# Thread: About some module over a DVR

1. ## About some module over a DVR

Let $R$ be a DVR and $M$ be a free $R$-module of rank $1$.
If $f:M\to M$ is a surjective $R$-linear map then $f$ is an isomorphism.

2. Originally Posted by KaKa
Let $R$ be a DVR and $M$ be a free $R$-module of rank $1$.
If $f:M\to M$ is a surjective $R$-linear map then $f$ is an isomorphism.
the problem is a very special case of this one: if $R$ is a ring (doesn't even have to be commutative) and $M$ is a (left or right) Noetherian $R$-module, then every $R$-linear surjective map $f : M \to M$ is an isomorphism. the proof is very easy:
look at the chain of submodules
$\ker f \subseteq \ker f^2 \subsetq \ldots$,
which has to stop at some point because $M$ is Noetherian. so
$\ker f^n = \ker f^{n+1}$,
for some positive integer $n$. now suppose that $f(x)=0$, for some $x \in M$. we have $x = f^n(a)$, for some $a \in M$, because $f$ is surjective. thus $0=f(x)=f^{n+1}(a)$ and hence $a \in \ker f^{n+1}=\ker f^n$. so $x = f^n(a)=0$. this proves that $f$ is injective, which is what we need.