# Galois group is isomorphic to group of characters. What can we say then?

Let $\overline{K}/K$ denote the separable closure of a finite field $K$ of characteristic $p$ and let $\mu_{n}$ denote the group of $n$-th roots of unity in $\overline{K}$, where $(n,p)=1$. Let us suppose that $\mu_{n}\in K$.

We have this result that says that the extension $L$ of $K$ obtained by adjoining to $K$ the $n$-th roots of elements of $K$ has Galois group $G(L/K)$ abelian, of exponent diving $n$, and furthermore, we have \begin{equation} G(L/K)\cong\operatorname{Hom}(K^{\times}/(K^{\times})^{n},\mu_{n}). \end{equation}

My question:

What is the motivation to consider such a group of homomorphisms (on the right side)?

I mean, I understand the proof, but if I was the one historically developing the theory, I would never think of considering such a group, so probably there is more going on behind the scenes.

I know that this thing has to do with character theory and maybe harmonic analysis, but I would appreciate some details on an intuitive level, or in other words, what we know about such groups and why they are worthy considering in this field- or number-theoretic context.

Do they somehow add more insight about what is going on in the Galois group?

Do they somehow add more insight about what is going on in the field?

Thanks.

• If you know about group cohomology and consider that to be a decent motivation, $H^1(G,M)\cong\operatorname{Hom}(G,M)$ when the action of $G$ on the $G$-module $M$ is trivial. Your Hom group might arise in particular when taking the long exact sequence in cohomology associated to the short exact sequence* $0\to \mu_n(K)\to K^\times\xrightarrow{\cdot n} K^\times\to 0$ (*when this is indeed exact) or a similar sequence. You might want to look for literature on Galois cohomology and/or Kummer theory. – Stahl Apr 14 '16 at 23:32
• I think your result is not correct, as stated, although it is similar to a correct result. Adjoining $\mu_n$ to $K$ will indeed give a Galois extension with abelian Galois group, but its order will be a divisor of $\phi(n)$, where $\phi$ is Euler's totient function. So if $n$ is prime, for instance, then the order of the Galois group is relatively prime to $n$, so not a divisor of $n$. What you should be adjoining are $n$th roots of elements of your finite field. And you need to assume that $K$ contains the $n$th roots of unity, and that the characteristic of the field does not divide $n$. – Barry Smith Apr 14 '16 at 23:47
• @BarrySmith You are correct, I screwed up and will edit it, thanks. – Shoutre Apr 14 '16 at 23:53
• @Stahl I'm not very confortable with these things yet, but I know the definitions. I can see how this group can arise in the long exact sequence, but the long exact sequence itself is something I'm not really familiar with it and I don't feel like I completely understand what it really captures (cohomology groups in general) . Maybe it is just lack of familiarity with these.. I'm still undergrad. – Shoutre Apr 14 '16 at 23:58
• @Stahl: yeah, sorry, I was confused. – Qiaochu Yuan Apr 15 '16 at 0:08

The point is not to know if it’s easier or smarter to look at the dual of G = Gal(L/K) instead of G itself. To understand the motivation, I think one should take the « experimental » point of view : given any Galois extension L/K, how does one describe its Galois group ? In exercises in Galois theory, generators of the extension L over K are usually given, and the student is asked to determine G using these generators. In this situation, one is « philosophically » convinced that one can do it, even if this could be quite non obvious, e.g. for L = $Q(\sqrt p_1, … , \sqrt p_n)$, where the $p_i$’s are distinct primes. But in a research problem, things do not happen this way. What is given is the base field K, and some desired properties of the extension L, and one must manage to describe G in order to go on. The setting of Kummer theory is a perfect example : K contains $\mu_n$, n prime to the characteristic of K (1), and L/K is an abelian extension of exponent dividing n. What Kummer’s main theorem says is that : first, L is generated over K by the $\sqrt [n] a$ of elements $a$ $\in K$* ; second, that the elements $\sigma$ of G are determined by their action on these $\sqrt [n] a$ , and this action is obviously given by $\sigma (\sqrt [n] a) /\sqrt [n] a$ = an n-th root of 1 (2).

It is striking that the answer is entirely « contained » in the base field, but the limitation of Kummer’s theory is the requirement that K must contain $\mu_n$, so one cannot catch abelian extensions of higher exponent. One could perhaps say that CFT was developped to overcome this difficulty : CFT gives a complete, much more elaborate description of the abelian extensions of a global field (number field and function field of positive characteristic), again with parameters entirely contained in the base field . However this is no longer abstract field theory, but number theory.

