Let $R$ be a Noetherian domain, let $\mathfrak{m}$ be a max ideal and $\phi:P\to Q$ an $R$-map between finitely generated modules. Suppose that $\forall f\not\in \mathfrak{m}$, the map $\phi_f:P_f\to Q_f$ is not surjective (in particular $\phi$ is not by setting $f=1$). Show that the map $P/\mathfrak{m}P\to Q/\mathfrak{m}Q$ is not surjective?
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I think this boils down to: if $M_f\neq 0$ for all $f\not\in \mathfrak{m}$, show that $\mathfrak{m}M\neq M$ – FunctionOfX Aug 27 '20 at 19:14
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I can show $Q/im(P)\mathfrak{m}\neq \mathfrak{m}Q/im(P)\mathfrak{m}$ – FunctionOfX Aug 27 '20 at 19:23
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I think then it is done by nakayama and surjectivity is a local property... – FunctionOfX Aug 27 '20 at 20:07
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Can you please move these comments to the body of the question? Also, if you believe you have an answer, you can post it here. – cqfd Aug 28 '20 at 13:55
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Let $M=Q/\phi(P)$. If $mM=M$ then there is $a\in m$ such that $(1-a)M=0$. Now set $f=1-a$. – user26857 Aug 28 '20 at 15:29