# If $E/F$ is finite divisible, then it is separable.

I want build a separable extension $E/F$. Suppose that $E/F$ is a finite divisible field extension. I want to prove that $E/F$ is separable in this method:

we know that if $Char(F)=0$, then $E/F$ is separable. So suppose that we are in case $Char(F)=p>0$ for a prime $p$ and $F$ is indivisible. We know that $F^*=N_{E/F}(E^*)\times C_m$ for $1\neq m|[E:F]=n$ and From ${F^*}^n‎\subseteq‎ N_{E/F}(E^*)$, every element of $C_m$ is a root of $x^n-1$. So if I can say $p\nmid m$, I can prove that ${F^*}^p=F^*$, and from it I can prove that $F$ is perfect and so $E/F$ is separable.

So my question is why $p\nmid m$?

• I have heard of divisible (abelian) groups and modules, but somebody pray tell me what is a divisible field extension? Are you referring to divisibility of the multiplicative group? – Jyrki Lahtonen Aug 14 '15 at 8:45
• A divisible group? – Bernard Aug 14 '15 at 8:57
• Yes, we say a group $G$ is radicable if for every $a\in G$ and every $n\in \mathbb{N}$, there exists $b\in G$ such that $b^n=a$. A ring $R$ is called radicable if its unit group is radicable. We say divisible when the group is abelian. A field $F$ is divisible when its multiplicative group is divisible. – MH.Fakharan Aug 14 '15 at 9:36
• Then “divisibility” is not a property of extensions, but of the individual fields. Did you mean for both $E$ and $F$ to be “divisible” or just $E$? – Lubin Aug 15 '15 at 16:58
• $F$ can be both of them, indivisible or divisible. – MH.Fakharan Aug 19 '15 at 3:23

As $C_m$ is a cyclic group of order $m$, so we have $m$th root of unity in $F$ like $a$. If $p=Char(F)|m$, then there exists natural number $r$ such that $rp=m$ we have :
$0=a^m-1=(a^r)^p-1=a^r-1$, so the order of $a$ can not be $m$ and it is contradiction. Therefore $p\nmid m$.