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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.

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

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  • $\begingroup$ 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? $\endgroup$ – zongxiang yi Jan 20 '18 at 13:43
  • $\begingroup$ In this case, does $f$ have $5$ distinct simple roots over $E$ ? $\endgroup$ – zongxiang yi Jan 20 '18 at 13:45
  • $\begingroup$ @zongxiangyi Yes, of course. $\endgroup$ – egreg Jan 20 '18 at 14:02
  • $\begingroup$ why $f$ has 5 distinct simple roots over $E$ ? Any hints? $\endgroup$ – zongxiang yi Jan 20 '18 at 14:07
  • $\begingroup$ @zongxiangyi A finite field is perfect. So irreducible polynomials have distinct roots in their splitting field. $\endgroup$ – egreg Jan 20 '18 at 14:12

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