同一条件的三副面孔
在第一门课程中你证明了 主理想整环 极其温顺：每个理想只需一个生成元。但要求 k[x,y] 也如此就太强了，那里的理想 (x,y) 确实需要两个生成元。正确的推广保留了精神——理想不会失控地膨胀——却不要求单一生成元。若交换 环 R 的每个理想都有限生成，则称 R 为 诺特环。
使该假设如此好用的，是三个看似不同的陈述彼此等价。要熟练地在它们之间切换；实践中你会挑最易验证的那个。
- 有限生成。 每个理想 I ⊆ R 都由有限多个元素生成。
- 升链条件 (ACC)。 每条理想升链 I_1 ⊆ I_2 ⊆ I_3 ⊆ … 最终稳定：存在 n 使 I_n = I_{n+1} = …。
- 极大条件。 每个非空理想集合在包含关系下都有极大元。
等价性的证明
循环 (1) ⇒ (2) ⇒ (3) ⇒ (1) 很短，值得亲手做一遍，因为同样的并集／升链技巧在整个学科中反复出现。唯一微妙之处以极大条件的形式用到了选择公理。
(1) => (2): Given a chain I_1 c I_2 c ... , let I = union of all I_n.
I is an ideal (a chain-union of ideals always is).
By (1), I = (a_1, ..., a_k) for finitely many generators.
Each a_j lies in some I_{n_j}; take N = max(n_1, ..., n_k).
Then all a_j in I_N, so I = I_N, and the chain stops at N.
(2) => (3): Suppose a nonempty set S of ideals has NO maximal element.
Pick I_1 in S. Not maximal => exists I_2 in S with I_1 (strictly) c I_2.
Repeat forever => a strictly ascending chain that never stabilizes,
contradicting (2). [uses dependent choice]
(3) => (1): Let I be any ideal. Let S = { finitely generated ideals c I }.
S is nonempty (0 is in it). By (3) it has a maximal element J = (a_1,...,a_m).
If J != I, pick x in I \ J. Then J + (x) is finitely generated, c I,
and strictly bigger than J -- contradicting maximality. So J = I.希尔伯特基定理：诺特性会传染
该条件无处不在的原因是 希尔伯特基定理：若 R 是诺特环，则 [[polynomial-ring|多项式环]] R[x] 也是。 迭代可得 R[x_1,…,x_n] 在任一域上或在 Z 上都是诺特的。由于仿射簇的每个坐标环都是这种环的商——而诺特环的商仍是诺特的——本质上代数几何中的每个环都自动是诺特的。
证明是一个关于首项系数的论证。提醒一下：这是那种第一次看像在变戏法的证明。诀窍是按理想中各元素的次数来组织 R[x] 的任意理想，并在每一层级倚靠 R 的诺特性。
Claim: R Noetherian => R[x] Noetherian.
Let J be an ideal of R[x]. For each d >= 0 set
L_d = { leading coeffs of degree-d elements of J } u {0} c R.
Each L_d is an ideal of R, and L_0 c L_1 c L_2 c ... (multiply by x).
By ACC in R this chain stabilizes at some N: L_d = L_N for all d >= N.
Each L_d (0 <= d <= N) is f.g.: L_d = (a_{d,1}, ..., a_{d,k_d}).
Choose f_{d,i} in J of degree d with leading coeff a_{d,i}.
Claim: these finitely many f_{d,i} generate J.
Given g in J of degree d, subtract an R[x]-combination of the f_{*,i}
matching its leading term (using L_d, or L_N times x^{d-N} if d > N).
This lowers deg g. Induct on degree => g lands in the ideal generated
by the f_{d,i}. Hence J is finitely generated. QED