$A\otimes_{\mathbb C}B$ is finitely generated as a $\mathbb C$-algebra. Does this imply that $A$ and $B$ are finitely generated? Consider $A$ and $B$ two $\mathbb C$-algebras such that $A\otimes_{\mathbb C}B$ is finitely generated as a $\mathbb C$-algebra. Does this imply that $A$ and $B$ are finitely generated?
I know that for general algebras, this is false. Indeed $\mathbb Q$ is infinitely generated over $\mathbb Z$ but the tensor product $ \mathbb Q\otimes_\mathbb Z \mathbb Z_2 =0$. For $\mathbb C$-algebras however, I just can't seem to find a counter-example. 
 A: Yes, it does, as long as $A$ and $B$ are both not the zero ring (obviously $A\otimes 0=0$ is finitely generated for any $A$).  Choose a finite set of generators of $A\otimes_\mathbb{C} B$; each of these is a finite sum of tensors $a\otimes b$.  Let $A_0\subseteq A$ be the subalgebra generated by all the $a$'s appearing in these tensors.  Then $A_0$ is finitely generated, and we see that the natural map $A_0\otimes_\mathbb{C} B\to A\otimes_\mathbb{C} B$ is surjective (since its image contains all of the tensors $a\otimes b$ in our generators).  This means $A/A_0\otimes_\mathbb{C} B=0$, so as long as $B\neq 0$, we must have $A/A_0=0$ and so $A_0=A$.  Thus $A$ is finitely generated.  By the same argument, $B$ is also finitely generated.
This argument clearly works with $\mathbb{C}$ replaced by any field.  Much more generally, a similar argument shows that if $R$ is any base ring and $A$ and $B$ are $R$-algebras such that $B$ is faithfully flat over $R$, then if $A\otimes_R B$ is finitely generated as a $B$-algebra (in particular, if it is finitely generated as an $R$-algebra), then $A$ is finitely generated as an $R$-algebra.
