Let $M$ be Noetherian $R$-module(where $R$ contains $1$) and $\phi:M \to M$ be $R$ -module homomorphism . Suppose $\phi$ is surjective, how do I show that $\phi$ is injective ?
Hints will suffice, thank you.
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Please attempt to use the search function first. Searching "Noetherian surjective injective " brings you to duplicates. – rschwieb Apr 07 '14 at 02:34
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Hint: $\ker \phi\subseteq \ker \phi^2\subseteq \dotsb$ is an ascending chain.
egreg
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$\text{ker}(\phi^n)$ is submodule of $M$ and since $M$ is Noetherian, the ascending chain will terminate. – Alexy Vincenzo Apr 06 '14 at 15:25
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@AlexyVincenzo Yes; if $\ker\phi\ne{0}$, there's $n$ such that $\ker\phi^{n-1}\ne\ker\phi^n=\ker\phi^{n+1}$ (where $\phi^0$ is the identity, of course). – egreg Apr 06 '14 at 15:29
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@AlexyVincenzo The sequence is stationary; but it starts, if $\ker\phi\ne{0}$, with ${0}=\ker\phi^0\subsetneq\ker\phi^1$, so you just pick the maximum $n$ such that $\ker\phi^{n-1}\subsetneq\ker\phi^n$. – egreg Apr 06 '14 at 15:37
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@AlexyVincenzo Why the minimum? The minimum such would be $1$ and would be useless, because you wouldn't be able to use the stationarity of the chain and the surjectivity of $\phi$. – egreg Apr 06 '14 at 15:45
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By the noetherianity we have $\ker(\phi^n) = \ker(\phi^{n+1})$ for some $n$.
$\phi$ is surjective, so $\phi^n$ is surjective; suppose $v \in \ker(\phi)$, so $\exists \ w $ such that $$ v= \phi^n(w) \Rightarrow 0 = \phi(v) = \phi^{n+1}(w) \Rightarrow w \in \ker(\phi^{n+1})$$ But $\ker(\phi^n) = \ker(\phi^{n+1})$ and so $$v= \phi^n(w) = 0 \Rightarrow \ker(\phi) = 0 $$ Then $\phi$ is injective
WLOG
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