(1) If char K divides n, the analog of Kummer theory is the Artin-Schreier-Witt theory

(2) To complete Stahl’s hint, I give here a sleek cohomological proof of Kummer’s theorem. Let L/K be Galois with group G, and take the cohomology of the exact sequence of G-modules 1 --> $\mu_n$--> $L^*$ --> $L^{*n}$ --> 1. This yields 1 --> $\mu_n (K)$ --> $K^*$ --> $K^* \cap L^{*n}$ --> $H^1(G, \mu_n )$ --> $H^1(G,L^*)$ . The last term is trivial by Hilbert 90, so the "Kummer radical" ($K^* \cap L^{*n}$) / $K^{*n}$ is isomorphic to $H^1(G,\mu_n )$, which is $Hom (G, \mu_n)$ if $K$ contains $\mu_n$. Note that these isomorphisms are explicit, they are exactly as described above. NB: in the formula given by Shoutre, one takes for L the maximal abelian extension of exponent dividing n .

The motivation is "Kummer theory". I think this map is also easier to understand initially in characteristic zero, where we can consider more general extensions than those obtained by adjoining roots of unity, so if you like, imagine we are in characteristic zero in what follows.

Suppose a field $K$ contains $n$ distinct roots of unity, with $n$ not divisible by the characteristic of $K$. If we choose an element $\alpha$ in $K$, then the extension field $M = K(\sqrt[n]{\alpha})$ is Galois over $K$ with cyclic Galois group. That it is Galois follows from the fact that $M$ is the splitting field of the polynomial $x^n - \alpha$ (the roots being the products of the form $\zeta \sqrt[n]{\alpha}$ with $\zeta$ an $n$th root of unity.

Let us see that $\mathrm{Gal}(M/K)$ is cyclic. If $\sigma$ is in $\mathrm{Gal}(M/K)$, then $\sigma \alpha = \zeta \alpha$ for some $\zeta \in \mu_n$. Rechoosing $\sigma$ if necessary, let us assume we chose $\sigma$ so that $\zeta$ is the smallest positive integer power of some fixed primitive $n$th root of unity. Then if $\tau \in \mathrm{Gal}(M/K)$, we must have $\tau \alpha = \zeta^k \alpha$ for some integer $k$. Thus, $\tau = \sigma^k$ in $\mathrm{Gal} (M/K)$, and so $\sigma$ generates $\mathrm{Gal} (M/K)$.

Now here's an important observation: if you pick a different $n$th root of $\alpha$ than that you used to form $L/K$, then $\sigma$ applied to that $n$th root will be the same root of unity $\zeta$ times your new $n$th root of unity. This follows from the assumption that $K$ contains $\mu_n$.

Now imagine that we adjoin the $n$th roots of every element of $K$ to $K$ and obtain the the extension $L$. Then we may define a very intuitive map giving the isomorphism in your post. Given a $\tau$ in $\mathrm{Gal}(L/K)$, we must define a map from $K/(K^{\times})^n$ to $\mu_n$. Given a representative $\alpha$ of a class in $K/(K^{\times})^n$, we define the map by sending the class to the root of unity $\zeta$ such that $\tau \sqrt[n]{\alpha} = \zeta \sqrt[n]{\alpha}$. The important observation above shows that it doesn't matter which root of unity we choose here. It is well defined on classes, since if $\alpha$ is an $n$th power in $K$, then the root of unity must be $1$. The check that this is a homomorphism is good practice. The fact that the map is one-to-one follows from the fact that different maps must act differently on some generator for $L/K$. The onto-ness is perhaps intuitive: some familiarity with extending isomorphisms will let you see that any action on the various generators can be extended to an isomorphism of $L$ -- except that you must be careful since some generators, for instance, may be powers of others.

A cleaner and more general version of this goes through group cohomology, and was one of the motivations for developing that theory in the first place. On the other hand, Kummer theory in this setting was one of the main tools for developing the fundamental theorems of class field theory. Kummer himself only thought in terms of "ideal numbers" with maps substituting for the notion of prime ideals. I can guess that Kummer extensions were already considered by Dedekind, but if not, then they are explicitly discussed shortly thereafter by Hilbert.

• In your third paragraph, you have $\alpha$ where I believe you should have $\sqrt[n]{\alpha}$. In the fifth paragraph, you have $K/(K^\times)^n$ instead of $K^\times/(K^\times)^n$. – Stahl Apr 15 '16 at 7:47