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A typical formulation of Hilbert's irreducibility theorem is like this (see [1]):

Let $k=\mathbb{Q}$ and $f\in k[x_1,\ldots,x_n,y_1,\ldots,y_m]$ be an irreducible polynomial. There exists a Zariski open dense subset $V\subseteq k^n$ such that for every $(a_1,\ldots,a_n)\in V$ the polynomial $f(a_1,\ldots,a_n,y_1,\ldots,y_m)$ is irreducible in $k[y_1,\ldots,y_m]$.

Versions of the above can be found for the field $k$ being a number field. My question is: can we generalize this to the prime ideals. More precisely, is the following true:

Let $k$ be a number field, let $k_0\subseteq k$ be a subfield, and let $\frak{p}$ be a prime ideal in $k[x_1,\ldots,x_n,y_1,\ldots,y_m]$ such that ${\frak p}\cap k[x_1,\ldots,x_n]=\{0\}$. There exists a Zariski open subset $\style{text-decoration:line-through}{V\subseteq k^n}$ an $(a_1,\ldots,a_n)\in k^n\setminus k_0^n$ such that the ideal $$ \bar{\frak p}=\{\omega(a_1,\ldots,a_n,y_1,\ldots,y_m)\mid \omega\in{\frak p}\} $$ is a prime ideal of $k[y_1,\ldots,y_m]$.

A reference would be greatly appreciated.

[1] M-D.Huang and Y.-Ch.Wong, Extended Hilbert irreducibility and its applications, J. Algorithms 37 (2000), no. 1, 121–145.

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    $\begingroup$ You need to add the hypothesis that $k[x_1,\dots,x_n]\to k[x_1,\dots,x_n,y_1,\dots,y_m]/\mathfrak{p}$ is an injective homomorphism. Assuming this, your formulation reduces to the usual formulation by applying the Noether Normalization Theorem to the integral closure, $A$, of $k[x_1,\dots,x_n]$ with respect to that morphism. There exists an element $z$ in $A$ such that the induced morphism $k[x_1,\dots,x_n,z]\to A$ becomes surjective with principal kernel $\langle f \rangle$ after inverting a nonzero element $g\in k[x_1,\dots,x_n]$. Now apply the usual theorem to $f\in k[x_1,\dots,x_n,z]$. $\endgroup$
    – Jason Starr
    Jul 2, 2018 at 8:54
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    $\begingroup$ @Adam Przeździecki Your `typical formulation' of Hilbert's irreducibility theorem is false. Take e.g. $m=n=1$ and $f=y_1^2-x_1$. Thus $f(y_1,a_1)$ is reducible for infinitely many rational $a_1$'s, so a Zariski open subset of the rationals avoiding these numbers is the empty set. $\endgroup$
    – Peter Mueller
    Jul 2, 2018 at 9:25
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    $\begingroup$ @Peter Mueller: The result stated in OP's question is true, just take $V$ to be empty. Presumably OP meant Zariski dense instead of Zariski open. $\endgroup$
    – js21
    Jul 2, 2018 at 9:45
  • $\begingroup$ @js21 The stated result is true, but it is not Hilbert's irreducibility theorem. $\endgroup$
    – Peter Mueller
    Jul 2, 2018 at 12:00
  • $\begingroup$ @PeterMueller Thanks, the "formulation" was copied from the paper cited. This also explains why I could not find this formulation in the Lang's book from which it was supposedly taken. $\endgroup$
    – Adam Przeździecki
    Jul 2, 2018 at 13:17

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