Is the contraction of a minimal prime of the extension always the original prime?

Question

Let $A\subseteq B$ be finite-type integral $k$-algebras, where $k$ is an algebraically closed field. Suppose we have a prime $\mathfrak{p}\subseteq A$ and $\mathfrak{q}\subseteq B$ a minimal prime of $\mathfrak{p}B$. Is it then always true that $\mathfrak{q}\cap A=\mathfrak{p}$? Only the inclusion $\mathfrak{q}\cap A\supseteq\mathfrak{p}$ is obvious. I suspect it is false, although I couldn't come up with a counter example.

Edit: Note that I don't consider an example where the extension $\mathfrak{p}B$ is the unit ideal $\mathfrak{p}B=(1)$ a counterexample. This is because then the set of minimal primes $\min(\mathfrak{p}B)$ of $\mathfrak{p}B$ is empty, and thus the statement $$ \forall \mathfrak{q}\in\min(\mathfrak{p}B):\ \mathfrak{q}\cap A=\mathfrak{p} $$ is vacuously true.

Answer 1

For an example where $\mathfrak{p}$ is not a contraction of a prime of $B$: take $A = k[x,xy] \subseteq k[x,y] = B$, and $\mathfrak{p} = xA$, so that $\mathfrak{p} B \cap A = (x, xy)A \supsetneq \mathfrak{p}$. (This is a classic example of a map of affine varieties with constructible but non-closed image).

For an example where $\mathfrak{p}$ is a contraction of a prime of $B$: take $A = k[x,y] \subseteq k[x,y,z,w]/(xw - yz) = B$, and $\mathfrak{p} = xA$, so that $\mathfrak{p} B = (x, xw - yz)B = (x, yz)B$, which has 2 minimal primes $\{ (x, y)B, (x, z)B \}$, and $(x, y)B \cap A = (x, y)A \supsetneq \mathfrak{p}$. (Note: this is the blowup of the affine plane at the origin.)

Answer 2

At least, without the hypothesis on $k$ being algebraically closed, you have the counter example with $A = \mathbb Z$ and $B = \mathbb Q$ :
$\mathfrak p = 3 \mathbb Z$.
Then $\mathfrak q = 3 \mathbb Q = \mathbb Q$ and $\mathfrak q \cap \mathbb Z = \mathbb Z \ne \mathfrak p$.

Adding the hypothesis i don't know.