Irreduciblity of a polynomial over a field extension

Posted: September 16, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: ,

Suppose that $p(x)$ is an irreducible polynomial over some field $F.$ Let $E/F$ be a finite field extension. The following problem investigates the irreducibility of $p(x)$ over $E.$

Problem. Suppose that $E/F$ is a finite field extension and $p(x) \in F[x].$ Prove that if $p(x)$ is irreducible over $F$ and $\gcd(\deg p(x), [E:F])=1,$ then $p(x)$ is irreducible over $E.$

Solution. We may assume, without loss of generality, that $p(x)$ is monic. Let $a$ be a root of $p(x)$ in some extension of $E$ and let $q(x)$ be the minimal polynomial of $a$ over $E.$ Clearly

$\deg q(x) \leq \deg p(x). \ \ \ \ \ \ \ (1)$

We also have

$[E(a):E)][E:F]=[E(a):F(a)][F(a):F]. \ \ \ \ \ \ (2)$

It now follows from $(2)$ and $\gcd(\deg p(x), [E:F]) = 1$ that $\deg p(x) \mid \deg q(x)$ and so by $(1)$

$\deg p(x)=\deg q(x). \ \ \ \ \ \ \ (3)$

Let $r(x)=q(x)-p(x) \in E[x].$ Then $\deg r(x) < \deg q(x)$ by $(3)$ and the fact that $p(x)$ and $q(x)$ are monic. We also have $r(a)=0$ and thus $r(x)=0$ by the minimality of $q(x).$ Hence $q(x)=p(x). \ \Box$

Note that it is possible for $p(x)$ to be irreducible over $E$ but $\gcd(\deg p(x), [E:F]) \neq 1.$ For example consider $p(x)=x^2+1, \ F = \mathbb{Q}$ and $E = \mathbb{Q}(\sqrt{2}).$