,6,4,3,5,2,1]],"a":7,"b":5}。請依 [a,b,ab,ba,equal] 欄序回答。

查看答案與解釋(7, 5, 5, 5, 是)

Wave4第二層核對=(7, 5, 5, 5, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0032/)

有限交換環 · qf-mod28-commring-product-swap · candidate





## 33. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-product-swap；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,5,7,4,0,6],[2,3,5,4,6,7,1,0],[3,5,4,7,0,6,2,1],[4,7,6,0,2,1,5,3],[5,4,7,6,1,0,3,2],[6,0,1,2,5,3,7,4],[7,6,0,1,3,2,4,5]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,7,2,0,7,5],[0,3,7,1,6,5,4,2],[0,4,2,6,1,5,3,7],[0,5,0,5,5,0,5,0],[0,6,7,4,3,5,1,2],[0,7,5,2,7,0,2,5]],"a":0,"b":6}。請依 [a,b,ab,ba,equal] 欄序回答。

查看答案與解釋(0, 6, 0, 0, 是)

Wave4第二層核對=(0, 6, 0, 0, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0033/)

有限交換環 · qf-mod28-commring-product-swap · candidate





## 34. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-product-swap；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,5,7,6,4,0],[2,3,5,6,0,4,7,1],[3,5,6,4,1,7,0,2],[4,7,0,1,5,2,3,6],[5,6,4,7,2,0,1,3],[6,4,7,0,3,1,2,5],[7,0,1,2,6,3,5,4]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,4,5,0,2,4],[0,3,4,1,2,5,7,6],[0,4,5,2,5,0,4,2],[0,5,0,5,0,0,5,5],[0,6,2,7,4,5,1,3],[0,7,4,6,2,5,3,1]],"a":1,"b":2}。請依 [a,b,ab,ba,equal] 欄序回答。

查看答案與解釋(1, 2, 2, 2, 是)

Wave4第二層核對=(1, 2, 2, 2, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0034/)

有限交換環 · qf-mod28-commring-product-swap · candidate





## 35. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-product-swap；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,5,0,6,7,4],[2,3,5,6,1,7,4,0],[3,5,6,7,2,4,0,1],[4,0,1,2,7,3,5,6],[5,6,7,4,3,0,1,2],[6,7,4,0,5,1,2,3],[7,4,0,1,6,2,3,5]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,7,7,0,2,5],[0,3,7,1,6,5,4,2],[0,4,7,6,1,5,3,2],[0,5,0,5,5,0,5,0],[0,6,2,4,3,5,1,7],[0,7,5,2,2,0,7,5]],"a":2,"b":6}。請依 [a,b,ab,ba,equal] 欄序回答。

查看答案與解釋(2, 6, 2, 2, 是)

Wave4第二層核對=(2, 6, 2, 2, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0035/)

有限交換環 · qf-mod28-commring-product-swap · candidate





## 36. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-product-swap；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,5,6,7,0,4],[2,3,5,7,0,4,1,6],[3,5,7,4,1,6,2,0],[4,6,0,1,5,2,7,3],[5,7,4,6,2,0,3,1],[6,0,1,2,7,3,4,5],[7,4,6,0,3,1,5,2]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,4,5,0,4,2],[0,3,4,1,2,5,7,6],[0,4,5,2,5,0,2,4],[0,5,0,5,0,0,5,5],[0,6,4,7,2,5,1,3],[0,7,2,6,4,5,3,1]],"a":3,"b":0}。請依 [a,b,ab,ba,equal] 欄序回答。

查看答案與解釋(3, 0, 0, 0, 是)

Wave4第二層核對=(3, 0, 0, 0, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0036/)

有限交換環 · qf-mod28-commring-product-swap · candidate





## 37. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-product-swap；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,5,0,7,4,6],[2,3,5,7,1,6,0,4],[3,5,7,6,2,4,1,0],[4,0,1,2,6,3,7,5],[5,7,6,4,3,0,2,1],[6,4,0,1,7,2,5,3],[7,6,4,0,5,1,3,2]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,6,6,0,5,2],[0,3,6,1,7,5,2,4],[0,4,6,7,1,5,2,3],[0,5,0,5,5,0,0,5],[0,6,5,2,2,0,5,6],[0,7,2,4,3,5,6,1]],"a":5,"b":3}。請依 [a,b,ab,ba,equal] 欄序回答。

查看答案與解釋(5, 3, 5, 5, 是)

Wave4第二層核對=(5, 3, 5, 5, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0037/)

有限交換環 · qf-mod28-commring-product-swap · candidate





## 38. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-product-swap；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,6,5,7,4,0],[2,3,6,4,7,0,5,1],[3,6,4,5,0,1,7,2],[4,5,7,0,2,3,1,6],[5,7,0,1,3,6,2,4],[6,4,5,7,1,2,0,3],[7,0,1,2,6,4,3,5]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,5,2,6,0,5],[0,3,5,1,7,2,6,4],[0,4,2,7,1,5,6,3],[0,5,6,2,5,6,0,2],[0,6,0,6,6,0,0,6],[0,7,5,4,3,2,6,1]],"a":4,"b":5}。請依 [a,b,ab,ba,equal] 欄序回答。

查看答案與解釋(4, 5, 5, 5, 是)

Wave4第二層核對=(4, 5, 5, 5, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0038/)

有限交換環 · qf-mod28-commring-product-swap · candidate





## 39. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-product-swap；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,6,7,0,4,5],[2,3,6,4,5,1,7,0],[3,6,4,7,0,2,5,1],[4,7,5,0,2,6,1,3],[5,0,1,2,6,7,3,4],[6,4,7,5,1,3,0,2],[7,5,0,1,3,4,2,6]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,7,2,7,0,6],[0,3,7,1,5,4,6,2],[0,4,2,5,1,3,6,7],[0,5,7,4,3,1,6,2],[0,6,0,6,6,6,0,0],[0,7,6,2,7,2,0,6]],"a":7,"b":1}。請依 [a,b,ab,ba,equal] 欄序回答。

查看答案與解釋(7, 1, 7, 7, 是)

Wave4第二層核對=(7, 1, 7, 7, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0039/)

有限交換環 · qf-mod28-commring-product-swap · candidate





## 40. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-product-swap；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,6,7,4,5,0],[2,3,6,5,0,7,4,1],[3,6,5,4,1,0,7,2],[4,7,0,1,6,3,2,5],[5,4,7,0,3,2,1,6],[6,5,4,7,2,1,0,3],[7,0,1,2,5,6,3,4]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,4,6,2,0,4],[0,3,4,1,2,7,6,5],[0,4,6,2,6,4,0,2],[0,5,2,7,4,1,6,3],[0,6,0,6,0,6,0,6],[0,7,4,5,2,3,6,1]],"a":7,"b":6}。請依 [a,b,ab,ba,equal] 欄序回答。

查看答案與解釋(7, 6, 6, 6, 是)

Wave4第二層核對=(7, 6, 6, 6, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0040/)

有限交換環 · qf-mod28-commring-product-swap · candidate





## 41. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-product-swap；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,6,0,7,5,4],[2,3,6,5,1,4,7,0],[3,6,5,7,2,0,4,1],[4,0,1,2,7,6,3,5],[5,7,4,0,6,2,1,3],[6,5,7,4,3,1,0,2],[7,4,0,1,5,3,2,6]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,7,7,2,0,6],[0,3,7,1,5,4,6,2],[0,4,7,5,1,3,6,2],[0,5,2,4,3,1,6,7],[0,6,0,6,6,6,0,0],[0,7,6,2,2,7,0,6]],"a":0,"b":4}。請依 [a,b,ab,ba,equal] 欄序回答。

查看答案與解釋(0, 4, 0, 0, 是)

Wave4第二層核對=(0, 4, 0, 0, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0041/)

有限交換環 · qf-mod28-commring-product-swap · candidate





## 42. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-product-swap；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,6,5,0,7,4],[2,3,6,7,0,1,4,5],[3,6,7,4,1,2,5,0],[4,5,0,1,6,7,2,3],[5,0,1,2,7,4,3,6],[6,7,4,5,2,3,0,1],[7,4,5,0,3,6,1,2]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,4,6,4,0,2],[0,3,4,1,2,7,6,5],[0,4,6,2,6,2,0,4],[0,5,4,7,2,1,6,3],[0,6,0,6,0,6,0,6],[0,7,2,5,4,3,6,1]],"a":1,"b":2}。請依 [a,b,ab,ba,equal] 欄序回答。

查看答案與解釋(1, 2, 2, 2, 是)

Wave4第二層核對=(1, 2, 2, 2, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0042/)

有限交換環 · qf-mod28-commring-product-swap · candidate





## 43. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-product-swap；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,6,0,4,7,5],[2,3,6,7,1,0,5,4],[3,6,7,5,2,1,4,0],[4,0,1,2,5,7,3,6],[5,4,0,1,7,6,2,3],[6,7,5,4,3,2,0,1],[7,5,4,0,6,3,1,2]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,5,5,6,0,2],[0,3,5,1,7,2,6,4],[0,4,5,7,1,2,6,3],[0,5,6,2,2,6,0,5],[0,6,0,6,6,0,0,6],[0,7,2,4,3,5,6,1]],"a":2,"b":7}。請依 [a,b,ab,ba,equal] 欄序回答。

查看答案與解釋(2, 7, 2, 2, 是)

Wave4第二層核對=(2, 7, 2, 2, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0043/)

有限交換環 · qf-mod28-commring-product-swap · candidate





## 44. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-product-swap；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,7,5,6,0,4],[2,3,7,4,6,0,1,5],[3,7,4,5,0,1,2,6],[4,5,6,0,2,3,7,1],[5,6,0,1,3,7,4,2],[6,0,1,2,7,4,5,3],[7,4,5,6,1,2,3,0]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,5,2,7,5,0],[0,3,5,1,6,2,4,7],[0,4,2,6,1,5,3,7],[0,5,7,2,5,7,2,0],[0,6,5,4,3,2,1,7],[0,7,0,7,7,0,7,0]],"a":3,"b":0}。請依 [a,b,ab,ba,equal] 欄序回答。

查看答案與解釋(3, 0, 0, 0, 是)

Wave4第二層核對=(3, 0, 0, 0, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0044/)

有限交換環 · qf-mod28-commring-product-swap · candidate





## 45. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-product-swap；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,7,6,0,5,4],[2,3,7,4,5,1,0,6],[3,7,4,6,0,2,1,5],[4,6,5,0,2,7,3,1],[5,0,1,2,7,6,4,3],[6,5,0,1,3,4,7,2],[7,4,6,5,1,3,2,0]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,6,2,6,7,0],[0,3,6,1,5,4,2,7],[0,4,2,5,1,3,6,7],[0,5,6,4,3,1,2,7],[0,6,7,2,6,2,7,0],[0,7,0,7,7,7,0,0]],"a":7,"b":3}。請依 [a,b,ab,ba,equal] 欄序回答。

查看答案與解釋(7, 3, 7, 7, 是)

Wave4第二層核對=(7, 3, 7, 7, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0045/)

有限交換環 · qf-mod28-commring-product-swap · candidate





## 46. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-product-swap；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,7,6,4,0,5],[2,3,7,5,0,6,1,4],[3,7,5,4,1,0,2,6],[4,6,0,1,7,3,5,2],[5,4,6,0,3,2,7,1],[6,0,1,2,5,7,4,3],[7,5,4,6,2,1,3,0]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,4,7,2,4,0],[0,3,4,1,2,6,5,7],[0,4,7,2,7,4,2,0],[0,5,2,6,4,1,3,7],[0,6,4,5,2,3,1,7],[0,7,0,7,0,7,7,0]],"a":5,"b":4}。請依 [a,b,ab,ba,equal] 欄序回答。

查看答案與解釋(5, 4, 4, 4, 是)

Wave4第二層核對=(5, 4, 4, 4, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0046/)

有限交換環 · qf-mod28-commring-product-swap · candidate





## 47. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-product-swap；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,7,0,6,4,5],[2,3,7,5,1,4,0,6],[3,7,5,6,2,0,1,4],[4,0,1,2,6,7,5,3],[5,6,4,0,7,2,3,1],[6,4,0,1,5,3,7,2],[7,5,6,4,3,1,2,0]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,6,6,2,7,0],[0,3,6,1,5,4,2,7],[0,4,6,5,1,3,2,7],[0,5,2,4,3,1,6,7],[0,6,7,2,2,6,7,0],[0,7,0,7,7,7,0,0]],"a":6,"b":1}。請依 [a,b,ab,ba,equal] 欄序回答。

查看答案與解釋(6, 1, 6, 6, 是)

Wave4第二層核對=(6, 1, 6, 6, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0047/)

有限交換環 · qf-mod28-commring-product-swap · candidate





## 48. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-product-swap；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,7,5,0,4,6],[2,3,7,6,0,1,5,4],[3,7,6,4,1,2,0,5],[4,5,0,1,7,6,3,2],[5,0,1,2,6,4,7,3],[6,4,5,0,3,7,2,1],[7,6,4,5,2,3,1,0]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,4,7,4,2,0],[0,3,4,1,2,6,5,7],[0,4,7,2,7,2,4,0],[0,5,4,6,2,1,3,7],[0,6,2,5,4,3,1,7],[0,7,0,7,0,7,7,0]],"a":5,"b":7}。請依 [a,b,ab,ba,equal] 欄序回答。

查看答案與解釋(5, 7, 7, 7, 是)

Wave4第二層核對=(5, 7, 7, 7, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0048/)

有限交換環 · qf-mod28-commring-product-swap · candidate





## 49. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-product-swap；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,7,0,4,5,6],[2,3,7,6,1,0,4,5],[3,7,6,5,2,1,0,4],[4,0,1,2,5,6,7,3],[5,4,0,1,6,7,3,2],[6,5,4,0,7,3,2,1],[7,6,5,4,3,2,1,0]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,5,5,7,2,0],[0,3,5,1,6,2,4,7],[0,4,5,6,1,2,3,7],[0,5,7,2,2,7,5,0],[0,6,2,4,3,5,1,7],[0,7,0,7,7,0,7,0]],"a":0,"b":4}。請依 [a,b,ab,ba,equal] 欄序回答。

查看答案與解釋(0, 4, 0, 0, 是)

Wave4第二層核對=(0, 4, 0, 0, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0049/)

有限交換環 · qf-mod28-commring-product-swap · candidate





## 50. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-product-swap；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,4,5,3,6,7,0],[2,4,3,6,5,7,0,1],[3,5,6,0,7,1,2,4],[4,3,5,7,6,0,1,2],[5,6,7,1,0,2,4,3],[6,7,0,2,1,4,3,5],[7,0,1,4,2,3,5,6]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,3,0,6,2,3,6],[0,3,0,0,3,3,0,3],[0,4,6,3,1,7,2,5],[0,5,2,3,7,1,6,4],[0,6,3,0,2,6,3,2],[0,7,6,3,5,4,2,1]],"a":1,"b":2}。請依 [a,b,ab,ba,equal] 欄序回答。

查看答案與解釋(1, 2, 2, 2, 是)

Wave4第二層核對=(1, 2, 2, 2, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0050/)

有限交換環 · qf-mod28-commring-product-swap · candidate





## 51. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-power-product；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2],[4,5,6,7,0,1,2,3],[5,6,7,0,1,2,3,4],[6,7,0,1,2,3,4,5],[7,0,1,2,3,4,5,6]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,6,0,2,4,6],[0,3,6,1,4,7,2,5],[0,4,0,4,0,4,0,4],[0,5,2,7,4,1,6,3],[0,6,4,2,0,6,4,2],[0,7,6,5,4,3,2,1]],"a":1,"b":2,"exponent":1}。請依 [a,b,exponent,ab,lhs,rhs,equal] 欄序回答。

查看答案與解釋(1, 2, 1, 2, 2, 2, 是)

Wave4第二層核對=(1, 2, 1, 2, 2, 2, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0051/)

有限交換環 · qf-mod28-commring-power-product · candidate





## 52. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-power-product；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,4,5,7,0,6],[2,3,4,5,7,6,1,0],[3,4,5,7,6,0,2,1],[4,5,7,6,0,1,3,2],[5,7,6,0,1,2,4,3],[6,0,1,2,3,4,7,5],[7,6,0,1,2,3,5,4]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,7,0,2,7,4],[0,3,7,1,4,6,5,2],[0,4,0,4,0,4,4,0],[0,5,2,6,4,1,3,7],[0,6,7,5,4,3,1,2],[0,7,4,2,0,7,2,4]],"a":2,"b":5,"exponent":2}。請依 [a,b,exponent,ab,lhs,rhs,equal] 欄序回答。

查看答案與解釋(2, 5, 2, 2, 4, 4, 是)

Wave4第二層核對=(2, 5, 2, 2, 4, 4, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0052/)

有限交換環 · qf-mod28-commring-power-product · candidate





## 53. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-power-product；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,4,6,7,5,0],[2,3,4,6,5,0,7,1],[3,4,6,5,7,1,0,2],[4,6,5,7,0,2,1,3],[5,7,0,1,2,4,3,6],[6,5,7,0,1,3,2,4],[7,0,1,2,3,6,4,5]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,5,0,4,2,5],[0,3,5,1,4,2,7,6],[0,4,0,4,0,0,4,4],[0,5,4,2,0,4,5,2],[0,6,2,7,4,5,1,3],[0,7,5,6,4,2,3,1]],"a":3,"b":0,"exponent":3}。請依 [a,b,exponent,ab,lhs,rhs,equal] 欄序回答。

查看答案與解釋(3, 0, 3, 0, 0, 0, 是)

Wave4第二層核對=(3, 0, 3, 0, 0, 0, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0053/)

有限交換環 · qf-mod28-commring-power-product · candidate





## 54. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-power-product；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,4,6,0,7,5],[2,3,4,6,7,1,5,0],[3,4,6,7,5,2,0,1],[4,6,7,5,0,3,1,2],[5,0,1,2,3,7,4,6],[6,7,5,0,1,4,2,3],[7,5,0,1,2,6,3,4]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,7,0,7,2,4],[0,3,7,1,4,6,5,2],[0,4,0,4,0,4,4,0],[0,5,7,6,4,1,3,2],[0,6,2,5,4,3,1,7],[0,7,4,2,0,2,7,4]],"a":4,"b":3,"exponent":4}。請依 [a,b,exponent,ab,lhs,rhs,equal] 欄序回答。

查看答案與解釋(4, 3, 4, 4, 0, 0, 是)

Wave4第二層核對=(4, 3, 4, 4, 0, 0, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0054/)

有限交換環 · qf-mod28-commring-power-product · candidate





## 55. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-power-product；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,4,7,6,0,5],[2,3,4,7,5,0,1,6],[3,4,7,5,6,1,2,0],[4,7,5,6,0,2,3,1],[5,6,0,1,2,4,7,3],[6,0,1,2,3,7,5,4],[7,5,6,0,1,3,4,2]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,5,0,4,5,2],[0,3,5,1,4,2,7,6],[0,4,0,4,0,0,4,4],[0,5,4,2,0,4,2,5],[0,6,5,7,4,2,1,3],[0,7,2,6,4,5,3,1]],"a":7,"b":5,"exponent":5}。請依 [a,b,exponent,ab,lhs,rhs,equal] 欄序回答。

查看答案與解釋(7, 5, 5, 5, 0, 0, 是)

Wave4第二層核對=(7, 5, 5, 5, 0, 0, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0055/)

有限交換環 · qf-mod28-commring-power-product · candidate





## 56. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-power-product；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,4,7,0,5,6],[2,3,4,7,6,1,0,5],[3,4,7,6,5,2,1,0],[4,7,6,5,0,3,2,1],[5,0,1,2,3,6,7,4],[6,5,0,1,2,7,4,3],[7,6,5,0,1,4,3,2]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,6,0,6,4,2],[0,3,6,1,4,7,2,5],[0,4,0,4,0,4,0,4],[0,5,6,7,4,1,2,3],[0,6,4,2,0,2,4,6],[0,7,2,5,4,3,6,1]],"a":6,"b":1,"exponent":6}。請依 [a,b,exponent,ab,lhs,rhs,equal] 欄序回答。

查看答案與解釋(6, 1, 6, 6, 0, 0, 是)

Wave4第二層核對=(6, 1, 6, 6, 0, 0, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0056/)

有限交換環 · qf-mod28-commring-power-product · candidate





## 57. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-power-product；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,5,6,4,7,0],[2,3,5,4,7,6,0,1],[3,5,4,6,0,7,1,2],[4,6,7,0,2,1,3,5],[5,4,6,7,1,0,2,3],[6,7,0,1,3,2,5,4],[7,0,1,2,5,3,4,6]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,6,2,0,5,6],[0,3,6,1,7,5,2,4],[0,4,2,7,1,5,6,3],[0,5,0,5,5,0,0,5],[0,6,5,2,6,0,5,2],[0,7,6,4,3,5,2,1]],"a":7,"b":5,"exponent":7}。請依 [a,b,exponent,ab,lhs,rhs,equal] 欄序回答。

查看答案與解釋(7, 5, 7, 5, 0, 0, 是)

Wave4第二層核對=(7, 5, 7, 5, 0, 0, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0057/)

有限交換環 · qf-mod28-commring-power-product · candidate





## 58. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-power-product；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,5,7,4,0,6],[2,3,5,4,6,7,1,0],[3,5,4,7,0,6,2,1],[4,7,6,0,2,1,5,3],[5,4,7,6,1,0,3,2],[6,0,1,2,5,3,7,4],[7,6,0,1,3,2,4,5]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,7,2,0,7,5],[0,3,7,1,6,5,4,2],[0,4,2,6,1,5,3,7],[0,5,0,5,5,0,5,0],[0,6,7,4,3,5,1,2],[0,7,5,2,7,0,2,5]],"a":0,"b":6,"exponent":1}。請依 [a,b,exponent,ab,lhs,rhs,equal] 欄序回答。

查看答案與解釋(0, 6, 1, 0, 0, 0, 是)

Wave4第二層核對=(0, 6, 1, 0, 0, 0, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0058/)

有限交換環 · qf-mod28-commring-power-product · candidate





## 59. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-power-product；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,5,7,6,4,0],[2,3,5,6,0,4,7,1],[3,5,6,4,1,7,0,2],[4,7,0,1,5,2,3,6],[5,6,4,7,2,0,1,3],[6,4,7,0,3,1,2,5],[7,0,1,2,6,3,5,4]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,4,5,0,2,4],[0,3,4,1,2,5,7,6],[0,4,5,2,5,0,4,2],[0,5,0,5,0,0,5,5],[0,6,2,7,4,5,1,3],[0,7,4,6,2,5,3,1]],"a":1,"b":2,"exponent":2}。請依 [a,b,exponent,ab,lhs,rhs,equal] 欄序回答。

