If $L/K$ is a finite field extension and $L$ is perfect, then $K$ is perfect.

Question

I'm aware that this question has been answered here. However, I wanna try and prove this directly:

Let $L/K$ be a finite field extension, suppose $L$ is perfect and $\text{char}(K) = p > 0$. Show that $L/K$ is separable. (Hint: Show that there exists an $n \in \mathbb{N}$ s.t $x^{p^n} \in K_s = \{a \in L\; |\; a\; \text{is separable over}\; K\}$ for every $x \in L$.)

Since $L$ is perfect, every $x \in L$ has a $p$-th root in $L$. I don't see how to prove the hint though. Any help is greatly appreciated!

Answer 1

In essence, this results from the fact that, over a field, $F$, of characteristic $p>0$, and any element, $\alpha$, that is algebraic over $F$, we have that $\alpha$ is separable over $F$ iff $F(\alpha)=F(\alpha^p)$. Put another way (as we trivially have $F(\alpha^p)\leq F(\alpha)$ regardless of whether or not $\alpha$ is separable and $F(\alpha)$ is obviously a finite extension of $F$), we have that an element $\alpha$ that is algebraic over $F$ is separable over $F$ iff $[F(\alpha):F]=[F(\alpha^p):F]$. The contrapositive of this is that an element, $\alpha$, that is algebraic over $F$ is inseparable over $F$ iff $[F(\alpha):F]>[F(\alpha^p):F]$. One more minor fact we need is that if $\alpha$ is separable so is $\alpha^p$ (this follows from, e.g., the fact that the separable closure is a field, or the fact that a simple extension by a separable element is separable).

Now addressing your question directly: as $L/K$ is finite, we may choose some integer $N\geq[L:K]-1$. I claim this is our desired integer. For suppose by way of contradiction that there exists some $y\in L$ such that $y^{p^N}$ is inseparable over $K$. By the minor fact stated above, this means $y^{p^i}$ is inseparable over $K$ $\forall\; 1\leq i\leq N$. So for each $1\leq i\leq N$, we have $[K(\alpha^{p^i}):K]>[K(\alpha^{p^{i+1}}):K]$ giving us the chain of inequalities:

$[K(\alpha):K]>[K(\alpha^p):K]>[K(\alpha^{p^2}):K]>\dots>[K(\alpha^{p^N}):K]>[K(\alpha^{p^{N+1}}):K]$

This is a sequence of $N+2\geq [L:K]+1$ integers. At the same time, as $L\geq K(\alpha)$ and $K(\alpha^{p^{N+1}})\geq K$, we have $[L:K]\geq [K(\alpha):K]$ and $[K(\alpha^{p^{N+1}}):K]\geq [K:K]=1$. So this is in fact a sequence of at least $[L:K]+1$ integers between $1$ and $[L:K]$, a contradiction by the pigeonhole principle. It follows that $x^{p^N}$ is separable over $K$ $\forall\; x\in L$.

There is another argument that proves the above that uses minimal polynomials of inseparable elements instead. That argument allows one to do some optimization and show that in fact any $N$ such that $p^N\geq [L:K]$ will work (and this is the best possible lower bound, I think).

Answer 2

For $K(a)/K$ a finite extension in characteristic $p$ let $$f(x)=\prod_j (x-a_j)^{p^{e_j}}\in K[x]$$ be its minimal polynomial and $F$ the splitting field. There is an automorphism $\sigma_j \in Aut(F/K)$ sending $a_1\to a_j$ thus $e_1=e_j$ so that $$f(x)=g(x)^{p^{e_1}}, \quad g(x)=\prod_j (x-a_j)=\sum_{n=0}^d c_n x^n$$ $$f(x)=\sum_{n=0}^d c_n^{p^{e_1}} x^{p^{e_1}n}, \qquad \sum_{n=0}^d c_n^{p^{e_1}} x^{n}=\prod_j (x-a_j^{p^{e_1}}) \in K[x]$$ Whence $K(a^{p^{e_1}})/K$ is separable.

Going back to your extension, if $n= [L:K]$ then $K(L^{p^n})/K$ is separable.

But $L=L^{p^n}$ so $K(L^{p^n})=L$ and $L/K$ is separable.