Let $\displaystyle R$ be a DVR and $\displaystyle M$ be a free $\displaystyle R$-module of rank $\displaystyle 1$.

If $\displaystyle f:M\to M$ is a surjective $\displaystyle R$-linear map then $\displaystyle f$ is an isomorphism.

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- Apr 11th 2011, 10:15 AMKaKaAbout some module over a DVR
Let $\displaystyle R$ be a DVR and $\displaystyle M$ be a free $\displaystyle R$-module of rank $\displaystyle 1$.

If $\displaystyle f:M\to M$ is a surjective $\displaystyle R$-linear map then $\displaystyle f$ is an isomorphism. - Apr 13th 2011, 12:48 AMNonCommAlg
the problem is a very special case of this one: if $\displaystyle R$ is a ring (doesn't even have to be commutative) and $\displaystyle M$ is a (left or right) Noetherian $\displaystyle R$-module, then every $\displaystyle R$-linear surjective map $\displaystyle f : M \to M$ is an isomorphism. the proof is very easy:

look at the chain of submodules

$\displaystyle \ker f \subseteq \ker f^2 \subsetq \ldots$,which has to stop at some point because $\displaystyle M$ is Noetherian. so

$\displaystyle \ker f^n = \ker f^{n+1}$,for some positive integer $\displaystyle n$. now suppose that $\displaystyle f(x)=0$, for some $\displaystyle x \in M$. we have $\displaystyle x = f^n(a)$, for some $\displaystyle a \in M$, because $\displaystyle f$ is surjective. thus $\displaystyle 0=f(x)=f^{n+1}(a)$ and hence $\displaystyle a \in \ker f^{n+1}=\ker f^n$. so $\displaystyle x = f^n(a)=0$. this proves that $\displaystyle f$ is injective, which is what we need.