Extension of scalars $M_B=B\otimes_A M$

Question

Most textbooks say that the $B$-module structure on $M_B$ (for $A\rightarrow B$ a ring morphism and $M$ an $A$-module) is "defined" by $b'(b\otimes m)=b'b\otimes m$. How is this a proper definition? After all not every element of $M_B$ is of the form $b\otimes m$, isn't it? Shouldn't the proper definition go as follows: fix $b'\in B$, define $b'\cdot:M_B\rightarrow M_B$ to be induced by the $A$-bilinear map $B\times M\rightarrow M_B$ sending $(b,m)\mapsto b'b\otimes m$.

Answer 1

Tensor products are defined via their universal property. This means that linear maps on tensor products are defined via their corresponding bilinear maps, which makes it possible to define them just on pure tensors (and check bilinearity of course). Thus, your interpretation is correct, but this doesn't mean that the other authors are wrong.

Here is a abstract-nonsense method to get the module structure:

An $A$-algebra $B$ is just an $A$-module with a homomorphism $m : B \otimes_A B \to B$ with nice properties (often also including a unit $A \to B$), which are just commutative diagrams. A left $B$-module is just an abelian group $N$ with a homomorphism $\alpha : B \otimes_A N \to N$ with nice properties. Now if $M$ is a left $A$-module and $B$ is an $A$-algebra, then $B \otimes_A M$ becomes an $B$-module via $$\alpha : B \otimes_A (B \otimes_A M) \cong (B \otimes_A B) \otimes_A M \xrightarrow{m \otimes \mathrm{id}_M} B \otimes_A M.$$ (This method has the advantage to be applicable in more abstract situations, too, where no elements are available anymore.)

Answer 2

Hint: show that for every $b' \in B$, the map

$$B \times M \to B \otimes_A M$$

given by $(b, m) \mapsto b'b \otimes m$ is $A$-bilinear, and therefore it factors through a map

$$B \otimes_A M \to B\otimes_A M$$

given on simple tensors by $b \otimes m \mapsto b'b\otimes m$. Show that these maps, for all $b' \in B$, give $B \otimes_A M$ the structure of a $B$-module.

(Oops, I just saw that this is exactly what you had in mind. Well, rest assured, you have the right idea.)