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

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.

• $A = k[x] \subseteq k[x,y]/(xy) = B$, $\mathfrak{p} = 0$, $\mathfrak{q} = (x)$ Nov 16, 2023 at 20:42
• @math54321 Thanks, but note that I'm looking for integral examples, and in your example $B$ is not an integral domain. Nov 16, 2023 at 21:51
• OK, then take e.g. $k[x] \subseteq k[x, x^{-1}]$ Nov 18, 2023 at 19:18
• @math54321 And what prime ideals? If we take $\mathfrak{p}=(x)$ then the extension of $\mathfrak{p}$ is $(1)$, which has no minimal prime. Hence for $\mathfrak{p}=(x)$, the statement is vacuously true. Nov 20, 2023 at 7:54
• Alright, let's see if 3rd time's the charm: $A = k[x,xy] \subseteq k[x,y] = B$, $\mathfrak{p} = xA$, so that $\mathfrak{p}B \cap A = (x, xy)$ Nov 20, 2023 at 16:44

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.)
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 $$\mathfrak q \cap \mathbb Z = \mathbb Z \ne \mathfrak p$$.
• Thanks for the input! I don't think that it works though. The extension $\mathfrak{p}^e$ is the unit ideal $\mathfrak{p}^e=(1)$, and by definition the unit ideal is not prime. Hence there are no minimal primes $\mathfrak{q}$ to test for, so the statement is vacuously true. Nov 20, 2023 at 8:05