同一條件的三副面孔
在第一門課程中你證明了 主理想整環 極其溫順：每個理想只需一個生成元。但要求 k[x,y] 也如此就太強了，那裡的理想 (x,y) 確實需要兩個生成元。正確的推廣保留了精神——理想不會失控地膨脹——卻不要求單一生成元。若交換 環 R 的每個理想都有限生成，則稱 R 為 諾特環。
使該假設如此好用的，是三個看似不同的陳述彼此等價。要熟練地在它們之間切換；實踐中你會挑最易驗證的那個。
- 有限生成。 每個理想 I ⊆ R 都由有限多個元素生成。
- 升鏈條件 (ACC)。 每條理想升鏈 I_1 ⊆ I_2 ⊆ I_3 ⊆ … 最終穩定：存在 n 使 I_n = I_{n+1} = …。
- 極大條件。 每個非空理想集合在包含關係下都有極大元。
等價性的證明
循環 (1) ⇒ (2) ⇒ (3) ⇒ (1) 很短，值得親手做一遍，因為同樣的聯集／升鏈技巧在整個學科中反覆出現。唯一微妙之處以極大條件的形式用到了選擇公理。
(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.希爾伯特基定理：諾特性會傳染
該條件無處不在的原因是 希爾伯特基定理：若 R 是諾特環，則 [[polynomial-ring|多項式環]] R[x] 也是。 迭代可得 R[x_1,…,x_n] 在任一域上或在 Z 上都是諾特的。由於仿射簇的每個坐標環都是這種環的商——而諾特環的商仍是諾特的——本質上代數幾何中的每個環都自動是諾特的。
證明是一個關於首項係數的論證。提醒一下：這是那種第一次看像在變戲法的證明。訣竅是按理想中各元素的次數來組織 R[x] 的任意理想，並在每一層級倚靠 R 的諾特性。
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