Kummer Theory - Example of Subgroup of $K^{*}$ containing $K^{*m}$ for global fields.

Question

I am trying to understand Kummer theory and I wish to apply it to global fields, so our field $K$ containing $\mu_m$ should be $\mathbb{Q}(\zeta_m)$.

Let $B$ be a subgroup of $K^{*}$ containing $K^{*m}$. We define $K_B:=K(B^{1/m})$ to be the compositum of all $K(a^{1/m})$ such that $a\in B$. Define $G=\text{Gal}(K_B/K)$.

(This is necessarily a Galois extension as $K_B$ is a splitting field for $x^n-a$ for any $a\in B$.)

We then have the Kummer Pairing, $$\kappa:G\:\text{x}\:B\rightarrow B$$

given by $\kappa(\sigma,a)=\frac{\sigma(\alpha)}{\alpha}$, for $\alpha^m=a$.

My question is this:

Can we think of an explicit example of $B$ such that $(B:K^{*m})$ is finite? I am struggling to see how this works

Answer 1

Let’s look at the case $m=2$ for simplicity. As you’ve seen, in this case your base field is $\Bbb Q$. Let’s not look at $\Bbb Q^\times$ but at $\Bbb Q^{>0}$. This simplifies some things slightly, and effectively allows us to consider fields gotten by taking square roots of positive integers only, i.e. real quadratic extensions of $\Bbb Q$ and their compositums.

Now, as a group, $\Bbb Q^{>0}$ is the direct sum of infinitely many infinite cyclic groups, one for each prime. This is the Fundamental Theorem of Arithmetic. You can write $\Bbb Q^{>0}=\bigoplus_pp^{\Bbb Z}$, if you accept the sumbol “$\oplus$” for groups written multiplicatively. Now, of course, we get $$ \bigl(\Bbb Q^{>0}\bigr)^2=\bigoplus_pp^{2\Bbb Z}\,, $$ equal to $(\Bbb Q^\times)^2$.

And if you take the quotient of $\Bbb Q^{>0}$ over this, i.e. if you take $\Bbb Q^{>0}\big/\bigl(\Bbb Q^{>0}\bigr)^2$, you get an infinite direct sum of cyclic groups of order $2$, one for each prime. I’m sure you know that Kummer Theory says that this is the dual of the Galois group of the compositum of all quadratic extensions of $\Bbb Q$, over $\Bbb Q$. Let’s denote this infinite extension of $\Bbb Q$ as $\mathcal Q$. Then the Galois group $G=\text{Gal}(\mathcal Q/\Bbb Q)$ is direct product of infinitely many cyclic groups of order $2$. Say we want to talk about a compositum $k$ of finitely many real quadratic extensions of $\Bbb Q$, gotten by adjoining square roots of the square-free positive integers $\{n_1,\cdots,n_r\}$. Then the Galois group of $k$ over $\Bbb Q$ will be $F\cong G/H$. Taking duals, the quotient becomes a subgroup and vice-versa, so we have $$ 1\longrightarrow H\longrightarrow G\longrightarrow F\longrightarrow 1\,, $$ giving rise to $$ 1\longleftarrow H^{\text D}\longleftarrow \Bbb Q^{>0}\big/\Bbb Q^2\longleftarrow F^{\text D}\longleftarrow 1\,. $$ So we’re talking about a finite subgroup of $\Bbb Q^{>0}\big/\Bbb Q^2 $, in other words, as you have clearly recognized from your posting, a subgroup $S\subset\Bbb Q^{>0}$ such that $(S :\Bbb Q^2)$ is finite. But now it’s clear which subgroup you’re talking about: it’s $\bigl\langle\Bbb Q^2,n_1,\cdots,n_r\bigr\rangle$, where the angle-bracket notation is meant to denote the group generated by the things inside. I think this answers your specific question.

I know I’ve gone into painful detail on things you have already understood. It was partly to make my own ideas clear and partly to go through everything step by step and try to be very explicit about how things fit together.

Answer 2

Why don't you take for B the multiplicative subgroup generated by K$^m$ and an element of K which is not in K$^m$ ? More "concretely", take m to be a prime p. If you think of the multiplication in K as an addition (group law, so the notation does not matter), then K*/K*$^p$ is a vector space over the prime field F$_p$. Any finite dimensional sub-vector space of K*/K*$^p$ will give you what you want. Of course, the case p = 2 covers the example given by Lubin .