# 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?

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.

• An imperfect field of characteristic $p$ has an extension of degree $p$. Commented Mar 31, 2016 at 20:29
• @Nas "...and it's known from field theory that every finite extension has degree a power of 2". What? When, how, where...? Did you mean some very particular case? Because this is blatantly false in general. Commented Mar 31, 2016 at 20:37
• @Joanpemo I think he's using that $F$ has no non-trivial odd extension. Commented Mar 31, 2016 at 20:45
• @Jake Thank you, that seems a reasonable assumption. Commented Mar 31, 2016 at 20:46
• It's not true that every algebraic extension of a perfect field is algebraically closed as you claim in your last sentence. Eg: Every number field is perfect, for example. Commented Mar 31, 2016 at 20:48

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.