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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?

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  • $\begingroup$ Did you mean $\phi(\hat{M})\subset\hat{\mathfrak{m}}\hat{N}$? If so, can you tell me what have you tried? $\endgroup$ – Mohan Nov 21 '16 at 17:26
  • $\begingroup$ @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 $\phi(\hat{M}) = \mathfrak{m}\hat{N}$ $\endgroup$ – user6565190 Nov 21 '16 at 18:15
  • $\begingroup$ Try $\phi=0$, clearly one such. $\endgroup$ – Mohan Nov 21 '16 at 18:18
  • $\begingroup$ @Mohan Well, but if $\phi$ is non zero? Sorry for insist but, I need the equality for another result. $\endgroup$ – user6565190 Nov 21 '16 at 18:35
  • $\begingroup$ 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. $\endgroup$ – Mohan Nov 21 '16 at 20:55

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