单位群的结构
O_K 中的元素 u 是单位当且仅当其范数 N(u) = ±1。在 Z 中唯一的单位是 ±1，但在更大的域中可以有无穷多个——例如在 Z[√2] 中，元素 1+√2 的范数为 -1，故其所有幂 (1+√2)^k 都是互异的单位。[[dirichlet-unit-theorem|狄利克雷单位定理]]将单位群 O_K* 完整地描述为一个有限生成阿贝尔群。
设 r₁ 为实嵌入 K → R 的个数，r₂ 为复嵌入的共轭对的个数，故 r₁ + 2r₂ = n。则 O_K* ≅ μ_K × Z^r，其中 μ_K 是 K 中单位根构成的有限循环群，而秩为 r = r₁ + r₂ − 1。证明通过单位的阿基米德绝对值的对数，将单位嵌入 R^(r₁+r₂) 中的一个超平面；其像是一个满秩格，同样依据数的几何。
Unit rank r = r1 + r2 - 1. Examples:
Q n=1, r1=1, r2=0 -> r = 0. Units = {±1}.
Q(i) n=2, r1=0, r2=1 -> r = 0. Units = {±1,±i} (finite!)
Q(sqrt(-5)) n=2, r1=0, r2=1 -> r = 0. Units = {±1}.
Q(sqrt(2)) n=2, r1=2, r2=0 -> r = 1. Units = ±(1+sqrt(2))^Z.
Q(sqrt(5)) n=2, r1=2, r2=0 -> r = 1. fund. unit = (1+sqrt(5))/2.
Q(cbrt(2)) n=3, r1=1, r2=1 -> r = 1. one fundamental unit.
Imaginary quadratic (r2=1, r1=0): rank 0, units are JUST roots of
unity -- finite. Real quadratic (r1=2): rank 1, one fundamental
unit u, every unit is ±u^k (this is the Pell equation engine).组装起来：x² + 5 = y³ 没有整数解
这是回报——一个真正的费马式丢番图方程，通过在 O_K = Z[√-5] 中工作而被攻破。我们用上了一切：在 O_K 中分解、通过类群（h_K = 2）控制理想、利用有限单位群（仅 ±1）、并读出分裂。看类数 2 如何起决定性作用。
Solve x^2 + 5 = y^3 in integers. Work in O_K = Z[sqrt(-5)].
Factor LHS: (x + sqrt(-5))(x - sqrt(-5)) = y^3. (*)
Step 1 (coprimality of ideals). Let p | gcd of the two factors.
Then p | their difference 2 sqrt(-5) and p | their sum 2x.
If y is even then x^2 ≡ -5 ≡ 3 (mod 4): impossible (squares are 0,1).
So y is odd, x is even -> x^2+5 odd, and one checks the ideals
(x+sqrt(-5)) and (x-sqrt(-5)) are COPRIME in O_K.
Step 2 (use unique factorization of IDEALS). Their product is
the cube (y)^3, and they are coprime, so each is itself a cube:
(x + sqrt(-5)) = a^3 for some ideal a.
Step 3 (use the class group, h_K = 2). In Cl(K) ≅ Z/2Z,
[a]^3 = [(x+sqrt(-5))] = [principal] = 1, so [a]^3 = 1.
But |Cl(K)| = 2, so [a]^2 = 1 too; gcd(3,2)=1 forces [a] = 1.
==> a IS principal: a = (a + b sqrt(-5)).
Step 4 (use the unit group, O_K* = {±1}). Then
x + sqrt(-5) = (unit)(a + b sqrt(-5))^3, unit = ±1 = (∓1)^3,
so WLOG x + sqrt(-5) = (a + b sqrt(-5))^3.
Expand: (a + b sqrt(-5))^3 = a(a^2 - 15 b^2) + b(3a^2 - 5 b^2) sqrt(-5).
Match sqrt(-5)-coefficients: b(3a^2 - 5b^2) = 1.
==> b = ±1 and 3a^2 - 5b^2 = ±1 ==> 3a^2 = 5 ± 1 = 6 or 4.
Neither gives integer a (a^2 = 2 or 4/3). CONTRADICTION.
Conclusion: x^2 + 5 = y^3 has NO integer solutions.