Splitting field and primitive element

Question

Let $F$ be a field of characteristic $p$

$f(x)$ a non-constant irreducible polynomial over $F$,

and $E$ a splitting field of $f(x)$.

Is it true that $E$ contains a primitive element? i.e., is $E=F(\alpha)$ for some $\alpha\in E$?

Answer 1

• There is no primitive element for $\Bbb{F}_p(x,y)/\Bbb{F}_p(x^p,y^p)$ the splitting field of $(T^p-x^p)(T^p-y^p)\in \Bbb{F}_p(x^p,y^p)[T]$ which is not irreducible.

  Proof: if $\Bbb{F}_p(x,y)=\Bbb{F}_p(x^p,y^p,a)$ then $a^p \in \Bbb{F}_p(x^p,y^p)$ so that $[\Bbb{F}_p(x,y):\Bbb{F}_p(x^p,y^p)]\le p$ contradicting that $[\Bbb{F}_p(x,y):\Bbb{F}_p(x^p,y^p)]=p^2$.

• Following the same idea there is no primitive element for $\Bbb{F}_p(x,y)/\Bbb{F}_p(x^py^p,x^p+y^p)$ the splitting field of $(T^p-x^p)(T^p-y^p)\in \Bbb{F}_p(x^py^p,x^p+y^p)[T]$ which is this time irreducible as $[\Bbb{F}_p(x,y^p):\Bbb{F}_p(x^py^p,x^p+y^p)]=2p$.

• If $f$ is separable so that $E/F$ is separable then there is a primitive element (follow the proof of the primitive element theorem).

• If $E/F$ is not separable but $F$ is algebraic over $\Bbb{F}_p(x)$ then there is a primitive element as well, because $F(E^{p^k})/F$ is separable for some $k$ so that $F(E^{p^k})=F(a)$, then $E=F(x^{1/p^m}a)$.