0
$\begingroup$

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?

$\endgroup$

1 Answer 1

2
$\begingroup$

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 $F(t)$.

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – Ofir
    Mar 24, 2012 at 14:50
  • $\begingroup$ 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. $\endgroup$
    – Ofir
    Mar 24, 2012 at 15:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .