# A class field theoretic phenomenon

In this post we discuss one example of what’s called a ‘class field theoretic phenomenon’. In particular, we focus on the application of trying to understand the property of when $X^3-2$ has three distinct roots modulo $p$, for various primes $p$.

# Motivation

Unfortunately, this post will not be a full discussion of the intuitive nature/desire for, class field theory. Instead, I’d just like to prove one particular incarnation of a class-field theoretic phenomenon.

So, what is a class field theoretic phenomenon? It’s a very loose, intuitive, touchy-feely phrase, but it can roughly be stated as the following meta-theorem:

Metatheorem: Let $K$ be a number field. Then, the fields $L/K$ for which all number theoretic properties can be described internal to $K$ are the abelian ones.

Let’s quickly describe what we mean by this. Recall first that an extension $L/K$ is called abelian if it is Galois, and if $\text{Gal}(L/K)$ is an abelian group. So, for, example, $\mathbb{Q}(\zeta_N)/\mathbb{Q}$ is abelian for all $N$ (in fact, they comprise a cofinal system of the most important abelian extensions of $\mathbb{Q}$– -the class fields). But, $\mathbb{Q}(\zeta_3,\sqrt[3]{7})$ is not.

The definition of ‘number theoretic properties’ is a vague one, and is best left to the Potter Stewart philosophy of identification. A similar statement can be made about things internal to $K$, but these are sort of easy to imagine- -they are objects/things/properties whose description only makes reference to elements/properties of $K$ (nothing ‘external’). As an example of these two ideas, consider the the Galois group $\text{Gal}(L/K)$ (clearly a number theoretic object). How could we describe it using ‘properties internal to $K$‘?

Well, if the metatheorem is to hold, we should, in particular be able to do so for $\text{Gal}(\mathbb{Q}(\zeta_N)/\mathbb{Q})$, since $\mathbb{Q}(\zeta_N)/\mathbb{Q}$ is abelian. And, indeed we can! Namely, we have an isomorphism

$\text{Gal}(\mathbb{Q}(\zeta_N)/\mathbb{Q})\cong I^{(N)}/P^{(N\cdot\infty)}$

(both of which are isomorphic to $(\mathbb{Z}/N\mathbb{Z})^\times$). Here $I^{(N)}$ denotes the free abelian group on the prime ideals $(p)$ of $\mathbb{Z}$, where $p\nmid N$, and

$P^{(N\cdot\infty)}=\left\{(\alpha):\alpha \in\mathbb{Q}, \alpha>0,\text{ and }v_p\left(\alpha-1\right)>v_p(N)\text{ for all }p\mid N\right\}$

To the uninitiated, this may seem highly unmotivated (it falls under the general study of ray class fields/ray class groups). That said, one thing is for sure, $I^{(N)}/P^{(N\cdot\infty)}$ is, in fact, an internal to $\mathbb{Q}$ object. Thus, at least in this case, the metatheorem holds.

Let’s give another example of the metatheorem with a, perhaps, less esoteric result. It’s a famous result of a first course in number theory that for a number $m\in\mathbb{Z}$, the set of primes $q$ for which $X^2-m\mod q$ has a solution, can be phrased in the form

$q$ is such a prime if and only if $q\equiv 1\mod N$

for some $N$. For example, $X^2+1$ has a root in $\mathbb{F}_q$ if and only if $q\equiv 1\mod 4$.

What does this have to do with the metatheorem? Well, the Dedekind-Kummer theorem tells us that $X^2+1$ has a root in in $\mathbb{F}_q$ if and only if the prime $q$ splits in $\mathcal{O}_{\mathbb{Q}(i)}=\mathbb{Z}[i]$. Thus, $X^2+1$ having a root in $\mathbb{F}_q$ is clearly a number theoretic property of $\mathbb{Q}(i)/\mathbb{Q}$. And, the previous paragraph tells us that this property has an entirely internal-to-$\mathbb{Q}$ description: $q$ is such a split prime if and only if $q\equiv 1\mod 4$. Taking the metatheorem into consideration, this shouldn’t be that big of a surprise- -the extension $\mathbb{Q}(i)/\mathbb{Q}$ is abelian! A similar analysis informs why we’d expect this to be true for $X^2-m$ for general $m$.

