# Is there an irreducible integral polynomial in two variables which is reducible for every value of one of the variables?

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.

• Hilbert's irreducibility theorem shows that this is impossible except for trivial counterexamples like A.Mortier's $f(x,y) = 2(x+y)$. – Noam D. Elkies Jun 27 '13 at 18:25
• Thanks @NoamD.Elkies! Embarrasingly, I didn't know Hilbert's irreducibility theorem. If you post that comment as an answer I'll accept it. – Omar Antolín-Camarena Jun 27 '13 at 18:29

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 $\mathbb{Z}$.
• That's right, argh! I meant over $\mathbb{Q}$, not $\mathbb{Z}$. – Omar Antolín-Camarena Jun 27 '13 at 18:25