查看答案與解釋(1, 2, 2, 2, 5, 5, 是)

Wave4第二層核對=(1, 2, 2, 2, 5, 5, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0059/)

有限交換環 · qf-mod28-commring-power-product · candidate





## 60. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-power-product；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,5,0,6,7,4],[2,3,5,6,1,7,4,0],[3,5,6,7,2,4,0,1],[4,0,1,2,7,3,5,6],[5,6,7,4,3,0,1,2],[6,7,4,0,5,1,2,3],[7,4,0,1,6,2,3,5]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,7,7,0,2,5],[0,3,7,1,6,5,4,2],[0,4,7,6,1,5,3,2],[0,5,0,5,5,0,5,0],[0,6,2,4,3,5,1,7],[0,7,5,2,2,0,7,5]],"a":2,"b":6,"exponent":3}。請依 [a,b,exponent,ab,lhs,rhs,equal] 欄序回答。

查看答案與解釋(2, 6, 3, 2, 0, 0, 是)

Wave4第二層核對=(2, 6, 3, 2, 0, 0, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0060/)

有限交換環 · qf-mod28-commring-power-product · candidate





## 61. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-power-product；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,5,6,7,0,4],[2,3,5,7,0,4,1,6],[3,5,7,4,1,6,2,0],[4,6,0,1,5,2,7,3],[5,7,4,6,2,0,3,1],[6,0,1,2,7,3,4,5],[7,4,6,0,3,1,5,2]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,4,5,0,4,2],[0,3,4,1,2,5,7,6],[0,4,5,2,5,0,2,4],[0,5,0,5,0,0,5,5],[0,6,4,7,2,5,1,3],[0,7,2,6,4,5,3,1]],"a":3,"b":0,"exponent":4}。請依 [a,b,exponent,ab,lhs,rhs,equal] 欄序回答。

查看答案與解釋(3, 0, 4, 0, 0, 0, 是)

Wave4第二層核對=(3, 0, 4, 0, 0, 0, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0061/)

有限交換環 · qf-mod28-commring-power-product · candidate





## 62. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-power-product；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,5,0,7,4,6],[2,3,5,7,1,6,0,4],[3,5,7,6,2,4,1,0],[4,0,1,2,6,3,7,5],[5,7,6,4,3,0,2,1],[6,4,0,1,7,2,5,3],[7,6,4,0,5,1,3,2]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,6,6,0,5,2],[0,3,6,1,7,5,2,4],[0,4,6,7,1,5,2,3],[0,5,0,5,5,0,0,5],[0,6,5,2,2,0,5,6],[0,7,2,4,3,5,6,1]],"a":5,"b":3,"exponent":5}。請依 [a,b,exponent,ab,lhs,rhs,equal] 欄序回答。

查看答案與解釋(5, 3, 5, 5, 0, 0, 是)

Wave4第二層核對=(5, 3, 5, 5, 0, 0, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0062/)

有限交換環 · qf-mod28-commring-power-product · candidate





## 63. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-power-product；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,6,5,7,4,0],[2,3,6,4,7,0,5,1],[3,6,4,5,0,1,7,2],[4,5,7,0,2,3,1,6],[5,7,0,1,3,6,2,4],[6,4,5,7,1,2,0,3],[7,0,1,2,6,4,3,5]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,5,2,6,0,5],[0,3,5,1,7,2,6,4],[0,4,2,7,1,5,6,3],[0,5,6,2,5,6,0,2],[0,6,0,6,6,0,0,6],[0,7,5,4,3,2,6,1]],"a":4,"b":5,"exponent":6}。請依 [a,b,exponent,ab,lhs,rhs,equal] 欄序回答。

查看答案與解釋(4, 5, 6, 5, 0, 0, 是)

Wave4第二層核對=(4, 5, 6, 5, 0, 0, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0063/)

有限交換環 · qf-mod28-commring-power-product · candidate





## 64. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-power-product；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,6,7,0,4,5],[2,3,6,4,5,1,7,0],[3,6,4,7,0,2,5,1],[4,7,5,0,2,6,1,3],[5,0,1,2,6,7,3,4],[6,4,7,5,1,3,0,2],[7,5,0,1,3,4,2,6]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,7,2,7,0,6],[0,3,7,1,5,4,6,2],[0,4,2,5,1,3,6,7],[0,5,7,4,3,1,6,2],[0,6,0,6,6,6,0,0],[0,7,6,2,7,2,0,6]],"a":7,"b":1,"exponent":7}。請依 [a,b,exponent,ab,lhs,rhs,equal] 欄序回答。

查看答案與解釋(7, 1, 7, 7, 0, 0, 是)

Wave4第二層核對=(7, 1, 7, 7, 0, 0, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0064/)

有限交換環 · qf-mod28-commring-power-product · candidate





## 65. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-power-product；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,6,7,4,5,0],[2,3,6,5,0,7,4,1],[3,6,5,4,1,0,7,2],[4,7,0,1,6,3,2,5],[5,4,7,0,3,2,1,6],[6,5,4,7,2,1,0,3],[7,0,1,2,5,6,3,4]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,4,6,2,0,4],[0,3,4,1,2,7,6,5],[0,4,6,2,6,4,0,2],[0,5,2,7,4,1,6,3],[0,6,0,6,0,6,0,6],[0,7,4,5,2,3,6,1]],"a":7,"b":6,"exponent":1}。請依 [a,b,exponent,ab,lhs,rhs,equal] 欄序回答。

查看答案與解釋(7, 6, 1, 6, 6, 6, 是)

Wave4第二層核對=(7, 6, 1, 6, 6, 6, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0065/)

有限交換環 · qf-mod28-commring-power-product · candidate





## 66. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-power-product；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,6,0,7,5,4],[2,3,6,5,1,4,7,0],[3,6,5,7,2,0,4,1],[4,0,1,2,7,6,3,5],[5,7,4,0,6,2,1,3],[6,5,7,4,3,1,0,2],[7,4,0,1,5,3,2,6]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,7,7,2,0,6],[0,3,7,1,5,4,6,2],[0,4,7,5,1,3,6,2],[0,5,2,4,3,1,6,7],[0,6,0,6,6,6,0,0],[0,7,6,2,2,7,0,6]],"a":0,"b":4,"exponent":2}。請依 [a,b,exponent,ab,lhs,rhs,equal] 欄序回答。

查看答案與解釋(0, 4, 2, 0, 0, 0, 是)

Wave4第二層核對=(0, 4, 2, 0, 0, 0, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0066/)

有限交換環 · qf-mod28-commring-power-product · candidate





## 67. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-power-product；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,6,5,0,7,4],[2,3,6,7,0,1,4,5],[3,6,7,4,1,2,5,0],[4,5,0,1,6,7,2,3],[5,0,1,2,7,4,3,6],[6,7,4,5,2,3,0,1],[7,4,5,0,3,6,1,2]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,4,6,4,0,2],[0,3,4,1,2,7,6,5],[0,4,6,2,6,2,0,4],[0,5,4,7,2,1,6,3],[0,6,0,6,0,6,0,6],[0,7,2,5,4,3,6,1]],"a":1,"b":2,"exponent":3}。請依 [a,b,exponent,ab,lhs,rhs,equal] 欄序回答。

查看答案與解釋(1, 2, 3, 2, 0, 0, 是)

Wave4第二層核對=(1, 2, 3, 2, 0, 0, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0067/)

有限交換環 · qf-mod28-commring-power-product · candidate





## 68. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-power-product；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,6,0,4,7,5],[2,3,6,7,1,0,5,4],[3,6,7,5,2,1,4,0],[4,0,1,2,5,7,3,6],[5,4,0,1,7,6,2,3],[6,7,5,4,3,2,0,1],[7,5,4,0,6,3,1,2]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,5,5,6,0,2],[0,3,5,1,7,2,6,4],[0,4,5,7,1,2,6,3],[0,5,6,2,2,6,0,5],[0,6,0,6,6,0,0,6],[0,7,2,4,3,5,6,1]],"a":2,"b":7,"exponent":4}。請依 [a,b,exponent,ab,lhs,rhs,equal] 欄序回答。

