A perfect field that is neither of characteristic $0$ nor algebraically closed

Question

Just recently I was given the task to find a non separable field extension. At first it seems like an easy task, but common fields you encounter are usually perfect. Eventually I found the example $\mathbb{F}_2(X)/\mathbb{F}_2(X^2)$, neither of characteristic $0$ nor of chracteristic $p>0$ where the Frobenius homomorphism is an automorphism, which would imply perfectness.

I then started to wonder what an infinite perfect field of characteristic $p>0$ would look like and only found algebraically closed fields, i.e. the algebraic closure of $\mathbb{F}_p(X)$. What would a non algebraically closed field look like where the characteristic isn't $0$, but it is perfect nonetheless, i.e. the Frobenius homomorphism is surjective. Can anyone give me an example?

Answer 1

Take your favorite infinite non-separably closed field $F$ of characteristic $p>0$ (e. g. $F = \Bbb F_p(X)$; the field $\Bbb F_{p^2}(X)$ is a separable non-trivial extension of $F$). Let $\overline F$ be an algebraic closure of $F$ and consider the perfect closure $F^{\rm perf}$ inside $\overline F$ given by $$ F^{\rm perf} := \{x\in \overline F \mid x^{p^n}\in F\text{ for some $n\in \Bbb N$}\}. $$ The field $F^{\rm perf}$ is perfect and purely inseparable over $F$. (The last claim follows from the fact that $x\in \overline F$ is separable over $F$ if and only if $F(x) = F(x^p)$.)
It is clear that $F^{\rm perf}$ is not algebraically closed, since for any separable (over $F$) $x\in \overline F$ with $x\notin F$ we necessarily have $x\notin F^{\rm perf}$.

In the case of $F = \Bbb F_p(X)$ we have $F^{\rm perf} = \bigcup_{n\in \Bbb N} \Bbb F_p(X^{p^{-n}})$.

Answer 2

Every subfield of $\overline{\mathbb{F}_p}$ is perfect. You can use that to produce many examples of the sort that you are after: the absolute Galois group of $\mathbb{F}_p$ is isomorphic to the profinite completion $\hat{\mathbb{Z}}$ of $\mathbb{Z}$, and that group has many infinite quotients (it is isomorphic to $\prod_l \mathbb{Z}_l$, where the product runs over all primes $l$, and $\mathbb{Z}_l$ denotes the $l$-adic integers), so $\mathbb{F}_p$ has many infinite degree extensions.

Answer 3

As you say, any char $0$ field is perfect. Apart from that, any finite field is perfect, and any algebraic extenstion of a perfect field is perfect. So, you can easily find infinite algebraic extensions of $\Bbb F_p$ for some $p$, apart from the algebraic closure, and it will be perfect. Say the quadratic closure of $\Bbb F_p$, for instance (the "constructible" elements over $\Bbb F_p$, if you'll excuse my abuse of terminology).