# Classifying Z/3 extensions of the rationals

The triumph of early twentieth century number theory is the formulation and proof of the main results of class field theory. In technical terms, class field theory classifies the abelian extensions of a number field and describes how primes split in such extensions. In more elementary (but less accurate) terms, class field theory answers two questions.

First, what are the polynomials $f$ such that $\mathbb{Q}[x]/f(x)$ is a Galois extension of $\mathbb{Q}$ with abelian Galois group? Remember that, if $g$ is a generic polynomial of degree $n$, then the splitting field of $\mathbb{Q}[x]/g(x)$ has Galois group $S_n$, and the field $\mathbb{Q}[x]/g(x)$ is fixed by $S_{n-1}$. So the polynomials $f$ which give Galois extensions are very rare, in that the Galois group is much smaller than the generic case.

Second, if $f$ is such a polynomial, how does $f$ factor modulo various primes $p$? I already wrote a post explaining what this has to do with Galois theory, so for now I’ll just refer you to there. Let me point out, though, that the description of that post is pretty impractical, if you actually want to answer a question like “for which primes $p$ does the polynomial $x^3-2$ have three roots modulo $p$?” That’s because these questions are really hard! The amazing thing about class field theory is that it will give you a completely explicit answer to the question “for which primes $p$ does the polynomial $x^3+x^2-2 x-1$ have three roots modulo $p$?”.

The first special case of these questions is the case where $f$ is a quadratic polynomial. In this case, the Galois group is always abelian, because it is contained in the abelian group $S_2$. The field $\mathbb{Q}[x]/f(x)$ is isomorphic to $\mathbb{Q}[\sqrt{D}]$ for some unique square-free integer $D$. The question of whether $f$ has roots modulo $p$ comes down to computing the Legendre symbol ${D \choose p}$ which, by quadratic reciprocity, comes down to evaluating the class of $p$ modulo $4D$.

In this post and the sequel, I will explain how elementary methods are good enough to get you through the next case, where $f$ has degree $3$. In this post, we will solely be concerned with the first question — what are the cubic extensions of $\mathbb{Q}$ with Galois group $\mathbb{Z}/3$?

One of the reasons I like this problem it that it shows you how far a little Galois theory can get you. Another reason is that it shows you that the main results of Class Field Theory don’t come out of nowhere. When I learned the subject, I knew very few examples of abelian Galois extensions, basically just the quadratic fields and the cycloctomic fields, and so the big classification results were baffling to me. Later, I realized that the number theorists who formulated these results knew tons of examples and could were very comfortable playing with them. In this post, I hope to get you comfortable playing with the $\mathbb{Z}/3$ extensions of $\mathbb{Q}$.

As I mentioned above, every $\mathbb{Z}/2$ extension is gotten by adjoining a square root. (In characteristic other than 2.) One might hope that every $\mathbb{Z}/3$-extension is gotten by adjoining a cube root, but this is wrong. The right generalization is Kummer’s Theorem:

Kummer’s Theorem: Let $K$ be a field whose characteristic does not divide $n$ and let $L$ be a $\mathbb{Z}/n$ extension of $K$. Assume also that all of the $n^{\textrm{th}}$ roots of unity lie in $K$. Then $L=K[a^{1/n}]$, for some $a \in K$. Moreover, if $a$ and $b$ are two elements of $K^*$, then $K[a^{1/n}] \cong K[b^{1/n}]$ if and only if $a$ and $b$ generate the same subgroup of $K^*/(K^*)^n$

Incidentally, most books that I’ve seen prove Kummer’s Theorem as a corollary of Hilbert’s Theorem 90, but the result is much easier than that. That might be a subject for a blog post at some other point…

Now, $\mathbb{Q}$ doesn’t contain the cube roots of unity. Let’s write $E$ for the field $\mathbb{Q}[e^{2 \pi i/3}]$. The Galois group of $E$ over $\mathbb{Q}$ is $\mathbb{Z}/2$. So, if $K$ is a $\mathbb{Z}/3$-extension of $E$, then the composite, call it $L$, of $E$ and $K$ is of the form $E[b^{1/3}]$ for some $b$ in $E$. We’ll write $x \mapsto \overline{x}$ for the involution of $E$.

