# The cubic Kronecker-Weber Theorem

This is the second in a sequence of posts where I look at $\mathbb{Z}/3$ extensions of the rational numbers in a very hands on, low tech, way. This time, we’re going to start checking the Kronecker-Weber theorem. The Kronecker-Weber Theorem states:

The Kronecker-Weber Theorem: If $K$ is an abelian Galois extension of $\mathbb{Q}$, then $K$ is contained in $\mathbb{Q}[e^{2 \pi i/n}]$ for some integer $n$.

So, when I said before that the only abelian extensions I was familiar with were the quadratic and the cyclotomic extensions, I had basically already listed them all! In the last post, I already listed all of the $\mathbb{Z}/3$ extensions of $\mathbb{Q}$; they were indexed by order three subgroups of $\lim_{\leftarrow} (\mathbb{Z}/N)^*$. Today, we’ll try to fit these inside cyclotomic fields.

By the way, I just discovered the very nice article Kronecker-Weber via Stickelberger by Franz Lemmermeyer; if you are enjoying this series of posts, you might want to work your way through that.

First of all, let’s talk about cyclotomic fields. Set $K_n=\mathbb{Q}[e^{2 \pi i/n}]$. From now on, we’ll write $\zeta_n$ in place of $e^{2 \pi i/n}$. The Galois group of $K_n$ is $(\mathbb{Z}/n)^*$, with the generator $\sigma_j : \zeta_n \to \zeta_n^j$. (It is obvious that every Galois automorphism is of this form. It requires an argument to see that these are all automorphisms; if you want to try it, I suggest first doing the case $n=p$, then $n=p^k$ and then deducing the case of general $n$ from the cases you’ve already done.) So, the cubic subfields of $K_n$ are in bijection with index three subgroups of $(\mathbb{Z}/n)^*$. If we put all of the cyclotomic fields together, we get that the $\mathbb{Z}/3$ extensions of $\mathbb{Q}$ which lie in cyclotomic fields are in bijection with the index three subgroups of $\lim_{\leftarrow} (\mathbb{Z}/n)^*$. Since these are the same objects that index the entire set of $\mathbb{Z}/3$ extensions of $\mathbb{Q}$, you should believe that all the $\mathbb{Z}/3$ extensions of $\mathbb{Q}$ lie in cyclotomic fields.

Of course, that is no proof at all, because we’ve just shown that two sets are both countable. For those who know a little algebraic number theory, here is how to make that argument rigorous: Recall that the $\mathbb{Z}/3$ extensions of $\mathbb{Q}$ are of the form $L=\mathbb{Q}[q^{1/3} \overline{q}^{2/3}+q^{2/3} \overline{q}^{1/3}]$. Here $q$ is an element of $\mathbb{Z}[\zeta_3]$ whose norm, $q \overline{q}$, is a square free product of primes that are $1$ modulo $3$. So the (integral) primes which ramify in $L$ are the primes which divide $N$, and possibly also $3$. Looking at the details of our classification, we see that, if $p_1$, $p_2$, …, $p_{r-1}$ is a list of primes that are $1$ modulo $3$, then there are precisely $2^{r-1}$ abelian cubic extensions of $\mathbb{Q}$ whose ramification is contained in the set $\{ 3, p_1, \ldots, p_{r-1} \}$. We can exhibit $2^{r-1}$ such extensions as subfields of $K_{3 p_1 \cdots p_{r-1}}$, so we see that every abelian cubic extension of $\mathbb{Q}$ lies in a cyclotomic field.

But the aim of this sequence of posts is to get explicit, and this argument isn’t. We show that finite counts match, but we don’t indicate how to fit a particular $\mathbb{Z}/3$ extension into a cyclotomic field. What we’d really like to do is, given an index $3$ subgroup of $(\mathbb{Z}/n)^*$, output an explicit formula for $q^{1/3} \overline{q}^{2/3}+q^{2/3} \overline{q}^{1/3}$ as an element of $K_n$. That is basically what we are going to do, but the details are different. What we are actually going to do is start with a character $\chi: (\mathbb{Z}/n)^* \to \{ 1, \zeta_3, \zeta_3^2 \}$ and give a formula for $q^{1/3} \overline{q}^{2/3}$ as an element of $K_{3n}$. We then just symmetrize the formula to project down to $K_n$. (This symmetrization makes $\chi$ and $\overline{\chi}$ yield the same element of $K_{n}$.)  We then use our classification of $\mathbb{Z}/3$ extensions to know that we have built all of them.

This seems like the point to make an admission. I thought I had a really slick way to do this step, but it is broken and, I think, unfixable. So I am going to use the standard argument, which goes through Gauss sums. Although it is sad to see my pretty argument go away, the Gauss sum method is probably better motivated and certainly more important. But, as a result, I’m going to have to pause our story soon and do a post about Gauss sums.

In the mean time, let me tell you where we are going, with a formula that you can check out for yourself.  Let $n$ be an integer which is a product of primes which are $1$ modulo $3$. Let $H$ be an index $3$ subgroup of $(\mathbb{Z}/n)^*$. So the fixed field of $H$ in $K_n$ is a $\mathbb{Z}/3$ extension of $\mathbb{Q}$. We’d like to find an element of this fixed field. The most obvious option is $\sum_{h \in H} \zeta^h$ — and this will work. Being a little more flexible, we could choose some function $f$ from  $(\mathbb{Z}/n)^* /H$ to $\mathbb{Q}$ and use $\sum_{i \in (\mathbb{Z}/n)^*)} f(i) \zeta^i$. What is the best choice of $f$? It turns out the best option is not to build an element of $K_n$, but to build an element of $K_{3n}$. We take $f$ to be a character $\chi$, mapping $(\mathbb{Z}/n)^*$ to $\{ 1, \zeta_3, \zeta_3^2 \}$ with kernel $H$. So we construct the element

$g:=\sum_{i \in (\mathbb{Z}/n)^*)} \chi(i) \zeta^i$

in $K_{3n}$. It is fixed by $\tilde{H}$, the preimage of $H$ in $(\mathbb{Z}/(3n))^*$. Now, the first fun fact to check is that $g^3$ is in $K_3=\mathbb{Q}[e^{2 \pi i/3}]$. And then comes the really fun fact: $g^3=q \overline{q}^2$, where $q$ is the element of $K_3$ associated to $H$

It’ll probably be about a week and a half for me to get the proof up; feel free to race me!