交换代数笔记

引言

本文以 Atiyah Macdonald 的 《Introduction to Commutative Algebra》为参考，总结了交换代数的知识点，并提供了一些自己的见解。

Noetherian Ring

Def
A ring A A A is called Noetherian if either

1. Every non-empty set of ideals of A A A has a maximal element.
2. Every ascending chain of ideals in A A A is stationary.
3. Every ideal in A A A is finitely generated.

Here we prove that 3 conditions above are equivalent.
(1) ⇒ \Rightarrow ⇒ (2): Suppose I = ( a 1 , a 2 , … ) I=(a_1,a_2,\ldots) I=(a1​,a2​,…) is the ascending chain, then I I I as a set has maximal element, say a n a_n an​. Then a n a_n an​ contains every ideal in I I I, and since a n ⊆ a n + k a_n \subseteq a_{n+k} an​⊆an+k​, we have a n = a n + k a_n=a_{n+k} an​=an+k​, for every k > 0 k > 0 k>0.
(2) ⇒ \Rightarrow ⇒ (3): Suppose a a a is a not-finitely-generated ideal, then ∃ x 1 , x 2 , … ∈ A \exists x_1,x_2,\ldots \in A ∃x1​,x2​,…∈A, such that the chain of ideals
( x 1 ) , ( x 1 , x 2 ) , … , ( x 1 , x 2 , … , x n ) , … (x_1), (x_1,x_2), \ldots , (x_1,x_2, \ldots,x_n), \ldots (x1​),(x1​,x2​),…,(x1​,x2​,…,xn​),…
is strictly ascending, which is not stationary.
(3) ⇒ \Rightarrow ⇒ (2): Suppose I = ( a 1 , a 2 , … ) I=(a_1,a_2,\ldots) I=(a1​,a2​,…) is ascending, let a = ∪ j = 1 ∞ a j a=\cup_{j=1}^\infty a_j a=∪j=1∞​aj​, then a a a is an ideal, hence a = ( x 1 , … , x n ) a=(x_1,\ldots,x_n) a=(x1​,…,xn​), and every x i ∈ a j i x_i\in a_{j_i} xi​∈aji​​ for some j i j_i ji​. Hence a ⊆ a max ⁡ ( j i ) a\subseteq a_{\max(j_i)} a⊆amax(ji​)​. Let n = m a x ( j i ) n=max(j_i) n=max(ji​), and for every k > 0 k>0 k>0, we have
a ⊆ a n + k ⊆ a ⊆ a n a\subseteq a_{n+k} \subseteq a \subseteq a_n a⊆an+k​⊆a⊆an​
hence a n + k = a n a_{n+k} = a_n an+k​=an​.
(2) ⇒ \Rightarrow ⇒ (1): Suppose I I I is a set of ideals without maximal element, then we can extract a strictly ascending chain of ideals one by one. Let x 1 ∈ I x_1\in I x1​∈I, since x 1 x_1 x1​ not maximal, ∃ x 2 > x 1 \exists x_2 > x_1 ∃x2​>x1​, then $\exists x 3 > x 2 x_3 > x_2 x3​>x2​, etc. In this manner, we have ( x n ) n = 1 ∞ (x_n)_{n=1}^\infty (xn​)n=1∞​ which is strictly ascending. Q . E . D \quad Q.E.D Q.E.D

We some easy facts of Noetherian rings without proof:

Fact

1. Every quotient ring of a Noetherian ring is Noetherian.
2. Every subring of a Noetherian ring is Noetherian.
3. Every localized ring of a Noetherian is Noetherian.

Thus Noetherian rings persists itself via plenty of operations. The next theorem is another support for this statement, although non-trivial to proof. It is called the Hilbert’s basis theorem.

Thm (Hilbert’s basis theorem)
If A A A is Noetherian, then A [ x ] A[x] A[x] is Noetherian.

sketch of proof.
We prove by show that any ideal of A [ x ] A[x] A[x] is finitely generated.
For a ⊂ A [ x ] a\subset A[x] a⊂A[x] to be an ideal, let l l l be the ideal generated by the lowest term of f f f for f ∈ a f\in a f∈a. Then l l l is an ideal of A A A hence is finitely generated. Let l = ( a 1 , … , a n ) l=(a_1,\ldots, a_n) l=(a1​,…,an​). For each a i a_i ai​, choose a f i ∈ a f_i\in a fi​∈a with
f i = a i x r i +  (lower terms) f_i = a_i x^{r_i} + \text{ (lower terms)} fi​=ai​xri​+ (lower terms)
Let r = m a x ( r i ) r = max(r_i) r=max(ri​). Let f ∈ a f\in a f∈a , then ∃ u i ∈ A [ x ] \exists u_i \in A[x] ∃ui​∈A[x] such that
f − ∑ u i f i f - \sum u_i f_i f−∑ui​fi​
has degree < r < r <r. Hence a a a is generated by f i f_i fi​ together with the generators of a ∩ ⨁ i = 0 r − 1 A x i a\cap \bigoplus_{i=0}^{r-1} Ax^i a∩⨁i=0r−1​Axi (as an A A A-module).

