If $E/F$ is finite, $F$ has no odd extensions, and $E$ has no extensions of degree $2$, then $F$ is perfect and $E$ is algebraically closed?

Question

I'm struggling to understand the claim that if $E/F$ is a finite field extension, $F$ has no odd extensions, and $E$ has no extensions of degree $2$, then $F$ is perfect and $E$ is algebraically closed.

I was thinking, since $F$ has no nontrivial odd extensions, then every extension is divisible by $2$, and it's known from field theory that in this case every finite extension has degree a power of $2$.

In particular, every nontrivial finite extension of $E$ must be have degree a power of $2$. If any one of these extensions is Galois, then choosing a fixed field of a subgroup of a suitable index would give a field extension of $E$ of degree $2$, a contradiction. Thus $E$ would have no nontrivial finite extensions, implying it's algebraically closed. Is there a way to make this work that I'm missing?

I'm blank on how to see that $F$ is perfect? If $F$ is imperfect of characteristic $p$, then $F\neq F^p$. If $a\in F\setminus F^p$, then I recall that $X^{p^e}-a$ is irreducible, so adjoining any root would give an extension of $F$ of degree $p^e$. Since $F$ only has extensions of degree $F$, necessarily $\operatorname{char}(F)=0,2$. In the former case, $F$ is perfect, but if $\operatorname{char}(F)=2$?

Furthermore, if $\operatorname{char}(F)=0$, let $K/E$ be a finite extension. Since we're working over characteristic $0$, we can enlarge this to a Galois extension, denoted $\tilde{K}$, to get a tower $\tilde{K}/E/F$. So $\tilde{K}/E$ is Galois as well, but then as noted above, if this is not proper, $E$ will have a quadratic extension. Thus $\tilde{K}=E$, so $K=E$. Thus $E$ is algebraically closed.

So I feel the only remaining difficulty is when $\operatorname{char}(F)=2$.

Edit: Maybe it's a sledgehammer, but I think showing $E$ is algebraically closed without assuming characteristic $0$ is sufficient. For if $[E:F]=1$, then $E=F$ is perfect as all algebraically closed fields are. If not, since then $1<[E:F]<\infty$, Artin-Shreier implies $E=F(\sqrt{-1})$ and $\operatorname{char}(F)=0$, so $F$ is again perfect.

Answer 1

If we know $F$ is perfect, then we can mimic the argument in Artin's beautiful proof of fundamental theorem of algebra using Galois correspondence to show $E$ is algebraically closed.

Now suppose $F$ is imperfect with $\mathrm{char}(F) = p > 0$. Take $I$ to be a nontrivial purely inseparable extension of $F$ (this is always possible by first taking an arbitrary inseparable extension $L$ over F and then taking the purely inseparable closure of $F$ in the normal closure of $L/F$). Let $\alpha \in I\setminus F$. Then $[F(\alpha):F] = p^k$ for some $k\ge 1$. Since $F$ has no nontrivial field extensions of odd degree, we can deduce $p = 2$. Next we will show in this case $E$ must perfect and algebraically closed. Suppose $E$ is imperfect and let $\beta$ be a purely inseparable element in an extension over $E$ . $[E(\beta):E] = 2^r$ for some $r\ge 1$ so $[E(\beta^{2^{r-1}}):E] = 2$, which contradicts the assumption that $E$ has no field extension of degree 2. So $E$ must be perfect. Let $K$ be a nontrivial finite extension over $E$. Since $E$ is perfect, $K/E$ is separable and let $N$ be the Galois closure of $K$ over $E$. $[K:E]$ is finite so $[N:E]$ is also finite and so is $[N:F]$. Every nontrivial finite extension of $F$ is of even degree, which implies the degree of every finite extension of $F$ is a power of 2 (a nontrivial fact). So $[N:F]$ is a power of 2 and so is $[N:E]$. Then $\mathrm{Gal}(N/E)$ is a 2-group. From the theory of $p$-group, any maximal subgroup $M$ of $\mathrm{Gal}(N/E)$ is of index 2 so the fixed field of $M$ in $N$ gives a quadratic extension of $E$, a contradiction. Hence $E$ must be algebraically closed.

At this point, we deduced that $E$ is algebraically closed whether $F$ is perfect or not. To prove that $F$ is perfect, I can't think of anything else other than the powerful Artin-Schreier theorem. Since $F$ is a subfield of the algebraically closed field $E$, by the Artin-Schreier theorem either $F = E$ or $E = F(i)$ and $\mathrm{char}(F) = 0$ . In either case, $F$ is perfect.

No sure if there is a better/shorter proof, but at least now I know the conclusion is true.