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Let $A$ be a UFD domain, i.e. integral and for any height one prime ${\frak p}$ of $A$, we have ${\frak p} = (u_{\frak p})$ for some $u_{\frak p} \in A$.

Once and for all, we fix the algebraic closure $\overline{K}$ and consider the integral closure $\overline{A}$ of $A$ in $\overline{K}$.

We consider, for a finite integer $d < \infty$, the following homomorphism of $A$-algebras $\colon$

\begin{equation}\label{P} f \colon A[X_1,\ldots,X_d] \to \overline{A}, \end{equation} where we define $X_i \mapsto a_i \in \overline{A}$.

We set ${\frak P} \colon= \mathrm{Kernel}(f)$, which is a prime ideal of $A[X_1,\ldots,X_d]$.

Q. Even if $A$ is $not$ noetherian, is ${\frak P}$ always finitely generated ?

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    $\begingroup$ Two questions to clarify what you're asking. First, where do the elements of $\overline A$ live? Maybe you're fixing an algebraic closure $\overline K$ of the fraction field $K$ of $A$ and taking $\overline A$ to be the elements of $\overline K$ that are integral over $A$. Second, you say that "$f$ is the following ring homomorphism", but you don't define $f$. Are you choosing elements $a_1,\ldots,a_d\in\overline A$ and taking $f$ to be the homomorphism defined by $X_i\to a_i$? If not, then what is $f$? $\endgroup$
    – Joe Silverman
    Commented Oct 4, 2016 at 20:55

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If $f$ is not required to be a morphism of $A$-algebras, a stupid counteraxmple exists. For instance, $A=k[x_1,x_2,\dots]$(infinitely many variables), $d=0$ and $$f:A\to A/(x_1,x_2,\dots)\cong k\subset A\subseteq \overline{A}$$

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    $\begingroup$ I think the OP meant $f$ to be a morphism of $A$-algebras. $\endgroup$
    – Fan Zheng
    Commented Oct 4, 2016 at 22:15
  • $\begingroup$ I am afraid that there might be some weird non-noetherain local ring $A$ such that ${\frak P}$ is infinitely generated. That is, it seems that the coherence of $A$ has much to do with my question. For example, if $A$ does not enjoy coherence, it could happen that ${\frak P}$ might be infinitely generated, although I cannot still give a concrete counter-example. Please just teach me any opinion. $\endgroup$
    – Pierre MATSUMI
    Commented Oct 5, 2016 at 17:17
  • $\begingroup$ @PierreMATSUMI Could you please clarify your comment, do you want $f$ to be a morphism of $A$-algebras?(with this asumption questoin becames much more interesting) I am asking because in my example $A$ is coherent. $\endgroup$
    – SashaP
    Commented Oct 5, 2016 at 21:41
  • $\begingroup$ Yes. I would like to have my $f$ as $A$-algebra homomorphism! $\endgroup$
    – Pierre MATSUMI
    Commented Oct 6, 2016 at 11:07
  • $\begingroup$ I am Pierre Matsumi. For the moment, I circumvented the above question for my researches. For readers, I express my gratitude. As for me, I guess there would be some weird non-neotherian ring $A$ which makes our ${\frak P}$ infinitely generated. I cannot give the concrete example, but I think it certainly exists. $\endgroup$
    – Pierre MATSUMI
    Commented Oct 7, 2016 at 7:50
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This is true at least for $d=1$.

In that case, $\mathfrak{P}=(m_{K,a_1}(X_1))$ is the principal ideal generated by the minimum polynomial of $a_1$ over $K$. Indeed, UFDs are integrally closed, so by Th. 9.2 of Matsumura's Commutative ring theory, $m_{K,a_1}(X)$ is in $A[X]$.

As $A[a_1]$ may not be integrally closed, using induction on $d$ seems to break down.

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