Finitely generated modules over a noetherian local ring whose completions are isomorphic

Question

This is part of an exercise in Eisenbud's commutative algebra, chapter 7, no. 5. Suppose $M$ and $N$ are finitely generated modules over a Noetherian local ring $R$ whose completions $\hat{M}$ and $\hat{N}$ are isomorphic. The goal is to show $M$ and $N$ are isomorphic. The problem is divided into four parts. I've handled the first three parts, which are the following:

1. $\text{Hom}_R(M,N)^\wedge \cong \text{Hom}_{\hat{R}}(\hat{M},\hat{N})$.

2. If $\hat{P}$ is the maximal ideal of $\hat{R}$ then $\hat{P} \text{Hom}_{\hat{R}}(\hat{M},\hat{N})$ consists of maps that take $\hat{M}$ to $\hat{P}\hat{N}$.

3. For $\phi, \phi' \in \text{Hom}_{\hat{R}}(\hat{M},\hat{N})$, if $\phi$ is an isomorphism and $\phi'$ differs from $\phi$ by an element of $\hat{P} \text{Hom}_{\hat{R}}(\hat{M},\hat{N})$ then $\phi'$ is surjective.

But I cannot solve the last part, which is:

4. There are elements $\phi'\in \text{Hom}_R(M,N)$, $\phi''\in \text{Hom}_R(N,M)$ that are surjective.

Actually part 2 is obvious, and it is used to prove 3. So I think I should handle part 4 by using 1 and 3, but I got stuck. Any hints?

Answer 1

I show only for $\phi’$. There is a $\hat{R}$-linear map $a: \hat{M} \rightarrow \hat{N}$ which is surjective. By 1, there is a map $b: M \rightarrow N$ such that $\hat{b}-a \in \hat{P}Hom(\hat{M},\hat{N})$, so by 3 $\hat{b}$ is surjective. In particular, $b(M)+PN=N$ thus by Nakayama $b(M)=N$.