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Let $R$ a noetherian local ring with maximal ideal $\mathfrak{m}$. Let $M$ and $N$ $R-$modules finitely generate and $\hat{M} \cong \hat{N}$ (the completion over the ideal $\mathfrak{m}$) . I have that $\widehat{\mbox{Hom}_R(M,N)} \cong \mbox{Hom}_{\hat{R}}(\hat{M},\hat{N})$, but I'm stuck with to prove that, if $\phi \in \hat{\mathfrak{m}} \mbox{Hom}_{\hat{R}}(\hat{M},\hat{N})$, then $\phi(\hat{M}) = \hat{\mathfrak{m}}\hat{N}$. (Considering that $\hat{\mathfrak{m}} = \mathfrak{m}$).

Some hint or idea?

ÝTAN
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  • Did you mean $\phi(\hat{M})\subset\hat{\mathfrak{m}}\hat{N}$? If so, can you tell me what have you tried? – Mohan Nov 21 '16 at 17:26
  • @Mohan I was trying using the fact that $\widehat{Hom_R(M,N)}$ is finitely generated and also is $\mathfrak{m}$. Now $\phi$ is a finite sum of the product of generator of the above objects. I think that is $\phi(\hat{M}) = \mathfrak{m}\hat{N}$ – ÝTAN Nov 21 '16 at 18:15
  • Try $\phi=0$, clearly one such. – Mohan Nov 21 '16 at 18:18
  • @Mohan Well, but if $\phi$ is non zero? Sorry for insist but, I need the equality for another result. – ÝTAN Nov 21 '16 at 18:35
  • No, still false and easy to construct them starting with the zero case. Think of $N=N_1\oplus N_2$ and $\phi$ zero in one of the factors. – Mohan Nov 21 '16 at 20:55
  • @Mohan Oh I got it thanks, I'm trying to solve the problem 7.5 of Eisenbud's Book(Commutative Alg.) and this say that I'm trying to prove, can you check it out please? To understand what is trying to say ? – ÝTAN Nov 22 '16 at 02:23

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