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

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?

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.
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.