-2
$\begingroup$

There is a correspondence between $k$-isomorphic finite separable normal extensions of $k$, $char(k) = p > 0$, and abelian p-groups.

Put it formally, $$\forall \, G = Gal(\Bbb F/k), |G| < \infty, \, char(k) = p > 0, \, \exists \, P \text{ - abelian p-group}, \exists \, \beta \text{ - bijection}: \beta(G) = P$$ up to $k$-isomorphism, and vise versa.

How to clearly define such a bijection (i.e. how to prove this)?

Well, since galois extension can be defined as automorphisms that fixes field $k$, and since $k$ has characteristic $p$, we can consider Frobenius endomorphism $\mathrm f: q \mapsto q^p$, which can be connected with p-groups as we talk about finite extensions (i.e. finite groups). $k$ is fixed by $G$, hence so do $\mathrm f(k), \mathrm f(\mathrm f(k))$ and so on. But one can generate Galois extension by Frobenius automorphisms only in case of finite fields, so the way is not so trivial.

$\endgroup$
  • $\begingroup$ Why do you think this is true? $\endgroup$ – Eric Wofsey Dec 3 '17 at 21:27
  • $\begingroup$ @EricWofsey I do not think this is true, but I consider. If not, one need to give a counterexample. $\endgroup$ – Evgeny Dec 3 '17 at 21:29
  • 1
    $\begingroup$ Well it's strange to ask people how to prove something if you don't think it's true... $\endgroup$ – Eric Wofsey Dec 3 '17 at 21:36
  • $\begingroup$ Abelian $p$-extensions of a field $k$ of characteristic $p$ can be characterized using the theory of Witt vectors. This generalizes the Artin-Schreier description of cyclic extensions of degree $p$. I recommend chapter 8 of Jacobson's Basic Algebra II for an account of this theory. $\endgroup$ – Jyrki Lahtonen Dec 5 '17 at 10:43
  • 1
    $\begingroup$ Of course, should it happen that the Galois group is elementary $p$-abelian, then $\Bbb{F}$ is a composition of cyclic extensions $\Bbb{F}=\prod_{i=1}^n K_i$, where $K_i$ is gotten as the splitting field of $x^p-x-a_i$ over $k$. $\endgroup$ – Jyrki Lahtonen Dec 5 '17 at 10:46
2
$\begingroup$

It's not entirely clear what precise statement you are asking to be true (what you have written does not make sense if taken literally), but any reasonable interpretation is false and I don't know why you would think it is true. For instance, if $k$ is algebraically closed, then there is only one finite extension of $k$. Or if $k$ is uncountable, there could be uncountably many non-isomorphic finite Galois extensions, whereas there are only countably many finite abelian $p$-groups.

$\endgroup$

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.