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

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 $$s'(s\otimes a) = (s's)\otimes a.$$
