Let $F$ be a field of characteristic $p \neq 0$ and $\alpha$ an element of an extension $E$. Show $\alpha$ is separable over $F$ iff $F(\alpha) = F(\alpha^p)$.
For the backward direction, I haven't got a clue where to start. For the forward direction, I've been thinking of letting $f$ be the minimal polynomial of $\alpha$. Then, if $\alpha^p$ can be shown to be separable, it may be possible to conclude something from the fact that $f(x) \mid g(x^p)$, where $g$ is the minimal polynomial of $\alpha^p$. This seems tough, though.
Looking through the internet, I found a few very lengthy proofs that $\alpha^p$ is separable, but these seem too involved for an exercise. I don't know where to go from here.
I should add that, from the structure of the textbook I'm reading, it seems that a proof may somehow be related to the notion of a perfect field, but it seems difficult (impossible?) to "extend" any given field to a perfect one, much less deduce any properties from that extension.
I now understand the backward direction. I'm having trouble, though, understanding the proof of the forward direction that was provided:
$F(\alpha)/F(\alpha^p)$ is purely inseparable, if $F(\alpha)/F$ is separable then it must be that $F(\alpha)/F(\alpha^p)$ is trivial ie. $F(\alpha)=F(\alpha^p)$.
I don't understand what is meant by $F(\alpha)/F(\alpha^p)$. I guess it's just that I've never seen this notation before (and I haven't been able to find it on the internet). Is it some sort of quotient field, as in $F(\alpha)$ modulo $F(\alpha^p)$? I assume it's not that because that field seems to be isomorphic to $F(\alpha)/(\alpha^p)$, but I might be wrong. Are we just trying to say that $F(\alpha)$ is inseparable in $F(\alpha^p)$? This also seems wrong – every polynomial over $F$ is also a polynomial over $F(\alpha^p$), but we know $\alpha$ to be separable over $F$.
Finally, I don't see how it follows that $F(\alpha)/F(\alpha^p)$ is trivial. I guess that's because I don't understand the notation, though.
Best Answer
$F(\alpha)/F(\alpha^p)$ is purely inseparable, if $F(\alpha)/F$ is separable then it must be that $F(\alpha)/F(\alpha^p)$ is trivial ie. $F(\alpha)=F(\alpha^p)$.
If $F(\alpha)/F$ is inseparable then the $F$-minimal polynomial of $\alpha$ is of the form $g(x^p)$ with $g\in F[x]$, so $\alpha^p$ is a root of $g$ and $[F(\alpha):F]=p[F(\alpha^p):F]$.