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 $v>1$.

  • $\begingroup$ 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

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 $\mathbb C[x_1,\ldots,x_n]$. $\square$

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

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