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Suppose $k$ is algebraically closed field. And $p(x_1,\ldots,x_n)\in k[x_1,\ldots,x_n]$ is an irreducible polynomial.

I wonder to show $p(x_1,\ldots,x_n)+z\in \overline{k(z)}[x_1,\ldots,x_n]$ is an irreducible polynomial, where overline denotes "the algebraic closure of" . (we are not allowed to use the result in this post, i.e., there are only finitely many value $a\in k$ such that $p+a$ is reducible over $k$.)

Then applying the following lemma we can show that there are only finitely many value $a\in k$ such that $p+a$ is reducible over $k$.(*)

Lemma Suppose $f:X\to Y$ is a morphism of finite type, with $Y$ irreducile. Suppose the generic fiber of $f$ is geometrically integral. Then there exists a nonempty open subset $U\subset Y$ such that the fiber $X_y$ is geometrically integral for all $y\in U$.

Another strong lemma stated like following(?): Suppose $X\to S$ is a morphism of schemes of finite presentation, and $s\in S$ is a point such that the fiber $X_s$ is geometrically integral then there is a open neighbourhood $U$ of $s$ such that for all $y\in U$ the fiber $X_y$ is geometrically integral.

If the strong lemma is true, then set $X=\operatorname{Spec} k[z][x_1,\ldots,x_n]/p(x_1,\ldots,x_n)+z$ and $S=\operatorname{Spec} k[z]$, $s=(z)\in S$, we can see (*) is true.

However, I cannot find a reference in the literature about this strong lemma, (so I donot know if it is true) could anyone give a reference or give a proof about it?

Thank you very much!

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Just being curious: why do you suspect that this "strong lemma" is true at all, if you cannot prove it yourself and can't find a reference in the literature? –  Nils Matthes Aug 13 '12 at 8:24

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As for the first version of the lemma - which is sufficient in your case - consult the free textbook of the Stacks Project at http://stacks.math.columbia.edu/.

In my version (which is not the newest one) you have to combine the lemmata 33.18.4 and 33.20.4 to get what you want. I guess that you can also derive it from some of the results in EGA IV/3, 15.5.

Comment: I am quite certain that the proof George gave in the related post is by far simpler than what you try now.

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