For $f:B\to A$, what can be said about $B/I\otimes _BA\overset{?}{\cong}A/ \left\langle fI \right\rangle $? Let $f:B\to A$, be a commutative unitary ring homomorphism and let $I$ be an ideal of $B$. When is it true that $B/I\otimes _BA\cong A/ \left\langle fI \right\rangle $?
My motivation is geometric: I'm asking whether the pullback of a closed subscheme of an affine schemes is the closed subscheme determined by corresponding ideal.
 A: I see that there is an answer in the comments, but I've already typed this, and it may be helpful, anyway:
They are always isomorphic as rings. Consider the short exact sequence of $B$-modules
$$ 0 \to I \to B \to B/I \to 0$$
Let us tensor the sequence over $B$ with the $B$-module $A$ (the map $f$ gives $A$ the structure of a $B$-module/algebra). Since this is a right-exact functor, we have that the sequence
$$ I\otimes_B A \to B \otimes_B A \to (B/I) \otimes_B A \to 0$$
is exact. It is easy to show that $I\otimes_B A$ is isomorphic to $f(I)A$, the ideal in $A$ generated by the image of $I$ under $f$, by the map extending $i\otimes a \mapsto f(i)a$. It is also easy to show that $B\otimes_B A$ is isomorphic to $A$ by the map extending $b\otimes a \mapsto f(b)a$. Finally, the diagram
$$\begin{array}{ccccccc}
I\otimes_B A & \to & B\otimes_B A & \to & (B/I)\otimes_B A & \to & 0 \\
\downarrow & & \downarrow \\
f(I)A & \to & A & \to & A/(f(I)A) & \to & 0
\end{array}$$
commutes and the rows are exact. You can fill in the last arrow $(B/I)\otimes_B A \to A/(f(I)A)$ by extending the map taking $\bar b \otimes a \mapsto \overline{f(b)a}$, where $\bar b$ is the class of $b$ in $B/I$ and $\overline{f(b)a}$ is the class of $f(b)a$ in $A/(f(I)A)$. You must show that this map is well-defined and then that it is an isomorphism, but this is a matter of following the diagram. 
However, this only shows that they are isomorphic as $B$-modules. This is something that's not addressed in the post referenced in the comments above because that post was only concerned with modules. You can check, however, that the map we constructed is actually a ring homomorphism, which finishes things off. 
