If for $A$ a commutative nonzero ring $A^m ≅ A^n$ as $A$-modules, then $m = n$ This is the problem I need to solve:

Let $A$ be a nonzero ring. Show that if $A^m ≅ A^n$, then $m = n$.

The book I got this problem from suggests using the following method to solve it:

Let $M$ be a maximal ideal of $A$ and let $f: A^m \rightarrow A^n$ be an isomorphism. Then $1\otimes  f:(A/M) \otimes A^m \rightarrow (A/M) \otimes  A^n$ is an isomorphism between vector spaces of dimensions $m$ and $n$ over the field $K = A/M$. Hence $m = n$.

The trouble is, I have no idea how to use this hint, mainly because I do not comprehend tensor products. We haven't gone over them in class due to some inclement weather closing it, and nothing I have read about them in the book or looking around on the internet, including a few answers to a question on this very website, makes sense to me. I just can't seem to get my head around what they are, how they're made from the modules they're made from, or what they're for.
Is there a way to solve this problem without using tensor products? And if there isn't, are there any proofs of other problems which use tensor products that I can read and maybe get an understanding of how to use this object in a proof?
 A: Yeah, I don't know why tesnor products are recommended here either.
Convince yourself that the existence of a module isomorphism between $R^n$ and $R^m$ is tantamount to the existence of an $n\times m $ matrix $A$ and an $m\times n$ matrix $B$ such that $AB=I_n$ and $BA=I_m$, all with entries in $R$. This is just basic linear algebra about what $Hom_R(R^n,R^m)$ looks like.
Now if you take a maximal ideal $M$ and apply the projection of $R$ onto $R/M$ to the entries of these matrices, you get matrices over a field with the same property. But we know this is not possible for fields since finite dimensional vector spaces have unique dimension.
A: Show that $f(M^n) = M^m$. This is harder than it sounds, but not too hard.
That means that $\bar f:A^n / M^n \to A^m / M^m : x + M^n \mapsto f(x) + M^m$ is well-defined and bijective; it is obviously $A$-linear.  However $A^n / M^n \cong (A/M)^n$ so we get an isomorphism $\hat f : (A/M)^n \to (A/M)^m$ that is $A/M$-linear. Since $A/M$ is a field, and dimension is well defined for vector spaces like $(A/M)^n$ and $(A/M)^m$ we get $m=n$.
