TITLE: Modules over a tensor product
QUESTION [9 upvotes]: Let $k$ be a field. Suppose $A$ and $B$ are two commutative $k$-algebras.
Let $M$ be a finite $A\otimes_k B$-module. 
Can one find a finite $A$-module $N$ and a finite $B$-module $L$ such that
$M \cong N\otimes_k L$ ?

REPLY [8 votes]: Here is a counterexample. Put 
$$
A=B=k^2.
$$
Let $(e_1,e_2)$ be the canonical basis of $k^2$, and define $M$ by 
$$
M:=\frac{k^2\otimes_k k^2}{(e_2\otimes e_2)}\quad.
$$ 
EDIT. Justification: 
A $k^2$-module is "the same thing as" a couple $(V_1,V_2)$ of $k$-vector spaces.
(More precisely, to the $k^2$-module $V$ we attach the pair $(e_1V,e_2V)$. The inverse functor is the obvious one.)
A $k^2\otimes_k k^2$-module is "the same thing as" a quadruple  $(V_{ij})_{i,j=1,2}$ of $k$-vector spaces. 
If $U$ is the $k^2$-module given by $(U_1,U_2)$ and $V$ is the $k^2$-module given by $(V_1,V_2)$, then $U\otimes_kV$ is the $k^2\otimes_k k^2$-module given by $(U_i\otimes_kV_j)_{i,j=1,2}$. 
This implies the claim.