Irreduciblity of a polynomial over a field extension

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

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Comments
  1. very helpful result…

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