I am reading the book “Fourier-Mulkai transforms in algebraic geometry” by Daniel Huybrechts. In the proof of Lemma 4.5, in page 92, it is written that if $M$ is a finite module over a local noetherian ring $(A,m)$ with $\operatorname{supp}(M)=\{m\}$, then there exists a surjection $M\twoheadrightarrow A/m$ and an injection $A/m\hookrightarrow M$. My question is that why is this true?
Why does a finite module over a Noetherian local ring supported only at the maximal ideal have the residue field as a submodule and a quotient
commutative-algebralocal-ringsmodules
Best Answer
The existence of a surjection $M\to A/m$ is easy and only requires $M$ to be nontrivial and finitely generated. In that case, by Noetherianness there is a maximal proper submodule $N\subset M$ so $M/N$ is a simple $A$-module, which is thus isomorphic to $A/m$ since $A$ is local.
To get an injection $A/m\to M$ you need to do more work and use the assumption that nothing except $m$ is in the support of $M$. Let $I\subset A$ be the annihilator of $M$. Since $\operatorname{supp}(M)=\{m\}$, the only prime ideal of $A$ that contains $I$ is $m$. That is, $A/I$ has only one prime ideal, so it is Artinian, so since $M$ is a finitely generated $A/I$-module it is also Artinian. Thus $M$ has a minimal nonzero submodule, which again must be isomorphic to $A/m$ since $A$ is local.