Inseparable extensions

Question

Given that $F/K$ is a finite extension of fields and the extension is not separable.

My question is whether we can always find a subfield $L$ such that $K\subset L \subset F$ and $[F:L]=p$, where $p=char(K)$.

Answer 1

You’re looking for a subfield $L$ that’ll have $[F:L]=p$. First we take the maximal separable extension $K^{\text{sep}}$ of $K$ in $F$, so that $F\supset K^{\text{sep}}$ is totally inseparable, and of finite degree $p^n$. So let’s go for an inductive argument, supposing that $p^m$ is the smallest degree of a proper subfield $F$ with $K^{\text{sep}}\subset F\subset L$ and $[L:F]=p^m$. We want $m=1$.

So we suppose that $[L:F]=p^m$ and take any $a\in L$ that’s not in $F$. Its minimal $F$-polynomial will be $X^{p^k}-b$ for some $k\le m$ and some $b\in F$. If $F(a)$ is a proper subfield of $L$, then we have $[L:F(a)]<[L:F]=p^m$, contradicting minimality of $m$. So we may take $F(a)=L$. If $k=1$ we’re done, but in any case, $F(a^p)$ is a proper subfield of $L$, ’cause if we in fact had $F(a^p)=L$, we’d also have $F(a^{p^i})=L$ for all $i$, and that’s not the case for $i=k$. Thus we take $F(a^p)$ for the desired subfield.