Now, $L=E[b^{1/3}]$ is supposed to be a $\mathbb{Z}/2 \times \mathbb{Z}/3$ extension of the rationals. In particular, $E[b^{1/3}]$ and $E[\overline{b}^{1/3}]$ are the same field. By Kummer’s Theorem, $b$ and $\overline{b}$ generate the same (order three) subgroup of $E^*/(E^*)^3$. So, either $\ [\overline{b}]=[b]$ or $\ [\overline{b}]=[b]^{-1}$. (Here square brackets denote classes in the group $E^*/(E^*)^3$.) Now, here is the fun part, which I’ll leave for you. Check that the first case, which looks prettier, corresponds to $S_3$ extensions of $\mathbb{Q}$. It is the second, peculiar, case which corresponds to $\mathbb{Z}/2 \times \mathbb{Z}/3$ extensions, and thus to the $\mathbb{Z}/3$ extensions. So our goal is to describe classes in $E^*/(E^*)^3$ which are inverted by $\ [b] \mapsto [\overline{b}]$.

Let $R$ be the ring $\mathbb{Z}[e^{2 \pi i/3}]$. This ring is known as the ring of Eisenstein integers. The most important fact about $R$ is that, like the ring of integers, $R$ has unique factorization into primes. Here is a detailed description of the primes of $R$. If $p$ is a (positive) prime of $\mathbb{Z}$ which is $1 \mod 3$ then $p$ factors as $p=\pi \overline{\pi}$ for a pair of distinct primes $(\pi, \overline{\pi})$ of $R$. Of course, $\pi$ is not unique, because we could always multiply by $\pm e^{2 \pi i k/3}$ for some choice of sign and of $k$. For future convenience, we adopt the convention that $\pi$ is always chosen to be $1 \mod 3$. This still leaves the ambiguity that we could switch $\pi$ and $\overline{\pi}$; we make the choice of which prime to call $\pi$ arbitrarily. If $p$ is a (positive) integer prime which is $2 \mod 3$, then $p$ is also prime in $R$. Finally, $3$ factors as $- \lambda^2$ where $\lambda$ is the prime $e^{2 \pi i/3}-e^{-2 \pi i/3}$. The units of $R$ are $\pm e^{2 \pi i k/3}$, for $k=0$, $1$, $2$.

In short, we can write $b$ as

$b=\pm e^{2 \pi i k/3} \lambda^c \prod_{p \equiv 1 \mod 3} \pi^{d_{\pi}} \overline{\pi}^{d_{\overline{\pi}}} \prod_{p \equiv 2 \mod 3} p^{e_p}.$

Since it is only the class of $b$ in $E^*/(E^*)^3$ that matters, we may assume that the sign is $+$ and that $c$, $d_{\pi}$, $d_{\overline{\pi}}$ and $e_p$ lie in $\{ 0,1,2 \}$. Then the condition that $\ [b]=[\overline{b}]^{-1}$ holds if and only if $c=0$, all the $e_p$ are zero, and, for each $p \equiv 1 \mod 3$ the ordered pair $(d_{\pi}, d_{\overline{\pi}})$ is one of $(0,0)$, $(1,2)$, $(2,1)$. Also, since it is only the group generated by $\ [b]$, and not $\ [b]$ itself that matters, we get the same cubic extension if we replace $b$ by $\overline{b}$.

In short, to describe a $\mathbb{Z}/3$ extension of $\mathbb{Q}$, we must choose a power of $e^{2 \pi i/3}$ and, for every prime which is $1 \mod 3$, we must choose whether $\pi \overline{\pi}^2$, $\pi^2 \overline{\pi}$, or neither, will appear in the factorization of $b$. All of these choices only count up to replacing $b$ by $\overline{b}$.

There are several directions to go from here. The first is to write down an explicit defining equation for the field $K$. Write $b=q \overline{q}^2$. Let $N=q \overline{q}$ and $T=q+\overline{q}$. These are both integers. Set $\theta =b^{1/3}+\overline{b}^{1/3}$.
Then

$\theta^3=3N \theta+NT$.

A different direction is to give other data classifying cubic extensions. We already saw two: the factorization of $b$ and the choice of $(N,T)$. Here is a better way of thinking of the former data. For each $p=1 \mod 3$, we can encode the choice of $(d_{\pi}, d_{\overline{\pi}})$ by choosing a morphism from $(\mathbb{Z}/p)^{\times}$ to the group $\{ 1, e^{2 \pi i}/3, e^{4 \pi i}/3 \}$. If $d_{\pi}=1$, we send $u \in \mathbb{Z}/p$ to whichever of these three units is congruent to $u^{(p-1)/3}$ modulo $\pi$. (Exercise: why must exactly one of these work?)
Similarly, if $d_{\pi}=2$, we do the same using $\overline{\pi}$ in place of $\pi$. If $d_{\pi}=0$, we use the trivial map. So, using the factorization of $b$, we get an order three character of $\prod_{d_{\pi} \neq 0} (\mathbb{Z}/p)^{\times}=(\mathbb{Z}/N)^{\times}$. We have not yet encoded the power of $e^{2 \pi i/3}$ in front of $b$; we can do this in a similar manner by giving a character of the unit group of $\mathbb{Z}/9$.

