Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

$M \cong N\otimes_k L$ ?

share|improve this question

1 Answer 1

up vote 4 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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.