If $U,V$ are left $A$-modules, then is $U \otimes V$ a left $A \otimes A$-module?

Question

Let $A$ be a unital associative algebra over the field $k$ and $U,V$ left $A$-modules. A book I'm reading claims that $U \otimes V$ is a left $A \otimes A$-module.

First, how is the tensor product taken? Since $k \subseteq A$, we have that $U$ and $V$ are $k$-vector spaces so we can consider $U \otimes_k V$ and I'm guessing this is what $U \otimes V$ denotes. Then, on this we want to put an $A \otimes A$-module structure.

I guess this is given by $$(a \otimes b). (u \otimes v)= (au) \otimes (bv)$$?

One then must check this is wel-defined and determines a unique action (we only said what it does to basis elements). Is my above interpretation correct?

Answer 1

Yes, that is what you need to prove. I agree that the book probably uses $\otimes$ to denote $\otimes_k$.

To show that the multiplication is well-defined, it is enough to show that $(a,b,u,v)\mapsto (au)\otimes(bv)$ is $k$-linear in each argument (which easily follows from the hypothesis that $U,V$ are $A$-modules). This will give you a $k$-bilinear map $(A\otimes_k A)\times (U\otimes_k V)\to (U\otimes_k V)$.

By bilinearity, to show that this is an action of $A\otimes_k A$, it is enough that you have mixed associativity for simple tensors. But this easily follows from mixed associativity for $U$ and $V$.

Answer 2

It seems to me that your interpretation is correct.

You should check conventions on what exactly $\otimes$ means, as it could also potentially mean tensor product over $\mathbb{Z}$. (I doubt that is the case, but I suppose it could happen.) But certainly, your interpretation is a reasonable one.