Let $k$ be an infinite perfect field in positive characteristic $p$, i.e. every element of $k$ is a $p$th power. I am interested in properties of finite fields that can be extended to $k$. For example:

  1. Let $L$ be a finite Galois extension of $k$. Is $\mathrm{Gal}(L/k)$ always cyclic?
  2. Let $L$ and $L'$ be two finite Galois extensions of $k$ of the same degree. Are $L$ and $L'$ always $k$-isomorphic fields?

The only example of such a field $k$ I have encountered is $\mathbb{F}_{q}(x,x^{p^{-1}},x^{p^{-2}},x^{p^{-3}},\ldots)$, where $q=p^d$ for some prime $p$, and the only finite Galois extensions I see are those of the form $\mathbb{F}_{q^e}(x,x^{p^{-1}},x^{p^{-2}},x^{p^{-3}},\ldots)$, which makes me expect a positive answer to both questions. (Edit: There are more (of course), see Emil Jeřábek's comment.)

Basically, I want to know whether Galois theory over infinite perfect fields in positive characteristic is as ''easy'' as over finite fields. In addition to an answer to the questions above, any properties supporting/contradicting this vague statement are apreciated.

  • 3
    $\begingroup$ Presumably, in (2), you want $L$ and $L'$ to be of the same degree? The property you are asking for seems closely related to being quasi-finite. $\endgroup$ – LSpice Apr 13 at 12:37
  • $\begingroup$ Yes, I meant same degree, thanks. I edited it. $\endgroup$ – JNS Apr 13 at 12:40
  • 2
    $\begingroup$ Your example field is in fact already a counterexample. For instance, where is the $l$-th root of $x$ for prime $l\ne p$? $\endgroup$ – Emil Jeřábek Apr 13 at 13:31
  • $\begingroup$ Galois theory of perfect fields is, in fact, as hard as the Galois theory of general fields: if $k$ is a file of characteristic $p$ then its finite separable extensions are in bijection with finite (necessarily separable) extensions of its perfection $k_{perf}=colim(k\xrightarrow{x\mapsto x^p} k\to \dots)$ $\endgroup$ – SashaP Apr 14 at 2:20

To produce an obvious counterexample to (1), fix a finite field $F$ of characteristic $p$, $n\ge 3$ and $L=\bigcup_m F[x_1^{p^{-m}},\dots,x_n^{p^{-m}}]$. This is a perfect field. The symmetric group $S_n$ acts by permuting the variables. Let $K$ be the fixed point field. Then $L$ is a finite Galois extension of $K$ with Galois group $S_n$.

A counterexample to (2), say with $p>2$: now the ground field is $L$. Let $F[u]$ be a quadratic extension of $F$. Then $L\subset L[u]$ and $L[x_1^{1/2}]$ are quadratic extensions of $L$, but are not isomorphic.

  • 1
    $\begingroup$ PS: the first construction, replacing $S_n$ with $G\subset S_n$ shows that for every prime $p$, every finite group is Galois group of a Galois extension of some perfect field of char. $p$. $\endgroup$ – YCor Apr 15 at 6:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.