$R^n \cong R^m$ iff $n=m$

How can i show that two $R$-modules of finite rank are isomorphic if and only if they have the same rank, i.e., $R^n \cong R^m$ iff $n=m$.

Solution 1:

This is true of rings that have the "IBN" (invariant basis number property). Among these are commutative rings--which is probably what you meant.

Use Krull's theorem to find a maximal ideal $\mathfrak{m}$ of $R$ and use the fact that since $R^n\cong R^m$ as $R$-modules that $(R/\mathfrak{m})\otimes_R R^n\cong (R/\mathfrak{m})\otimes_R R^m$ as $R/\mathfrak{m}$-modules. But, from basic module theory this is just $(R/\mathfrak{m})^n\cong(R/\mathfrak{m})^m$ and so we've reduced the problem from general rings to fields--something you should be familiar with.

Solution 2:

If localization is not your thing, there is a different elementary proof.

Just as in linear algebra, you can prove that the set of R linear transformations from $R^n\to R^m$ is isomorphic to the set of $n\times m$ matrices over R. An isomorphism would amount to an $n\times m$ matrix A and an $m\times n$ matrix B such that AB and BA are identity matrices.

Now pick a maximal ideal M of R, and apply the quotient map from R to the field R/M entry wise to the matrix. Now you have two matrices over a field which multiply to identities in either order, giving an isomorphism of vector spaces. Since we know two isomorphic vector spaces have equal dimension, m=n.