I am reading Algebraic Number Fields by Janusz. In the middle of one of his proofs on page 23, he makes the following claim:
Suppose $K$ is a field and $L$ a finite field extension that is not separable over $K$. Then $K$ is a field of characteristic $p$ and there is a subfield $F$ of $L$ with $K\subseteq F\subset L$ and
- $\dim_FL=[L:F]=p^m\neq 1$, for some $m\geq 1$;
- for each $x\in L$, $x^p\in F$.
I know how to prove that $K$ is a field of characteristic $p$, and I have tried various fields for $F$ [for example, I tried one given $L=K(a_1,\ldots,a_n)$ where I took $F=K(a_1^p,\ldots,a_n^p)$]; however, I always have trouble proving that either $x^p\in F$ or $K\subseteq F\subset L$. For the example given above, I couldn't understand why $F\neq L$ is necessarily true.
First take the separable closure $K^s$ of $K$ in $L$, so the subfield generated by those elements in $L$ which are separable over $K$. Then $K^s/K$ is separable, so $K^s$ is a proper subfield of $L$ since $L/K$ is not separable. Also, $L/K^s$ is purely inseparable, so every element $a\in L$ satisfies $a^q\in K^s$ for some power $q$ of $p$. Now write $L=K^s(a_1,\ldots,a_n)$ with $a_1$ not in $E=K^s(a_2,\ldots,a_n)$. The minimal polynomial of $a_1$ over $E$ is of the form $x^{p^m}-b$ for some $b\in E$ and some $m\geq1$, so the minimal polynomial of $a_1^p$ over $E$ is $x^{p^{m-1}}-b$. We can now set $F=E(a_1^p)$. We have $[L:F]=p$ and every element $a\in L$ satisfies $a^p\in F$.