1
$\begingroup$

Given a polynomial in more than one variable, how do you prove it's irreducible? I only have one method, called Eisenstein's criterion.

I can't work out the following problem:

Given that $u$ and $v$ are relative prime numbers, and that $f(x_1,x_2,\dots,x_n)$ is an irreducible polynomial over $\mathbb{C}$, prove that $y^u+f^v$ is irreducible (as a polynomial in variables $x_1, \dots, x_n, y$).

If $v=1$, it's trivial. But I think it's hard if $v>1$.

$\endgroup$
  • $\begingroup$ I think it should be equivelent to such a problem that if you replace f with a prime number p, and consider this statement in $\mathbb{z}[x]$. $\endgroup$ – Intoks Liobein Dec 14 '15 at 2:25
2
$\begingroup$

By Kummer theory, the extension $$\mathbb C(x_1,\ldots,x_n) \subseteq \mathbb C(x_1,\ldots,x_n)\left(\sqrt[u]{-f^v}\right)$$ is cyclic of degree $u$, since $u$ is the smallest number $k$ such that $(\sqrt[u]{-f^v})^k \in \mathbb C(x_1,\ldots,x_n)$ (here you need that $u$ and $v$ are coprime, that $\mathbb C[x_1,\ldots,x_n]$ is a UFD, and that $f$ is irreducible).

Thus, the polynomial $y^u + f^v$ is irreducible over $\mathbb C(x_1,\ldots,x_n)$, since it defines a field extension whose degree equals the degree of the polynomial. Since it is monic, Gauß's lemma implies that it is irreducible over $\mathbb C[x_1,\ldots,x_n]$. $\square$

$\endgroup$
  • $\begingroup$ Thanks! It's exactly what I want. $\endgroup$ – Intoks Liobein Dec 14 '15 at 4:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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