第 15948 題：有限商環

原題與已知條件

Z_n與proper雙邊理想的全部加法cosets。操作=quotring-classes；資料={"n":24,"d":4}。請依 [ideal,cosets,quotient_size,certificate] 欄序回答。

• G001 — Z_n與proper雙邊理想的全部加法cosets。操作=quotring-classes；資料={"n":24,"d":4}。請依 [ideal,cosets,quotient_size,certificate] 欄序回答。

定義與適用範圍

沿用原題明示的論域、假設與來源 MKO 定義。本題已完成第二對話數理與中英語義覆核，並由原始參數逐題獨立重算模型、答案與每個算術步驟；原題仍為 candidate。

• mko-finite-quotient-ring

逐步推導

quotient ring parent modulus divided by ideal generator := div(24, 4)

quotient ring parent modulus divided by ideal generator = 6

same declared parent Z_24;generator d=4;n/d=6;ideal multiples=[0, 4, 8, 12, 16, 20]

I=kernel of residue map Z_24 to Z_4;proper two-sided ideal=是;zero ideal allowed when d=n

complete parent residues=[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23];same ideal I=[0, 4, 8, 12, 16, 20]

all additive cosets in representative order0..d-1=[[0, 4, 8, 12, 16, 20], [1, 5, 9, 13, 17, 21], [2, 6, 10, 14, 18, 22], [3, 7, 11, 15, 19, 23]];quotient size=4;no field conclusion

逐步反向核對

• S003 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S002 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S001 · premise_and_result_audit — 以精確有理數重算 quotient ring parent modulus divided by ideal generator，不以小數近似替代。

結論

([0, 4, 8, 12, 16, 20], [[0, 4, 8, 12, 16, 20], [1, 5, 9, 13, 17, 21], [2, 6, 10, 14, 18, 22], [3, 7, 11, 15, 19, 23]], 4, proper_two_sided_ideal_in_commutative_Zn)

依上列完整代值及條件核對得到原題答案；原題編號、格式與狀態保持不變。

本題覆核與來源綁定

覆核日期：2026-10-26

原題、答案、步驟表達式、數值 AST 與既有歷史保持原樣；原來的候選狀態屬於來源歷史，不覆寫為形式化證明。