If char(K) is prime and $x\longmapsto x^p$ is bijective, then $L/K$ is separable.

Question

Let $K$ a filed of characteristic $p$ a prime and $L/K$ a field extension. If $\varphi:x\longmapsto x^p$ is bijective, then $L/K$ is separable.

Q1) First of all, where is $\varphi$ defined ? on $L$ ? On an other extension ?

I will suppose $\varphi$ is defined on $L$ (but maybe it's only in $K$ ?)

Suppose it's not separable. Then, there is an $a\in L$ s.t. the minimal polynomial of $a$ (let denote it $P$) has multiple root. In particular, $$P(x)=(x-a)^nG(x)\in L[x],\quad n\geq 2.$$

Q2) I'm not sure if $P(x)$ is really of the form $(x-a)^nG(x)\in L[x]$, may be it's only of the form $(x-a)G(x)$ in spite of the fact that $a$ is a multiple root. Or maybe it's an it's an other root of the minimal polynomial that is not separable in an other extension, so it's maybe of the form $(x-a)(x-b)^n G(x)$ in an other field $F\supset L$. What do you think ?

But I will follow my (probably wrong) logic. I would say (I don't know why, but I think it's this) that $L$ is a finite field with $p^d$ elements. In particular, I would say (I don't know why, but it looks helpful) $n=p^d$, and thus $$(x-a)^n=x^{p^d}-a^{p^d}.$$

I'm lost here... (but I think that every thing looks wrong). I really need help.

Answer 1

Beware that a field of characteristic $p$ is not necessarily finite !

Question 1 : $\varphi:x\mapsto x^p$ is the Frobenius automorphism. It is defined on any field of characteristic $p$. So in particular in $L$ and $K$. But what is important here is that $\varphi:K\rightarrow K$ is bijective (as an automorphism of $K$).

So in fact, you are trying to prove that if $\varphi$ is bijective, then any field extension is separable. (If $K$ is of characteristic $p$, this is in fact equivalent, and a field satisfying the second property is called perfect).

Question 2 : It depends a bit on your definition of inseparable extension, but I guess you mean that there is $a\in L$ such that the minimal polynomial of $a$ has multiple roots.

Let us show that if $P$ is an irreducible polynomial (as any minimal polynomial), then $P$ is of the form $\sum a_{ip} x^{ip}$. Note that $P$ and $P'$ have a common factor, because $P$ has a multiple root. But then $\gcd(P,P')$ would divide $P$. Since $P$ is irreducible, the only possibility for this to happen is that $P'=0$. So, if $P=\sum a_i x^i$, its derivative is $P'=\sum ia_i x^{i-1}=0$. This implies that the only non zero coefficients of $P$ are the $a_{ip}$.

Now if $P=\sum a_{ip} x^{ip}$ and $\varphi$ is bijective, then any $a_{ip}$ is of the form $a_{ip}=b_i^p$ for some $p$. So $P=\sum b_i^p x^{ip}=(\sum b_i x^i)^p$ which contradict that $P$ is irreducible.