查看答案與解釋(2, 7, 4, 2, 0, 0, 是)

Wave4第二層核對=(2, 7, 4, 2, 0, 0, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0068/)

有限交換環 · qf-mod28-commring-power-product · candidate





## 69. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-power-product；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,7,5,6,0,4],[2,3,7,4,6,0,1,5],[3,7,4,5,0,1,2,6],[4,5,6,0,2,3,7,1],[5,6,0,1,3,7,4,2],[6,0,1,2,7,4,5,3],[7,4,5,6,1,2,3,0]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,5,2,7,5,0],[0,3,5,1,6,2,4,7],[0,4,2,6,1,5,3,7],[0,5,7,2,5,7,2,0],[0,6,5,4,3,2,1,7],[0,7,0,7,7,0,7,0]],"a":3,"b":0,"exponent":5}。請依 [a,b,exponent,ab,lhs,rhs,equal] 欄序回答。

查看答案與解釋(3, 0, 5, 0, 0, 0, 是)

Wave4第二層核對=(3, 0, 5, 0, 0, 0, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0069/)

有限交換環 · qf-mod28-commring-power-product · candidate





## 70. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-power-product；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,7,6,0,5,4],[2,3,7,4,5,1,0,6],[3,7,4,6,0,2,1,5],[4,6,5,0,2,7,3,1],[5,0,1,2,7,6,4,3],[6,5,0,1,3,4,7,2],[7,4,6,5,1,3,2,0]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,6,2,6,7,0],[0,3,6,1,5,4,2,7],[0,4,2,5,1,3,6,7],[0,5,6,4,3,1,2,7],[0,6,7,2,6,2,7,0],[0,7,0,7,7,7,0,0]],"a":7,"b":3,"exponent":6}。請依 [a,b,exponent,ab,lhs,rhs,equal] 欄序回答。

查看答案與解釋(7, 3, 6, 7, 0, 0, 是)

Wave4第二層核對=(7, 3, 6, 7, 0, 0, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0070/)

有限交換環 · qf-mod28-commring-power-product · candidate





## 71. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-power-product；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,7,6,4,0,5],[2,3,7,5,0,6,1,4],[3,7,5,4,1,0,2,6],[4,6,0,1,7,3,5,2],[5,4,6,0,3,2,7,1],[6,0,1,2,5,7,4,3],[7,5,4,6,2,1,3,0]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,4,7,2,4,0],[0,3,4,1,2,6,5,7],[0,4,7,2,7,4,2,0],[0,5,2,6,4,1,3,7],[0,6,4,5,2,3,1,7],[0,7,0,7,0,7,7,0]],"a":5,"b":4,"exponent":7}。請依 [a,b,exponent,ab,lhs,rhs,equal] 欄序回答。

查看答案與解釋(5, 4, 7, 4, 0, 0, 是)

Wave4第二層核對=(5, 4, 7, 4, 0, 0, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0071/)

有限交換環 · qf-mod28-commring-power-product · candidate





## 72. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-power-product；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,7,0,6,4,5],[2,3,7,5,1,4,0,6],[3,7,5,6,2,0,1,4],[4,0,1,2,6,7,5,3],[5,6,4,0,7,2,3,1],[6,4,0,1,5,3,7,2],[7,5,6,4,3,1,2,0]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,6,6,2,7,0],[0,3,6,1,5,4,2,7],[0,4,6,5,1,3,2,7],[0,5,2,4,3,1,6,7],[0,6,7,2,2,6,7,0],[0,7,0,7,7,7,0,0]],"a":6,"b":1,"exponent":1}。請依 [a,b,exponent,ab,lhs,rhs,equal] 欄序回答。

查看答案與解釋(6, 1, 1, 6, 6, 6, 是)

Wave4第二層核對=(6, 1, 1, 6, 6, 6, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0072/)

有限交換環 · qf-mod28-commring-power-product · candidate





## 73. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-power-product；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,7,5,0,4,6],[2,3,7,6,0,1,5,4],[3,7,6,4,1,2,0,5],[4,5,0,1,7,6,3,2],[5,0,1,2,6,4,7,3],[6,4,5,0,3,7,2,1],[7,6,4,5,2,3,1,0]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,4,7,4,2,0],[0,3,4,1,2,6,5,7],[0,4,7,2,7,2,4,0],[0,5,4,6,2,1,3,7],[0,6,2,5,4,3,1,7],[0,7,0,7,0,7,7,0]],"a":5,"b":7,"exponent":2}。請依 [a,b,exponent,ab,lhs,rhs,equal] 欄序回答。

查看答案與解釋(5, 7, 2, 7, 0, 0, 是)

Wave4第二層核對=(5, 7, 2, 7, 0, 0, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0073/)

有限交換環 · qf-mod28-commring-power-product · candidate





## 74. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-power-product；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,7,0,4,5,6],[2,3,7,6,1,0,4,5],[3,7,6,5,2,1,0,4],[4,0,1,2,5,6,7,3],[5,4,0,1,6,7,3,2],[6,5,4,0,7,3,2,1],[7,6,5,4,3,2,1,0]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,5,5,7,2,0],[0,3,5,1,6,2,4,7],[0,4,5,6,1,2,3,7],[0,5,7,2,2,7,5,0],[0,6,2,4,3,5,1,7],[0,7,0,7,7,0,7,0]],"a":0,"b":4,"exponent":3}。請依 [a,b,exponent,ab,lhs,rhs,equal] 欄序回答。

查看答案與解釋(0, 4, 3, 0, 0, 0, 是)

Wave4第二層核對=(0, 4, 3, 0, 0, 0, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0074/)

有限交換環 · qf-mod28-commring-power-product · candidate





## 75. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-power-product；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,4,5,3,6,7,0],[2,4,3,6,5,7,0,1],[3,5,6,0,7,1,2,4],[4,3,5,7,6,0,1,2],[5,6,7,1,0,2,4,3],[6,7,0,2,1,4,3,5],[7,0,1,4,2,3,5,6]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,3,0,6,2,3,6],[0,3,0,0,3,3,0,3],[0,4,6,3,1,7,2,5],[0,5,2,3,7,1,6,4],[0,6,3,0,2,6,3,2],[0,7,6,3,5,4,2,1]],"a":1,"b":2,"exponent":4}。請依 [a,b,exponent,ab,lhs,rhs,equal] 欄序回答。

查看答案與解釋(1, 2, 4, 2, 0, 0, 是)

Wave4第二層核對=(1, 2, 4, 2, 0, 0, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0075/)

有限交換環 · qf-mod28-commring-power-product · candidate





## 76. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-binomial-square；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,4,5,6,7,0],[2,3,4,5,6,7,0,1],[3,4,5,6,7,0,1,2],[4,5,6,7,0,1,2,3],[5,6,7,0,1,2,3,4],[6,7,0,1,2,3,4,5],[7,0,1,2,3,4,5,6]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,6,0,2,4,6],[0,3,6,1,4,7,2,5],[0,4,0,4,0,4,0,4],[0,5,2,7,4,1,6,3],[0,6,4,2,0,6,4,2],[0,7,6,5,4,3,2,1]],"a":1,"b":2}。請依 [a,b,sum,lhs,a2,ab,b2,rhs,equal] 欄序回答。

查看答案與解釋(1, 2, 3, 1, 1, 2, 4, 1, 是)

Wave4第二層核對=(1, 2, 3, 1, 1, 2, 4, 1, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0076/)

