Help understand isomorphism of tensor product and connection to vector spaces well I'm having a hard time understanding the tensor product. Here's a problem from Atiyah and Macdonald's book:
Let $A$ be a non-trivial ring and let $m,n$ be positive integers. Let $f: A^{n} \rightarrow A^{m}$ be an isomorphism of $A$-modules. Show this implies that $n=m$. 
Well the solution is at follows:
Let $m$ be a maximal ideal of $A$. Then we have an induced isomorphism:
$(A/m) \otimes_{A} A^{n} \rightarrow (A/m) \otimes_{A} A^{m}$.
Now it says that this is an isomorphism between vector spaces of dimension $n$ and $m$.
My questions are:
1) How do we know that $(A/m) \otimes_{A} A^{n}$ is a vector space over $A/m$ ?
2) How do we know it has exactly dimension $n$?
Is there some "standard" theorem that tells us this? Can you please explain this in detail?
Thanks
 A: 1) Let $f: A \rightarrow B$ be any homomorphism of commutative rings and let $M$ be any $A$-module.  Then one may view $B \otimes_A M$ as a $B$-module via $b \cdot (\sum_i b_i \otimes x_i) := \sum_i b b_i \otimes x_i$.  This process of starting with an $A$-module and getting a $B$-module is called base change.
In your particular case the map is $f: A \rightarrow A/\mathfrak{m}$, so $(A/\mathfrak{m}) \otimes_A M$ is an $A/\mathfrak{m}$-module.  But since $\mathfrak{m}$ is maximal, $A/\mathfrak{m}$ is a field, and a module over a field is simply a vector space over that field.  This answers your first question.
2) You should establish the following two properties of tensor products:
(i) For any $A$-module $M$, there is a canonical isomorphism $M \otimes_A A \stackrel{\sim}{\rightarrow} M$.  (Thus "changing the base from $A$ to $A$" has no effect.)
(ii) For any family $M_i$ of $A$-modules indexed by a set $I$ and any $A$-module $M$, there is a canonical isomorphism $(\bigoplus_{i \in I} M_i) \otimes_A N \stackrel{\sim}{\rightarrow} \bigoplus_{i \in I} (M_i \otimes_A N)$.  
What you want follows easily from these facts (which are themselves ubiquitously useful).  Try it yourself and ask again if you have further questions.
A: The original question you are trying to solve (that $A^m \cong A^n$ as $A$-modules implies $m = n$) does not need tensor products.  If $M$ and $N$ are isomorphic $A$-modules then, for any ideal $I$ in $A$, $M/IM$ and $N/IN$ are isomorphic $A/I$-modules. More precisely, if $f \colon M \rightarrow N$ is an $A$-module isomorphism with inverse $g$, then the induced map $M/IM \rightarrow N/IN$ given by $x \bmod IM \mapsto f(x) \bmod IN$ is well-defined and $A/I$-linear, and it's an $A/I$-module isomorphism because the map $y \bmod IN \mapsto g(y) \bmod IM$ going the other way is an inverse (just check on each element that these two maps undo each other). Taking $M = A^m$, $N = A^n$, and $I = \mathfrak m$ to be a maximal ideal in $A$, we get from an $A$-module isom. of $A^m$ with $A^n$ that the $A/\mathfrak m$-vector spaces $A^m/{\mathfrak m}A^m$ and $A^n/{\mathfrak m}A^n$ are isomorphic. Now it remains to show $A^m/IA^m \cong (A/IA)^m$ for any ideal $I$, and this is left to you. No tensor products are needed. I think this is useful because it makes this result available if you are teaching an abstract algebra class without having to introduce tensor products already. Tensor products are of course very important, but they need not be used here. In fact, later on in the course when you have tensor products you can come back and reprove this result on isomorphisms with them and observe that the argument is basically the same as the one that didn't use tensor products.
After learning about tensor products, and then exterior powers, you can prove finer properties of $A$-linear maps $A^m \rightarrow A^n$. See Corollary 5.11 in 
http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/extmod.pdf, which incidentally provides a third way to handle the original question via exterior powers without using maximal ideals in $A$.
