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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.

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  • $\begingroup$ $A = k[x] \subseteq k[x,y]/(xy) = B$, $\mathfrak{p} = 0$, $\mathfrak{q} = (x)$ $\endgroup$
    – math54321
    Commented Nov 16, 2023 at 20:42
  • $\begingroup$ @math54321 Thanks, but note that I'm looking for integral examples, and in your example $B$ is not an integral domain. $\endgroup$
    – imtrying46
    Commented Nov 16, 2023 at 21:51
  • $\begingroup$ OK, then take e.g. $k[x] \subseteq k[x, x^{-1}]$ $\endgroup$
    – math54321
    Commented Nov 18, 2023 at 19:18
  • $\begingroup$ @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. $\endgroup$
    – imtrying46
    Commented Nov 20, 2023 at 7:54
  • $\begingroup$ Alright, let's see if 3rd time's the charm: $A = k[x,xy] \subseteq k[x,y] = B$, $\mathfrak{p} = xA$, so that $\mathfrak{p}B \cap A = (x, xy)$ $\endgroup$
    – math54321
    Commented Nov 20, 2023 at 16:44

2 Answers 2

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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.)

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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.

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  • $\begingroup$ 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. $\endgroup$
    – imtrying46
    Commented Nov 20, 2023 at 8:05

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