So, while this is a famous theorem, covered in most first introductions to number theory, making even a slight modification leads to a significantly harder question. For example, consider the polynomial $X^3-2$. Analogizing to the case of, say, $X^2-2$, one might be led to conjecture that for some $N$ and for some $S\subseteq\mathbb{Z}/N\mathbb{Z}$ we might have:

$X^3-2$ has three distinct roots in $\mathbb{F}_q$ if and only if $(q\mod N)\in S$

For example, you might guess that $X^3-2$ has a root in $\mathbb{F}_q$ if and only if $q\equiv 1,2,3\mod 4$, or perhaps, if and only if $q\equiv 1,5,7\mod 8$.

One may try and solve this by elementary means, but quickly one realizes that the question is deceivingly difficult. That said, if we take our metatheorem into account again, then we’d expect the answer to this question to be a resounding no: there does not exist such an $N$ and such an $S$. Why? Begin again by considering the Dedekind-Kummer theorem, this tells us that $X^3-2$ having three distinct roots in $\mathbb{F}_q$ is equivalent to $q$ splitting in $\mathcal{O}_{\mathbb{Q}(\sqrt[3]{2})}=\mathbb{Z}\left[\sqrt[3]{2}\right]$. Thus, $X^3-2$ having a root in $\mathbb{F}_q$ is a number theoretic property of $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$.

So, if we could find such an $N$ and $S$, this would give an internal-to-$\mathbb{Q}$ description of this number theoretic property. But, the metatheorem predicts that this shouldn’t happen- -the extension $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ is non-abelian!

Now, while all of the above is good intuition about what class field theory says (in the form of the metatheorem) it is, after all, just intuition. In particular, the fairly basic question of whether $X^3-2$ having three distinct roots in $\mathbb{F}_q$ can be determined modulo some integer $N$ is left unanswered. It is the goal of this post to prove this particular case of the metatheorem.

# The actual result

So, let’s rigorously state what we want to prove. But, first, let’s recall some basic definitions. For a number field $L/K$, we define the set $\text{Spl}(L/K)$ to be the set of primes $\mathfrak{p}$ of $K$ which split completely in $L$. So, for example, our discussion in the motivation implies

$\text{Spl}\left(\mathbb{Q}(i)/\mathbb{Q}\right)=\left\{p:p\equiv 1\mod 4\right\}$

The goal then of this post is the following:

Theorem: Let $K/\mathbb{Q}$ be a number field. If there exists an integer $N$, and a set $S\subseteq\mathbb{Z}/N\mathbb{Z}$ such that

$\text{Spl}(K/\mathbb{Q})=\left\{p:(p\mod N)\in S\right\}$

then, $K/\mathbb{Q}$ is abelian. In fact, $K\subseteq\mathbb{Q}(\zeta_N)$.

As a corollary, we obtain the following very pleasant result. Call an irreducible monic polynomial $f(X)\in\mathbb{Z}[X]$ monogenic if $\mathcal{O}_K=\mathbb{Z}[X]/(f(X))$, where $K=\mathbb{Q}[X]/f(X)$. Furthermore, say that $f(X)$‘s splitting behavior can be determined modulo $N$ if for some subset $S\subseteq \mathbb{Z}/N\mathbb{Z}$, $f(X)$ splits into $\deg f$ linear factors in $\mathbb{F}_q$ if and only if $(q\mod N)\in S$.

Corollary 1: A monogenic polynomial $f(X)\in\mathbb{Z}[X]$ has its splitting behavior determined modulo $N$ if and only if $K/\mathbb{Q}$ is abelian, where $K=\mathbb{Q}[X]/(f(X))$.

Moreover, if $\mathbb{Q}[X]/(f(X))$ is Galois (perhaps non-abelian!), then $f(X)$ having a root in $\mathbb{F}_q$ can be determined modulo $N$ if and only if $\mathbb{Q}[X]/f(X)$ is Galois! This is because having a root in $\mathbb{F}_q$ is equivalent to splitting completely, by first principles.