It follows that

• If A A A is Noetherian ring then A [ x 1 , … , x n ] A[x_1,\ldots, x_n] A[x1​,…,xn​] is Noetherian ring.
• Every finitely generated algebra B B B of A A A is Noetherian ring.

Next we will do some exercise. First we investigate the behavior of a non-Noetherian ring.

Exercise 1
If A A A is non-Noetherian, let Σ \Sigma Σ be the not-finitely-generated ideals. Hence Σ \Sigma Σ is non-empty. By Zorn lemma Σ \Sigma Σ has maximal elements. Prove that the maximal element is a prime ideal.

proof.
Let p p p be the maximal element of Σ \Sigma Σ. Suppose x ∉ p , y ∉ p x\notin p,y\notin p x∈/​p,y∈/​p, and x y ∈ p xy\in p xy∈p. Then since p + ( x ) p + (x) p+(x) is finitely generated, let p + ( x ) = ( x 1 , … , x n ) p+(x) = (x_1,\ldots,x_n) p+(x)=(x1​,…,xn​). W.L.O.G let x 1 , … , x r ∉ ( x ) x_1, \ldots, x_r \notin (x) x1​,…,xr​∈/​(x), and let x ~ 1 , … , x ~ r ∈ p \tilde {x}_1, \ldots ,\tilde {x}_r\in p x~1​,…,x~r​∈p such that
x i = x ~ i + λ i λ i ∈ ( x ) i = 1 , … , r x_i = \tilde {x}_i + \lambda_i \quad \lambda_i\in (x) \quad i=1,\ldots, r xi​=x~i​+λi​λi​∈(x)i=1,…,r
Hence p 0 : = ( x ~ 1 , … , x ~ r ) ⊂ p p_0 := (\tilde {x}_1, \ldots, \tilde{x}_r)\subset p p0​:=(x~1​,…,x~r​)⊂p and p 0 + ( x ) = p + ( x ) p_0+(x) = p+(x) p0​+(x)=p+(x). Moreover, p = p 0 + x ⋅ ( p : x ) p = p_0 + x \cdot (p:x) p=p0​+x⋅(p:x), since is c ∈ p c\in p c∈p, then
d + m x = c + n x d + mx = c + nx d+mx=c+nx ⇒ c = d + ( m − n ) x \Rightarrow c = d + (m-n)x ⇒c=d+(m−n)x
For d ∈ p 0 d\in p_0 d∈p0​, and m , n ∈ A m,n \in A m,n∈A. But since ( m − n ) x = c − d ∈ p (m-n)x = c-d\in p (m−n)x=c−d∈p we have m − n ∈ ( p : x ) m-n\in (p:x) m−n∈(p:x). Therefore, since y ∈ ( p : x ) y\in (p:x) y∈(p:x), ( p : x ) (p:x) (p:x) is strictly larger thant p p p, hence is finitely generated, hence so is p p p. Contradiction. Q . E . D . \quad Q.E.D. Q.E.D.

We have demonstrated the reproductivity of Noetherian ring. Now we turn to the its primary decomposition. The following two simple lemmas show that every ideal in a Noetherian ring has primary decomposition.

In a Noetherian ring A A A:

Lemma
Every ideal is finite intersection of irreducible ideals.

Lemma
Every ideal is primary.

Hence, every ideal has primary decomposition.

Perhaps the very important consequence of a Noetherian ring is that the power of its nilradical is zero. (suppose a i a_i ai​ generates r ( a ) r(a) r(a), and a i r i a_i^{r_i} airi​​=0. Let m = ∑ ( r i − 1 ) + 1 m = \sum (r_i - 1) + 1 m=∑(ri​−1)+1, then r ( a ) m = 0 r(a)^m=0 r(a)m=0.) The next exercise is an application.

Exercise 2
Let A A A be an Noetherian ring and f = ∑ n = 0 ∞ a n x n ∈ A [ [ x ] ] f=\sum_{n=0}^\infty a_nx^n \in A[[x]] f=∑n=0∞​an​xn∈A[[x]]. Show that f f f is nilpotent if and only if all a n a_n an​ are nilpotent.

Proof.
Suppose f n = 0 f^n=0 fn=0, then a 0 n = 0 a_0^n=0 a0n​=0, then f − a 0 f-a_0 f−a0​ is nilpotent, hence a 1 m = 0 a_1^m=0 a1m​=0 for some m m m. By induction we know that a n = 0 a_n=0 an​=0 for all n ≥ 0 n\ge 0 n≥0.
Now suppose a n a_n an​ are all nilpotent. we need to show that N A [ [ x ] ] ⊂ N A [ [ x ] ] \mathfrak{N}_A[[x]] \subset \mathfrak{N}_{A[[x]]} NA​[[x]]⊂NA[[x]]​. By Noetherian condition we know that ∃ n \exists n ∃n, ( N A ) n = 0 (\mathfrak{N}_A)^n = 0 (NA​)n=0. Hence if f ∈ N A [ [ x ] ] f\in \mathfrak{N}_A[[x]] f∈NA​[[x]], then each coefficient of f n f^n fn lies in ( N A ) n (\mathfrak{N}_A)^n (NA​)n, hence is 0, hence f n = 0 f^n=0 fn=0. Q . E . D . \quad Q.E.D. Q.E.D.