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).

  • $\begingroup$ If $a_1,...a_n$ are such that $a_1u_1+...a_nu_n=0$, then all $a_i$ belongs to $m$. How proceed? $\endgroup$ Jul 25 '18 at 23:03
  • $\begingroup$ 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 $K=0$. $\endgroup$
    – Bernard
    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 $g$.


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