So, this corollary solves entirely our simple problem. Since $X^3-2$ is monogenic, but $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ is non-abelian, we cannot determine its splitting behavior modulo $N$ for any $N$.

Before we prove our main theorem, we’ll need two lemmas. The first is trivial, but the second requires a serious amount of work:

Lemma 1: Let $K/\mathbb{Q}$ be a number field. Then, for a Galois closure $M$ of $K$ in $\overline{\mathbb{Q}}$ the following holds:

$\text{Spl}(K/\mathbb{Q})=\text{Spl}(M/\mathbb{Q})$

This is a simple exercise in ‘Hilbert theory’ (the study of decomposition and inertia groups in Galois extension).

Proof: First assume that $p\in\text{Spl}(M/\mathbb{Q})$. Take any prime $\mathfrak{p}$ of $K$ lying over $p$, and take a prime $\mathfrak{P}$ of $M$ lying  over $\mathfrak{p}$. By basic number theory, we have the following identities:

$e\left(\mathfrak{P}\mid (p)\right)=e\left(\mathfrak{P}\mid\mathfrak{p}\right)e\left(\mathfrak{p}\mid(p)\right)\qquad f\left(\mathfrak{P}\mid(p)\right)=f\left(\mathfrak{P}\mid\mathfrak{p}\right)f\left(\mathfrak{p}\mid(p)\right)$

Since, by assumption,

$e\left(\mathfrak{P}\mid(p)\right)=f\left(\mathfrak{P}\mid(p)\right)=1$

we may conclude that

$e\left(\mathfrak{p}\mid(p)\right)=f\left(\mathfrak{p}\mid(p)\right)=1$

And, so, the identity

$\displaystyle \sum_{\mathfrak{p}\mid p}e\left(\mathfrak{p}\mid (p)\right) f\left(\mathfrak{p}\mid (p)\right)=[K:\mathbb{Q}]$

implies that the number of primes over $(p)$ in $K$ is $[K:\mathbb{Q}]$, which is the definition of $p$ splitting completely in $K$.

Conversely, suppose that $p$ splits completely in $K$. Then, for every Galois conjugate $\sigma(K)$ (with $\sigma\in \text{Gal}(M\mathbb{Q})$) we have that $p$ splits completely in $\sigma(K)$. Indeed, if

$p\mathcal{O}_K=\mathfrak{p}_1\cdots\mathfrak{p}_n$

where $n=[K:\mathbb{Q}]$, then by  the nature of $\sigma$,

$p\mathcal{O}_{\sigma(K)}=\sigma\left(p\mathcal{O}_K\right)=\sigma(\mathfrak{p}_1)\cdots\sigma\left(\mathfrak{p}_n\right)$

Each $\sigma(\mathfrak{p}_i)$ is a distinct prime, and thus we see that $p$ splits completely in $\sigma(K)$.

So, now choose a prime $\mathfrak{P}$ of $M$ lying over $(p)$. Let $D:= D(\mathfrak{P}\mid \mathfrak{p})$ be its decomposition group, and consider the associated decomposition field $L^D$. Then, by basic theory, $p$ splitting in $\sigma(K)$ implies that $\sigma(K)\subseteq L^D$ for all $\sigma$. So,

$\displaystyle M=\prod_\sigma \sigma(K)\subseteq L^D$

Thus, $M=L^D$, and so, once again by the theory of decomposition fields, we conclude that $(p)$ is totally split in $M$. $\blacksquare$

The second lemma requires quite a bit of analytic work. In fact, it’s a corollary of the Chebotarev density theorem which requires a detailed study of the special values of $L$-functions. For a proof of this lemma, see Corollary 3.5 of Chapter VI of Milne’s Class Field Theory.

Lemma 2: Let $L,L'/K$ be Galois extensions of number fields. Then, $L\subseteq L'$ if and only if $\text{Spl}(L'/K)\subseteq \text{Spl}(L/K)$.

Note that while this seems like it contradicts our metatheorem, it really doesn’t. While this relates number theoretic properties of extensions of $K$ (their containment), the relation which guides this (primes splitting in the extension) is NOT internal to $K$.

