# Correspondence between finite Galois extensions and abelian p-groups

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.

• Why do you think this is true? – Eric Wofsey Dec 3 '17 at 21:27
• @EricWofsey I do not think this is true, but I consider. If not, one need to give a counterexample. – Evgeny Dec 3 '17 at 21:29
• Well it's strange to ask people how to prove something if you don't think it's true... – Eric Wofsey Dec 3 '17 at 21:36
• 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. – Jyrki Lahtonen Dec 5 '17 at 10:43
• 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 $k$. – Jyrki Lahtonen Dec 5 '17 at 10:46

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.