有限交換環 · qf-mod28-commring-binomial-square · candidate





## 77. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-binomial-square；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,4,5,7,0,6],[2,3,4,5,7,6,1,0],[3,4,5,7,6,0,2,1],[4,5,7,6,0,1,3,2],[5,7,6,0,1,2,4,3],[6,0,1,2,3,4,7,5],[7,6,0,1,2,3,5,4]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,7,0,2,7,4],[0,3,7,1,4,6,5,2],[0,4,0,4,0,4,4,0],[0,5,2,6,4,1,3,7],[0,6,7,5,4,3,1,2],[0,7,4,2,0,7,2,4]],"a":2,"b":5}。請依 [a,b,sum,lhs,a2,ab,b2,rhs,equal] 欄序回答。

查看答案與解釋(2, 5, 6, 1, 4, 2, 1, 1, 是)

Wave4第二層核對=(2, 5, 6, 1, 4, 2, 1, 1, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0077/)

有限交換環 · qf-mod28-commring-binomial-square · candidate





## 78. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-binomial-square；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,4,6,7,5,0],[2,3,4,6,5,0,7,1],[3,4,6,5,7,1,0,2],[4,6,5,7,0,2,1,3],[5,7,0,1,2,4,3,6],[6,5,7,0,1,3,2,4],[7,0,1,2,3,6,4,5]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,5,0,4,2,5],[0,3,5,1,4,2,7,6],[0,4,0,4,0,0,4,4],[0,5,4,2,0,4,5,2],[0,6,2,7,4,5,1,3],[0,7,5,6,4,2,3,1]],"a":3,"b":0}。請依 [a,b,sum,lhs,a2,ab,b2,rhs,equal] 欄序回答。

查看答案與解釋(3, 0, 3, 1, 1, 0, 0, 1, 是)

Wave4第二層核對=(3, 0, 3, 1, 1, 0, 0, 1, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0078/)

有限交換環 · qf-mod28-commring-binomial-square · candidate





## 79. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-binomial-square；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,4,6,0,7,5],[2,3,4,6,7,1,5,0],[3,4,6,7,5,2,0,1],[4,6,7,5,0,3,1,2],[5,0,1,2,3,7,4,6],[6,7,5,0,1,4,2,3],[7,5,0,1,2,6,3,4]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,7,0,7,2,4],[0,3,7,1,4,6,5,2],[0,4,0,4,0,4,4,0],[0,5,7,6,4,1,3,2],[0,6,2,5,4,3,1,7],[0,7,4,2,0,2,7,4]],"a":4,"b":3}。請依 [a,b,sum,lhs,a2,ab,b2,rhs,equal] 欄序回答。

查看答案與解釋(4, 3, 5, 1, 0, 4, 1, 1, 是)

Wave4第二層核對=(4, 3, 5, 1, 0, 4, 1, 1, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0079/)

有限交換環 · qf-mod28-commring-binomial-square · candidate





## 80. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-binomial-square；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,4,7,6,0,5],[2,3,4,7,5,0,1,6],[3,4,7,5,6,1,2,0],[4,7,5,6,0,2,3,1],[5,6,0,1,2,4,7,3],[6,0,1,2,3,7,5,4],[7,5,6,0,1,3,4,2]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,5,0,4,5,2],[0,3,5,1,4,2,7,6],[0,4,0,4,0,0,4,4],[0,5,4,2,0,4,2,5],[0,6,5,7,4,2,1,3],[0,7,2,6,4,5,3,1]],"a":7,"b":5}。請依 [a,b,sum,lhs,a2,ab,b2,rhs,equal] 欄序回答。

查看答案與解釋(7, 5, 3, 1, 1, 5, 4, 1, 是)

Wave4第二層核對=(7, 5, 3, 1, 1, 5, 4, 1, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0080/)

有限交換環 · qf-mod28-commring-binomial-square · candidate





## 81. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-binomial-square；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,4,7,0,5,6],[2,3,4,7,6,1,0,5],[3,4,7,6,5,2,1,0],[4,7,6,5,0,3,2,1],[5,0,1,2,3,6,7,4],[6,5,0,1,2,7,4,3],[7,6,5,0,1,4,3,2]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,4,6,0,6,4,2],[0,3,6,1,4,7,2,5],[0,4,0,4,0,4,0,4],[0,5,6,7,4,1,2,3],[0,6,4,2,0,2,4,6],[0,7,2,5,4,3,6,1]],"a":6,"b":1}。請依 [a,b,sum,lhs,a2,ab,b2,rhs,equal] 欄序回答。

查看答案與解釋(6, 1, 5, 1, 4, 6, 1, 1, 是)

Wave4第二層核對=(6, 1, 5, 1, 4, 6, 1, 1, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0081/)

有限交換環 · qf-mod28-commring-binomial-square · candidate





## 82. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-binomial-square；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,5,6,4,7,0],[2,3,5,4,7,6,0,1],[3,5,4,6,0,7,1,2],[4,6,7,0,2,1,3,5],[5,4,6,7,1,0,2,3],[6,7,0,1,3,2,5,4],[7,0,1,2,5,3,4,6]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,6,2,0,5,6],[0,3,6,1,7,5,2,4],[0,4,2,7,1,5,6,3],[0,5,0,5,5,0,0,5],[0,6,5,2,6,0,5,2],[0,7,6,4,3,5,2,1]],"a":7,"b":5}。請依 [a,b,sum,lhs,a2,ab,b2,rhs,equal] 欄序回答。

查看答案與解釋(7, 5, 3, 1, 1, 5, 0, 1, 是)

Wave4第二層核對=(7, 5, 3, 1, 1, 5, 0, 1, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0082/)

有限交換環 · qf-mod28-commring-binomial-square · candidate





## 83. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-binomial-square；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,5,7,4,0,6],[2,3,5,4,6,7,1,0],[3,5,4,7,0,6,2,1],[4,7,6,0,2,1,5,3],[5,4,7,6,1,0,3,2],[6,0,1,2,5,3,7,4],[7,6,0,1,3,2,4,5]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,7,2,0,7,5],[0,3,7,1,6,5,4,2],[0,4,2,6,1,5,3,7],[0,5,0,5,5,0,5,0],[0,6,7,4,3,5,1,2],[0,7,5,2,7,0,2,5]],"a":0,"b":6}。請依 [a,b,sum,lhs,a2,ab,b2,rhs,equal] 欄序回答。

查看答案與解釋(0, 6, 6, 1, 0, 0, 1, 1, 是)

Wave4第二層核對=(0, 6, 6, 1, 0, 0, 1, 1, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0083/)

有限交換環 · qf-mod28-commring-binomial-square · candidate





## 84. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-binomial-square；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,5,7,6,4,0],[2,3,5,6,0,4,7,1],[3,5,6,4,1,7,0,2],[4,7,0,1,5,2,3,6],[5,6,4,7,2,0,1,3],[6,4,7,0,3,1,2,5],[7,0,1,2,6,3,5,4]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,4,5,0,2,4],[0,3,4,1,2,5,7,6],[0,4,5,2,5,0,4,2],[0,5,0,5,0,0,5,5],[0,6,2,7,4,5,1,3],[0,7,4,6,2,5,3,1]],"a":1,"b":2}。請依 [a,b,sum,lhs,a2,ab,b2,rhs,equal] 欄序回答。

查看答案與解釋(1, 2, 3, 1, 1, 2, 5, 1, 是)

Wave4第二層核對=(1, 2, 3, 1, 1, 2, 5, 1, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0084/)

