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 has three distinct roots modulo , for various primes .
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 be a number field. Then, the fields for which all number theoretic properties can be described internal to are the abelian ones.
Let’s quickly describe what we mean by this. Recall first that an extension is called abelian if it is Galois, and if is an abelian group. So, for, example, is abelian for all (in fact, they comprise a cofinal system of the most important abelian extensions of – -the class fields). But, 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 , but these are sort of easy to imagine- -they are objects/things/properties whose description only makes reference to elements/properties of (nothing ‘external’). As an example of these two ideas, consider the the Galois group (clearly a number theoretic object). How could we describe it using ‘properties internal to ‘?
Well, if the metatheorem is to hold, we should, in particular be able to do so for , since is abelian. And, indeed we can! Namely, we have an isomorphism
(both of which are isomorphic to ). Here denotes the free abelian group on the prime ideals of , where , and
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, is, in fact, an internal to 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 , the set of primes for which has a solution, can be phrased in the form
“ is such a prime if and only if “
for some . For example, has a root in if and only if .
What does this have to do with the metatheorem? Well, the Dedekind-Kummer theorem tells us that has a root in in if and only if the prime splits in . Thus, having a root in is clearly a number theoretic property of . And, the previous paragraph tells us that this property has an entirely internal-to- description: is such a split prime if and only if . Taking the metatheorem into consideration, this shouldn’t be that big of a surprise- -the extension is abelian! A similar analysis informs why we’d expect this to be true for for general .
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 . Analogizing to the case of, say, , one might be led to conjecture that for some and for some we might have:
“ has three distinct roots in if and only if “
For example, you might guess that has a root in if and only if , or perhaps, if and only if .
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 and such an . Why? Begin again by considering the Dedekind-Kummer theorem, this tells us that having three distinct roots in is equivalent to splitting in . Thus, having a root in is a number theoretic property of .
So, if we could find such an and , this would give an internal-to- description of this number theoretic property. But, the metatheorem predicts that this shouldn’t happen- -the extension 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 having three distinct roots in can be determined modulo some integer 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 , we define the set to be the set of primes of which split completely in . So, for example, our discussion in the motivation implies
The goal then of this post is the following:
Theorem: Let be a number field. If there exists an integer , and a set such that
then, is abelian. In fact, .
As a corollary, we obtain the following very pleasant result. Call an irreducible monic polynomial monogenic if , where . Furthermore, say that ‘s splitting behavior can be determined modulo if for some subset , splits into linear factors in if and only if .
Corollary 1: A monogenic polynomial has its splitting behavior determined modulo if and only if is abelian, where .
Moreover, if is Galois (perhaps non-abelian!), then having a root in can be determined modulo if and only if is Galois! This is because having a root in is equivalent to splitting completely, by first principles.
So, this corollary solves entirely our simple problem. Since is monogenic, but is non-abelian, we cannot determine its splitting behavior modulo for any .
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 be a number field. Then, for a Galois closure of in the following holds:
This is a simple exercise in ‘Hilbert theory’ (the study of decomposition and inertia groups in Galois extension).
Proof: First assume that . Take any prime of lying over , and take a prime of lying over . By basic number theory, we have the following identities:
Since, by assumption,
we may conclude that
And, so, the identity
implies that the number of primes over in is , which is the definition of splitting completely in .
Conversely, suppose that splits completely in . Then, for every Galois conjugate (with ) we have that splits completely in . Indeed, if
where , then by the nature of ,
Each is a distinct prime, and thus we see that splits completely in .
So, now choose a prime of lying over . Let be its decomposition group, and consider the associated decomposition field . Then, by basic theory, splitting in implies that for all . So,
Thus, , and so, once again by the theory of decomposition fields, we conclude that is totally split in .
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 -functions. For a proof of this lemma, see Corollary 3.5 of Chapter VI of Milne’s Class Field Theory.
Lemma 2: Let be Galois extensions of number fields. Then, if and only if .
Note that while this seems like it contradicts our metatheorem, it really doesn’t. While this relates number theoretic properties of extensions of (their containment), the relation which guides this (primes splitting in the extension) is NOT internal to .
Ok, so with both of these lemmas, we are prepared to prove our theorem:
Proof(of theorem): Suppose first that is abelian. Then, by class field theory, for some minimal (this is the Kronecker-Weber theorem). Moreover, ramifies in if and only if . Now, let be the subgroup corresponding the subextension . Then, the kernel of map
That said, the isomorphism is dictated by , the unique Frobenius element of correspond to . So, the map sends to the restriction of to . But, since Frobenii are functorial, this is just the Frobenius element . Thus, we see that is in if and only if is trivial. But, by basic theory, this is equivalent to splitting completely in . Thus, we see that is our desired set !
Conversely, suppose that is such that
Then, by Lemma 1, if is a Galois closure of , we have
Suppose for a second that . Then, we claim that . Indeed, basic number theory shows that
and thus if we have . But, since both and are Galois over , this implies by Lemma 2 that .
Thus, it remains to show that . To see this, suppose not. Then, consider the compositum of and . Note that if split completely in , then would have to split completely in and . But, by assumption, . So, no prime splits in . But, this is impossible (by an application of Chebotarev density).
Note a sort of strange fact that comes out of the above proof. Let be as in the second part of the proof, with a given . Then, since we can emulate the techniques in the first part of the proof to actually show that needs to be a subgroup of ! This shows a very non-obvious restriction on what sets of primes occur as for some !