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Math Help - About some module over a DVR

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by KaKa View Post
    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.
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