醜聞：在 Z[√-5] 中 6 有兩種分解
Z 是唯一分解整環：每個整數本質上以唯一方式分解為質數。我們當然希望 O_K 也是 UFD。但它通常不是。標準的警示例子是 O_K = Z[√-5]（這裡 -5 ≡ 3 mod 4，所以這就是完整的整數環）。數 6 有兩種分解為不可約元的方式。
In O_K = Z[sqrt(-5)], norm N(a+b sqrt(-5)) = a^2 + 5 b^2.
N is multiplicative; an element is a unit iff N = 1, so units = {+1,-1}.
6 = 2 * 3 = (1 + sqrt(-5)) * (1 - sqrt(-5)).
Norms:
N(2) = 4
N(3) = 9
N(1 + sqrt(-5)) = 1 + 5 = 6
N(1 - sqrt(-5)) = 6
Are these irreducible? A proper factor would have norm 2 or 3.
But a^2 + 5 b^2 = 2 and = 3 have NO integer solutions.
So 2, 3, 1±sqrt(-5) are all irreducible, and none is an associate
of another (their norms differ / units are only ±1).
==> 6 = 2*3 = (1+sqrt(-5))(1-sqrt(-5)) are TWO distinct
factorizations into irreducibles. UFD FAILS.戴德金的修復：分解理想，而非元素
庫默爾的洞見，由戴德金嚴格化：停止分解元素，改為分解理想。元素 2 無法進一步分裂，但理想 (2) 可以——分裂為一個質理想的平方。設 p = (2, 1+√-5)。可驗證 p² = (2)，類似地，3 之上的理想給出 q = (3, 1+√-5)、q' = (3, 1-√-5)，且 (3) = qq'。於是 6 的兩種分解都細化到*同一個*理想分解。
Set p = (2, 1+sqrt(-5))
q = (3, 1+sqrt(-5))
q' = (3, 1-sqrt(-5))
Claim: (2) = p^2, (3) = q q'.
Ideal factorizations of the two sides of 6 = 2*3 = (1+s)(1-s):
(6) = (2)(3) = p^2 q q'
(6) = (1+s)(1-s) = (p q)(p q')
because (1+sqrt(-5)) = p q, (1-sqrt(-5)) = p q'.
Both give (6) = p^2 q q' -- the SAME prime-ideal factorization.
Uniqueness is restored: the ambiguity was only in how the prime
ideals clumped together into principal ideals (elements).為何總是奏效：O_K 是戴德金整環
這種拯救並非 Z[√-5] 的僥倖技巧；它是結構性的。戴德金整環是滿足以下條件的整環：(1) 諾特環，(2) 在其分式域中整閉，(3) 維數為一——每個非零質理想都是極大理想。任何數域的整數環 O_K 都滿足這三條，所以 O_K 總是戴德金整環。
回報定理：在戴德金整環中，每個非零理想都唯一地分解為質理想之積。這就是理想的唯一分解——本學科的核心結構性結果。局部地，每個質理想給出一個離散賦值環，因此分解由每個質理想處一個乾淨的消沒階來支配。下一份指南將理想與元素之間的*差距*轉化為一個有限不變量：類群。