# Find the degree of splitting field of a separable polynomial over finite field

Question:

Let $f$ be a polynomial of degree $n$ over finite field $F=\mathbb{F}_q$ and $E$ be the splitting field of $f$. If $f$ has $n$ simple roots over $E$, can we say that $[E:F] \mid n$ ? If it's true, show it.

Example:$F=\mathbb{F}_3$, $f(x)=x^4+1$, $E=F(\alpha)$, where $\alpha \in \mathbb{C}$ and $f(\alpha)=0$. Then $[E:F]=2$ and $\deg(f)=4$. It has $[E:F]\mid \deg(f)$.

Example:$g=\mathbb{F}_5$, $f(x)=x(x-1)$, $E=F(\beta)$, where $\beta\in \mathbb{C}$ and $g(\beta)=0$. Then $[E:F]=1$ and $\deg(g)=2$. It also has $[E:F]\mid \deg(g)$.

Thanks for any replies.

• 1 2 are true, but 3 and "$[E:F]=n$" is false – Kenny Lau Jan 20 '18 at 12:28
• Consider $f(x) = x^4+1$ over $\Bbb F_3$ – Kenny Lau Jan 20 '18 at 12:30
• @KennyLau is "$[E:F] \mid n$" true ? – zongxiang yi Jan 20 '18 at 12:32
• Even simpler: consider $\;f(x)=x(x-1)\;$ over any field... – DonAntonio Jan 20 '18 at 12:32
• @DonAntonio He asked "if $f$ is irreducible" in his now deleted comment – Kenny Lau Jan 20 '18 at 12:32

Let $f=gh$, where $g$ is irreducible of degree $2$ and $h$ is irreducible of degree $3$. If $a$ is a root of $g$, then the splitting field of $f$ is the same as the splitting field of $h$ over $F(a)$ (which is the splitting field of $g$).

Can you show that $[E:F(a)]=3$, that is, $h$ is irreducible over $F(a)$?

• If $h$ is irreducible over $F(a)$, then $[E:F]=[E:F(a)]*[F(a):F]=2*3=6$, but $\deg(f)=2+3=5$, So $[E:F] \nmid deg(f)$. Am I right? – zongxiang yi Jan 20 '18 at 13:43
• In this case, does $f$ have $5$ distinct simple roots over $E$ ? – zongxiang yi Jan 20 '18 at 13:45
• @zongxiangyi Yes, of course. – egreg Jan 20 '18 at 14:02
• why $f$ has 5 distinct simple roots over $E$ ? Any hints? – zongxiang yi Jan 20 '18 at 14:07
• @zongxiangyi A finite field is perfect. So irreducible polynomials have distinct roots in their splitting field. – egreg Jan 20 '18 at 14:12