How does $S \otimes_R A$ become an $S$-algebra?

Question

Let $f : R \longrightarrow S$ is a ring homomorphism and $A$ is an $R$-algebra then in our lecture note it has been stated that its scalar extension $S \otimes_R A$ is an $S$-algebra. How is that possible? I know that the scalar extension is a $S$-module given by the well defined operation $s(s_1 \otimes x):=ss_1 \otimes x$. Also I know that if $A$ and $B$ are two $R$-algebra then their tensor product $A \otimes B$ is again an $R$-algebra with the well-defined operation given by $(a \otimes b)(a' \otimes b'):=aa' \otimes bb'$.

Now from these two facts how can I reach at the desired conclusion? Please help me in this regard.

Thank you in advance.

Answer 1

You have to be careful about actions being left or right if the rings are not commutative. In general, if you have $AB$-bimodule $M$ (i.e. left $A$-action and right $B$-action) and $BC$-bimodule $N$ (left $B$-action and right $C$-action), then the tensor product $M\otimes_B N$ is $AC$-module:

$$a(m\otimes n) = (am)\otimes n,\ (m\otimes n)c = m\otimes(nc)$$ and if $M$ and $N$ are moreover algebras, then the tensor product will again be algebra.

In your case you have $S\otimes_R A$. To make it into $S$ module, you need to have

• left $S$-action on $S$ - this is just multiplication in $S$,
• right $R$-action on $S$ - this is given by $f$, $s.r := sf(r)$,
• left $R$-action on $A$ that you have since $A$ is an algebra.

Algebra multiplication is defined as you wrote it. The left $S$-action is defined as $$s'(s\otimes a) = (s's)\otimes a.$$