有限交換環 · qf-mod28-commring-binomial-square · candidate





## 85. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-binomial-square；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,5,0,6,7,4],[2,3,5,6,1,7,4,0],[3,5,6,7,2,4,0,1],[4,0,1,2,7,3,5,6],[5,6,7,4,3,0,1,2],[6,7,4,0,5,1,2,3],[7,4,0,1,6,2,3,5]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,7,7,0,2,5],[0,3,7,1,6,5,4,2],[0,4,7,6,1,5,3,2],[0,5,0,5,5,0,5,0],[0,6,2,4,3,5,1,7],[0,7,5,2,2,0,7,5]],"a":2,"b":6}。請依 [a,b,sum,lhs,a2,ab,b2,rhs,equal] 欄序回答。

查看答案與解釋(2, 6, 4, 1, 5, 2, 1, 1, 是)

Wave4第二層核對=(2, 6, 4, 1, 5, 2, 1, 1, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0085/)

有限交換環 · qf-mod28-commring-binomial-square · candidate





## 86. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-binomial-square；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,5,6,7,0,4],[2,3,5,7,0,4,1,6],[3,5,7,4,1,6,2,0],[4,6,0,1,5,2,7,3],[5,7,4,6,2,0,3,1],[6,0,1,2,7,3,4,5],[7,4,6,0,3,1,5,2]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,4,5,0,4,2],[0,3,4,1,2,5,7,6],[0,4,5,2,5,0,2,4],[0,5,0,5,0,0,5,5],[0,6,4,7,2,5,1,3],[0,7,2,6,4,5,3,1]],"a":3,"b":0}。請依 [a,b,sum,lhs,a2,ab,b2,rhs,equal] 欄序回答。

查看答案與解釋(3, 0, 3, 1, 1, 0, 0, 1, 是)

Wave4第二層核對=(3, 0, 3, 1, 1, 0, 0, 1, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0086/)

有限交換環 · qf-mod28-commring-binomial-square · candidate





## 87. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-binomial-square；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,5,0,7,4,6],[2,3,5,7,1,6,0,4],[3,5,7,6,2,4,1,0],[4,0,1,2,6,3,7,5],[5,7,6,4,3,0,2,1],[6,4,0,1,7,2,5,3],[7,6,4,0,5,1,3,2]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,5,6,6,0,5,2],[0,3,6,1,7,5,2,4],[0,4,6,7,1,5,2,3],[0,5,0,5,5,0,0,5],[0,6,5,2,2,0,5,6],[0,7,2,4,3,5,6,1]],"a":5,"b":3}。請依 [a,b,sum,lhs,a2,ab,b2,rhs,equal] 欄序回答。

查看答案與解釋(5, 3, 4, 1, 0, 5, 1, 1, 是)

Wave4第二層核對=(5, 3, 4, 1, 0, 5, 1, 1, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0087/)

有限交換環 · qf-mod28-commring-binomial-square · candidate





## 88. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-binomial-square；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,6,5,7,4,0],[2,3,6,4,7,0,5,1],[3,6,4,5,0,1,7,2],[4,5,7,0,2,3,1,6],[5,7,0,1,3,6,2,4],[6,4,5,7,1,2,0,3],[7,0,1,2,6,4,3,5]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,5,2,6,0,5],[0,3,5,1,7,2,6,4],[0,4,2,7,1,5,6,3],[0,5,6,2,5,6,0,2],[0,6,0,6,6,0,0,6],[0,7,5,4,3,2,6,1]],"a":4,"b":5}。請依 [a,b,sum,lhs,a2,ab,b2,rhs,equal] 欄序回答。

查看答案與解釋(4, 5, 3, 1, 1, 5, 6, 1, 是)

Wave4第二層核對=(4, 5, 3, 1, 1, 5, 6, 1, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0088/)

有限交換環 · qf-mod28-commring-binomial-square · candidate





## 89. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-binomial-square；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,6,7,0,4,5],[2,3,6,4,5,1,7,0],[3,6,4,7,0,2,5,1],[4,7,5,0,2,6,1,3],[5,0,1,2,6,7,3,4],[6,4,7,5,1,3,0,2],[7,5,0,1,3,4,2,6]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,7,2,7,0,6],[0,3,7,1,5,4,6,2],[0,4,2,5,1,3,6,7],[0,5,7,4,3,1,6,2],[0,6,0,6,6,6,0,0],[0,7,6,2,7,2,0,6]],"a":7,"b":1}。請依 [a,b,sum,lhs,a2,ab,b2,rhs,equal] 欄序回答。

查看答案與解釋(7, 1, 5, 1, 6, 7, 1, 1, 是)

Wave4第二層核對=(7, 1, 5, 1, 6, 7, 1, 1, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0089/)

有限交換環 · qf-mod28-commring-binomial-square · candidate





## 90. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-binomial-square；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,6,7,4,5,0],[2,3,6,5,0,7,4,1],[3,6,5,4,1,0,7,2],[4,7,0,1,6,3,2,5],[5,4,7,0,3,2,1,6],[6,5,4,7,2,1,0,3],[7,0,1,2,5,6,3,4]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,4,6,2,0,4],[0,3,4,1,2,7,6,5],[0,4,6,2,6,4,0,2],[0,5,2,7,4,1,6,3],[0,6,0,6,0,6,0,6],[0,7,4,5,2,3,6,1]],"a":7,"b":6}。請依 [a,b,sum,lhs,a2,ab,b2,rhs,equal] 欄序回答。

查看答案與解釋(7, 6, 3, 1, 1, 6, 0, 1, 是)

Wave4第二層核對=(7, 6, 3, 1, 1, 6, 0, 1, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0090/)

有限交換環 · qf-mod28-commring-binomial-square · candidate





## 91. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-binomial-square；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,6,0,7,5,4],[2,3,6,5,1,4,7,0],[3,6,5,7,2,0,4,1],[4,0,1,2,7,6,3,5],[5,7,4,0,6,2,1,3],[6,5,7,4,3,1,0,2],[7,4,0,1,5,3,2,6]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,7,7,2,0,6],[0,3,7,1,5,4,6,2],[0,4,7,5,1,3,6,2],[0,5,2,4,3,1,6,7],[0,6,0,6,6,6,0,0],[0,7,6,2,2,7,0,6]],"a":0,"b":4}。請依 [a,b,sum,lhs,a2,ab,b2,rhs,equal] 欄序回答。

查看答案與解釋(0, 4, 4, 1, 0, 0, 1, 1, 是)

Wave4第二層核對=(0, 4, 4, 1, 0, 0, 1, 1, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0091/)

有限交換環 · qf-mod28-commring-binomial-square · candidate





## 92. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-binomial-square；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,6,5,0,7,4],[2,3,6,7,0,1,4,5],[3,6,7,4,1,2,5,0],[4,5,0,1,6,7,2,3],[5,0,1,2,7,4,3,6],[6,7,4,5,2,3,0,1],[7,4,5,0,3,6,1,2]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,4,6,4,0,2],[0,3,4,1,2,7,6,5],[0,4,6,2,6,2,0,4],[0,5,4,7,2,1,6,3],[0,6,0,6,0,6,0,6],[0,7,2,5,4,3,6,1]],"a":1,"b":2}。請依 [a,b,sum,lhs,a2,ab,b2,rhs,equal] 欄序回答。

查看答案與解釋(1, 2, 3, 1, 1, 2, 6, 1, 是)

Wave4第二層核對=(1, 2, 3, 1, 1, 2, 6, 1, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0092/)

有限交換環 · qf-mod28-commring-binomial-square · candidate





## 93. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-binomial-square；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,6,0,4,7,5],[2,3,6,7,1,0,5,4],[3,6,7,5,2,1,4,0],[4,0,1,2,5,7,3,6],[5,4,0,1,7,6,2,3],[6,7,5,4,3,2,0,1],[7,5,4,0,6,3,1,2]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,6,5,5,6,0,2],[0,3,5,1,7,2,6,4],[0,4,5,7,1,2,6,3],[0,5,6,2,2,6,0,5],[0,6,0,6,6,0,0,6],[0,7,2,4,3,5,6,1]],"a":2,"b":7}。請依 [a,b,sum,lhs,a2,ab,b2,rhs,equal] 欄序回答。

