丑闻：在 Z[√-5] 中 6 有两种分解
Z 是唯一分解整环：每个整数本质上以唯一方式分解为素数。我们当然希望 O_K 也是 UFD。但它通常不是。标准的警示例子是 O_K = Z[√-5]（这里 -5 ≡ 3 mod 4，所以这就是完整的整数环）。数 6 有两种分解为不可约元的方式。
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.戴德金的修复：分解理想，而非元素
库默尔的洞见，由戴德金严格化：停止分解元素，改为分解理想。元素 2 无法进一步分裂，但理想 (2) 可以——分裂为一个素理想的平方。设 p = (2, 1+√-5)。可验证 p² = (2)，类似地，3 之上的理想给出 q = (3, 1+√-5)、q' = (3, 1-√-5)，且 (3) = qq'。于是 6 的两种分解都细化到*同一个*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).为何总是奏效：O_K 是戴德金整环
这种拯救并非 Z[√-5] 的侥幸技巧；它是结构性的。戴德金整环是满足以下条件的整环：(1) 诺特环，(2) 在其分式域中整闭，(3) 维数为一——每个非零素理想都是极大理想。任何数域的整数环 O_K 都满足这三条，所以 O_K 总是戴德金整环。
回报定理：在戴德金整环中，每个非零理想都唯一地分解为素理想之积。这就是理想的唯一分解——本学科的核心结构性结果。局部地，每个素理想给出一个离散赋值环，因此分解由每个素理想处一个干净的消没阶来支配。下一份指南将理想与元素之间的*差距*转化为一个有限不变量：类群。