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Let $A \subsetneq B$ be a ring extension, where $A$ and $B$ are commutative $k$-algebras, $k$ is a field of characteristic zero. Let $P$ be a prime ideal of $B$. Denote the contraction of $P$ to $A$ by $P^{c}=P \cap A$, and its extension to $B$ by $P^{ce}=(P^{c})^{e}=(P \cap A)^{e}=(P \cap A)B$.

I am looking for special cases in which $P^{ce}=P$.

Remarks:

(1) If $f: U \to V$ is a surjective ring homomorphism, then for every ideal of $U$ (prime or not) we have $I^{ce}=I$, see here.

(2) If $A \subset B$ is faithfully flat, then for every prime ideal $p$ of $A$, $p^{ec}=p$, see the answer to this question.

(3) Is it true that if $B$ is a finitely generated free $A$-module, then my question has a positive answer? (the condition "finitely generated" is necessary, since, for example, my question has a negative answer for $A=k$, $B=k[x,y]$).

user26857
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user237522
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    If I'm not mistaken (3) fails in characteristic zero, too: set $A=\mathbb Q[X]$, $B=\mathbb Q[\sqrt X]$, and $P=\sqrt X B$. Then $P^c=XA$ and $P^{ce}=XB\subsetneq P$. – user26857 Jun 05 '17 at 22:44
  • If I am not wrong, another example for a negative answer for (3) is: $A=\mathbb{C}[x^2]$, $B=\mathbb{C}[x^2,x^3]$. – user237522 Jun 05 '17 at 23:14
  • In (2) I meant to bring an example where a similar property ($p^{ec}=p$) holds for a 'nice' ring extension, namely, a faithfully flat ring extension. My hope was that, for example, a free ring extension will satisfy $P^{ce}=P$, but unfortunately this is not true. It still interests me if another 'nice' extension satisfies $P^{ce}=P$. – user237522 Jun 05 '17 at 23:25
  • Slightly (but not really) changing your example to: $A=\mathbb{C}[x^2]$, $B=\mathbb{C}[x]$, $P=Bx$; I am not sure, but think that I can show that if: (i) $A$ is a UFD. (ii) $A \subset B$ is integral. (iii) $P^{ce}=P$ for every prime ideal $P$ of $B$, then $B$ is also a UFD. Your example shows that $B$ can be a UFD without requiring (iii). – user237522 Jun 05 '17 at 23:48
  • If $B$ is also an integral domain, then you are right. This follows easily from Kaplansky's Theorem for UFDs. – user26857 Jun 06 '17 at 08:50
  • Yes, I had in mind a Noetherian integral domain $B$ and the following argument, based on the results brought in http://www2.gsu.edu/~matfxe/commalglectures/lect16.pdf: By Theorem 1.1 we need to show that every height one prime is principal. Let $P$ be a height one prime in $B$. then by Lemma 1.2 $P \cap A= aA$ for some $a \in A$. By my assumption (iii) $aB=(aA)B=P$, so $P$ is principal, and we are done by Theorem 1.1, or by Kaplansky's Theorem, since $a \in aB=P$ is a prime element of $B$ (If I am not wrong,Theorem 47 in Kaplansky's book "commutative rings" guarantees that $P \cap A \neq 0$). – user237522 Jun 06 '17 at 10:17
  • No need to assume $B$ Noetherian. The condition (iii) guarantees that for $P\ne(0)$ we have $P\cap A\ne(0)$. – user26857 Jun 06 '17 at 10:31
  • Thanks. Please do you have any ideas when condition (iii) is satisfied? – user237522 Jun 06 '17 at 17:35
  • Well, the localization is an obvious example, but if you want to deal with finitely generated algebras over a field, then this is not useful. – user26857 Jun 06 '17 at 20:41
  • Can you please elaborate on localizations (in an answer)? This idea sounds interesting for me. – user237522 Jun 06 '17 at 21:09
  • If $P$ is a prime in $S^{-1}R$ then $P=S^{-1}p$ for $p$ a prime of $R$ with $S\cap p=\emptyset$. Then $P^c=p$ and $p^e=S^{-1}p$. That's all. – user26857 Jun 06 '17 at 21:14
  • Perhaps (I am not sure yet, I might be wrong) from ideas discussed in https://math.stackexchange.com/questions/140584/about-the-localization-of-a-ufd it will be possible to show that $B$ (as in one of the above comments) is a UFD. – user237522 Jun 06 '17 at 21:16
  • I hoped that something can be done when $A \subset B$ is an integral extension of integral domains with $A$ UFD, but this is not possible, as $k[x^2] \subset k[x^2,x^3]$ shows. – user237522 Jun 06 '17 at 21:35

1 Answers1

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I know it has been a while since the question was posted, anyway (if I'm right) the following should be an example of $\mathfrak p^{ce}=\mathfrak p.$

Let $f\colon A\longrightarrow B$ be an injective, flat ring homomorphism. Let $\mathfrak p\in\mathsf{Spec}(B)$ and $\mathfrak q:=f^{-1}(\mathfrak p)$. I claim that $$\mathfrak qB=\mathfrak p.$$

The ring map $B/\mathfrak qB\longrightarrow B/\mathfrak p$ is onto. We show that it is injective too. As $f$ is flat and $A/\mathfrak q\longrightarrow B/\mathfrak p$ is injective, tensorizing by -$\otimes_AB$ we get an injective ring homomorphism $$\phi\colon B/\mathfrak qB=B\otimes_A A/\mathfrak q\longrightarrow B\otimes_AB/\mathfrak p.$$ Now just notice that $\phi$ factorizes as $$B/\mathfrak qB\longrightarrow B/\mathfrak p \longrightarrow B\otimes_AB/\mathfrak p.$$

Vincenzo Zaccaro
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    It's hard to believe that for $K\subseteq K[X]$ we have $(X)^{ce}=(X)$, so you are not right. – user26857 May 28 '19 at 07:54
  • @user26857 You are right. Did you see the mistake in my proof? – Vincenzo Zaccaro May 28 '19 at 13:54
  • @Vincenzo Zaccaro The mistake you made is in the last line. In general your putative factorization of $\phi$ is broken. Informally you can see that $\phi$ send the $B$ data in $B/\mathfrak{q}B$ to the left-hand tensor in $B \otimes_A B/\mathfrak{p}$, whereas in $B/\mathfrak{q}B \rightarrow B/\mathfrak{p} \rightarrow B \otimes_A B/\mathfrak{p}$ the $B$ data is sent to the right-hand tensor. However, with the additional assumption that $A \rightarrow B$ is an epimorphism your proof goes through fine! – Badam Baplan Nov 23 '20 at 06:44