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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

$M \cong N\otimes_k L$ ?

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up vote 6 down vote accepted

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.

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