Noetherian Rings and Modules
This chapter may serve as an introduction to the methods of algebraic geometry rooted in commutative algebra and the theory of modules, mostly over a Noetherian ring.
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Noetherian Rings and Modules
This chapter may serve as an introduction to the method s of algebraic geometry rooted in commutative algebra and the theory of modules, mostly over a Noetherian ring .
§1.
BASIC CRITERIA
Let A be a ring and M a module (i.e., a left A-module). We shall say that M is Noetherian if it satisfies anyone of the following three cond itions : (1) Every submodule of M is finitely generated.
(2) Every ascending sequence of submodules of M,
such that M, =I: M j + I is finite. (3) Every non-empty set S of submodules of M has a maximal element (i.e., a submodule M 0 such that for any element N of S which contains M o we have N = M o). We shall now prove that the above three conditions are equivalent. (I) = (2) Suppose we have an ascending sequence of submodules of M as above. Let N be the union of all the M, (i = 1,2, ...). Then N is finitely generated, say by elements XI" ' " X" and each generator is in some M j • Hence there exists an index j such that
413 S. Lang, Algebra © Springer Science+Business Media LLC 2002
414
NOETHERIAN RINGS AND MODULES
X, §1
Then
whence eq ua lity holds and o ur implica tion is pr oved. (2) = (3) Let No be an element of S. If No is not max imal, it is properly contained in a submodule N i- If N, is not maximal, it is properly contained in a submod ule N 2 ' Inductively, if we ha ve found N, which is not ma ximal, it is contained properly in a submod ule N i + , . In thi s way we could construct an infinite chain, which is impossible. (3) = (1) Let N be a submodule of M . Let ao E N . If N # , then there exists an element a, EN which does not lie in M' --> M --> M " --> 0, M is Noetherian if and only if M ' and M" are Noetherian.
°
Corollary 1.3. Let M be a module, and let N , N ' be submodules. If M = N + N ' and if both N, N ' are Noetherian , then M is Noetherian. A finite direct sum of Noetherian modules is Noetherian. Proof. We first observe that the direct product N x N' is Noetherian since it contains N as a submodule whose factor module is isomorphic to N', and Proposition 1.2 applies. We have a surjective homomorphism
N x N'
-->
M
such that the pair (x, x ') with x E N and x ' EN' maps on x + x' . By Proposition 1.1, it follows that M is Noetherian. Finite products (or sums) follow by induction. A ring A is called Noetherian if it is Noetherian as a left module over itself. This means that every left ideal is finitely generated. Proposition 1.4. Let A be a Noetherian ring and let M be afinitely generated module. Then M is Noetherian . Proof.
Let
Xl"
' "
x, be generators of M. There exists a homomorphism f : A x A x .. . x A
-->
M
of the product of A with itself n times such that
This homomorphism is surjective. By the corollary ofthe preceding proposition, the product is Noetherian, and hence M is Noetherian by Proposition 1.1. Proposition 1.5. Let A be a ring which is Noetherian, and let qJ : A a surjective ring-homomorphism. Then B is Noetherian.
-->
B be
Proof. Let b 1 C . .. c b, c . .. be an ascending chain of left ideals
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