查看答案與解釋(2, 7, 4, 1, 6, 2, 1, 1, 是)

Wave4第二層核對=(2, 7, 4, 1, 6, 2, 1, 1, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0093/)

有限交換環 · qf-mod28-commring-binomial-square · candidate





## 94. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-binomial-square；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,7,5,6,0,4],[2,3,7,4,6,0,1,5],[3,7,4,5,0,1,2,6],[4,5,6,0,2,3,7,1],[5,6,0,1,3,7,4,2],[6,0,1,2,7,4,5,3],[7,4,5,6,1,2,3,0]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,5,2,7,5,0],[0,3,5,1,6,2,4,7],[0,4,2,6,1,5,3,7],[0,5,7,2,5,7,2,0],[0,6,5,4,3,2,1,7],[0,7,0,7,7,0,7,0]],"a":3,"b":0}。請依 [a,b,sum,lhs,a2,ab,b2,rhs,equal] 欄序回答。

查看答案與解釋(3, 0, 3, 1, 1, 0, 0, 1, 是)

Wave4第二層核對=(3, 0, 3, 1, 1, 0, 0, 1, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0094/)

有限交換環 · qf-mod28-commring-binomial-square · candidate





## 95. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-binomial-square；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,7,6,0,5,4],[2,3,7,4,5,1,0,6],[3,7,4,6,0,2,1,5],[4,6,5,0,2,7,3,1],[5,0,1,2,7,6,4,3],[6,5,0,1,3,4,7,2],[7,4,6,5,1,3,2,0]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,6,2,6,7,0],[0,3,6,1,5,4,2,7],[0,4,2,5,1,3,6,7],[0,5,6,4,3,1,2,7],[0,6,7,2,6,2,7,0],[0,7,0,7,7,7,0,0]],"a":7,"b":3}。請依 [a,b,sum,lhs,a2,ab,b2,rhs,equal] 欄序回答。

查看答案與解釋(7, 3, 5, 1, 0, 7, 1, 1, 是)

Wave4第二層核對=(7, 3, 5, 1, 0, 7, 1, 1, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0095/)

有限交換環 · qf-mod28-commring-binomial-square · candidate





## 96. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-binomial-square；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,7,6,4,0,5],[2,3,7,5,0,6,1,4],[3,7,5,4,1,0,2,6],[4,6,0,1,7,3,5,2],[5,4,6,0,3,2,7,1],[6,0,1,2,5,7,4,3],[7,5,4,6,2,1,3,0]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,4,7,2,4,0],[0,3,4,1,2,6,5,7],[0,4,7,2,7,4,2,0],[0,5,2,6,4,1,3,7],[0,6,4,5,2,3,1,7],[0,7,0,7,0,7,7,0]],"a":5,"b":4}。請依 [a,b,sum,lhs,a2,ab,b2,rhs,equal] 欄序回答。

查看答案與解釋(5, 4, 3, 1, 1, 4, 7, 1, 是)

Wave4第二層核對=(5, 4, 3, 1, 1, 4, 7, 1, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0096/)

有限交換環 · qf-mod28-commring-binomial-square · candidate





## 97. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-binomial-square；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,7,0,6,4,5],[2,3,7,5,1,4,0,6],[3,7,5,6,2,0,1,4],[4,0,1,2,6,7,5,3],[5,6,4,0,7,2,3,1],[6,4,0,1,5,3,7,2],[7,5,6,4,3,1,2,0]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,6,6,2,7,0],[0,3,6,1,5,4,2,7],[0,4,6,5,1,3,2,7],[0,5,2,4,3,1,6,7],[0,6,7,2,2,6,7,0],[0,7,0,7,7,7,0,0]],"a":6,"b":1}。請依 [a,b,sum,lhs,a2,ab,b2,rhs,equal] 欄序回答。

查看答案與解釋(6, 1, 4, 1, 7, 6, 1, 1, 是)

Wave4第二層核對=(6, 1, 4, 1, 7, 6, 1, 1, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0097/)

有限交換環 · qf-mod28-commring-binomial-square · candidate





## 98. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-binomial-square；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,7,5,0,4,6],[2,3,7,6,0,1,5,4],[3,7,6,4,1,2,0,5],[4,5,0,1,7,6,3,2],[5,0,1,2,6,4,7,3],[6,4,5,0,3,7,2,1],[7,6,4,5,2,3,1,0]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,4,7,4,2,0],[0,3,4,1,2,6,5,7],[0,4,7,2,7,2,4,0],[0,5,4,6,2,1,3,7],[0,6,2,5,4,3,1,7],[0,7,0,7,0,7,7,0]],"a":5,"b":7}。請依 [a,b,sum,lhs,a2,ab,b2,rhs,equal] 欄序回答。

查看答案與解釋(5, 7, 3, 1, 1, 7, 0, 1, 是)

Wave4第二層核對=(5, 7, 3, 1, 1, 7, 0, 1, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0098/)

有限交換環 · qf-mod28-commring-binomial-square · candidate





## 99. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-binomial-square；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,3,7,0,4,5,6],[2,3,7,6,1,0,4,5],[3,7,6,5,2,1,0,4],[4,0,1,2,5,6,7,3],[5,4,0,1,6,7,3,2],[6,5,4,0,7,3,2,1],[7,6,5,4,3,2,1,0]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,7,5,5,7,2,0],[0,3,5,1,6,2,4,7],[0,4,5,6,1,2,3,7],[0,5,7,2,2,7,5,0],[0,6,2,4,3,5,1,7],[0,7,0,7,7,0,7,0]],"a":0,"b":4}。請依 [a,b,sum,lhs,a2,ab,b2,rhs,equal] 欄序回答。

查看答案與解釋(0, 4, 4, 1, 0, 0, 1, 1, 是)

Wave4第二層核對=(0, 4, 4, 1, 0, 0, 1, 1, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0099/)

有限交換環 · qf-mod28-commring-binomial-square · candidate





## 100. 有限交換環

同一完整有限含一環的加法表與乘法表。操作=commring-binomial-square；資料={"n":8,"addition":[[0,1,2,3,4,5,6,7],[1,2,4,5,3,6,7,0],[2,4,3,6,5,7,0,1],[3,5,6,0,7,1,2,4],[4,3,5,7,6,0,1,2],[5,6,7,1,0,2,4,3],[6,7,0,2,1,4,3,5],[7,0,1,4,2,3,5,6]],"multiplication":[[0,0,0,0,0,0,0,0],[0,1,2,3,4,5,6,7],[0,2,3,0,6,2,3,6],[0,3,0,0,3,3,0,3],[0,4,6,3,1,7,2,5],[0,5,2,3,7,1,6,4],[0,6,3,0,2,6,3,2],[0,7,6,3,5,4,2,1]],"a":1,"b":2}。請依 [a,b,sum,lhs,a2,ab,b2,rhs,equal] 欄序回答。

查看答案與解釋(1, 2, 4, 1, 1, 2, 3, 1, 是)

Wave4第二層核對=(1, 2, 4, 1, 1, 2, 3, 1, 是)；依完整明示資料重算；完整有限／命名精確切片不自動升格為任意無限定理；乘法交換律逐全部元素對核對；交換環不自動是整環或域且可有零因子。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261028-r01-finite-commutative-ring-0100/)

有限交換環 · qf-mod28-commring-binomial-square · candidate



[原始題目 JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations/questions.jsonl) · [驗收報告](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261028-r01-finite-commutative-ring-foundations/validation-report.json)