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.

  • 9
    $\begingroup$ Hilbert's irreducibility theorem shows that this is impossible except for trivial counterexamples like A.Mortier's $f(x,y) = 2(x+y)$. $\endgroup$ Jun 27, 2013 at 18:25
  • 1
    $\begingroup$ Thanks @NoamD.Elkies! Embarrasingly, I didn't know Hilbert's irreducibility theorem. If you post that comment as an answer I'll accept it. $\endgroup$ Jun 27, 2013 at 18:29

1 Answer 1


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 $\mathbb{Z}$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.