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Is there a polynomial $f(x,y)$ in two variables, with integer coefficients, such that $f$ is irreducible over the complex numbers (i.e., in $\mathbb{C}[x,y]$), but for every integer $n$, the polynomial in one variable $f(x,n)$ is reducible over $\mathbb{Q}$?

For comparison, there are polynomials in one variable which are irreducible, but reducible mod $n$ for every $n$. See the question Polynomial reducible modulo every integer.

EDIT: As Arnaud Mortier's example showed, I should have said reducible over the rationals, not the integers, so I edited the question.

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    $\begingroup$ Hilbert's irreducibility theorem shows that this is impossible except for trivial counterexamples like A.Mortier's $f(x,y) = 2(x+y)$. $\endgroup$
    – Noam D. Elkies
    Commented Jun 27, 2013 at 18:25
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    $\begingroup$ Thanks @NoamD.Elkies! Embarrasingly, I didn't know Hilbert's irreducibility theorem. If you post that comment as an answer I'll accept it. $\endgroup$
    – Omar Antolín-Camarena
    Commented Jun 27, 2013 at 18:29

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I would go for $2(X+Y)$. It is irreducible over the complex numbers for degree reasons, and for every integer $n$, $f(x,n)$ is a non-zero polynomial divisible by 2 and another polynomial of positive degree, hence reducible over $\mathbb{Z}$.

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  • $\begingroup$ That's right, argh! I meant over $\mathbb{Q}$, not $\mathbb{Z}$. $\endgroup$
    – Omar Antolín-Camarena
    Commented Jun 27, 2013 at 18:25

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