# Can we take a tensor product of algebra and module?

I'm trying to learn tensor product and I found that there are at least two different tensor products, tensor product of modules and tensor product of algebras. But can we mix them? Like if $M$ is an $A$-module and $N$ is a $A$-algebra, does $M\otimes_A N$ make ever sense?

• how do you define "tensor product of algebras" and from which reference? Commented Jan 10, 2015 at 17:47
• Have a look at scalar extensions, maybe you will find there what you are looking for. Commented Jan 10, 2015 at 17:50
• From Liu's "Algebraic Geometry and Arithmetic Curves" 1.1.3. k.stm: That was a good explanation. Thanks! Commented Jan 10, 2015 at 17:52

## 2 Answers

Sure; one does this all the time, in fact. Let me use $B$ for an $A$-algebra instead of $N$. It is, in particular, an $A$-module so we can at least form the $A$-module $M \otimes_A B$. The additional feature is that $M \otimes_A B$ is also a $B$-module in a natural way — to see this, you can use the $A$-trilinear map \begin{align*} B \times M \times B &\to M \otimes_A B \\ (b, m, b') &\mapsto m \otimes bb' \end{align*} To give you some keywords: this is often called extension of the base or of scalars. In algebraic geometry this corresponds to the operation of pulling back a quasicoherent sheaf, so it's important to study how the properties of $M$ relate to those of $M \otimes_A B$.

An example you may have seen before: to study a real vector space $V$ one forms the complexification $V \otimes_{\mathbf{R}} \mathbf{C}$ and enjoys an abundance of eigenvalues; the game is now to relate what you've learned back to $V$. See almost any text in Lie theory, eg, Bump's, for applications.

There is only one definition; in fact when you want to define "tensor product of algebras", you take the module structure of algebra via homomorphism