# Question about homomorphisms between free modules

Let $(A,m,k)$ be a Noetherian local ring and let $g:L\rightarrow L'$ be a homomorphism between free finitely generated $A$-modules. I want to prove that $g$ is inversible to the left if, and only if, the induced homomorphism $h:L/mL\rightarrow L'/mL'$ is injective. It is clear that if $g$ has an inverse to the left, then $h$ is injective, but the converse is hard to me.

Working in the converse I note that if I find basis to $L$ ad $L'$ then is easy define an inverse to the left to $g$. So other question: basis of $L/mL$ (like a $k$-vector space) induces a basis of $L$?

Yes a basis of $$L/\mathfrak mL$$ can be lifted to a basis of $$L$$ by Nakayama's lemma:

Let $$R$$ be a commutative ring, $$I$$ an ideal of $$R$$, $$M$$ a finitely generated $$R$$-module, $$N\subset M$$ a submodule.

If $$M\subset N+IM$$, there exists an element $$a\in I$$ such that $$(1+a)M\subset N$$.

In the present; you consider vectors $$u_1, \dots u_n\in L$$ such that their images in $$L/\mathfrak mL$$ are a basis of this $$A/\mathfrak m$$-vector space, and denote $$N=\langle u_1, \dots u_n\rangle$$. By hypothesis, we have $$L\subset N+\mathfrak mL,$$ so $$(1+a)L\subset N$$ for some $$a\in\mathfrak m$$? As $$A$$ is local with maximal ideal $$\mathfrak m$$, $$1+a$$ is a unit in $$A$$, so actually $$L=N$$.

Checking $$u_1, \dots u_n$$ are linearly independent is easy (always with Nakayama).

• If $a_1,...a_n$ are such that $a_1u_1+...a_nu_n=0$, then all $a_i$ belongs to $m$. How proceed? Jul 25 '18 at 23:03
• Map $A^n$ onto $L$ (via the basis) and consider the submodule $K$ of $A^n$ made up of the linear relations between the generators. You can show $K=\mathfrak m K$, and as it's finitely generated, it implies $K=0$. Jul 25 '18 at 23:50

First let's prove that if $L$ has rank $n$ then any set with $n$ generators of $L$ is a basis of $L$. For this, let $\{x_1,..., x_n\}$ be a set of generators of $L$ and consider the canonical exact sequence $0\rightarrow K\rightarrow L\rightarrow L\rightarrow0$ where $L\rightarrow L$ maps $e_i$ into $x_i$ and $\{e_i\}$ is the canonical basis of $L$. Since $L$ is free, we have exact sequence between $k$-vector spaces of finite dimension.

$$0\rightarrow K/mK\rightarrow L/mL\rightarrow L/mL\rightarrow0$$

Therefore $L/mL\rightarrow L/mL$ is an isomorphism and $K=mK$.By Nakayama's lemma we have $K=0$ and we conclude that $L\rightarrow L$ is an isomorphism, i.e., $\{x_1,...,x_n\}$ is a basis of $L$.

Now, if $h:L/mL\rightarrow L'/mL'$ is injective, let $\{\overline{x_1},...,\overline{x_n}\}$ be a basis of $L/mL$ and let $\{\overline{g(x_1)},...,\overline{g(x_n)},\overline{y_1},...,\overline{y_m}\}$ be a basis of $L'/mL'$. By Nakayama these sets span $L$ and $L'$ and we see above that $\{x_1,...,x_n\}$ and $\{g(x_1),...,g(x_n),y_1,...,y_m\}$ are basis of $L$ and $L'$, respectively. Thus the map $f:L'\rightarrow L$ defined by $f(g(x_i))=x_i$ and $f(y_j)=0$ is an inverse to the left to $g$.