Nakayama's Lemma for Finitely Generated Modules over Local Rings

Theorem

Let R be a local ring with maximal ideal m. Let M be a finitely generated R-module. If mM=M, then M=0.

Proof

We are given a local ring ( R , m ) and a finitely generated R -module M such that mM = M . We assume for the sake of contradiction that M ≠ 0 . Let S be the set of all proper submodules of M . Since M ≠ 0 , the zero submodule { 0 } is in S , so S is non-empty. The set S is partially ordered by set inclusion ⊆ . Let C be a chain of submodules in S . The union U = ⋃ L ∈ C L is a submodule of M . Since M is finitely generated, any chain of proper submodules has an upper bound that is also a proper submodule. By Zorn's Lemma, S has a maximal element. Let N be such a maximal proper submodule of M . Consider the quotient module M / N . By the choice of N as a maximal submodule, there are no submodules strictly between N and M . This implies that M / N is a simple module, meaning its only submodules are { 0 } and M / N itself. Now consider the submodule m ( M / N ) . Since M / N is simple, we have two cases: 1) m ( M / N ) = M / N or 2) m ( M / N ) = { 0 } . If case 1 holds, then since M / N is finitely generated (as it is cyclic), we would have M / N = 0 , which implies M = N . This contradicts that N is a proper submodule. Therefore, we must have case 2: m ( M / N ) = { 0 } . This means the maximal ideal m annihilates M / N . This implies mM ⊆ N . But we started with the assumption that mM = M . Substituting this into the previous result gives M ⊆ N . This is a contradiction, as N was chosen to be a proper submodule of M (so N ⊊ M ). Our initial assumption that M ≠ 0 must be false. Thus, we conclude that M = 0 . ■