Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to understand irreducibility of polynomials in 2 variables. Specifically, I want to prove that if $f(t)$ is irreducible in $F[t]$ (where $F$ is a field), then $f(s-t)$ is irreducible in $F(t)[s]$.

Is there a criterion\theorem which helps to prove this result? Is it even true?

It feels to me like an opposite direction of Hilbert's irreducibility theorem, i.e. turning an irreducible element of a polynomial ring with 1 variable, into an irreducible element of a polynomial ring with 2 variables. I hope it is not hard, I actually feel it is easy, but I don't have enough experience with polynomial rings over 2 variables.

EDIT: What about the following generalization? If $f(t)$ is irreducible in $F[t]$ and $h(s,t)$ is irreducible in $F[s,t]$, is $f(h(s,t))$ irreducible in $F(t)[s]$? Is it true?

share|cite|improve this question
up vote 2 down vote accepted

It seems to me that since $t$ is in the coefficient field for $f(s-t)$ it can't affect factorization; $f(s-t)$ is irreducible over $F(t)$ if and only if $f(s)$ is. And the factorization of $f(s)$ can't involve the indeterminate $t$; if it's irreducible over $F$, it's still irreducible over $F(t)$.

share|cite|improve this answer
Can you explain the last point ("the factorization of f(s) can't involve the indeterminate t")? It sounds intuitive, but when I try to prove it I get many different cases and none of them seems easy or simple. – Ofir Mar 24 '12 at 14:50
OK, I managed to prove it. Gauss's Lemma reduced it to proving that $f(s)$ is irreducible in $F[s][t]$. Then I used the fact that the only invertible polynomials in an integral domain are constant. – Ofir Mar 24 '12 at 15:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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