Ok, so with both of these lemmas, we are prepared to prove our theorem:

Proof(of theorem): Suppose first that $K/\mathbb{Q}$ is abelian. Then, by class field theory, $K\subseteq\mathbb{Q}(\zeta_N)$ for some minimal $N$ (this is the Kronecker-Weber theorem). Moreover, $p$ ramifies in $K$ if and only if $p\mid N$. Now, let $H\subseteq (\mathbb{Z}/N\mathbb{Z})^\times=\text{Gal}(\mathbb{Q}(\zeta_N)/\mathbb{Q})$ be the subgroup corresponding the subextension $K/\mathbb{Q}$. Then, the kernel of map

$(\mathbb{Z}/N\mathbb{Z})^\times\to \text{Gal}(K/\mathbb{Q})\to \text{Gal}(K/\mathbb{Q})$

is $H$.

That said, the isomorphism $(\mathbb{Z}/N\mathbb{Z})^\times\to\text{Gal}(\mathbb{Q}(\zeta_N)/\mathbb{Q})$ is dictated by $p\mapsto \text{Frob}_p$, the unique Frobenius element of $\text{Gal}(\mathbb{Q}(\zeta_N)/\mathbb{Q}$ correspond to $p$. So, the map $(\mathbb{Z}/N\mathbb{Z})^\times\to\text{Gal}(K/\mathbb{Q})$ sends $p$ to the restriction of $\text{Frob}_p$ to $K$. But, since Frobenii are functorial, this is just the Frobenius element $\text{Frob}_p\in\text{Gal}(K/\mathbb{Q})$. Thus, we see that $p\in(\mathbb{Z}/N\mathbb{Z})^\times$ is in $H$ if and only if $\text{Frob}_p\in\text{Gal}(K/\mathbb{Q})$ is trivial. But, by basic theory, this is equivalent to $p$ splitting completely in $K$. Thus, we see that $H$ is our desired set $S$!

Conversely, suppose that $K/\mathbb{Q}$ is such that

$\text{Spl}(K/\mathbb{Q})=\left\{p:(p\mod N)\in S\right\}$

Then, by Lemma 1, if $M$ is a Galois closure of $K$, we have

$\text{Spl}\left(M/\mathbb{Q}\right)=\left\{p:(p\mod N)\in S\right\}$

Suppose for a second that $1\in S$. Then, we claim that $M\subseteq \mathbb{Q}(\zeta_N)$. Indeed, basic number theory shows that

$\text{Spl}(\mathbb{Q}(\zeta_N)/\mathbb{Q})=\left\{p:p\equiv 1\mod N\right\}$

and thus if $1\in S$ we have $\text{Spl}(\mathbb{Q}(\zeta_N)/\mathbb{Q})\subseteq\text{Spl}(M/\mathbb{Q})$. But, since both $\mathbb{Q}(\zeta_N)$ and $M$ are Galois over $\mathbb{Q}$, this implies by Lemma 2 that $M\subseteq\mathbb{Q}(\zeta_N)$.

Thus, it remains to show that $1\in S$. To see this, suppose not. Then, consider the compositum $F$ of $M$ and $\mathbb{Q}(\zeta_N)$. Note that if $p$ split completely in $F$, then $p$ would have to split completely in $M$ and $\mathbb{Q}(\zeta_N)$. But, by assumption, $\text{Spl}(\mathbb{Q}(\zeta_N)/\mathbb{Q})\cap \text{Spl}(M/\mathbb{Q})=\varnothing$. So, no prime splits in $F$. But, this is impossible (by an application of Chebotarev density). $\blacksquare$

Note a sort of strange fact that comes out of the above proof. Let $K/\mathbb{Q}$ be as in the second part of the proof, with a given $S\subseteq\mathbb{Z}/N\mathbb{Z}$. Then, since $K\subseteq\mathbb{Q}(\zeta_N)$ we can emulate the techniques in the first part of the proof to actually show that $S$ needs to be a subgroup of $(\mathbb{Z}/N\mathbb{Z})^\times$! This shows a very non-obvious restriction on what sets of primes occur as $\text{Spl}(K/\mathbb{Q})$ for some $K/\mathbb{Q}$!