Module over a tensorial product of algebras

Question

Consider $A$ and $B$ $\mathbb{K}$-algebras. If $V$ is an $A$-module and simutaneously an $B$-module, then $V$ is an $A\otimes B$-module?

I can't prove or disprove this one. Someone can help me? Thank you in advance.

Answer 1

In general this won't be true. The point is that in $A\otimes_K B$, the elements in $1\otimes B$ and $A\otimes 1$ commute with each other. So actually a structure of $(A\otimes_K B)$-module on a $K$-vector space $V$ is the same as a structure of $A$-module and $B$-module such that the actions commute.

For a counter-example, you can take $A=B=V$, with the action of $A$ on itself by product on the left. Then if $A$ is not commutative, in general it won't be a $(A\otimes_K A)$-module. Note that it might have some other $(A\otimes_K A)$-module structure, but it won't be induced by the multiplication on the left.

You can actually find $A$ such that it cannot have any $(A\otimes_K A)$-module structure, but it's a little more involved. For the sake of completeness: you can take $A$ to be a finite-dimensional division algebra of odd exponent, say of $K$-dimension $d$. Then $A\otimes_K A$ will still be a division algebra, and any non-trivial module over it needs to have $K$-dimension at least $d^2$, so $A$ cannot be such a module.