Isomorphism of (tensored) algebras by restricting/extending scalars Let $A, B$ be commutative rings with identities, the ring map $f: A \rightarrow B$ gives $B$ the $A$-algebra structure. Let $S=f(A)$, $C$ an $S$-algebra (hence is also an $A$-algebra). Then is the map
$$\phi:B\otimes_A C \rightarrow B\otimes_S C, b\otimes c\mapsto b\otimes c$$
an $A$-algebra isomorphism?
 A: Yes, this is a particular case of the following result:    
Given two $A$-modules $B,C$ (not necessarily algebras) killed by an ideal $I\subset A$, the canonical $A$-module morphism $$\phi:B\otimes_A C \rightarrow B\otimes_{A/I} C: b\otimes c\mapsto b\otimes c$$  is an isomorphism.
This is proved by going back to the construction of a tensor product as a quotient of a free group.    
Now, just apply this to $I=\text{ker}(\phi:A\to B)$ and $A/I=S$.  
Edit
As usual in these more advanced questions Bourbaki is the best (only?) reference: Algebra, Chapter II, §3, Corollary to Proposition 2, page 246.
A: Yes, the formula $b \otimes c \mapsto b \otimes c$ defines maps $B \otimes_A C \to B \otimes_S C$ and $B \otimes_S C \to B \otimes_A C$ which, if they are well defined, are clearly inverse to each other.
To check that these maps are well defined you just have to prove that they respect the bilinear relations between simple tensors in the domain.  Additivity is obvious and balanced with respect to the action of $A$ and $S$ follows from the fact that $A$ acts through $f$.
