If $R{^m}$ is isomorphic to $R{^n}$ as $R$-modules and if $M$ is a maximal ideal of $R$ then how can I show that image of $M{^m}$ is $M{^n}$?

Question

If $R{^m}$ is isomorphic to $R{^n}$ as $R$-modules and if $M$ is a maximal ideal of $R$ then how can I show that image of $M{^m}$ is $M{^n}$?

Background: I was trying to prove that if $R{^m}$ is isomorphic to $R{^n}$ as $R$-modules then $m=n$. For this I need a step like $R{^m/M{^m}}$ isomorphic to $R{^n/M{^n}}$.

I need some help. Thank You.

Answer 1

You should look at a much more general statement: For any ideal $I\subset R$, and any homomorphism $f: N\to N'$ of $R$-modules, $f(IN) \subset IN'$. This statement is immediate from the definition of a module homomorphism.

If $f$ is an isomorphism, then we can apply the same claim to $f^{-1}$, and see that $f$ restricts to an isomorphism $IN\to IN'$. Setting $I=M$, $N=R^m$, $N'=R^n$ gives what you want.

Answer 2

Let $f:R^n\rightarrow R^m$ be an isomorphism. Consider $\overline{f}:(R/M)^n\rightarrow (R/N)^m$ given by $\overline{f}([r])=f(r)$ (the map is clearly well-defined). Proceeding in an analogue way we can construct $\overline{f^{-1}}$. It is clear that $\overline{f^{-1}}$ is the inverse of $\overline{f}$.