TITLE: If $L\mid K$ is a finite extension of fields then K is perfect iff L is perfect
QUESTION [5 upvotes]: Problem: Let $L\mid K$ be a finite extension of fields. Then $K$ is perfect $\iff$ $L$ is perfect.

The implication $\implies$ is quite easy and it has already been discussed here.
I'm interested in the implication $\impliedby$ because of the following application:

Suppose $\text{char}(k)=p>0$. Then any finitely generated field extension $K\mid k$ that is perfect has $\text{tr.deg}_k(K)=0$.

For the proof of this recall that a purely trascendental extension is never perfect as $t_1$ is never a $p$-power in the field $k(t_1,\dots,t_n)$. Then if $K$ is a finite extension of a purely trascendental extension it wouldn't be perfect as well.
That can be restated in a neat way in the language of algebraic geometry:

If $\text{char}(k)>0$ then any $k$-variety with perfect function field must have dimension $0$.

Also it would be interesting to see if the equivalence in the problem is true if we change finite extension by algebraic extension. Again the implication $\implies$ is not so hard and has been discussed here.

REPLY [5 votes]: In the answer of darij grinberg the following equality is given $[L:K]=[L^p:K^p]$ (for the proof notice that $K$-basis of the vector space $L$ correspond to $K^p$-basis of $L^p$ under Frobenius). Using this I think we can answer the problem in a really short way:
As we have the tower of extensions
$$K^p\hookrightarrow K\hookrightarrow L=L^p$$
we have that $[L^p:K^p]=[L:K][K:K^p]$ but as $[L:K]=[L^p:K^p]$ we conclude $[K:K^p]=1$.