Do tensor products support cancellation? If $A\otimes B\simeq A\otimes C$, is it true that $B\simeq C$? If not, under what conditions might this be true?
EDIT:
More specifically, I'm working on a problem where I have a field $K$ and an ideal $I\subseteq A$, where $A$ is an algebra over $K$. I also have that $I\otimes_K K\simeq A\otimes_K K$. I would like to conclude that $I=A$. Can I do this? (This seems like it has to be true.)
 A: Let $K$ be the field of rational numbers, $A=K[x]$ be the ring of polynomials with coefficients from $K$, and $I=(x)$ be the ideal of polynomials $f$ with $f(0)=0$.
As $A$-modules, $I \cong A \cong I \otimes_K K \cong A \otimes_K K$, but $I \neq A$.
One of course has cancellation up to isomorphism: if $M$ is an $A$-module, then $M \otimes_K K \cong M$ as $A$-modules, so if $I \otimes_K K \cong A \otimes_K K$, then $I \cong I \otimes_K K \cong A \otimes_K K \cong A$.
If $A$ is a finite dimensional algebra, then things are better:  if $I \otimes_K K \cong A \otimes_K K$ then both $I$ and $A$ have the same dimension as $K$-vector spaces, and since $I \leq A$ and $A$ is finite dimensional, one must actually have equality: $I=A$.
If $A$ is actually an algebra over a subfield of $L$, and you want to change fields to $K$, then look for "separable algebra" to see cases where things behave reasonably.
A: Consider $\mathbb{Q} \otimes_\mathbb{Z} \mathbb{Z}_n$. Since we have 
$$\frac{a}{b} \otimes k = \frac{a}{nb} \otimes nk = 0,$$
all the pure tensors vanish, so the tensor product does too. Comparing this with $\mathbb{Q} \otimes_\mathbb{Z} \mathbb{Z}$ shows that cancelation need not hold.
A: I think I should point out the following in an answer.
As I said in the comment above, if you take a divisible group $D$ and a finite abelian group $G$ then $D\otimes_\mathbb{Z} G=0$ for basically the same reason why the above user pointed out $\mathbb{Q}\otimes_\mathbb{Z}\mathbb{Z}_n=0$--of course neither of the factors is zero.
Moreover, it's clear that there are problems even with "nice" modules. For example, if you take something really nice, say some field $k$, then $k^m\otimes_k k^n\cong k^\ell\otimes_k k^j$ whenever $mn=\ell j$. That said, it's clear that since fields have the IBN property if we choose $m\ne \ell$ that $k^m\not\cong k^\ell$.
