Irreducible polynomials in extensions of the wrong degree
by math_explorer, Apr 22, 2016, 4:32 PM
Theorem. Let 
be a field, let 
be a degree-
irreducible polynomial in ![$\mathbb{F}[x]$](//latex.artofproblemsolving.com/2/1/e/21efdb2c5a47c516116b846c750a130557395cbe.png)
, and let 
be a degree-
extension of . If and are coprime, then is still irreducible over ![$\mathbb{K}[x]$](//latex.artofproblemsolving.com/9/a/4/9a4ac7b5d11ac5e92e9ad4bc2aa3de7fafdbd91c.png)
.
(The condition is tight; if
, then 
has the factor ![$x - \sqrt[g]{2}$](//latex.artofproblemsolving.com/3/3/3/3333747fc3152c87deba8511f891661c2881b7a6.png)
in ![$\mathbb{K}(\sqrt[q]{2})$](//latex.artofproblemsolving.com/5/3/9/5396dde60d65b2351c33a92376c2d1f9f20c2892.png)
.
Proof. Suppose not, and let
be a (proper, nontrivial) irreducible polynomial factor of over . Let 
be the degree of . Adjoin a root 
of to , producing the field 
, which has degree over and thus degree 
over .
Then note that, since is a factor of , we have that is also a root of , which means 
has degree . Furthermore, 
. But since the former has degree and the latter has degree over , ![\[ p \mid dq \Longrightarrow p \mid d \Longrightarrow d \in \{0, p\}, \]](//latex.artofproblemsolving.com/b/d/8/bd83f9995c7e2262ef2b4d9bd3fe5eac21fbb0a3.png)
contradicting the claim that is a proper nontrivial factor.
be a field, let 
be a degree-
irreducible polynomial in ![$\mathbb{F}[x]$](http://latex.artofproblemsolving.com/2/1/e/21efdb2c5a47c516116b846c750a130557395cbe.png)
, and let 
be a degree-
extension of . If and are coprime, then is still irreducible over ![$\mathbb{K}[x]$](http://latex.artofproblemsolving.com/9/a/4/9a4ac7b5d11ac5e92e9ad4bc2aa3de7fafdbd91c.png)
.(The condition is tight; if
, then 
has the factor ![$x - \sqrt[g]{2}$](http://latex.artofproblemsolving.com/3/3/3/3333747fc3152c87deba8511f891661c2881b7a6.png)
in ![$\mathbb{K}(\sqrt[q]{2})$](http://latex.artofproblemsolving.com/5/3/9/5396dde60d65b2351c33a92376c2d1f9f20c2892.png)
.Proof. Suppose not, and let
be a (proper, nontrivial) irreducible polynomial factor of over . Let 
be the degree of . Adjoin a root 
of to , producing the field 
, which has degree over and thus degree 
over .Then note that, since
is a factor of , we have that is also a root of , which means 
has degree . Furthermore, 
. But since the former has degree and the latter has degree over , ![\[ p \mid dq \Longrightarrow p \mid d \Longrightarrow d \in \{0, p\}, \]](http://latex.artofproblemsolving.com/b/d/8/bd83f9995c7e2262ef2b4d9bd3fe5eac21fbb0a3.png)
contradicting the claim that is a proper nontrivial factor.
May 2020
January 2020