If $M$ is a finitely generated module over a local ring $A$, then there is a free $A$-module $L$ such that $L/mL\simeq M/mM$.

Question

I want to show the title.

If $M$ is a finitely generated module over a local ring $A$, then there is a free $A$-module $L$ such that $L/mL\simeq M/mM$ where $m$ is a unique maximal ideal of $A$.

My attempt: Let $M$ is a finitely generated $A$-module where $A$ is a local ring with maximal ideal $m$. Then $M/mM$ is a $A/m$-vector space so there is a basis $x_1+mM,...,x_n+mM$ of $M/mM$. Now let $L$ be a free $A$-module generated by $x_1,...,x_n$. Define $\phi:L\to M/mM$ by $x_i\mapsto x_i+mM$. Then I want to show the kernel $\{\sum_{i=1}^na_ix_i| \sum_{i=1}^na_ix_i \in mM\}=mL$. $\supset$ is clear but how can I prove the reverse inclusion?

Answer 1

We know that $\phi$ is surjective, and $mL\subseteq\ker\phi$. Then one can define a surjective homomorphism of $A/m$-vector spaces $\overline\phi:L/mL\to M/mM$, $\overline\phi(x+mL)=\phi(x),\ \forall x\in L$, and by dimension reasons this is an isomorphism.

Answer 2

The idea is right. The kernel of $\phi$ certainly contains $mL$.

Suppose $x=a_1x_1+\dots+a_nx_n\in\ker\phi$. This means $$ (a_1+m)(x_1+mM)+(a_2+m)(x_2+mM)+\dots+(a_n+m)(x_n+mM)=0+mM $$ and, since $\{x_1+mM,\dots,x_n+mM\}$ is a basis of $M/mM$ as a vector space over $A/m$, we obtain $a_1+m=a_2+m=\dots=a_n+m=0+m$.

Therefore $a_1,a_2,\dots,a_n\in m$ and $x\in mL$.

Now you have to prove that $\phi$ is surjective, which follows from $\{x_1+mM,\dots,x_n+mM\}$ being a spanning set.