Question: Let $f: A\rightarrow B$ be a homomorphism of rings which makes $B$ an $A$-algebra. Suppose $\mathfrak{q}$ is a prime ideal of $B$ and $f^{-1}(\mathfrak{q})=\mathfrak{p}$, then I want to know if we have $A_{\mathfrak{p}}\otimes_AB\cong B_\mathfrak{q}$ as rings?
I don't assume that $f$ is surjctive. I can consturct a map $A_{\mathfrak{p}}\otimes_AB\rightarrow B_\mathfrak{q}$ by $\frac{a}{s}\otimes b\rightarrow \frac{f(a)b}{f(s)}$. To prove the isomorphism, I want to construct the inverse map, but it seems to me there is no obvious map, since $f$ is not assumed to be surjcetive.
Motivation: This result is used to prove $\mathcal{O}_{X, x}/\mathfrak{m}y \, \mathcal{O}_{X,x} \simeq \mathcal{O}_{X_y, x}$ in this question, User "nowhere dense" considered the map $B\xrightarrow{g} B\otimes_A A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$ in that question, and he got $(B\otimes_A A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p})_\tilde{\mathfrak{q}}\cong ( A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}\otimes_A B)\otimes_B B_\mathfrak{q}$. I don't know why this is true.
Edit. I don't think this question is just $S^{-1}A\otimes_A B\cong f(S)^{-1}B$, since we don't know $A-\mathfrak{p}$ doesn't map to $B-\mathfrak{q}$.
Any help is appreciated.
This is not true. Suppose $A= \mathbb{C}, p=0$ and $B=\mathbb{C}[t], q=(t) $. The map $f: \mathbb{C} \to \mathbb{C}[t] $ is the inclusion. Note that $f^{-1}(q) = q \cap \mathbb{C} = 0$.
You have that $A_p = A$ because $\mathbb{C}$ is a field and everything beside zero is already invertible. Now $$ A_p \otimes_A B = A \otimes_A B \simeq B$$
but $$ B_q = \mathbb{C}[t]_{(t) }$$
is not isomorphic to $B$, because it is a local ring while $B$ has infinitely many maximal ideals.