Right now it seems arbitrary that we have to give a choice of $N$ and a character; the better way to phrase this that cubic extensions are parameterized by order three characters of $\lim_{\leftarrow M} (\mathbb{Z}/M)^{\times}$. A nice feature of this presentation is that we are now taking the inverse limit over all integers $M$; if $M$ has factors which are $2 \mod 3$, or if $M$ is divisible by a prime (other than $3$) more than once, these simply will not contribute any order three characters in the limit. Finally, we must remember that we don’t want the trivial character, and we want to forget about interchanging $b$ and $\overline{b}$. We can do both of these at once by, instead of bringing up characters, simply saying that we are interested in index three subgroups of $\lim_{\leftarrow M} (\mathbb{Z}/M)^{\times}$.

We have now, by fairly elementary means, reached the cubic case of one of the main theorems of Class Field Theory: cyclic extensions of $\mathbb{Q}$ of order $c$ are in bijection with index $c$ subgroups of $\lim_{\leftarrow M} (\mathbb{Z}/M)^{\times}$. It is surely worth attempting to prove the general result by just such an elementary attack. Be warned, there are two major difficulties. The ring analogous to $R$ does not have unique factorization for general $c$, and its unit group is very large.

Next time, we will see how to prove the cubic case of the Kronecker-Weber theorem. This will also give us a far better way of understanding these mysterious characters.

## 12 thoughts on “Classifying Z/3 extensions of the rationals”

1. “Later, I realized that the number theorists who formulated these results knew tons of examples and could were very comfortable paying with them.”

I’d hate to be their waiter!

2. Z says:

Very nice, as usual.

3. Martin says:

(I am not an expert, but I teach the basics to undergrads this semester…)

Do you want to prove Kummer’s theorem with linear algebra (eigenvalues)?
(That was a proof I found and gave as a homework.)

4. David Speyer says:

Martin: yup, that’s exactly the way I want to prove it. If that blog post every gets written, it will also explain how the standard proof of Hilbert 90 is a (rather large) generalization of this argument.

Scott: You know, I was going to fix that sentence, and now I can’t. :)

5. Rolando says:

Could you please help me to find a way to print this interesting article? I have viewed it in the browsers K-Meleon, Firefox and Internet Explorer, but it won’t print properly after the first page.

6. David Speyer says:

I have the same problem (using Firefox on Fedora). I think that the browsing is sending the printer the whole document as one page, rather than paginating it correctly. The HTML is very simple, so I’m not sure why all of our browsers should fail this way.

My only thought is a hack. Copy the HTML source into your favorite editor, replace every occurrence of “[dollarsign]latex” with just “[dollarsign]”, and strip out the links. (Here “[dollarsign]” is the character you get by typing SHIFT-4.) At that point, you should have functional latex code. Throw in your favorite LaTeX preamble and you should be able to compile it to a PDF.

Obviously, that is a terrible solution. Better ideas are welcome!

7. Rolando says:

The printing works perfectly on Opera.

8. Lalit Jain says:

Could you explain your comment on Kummer’s theorem or perhaps give a reference? As a student learning class field theory for the first time I am quite intrigued….

“Incidentally, most books that I’ve seen prove Kummer’s Theorem as a corollary of Hilbert’s Theorem 90, but the result is much easier than that. That might be a subject for a blog post at some other point…

9. David Speyer says:

Kummer’s theorem: Let $L/K$ be cyclic with Galois group $\mathbb{Z}/n$, with $K$ containing $n$-th roots of unity and the characteristic of $K$ relatively prime to $n$. Let $\sigma$ be the order $n$ automorphism of $L$. As a $K$ linear map, $\sigma$ obeys $\latex \sigma^n=1$. So, as a $K$-vector space, $L$ breaks up as a direct sum of $\zeta^j$-eigenspaces for various values of $j$.

Suppose that $u \in L^*$ with $\sigma(u) = \zeta^j u$ and $v \in L^*$ with $\sigma(v) = \zeta^k v$. Then $\sigma(uv) = \zeta^{j+k} uv$. So the set of $j$ such that $\zeta^j$ is an eigenvalue of $\sigma$ is a subgroup of $\mathbb{Z}/n$. If this subgroup is not all of $\mathbb{Z}/n$, then $\sigma$ has order less than $n$. So we conclude that there is some $v$ in $L$ with $\sigma(v) = \zeta v$.

Then $\sigma(v^n) = \zeta^n v^n = v^n$, and we conclude that $v^n \in K$. Let $v^n=w$. Then $L=K(w^{1/n})$.