Tuesday, April 01, 2008

Factorization of prime ideals in extension fields

In this installment of our series on algebraic number theory, we're going to do two things. First, we'll look at how a prime ideal of one ring of algebraic integers factors into multiple prime ideals in the ring of integers of a larger field. This is an absolutely central concern of the theory. We'll look at some simple examples. Second, we'll make some general definitions that are used to describe this factorization. These definitions will be needed in most of what follows.

In order to understand what's discussed here, you'll need to recall a great deal of what's been discussed before. You can find a list of all previous installments here if you need to review.

Suppose E⊇F is a field extension and OE, OF are the corresponding rings of integers. We have OF ⊆ OE, and if I is any ideal of OF then of course I ⊆ OE. I isn't an ideal of OE, because it isn't closed under multiplication by elements of OE, but I⋅OE is an ideal of OE.

If P is a prime ideal of OF, we have no good reason to expect that P⋅OE is prime in OE, and in fact it generally is not. For example, let's look at quadratic extensions of ℚ. So suppose F=ℚ, d is a square-free integer, positive or negative, and d ≠ 0 or 1. Let p∈ℤ be a positive prime. We can ask whether (the ideal corresponding to) p is still a prime ideal of the ring of integers of E=ℚ(√d). That is, we can ask what happens to the prime, principal ideal (p) in Oℚ(√d). (With principal ideals, we can usually get away without specifying what ring they are contained in.)

Consider the Diophantine equation ±p=a2-b2d. If we can solve it by finding a,b∈ℤ that make the equation true, then we can verify that we have a factorization of the ideal (p) in Oℚ(√d) expressed by the equation of principal ideals (p)=(a+b√d)(a-b√d). (Proof: p∈(a+b√d)(a-b√d) since ±p=a2-b2d, so (p)⊆(a+b√d)(a-b√d). But by the definition of a product of ideals, all members of (a+b√d)(a-b√d) are a product of an element of Oℚ(√d) and the number [a+b√d][a-b√d]=±p, and so (a+b√d)(a-b√d)⊆(p).)

So solutions of a certain Diophantine equation tell us about how an ideal (p) of ℤ factors in the integers of a quadratic extension. And in fact, if the equation can be solved, then the prime ideal (p) of ℤ is not also a prime ideal of Oℚ(√d). Is there a converse, that is, can we infer solutions of a Diophantine equation from how ideals factor in extensions? Unfortunately, the situation is more complicated. For instance, if Oℚ(√d) is not a PID, then it is possible for (p) to not be a prime ideal of Oℚ(√d), yet there may be no solutions of ±p=a2-b2d. One of the reasons that algebraic number theory is useful and interesting is that it helps get a handle on the complications, as we shall see.

For simplicity, suppose d≡3 (mod 4). Then we have remarked that the integers of ℚ(√d) are given by Oℚ(√d) = {a+b√d | a,b∈ℤ}. Suppose we could find some integer α∈ℚ(√d) such that the norm Nℚ(√d)/ℚα = ±p. (For short, write Nα for the norm.) So if α=a+b√d with a,b∈ℤ, we have ±p=Nα=a2-b2d. When this happens, we can write ±p=(a+b√d)(a-b√d). Since Nα is a prime (in ℤ), α=a+b√d must be a prime of Oℚ(√d) – and p is not a prime in that ring. To make things really easy for showing examples, let d=3. Trying a few values for a and b, let α=4+√3, so Nα=13=(4+√3)(4-√3). Or if α=8+5√3, then Nα=-11=(8+5√3)(8-5√3). So both 11 and 13 are not primes in Oℚ(√3).

This translates directly to a factorization of the principal ideal (13) in Oℚ(√d), namely (13)=(4+√3)(4-√3). Clearly (13) isn't a prime ideal, since it has two distinct nontrivial prime factors. (11)=(-11) is likewise not a prime ideal. (ℚ(√3) happens to be a PID, but that isn't crucial here.) In general, then, if p∈ℤ is a prime, the principal ideal (p) is a prime ideal of ℤ, but for any square-free d∈ℤ (with d≠0,1), (p) splits into the product of two distinct ideals in the integers of the quadratic extension ℚ(√d) if we can find a,b∈ℤ, with a≠0, such that ±p = N(a±b√d) = a2-b2q. In that case, we have a factorization of (p) in Oℚ(√d) into principal ideals such that (p)=(a+b√d)(a-b√d), and the factors are prime ideals. If a≠0, these will be distinct ideals. (Since p is assumed to be prime in ℤ, if a=0, we must have b=±1, and the two ideals will be the same. In this special case, one actually says that p "ramifies" in the extension field.)

Conversely, what if, for the given d, there are no integers a,b∈ℤ that satisfy the equation ±p = a2-b2d – can we conclude that p doesn't split in Oℚ(√d)? Actually, no, we can't in general. We could if Oℚ(√d) happens to be a PID, since, because d≡3 (mod 4), the factor ideals would have to have the form (a±b√d) for some a,b∈ℤ, which would yield a solution of the equation, contrary to supposition. (And remember, by the general theory of Dedekind rings, the same is true if Oℚ(√d) has unique factorization, since a ring of integers of a field is a Dedekind ring, and therefore is a PID, if and only if it has unique factorization.)

However, and this is a major "however", if Oℚ(√d) isn't a PID, or equivalently if it doesn't have unique factorization of elements, then we could have a factorization of ideals where (p)=AB into prime ideals that are not principal. (The prime ideal factors themselves are uniquely determined, which, do not forget, is always true in Dedekind rings.) In that case, the factorization doesn't necessary give us a solution of ±p = a2-b2q. Thus it's possible to have primes p where (p) splits in Oℚ(√d) even if the equation has no integer solutions. We might even have A=B, in which case (p) ramifies.

So one key thing this discussion illustrates is that there is a close, though not entirely simple, connection between solving simple Diophantine equations (like ±p=a2-b2q) and determining how prime ideals of a subfield split into prime ideals in an extension field. Further, these equations arise from asking what numbers can occur as norms of some integer of an extension field. But the situation is more complicated if the ring of integers of the extension field doesn't have unique factorization (or equivalently, if it's not a principal ideal domain). This is another reason uniqueness of factorization (or the lack of it) is rather important.

To give some idea of where this leads, we'll just say that class field theory, which we will be coming to shortly, originated in trying to deal with such questions, in the form of saying someting about how prime ideals split in extension fields, and also about when nonprincipal ideals can become principal ideals in extension fields. It helps us deal with the complexities that result from failure of unique factorization of numbers, and from the equalivalent untidiness of having ideals that are not principal ideals.

But before proceeding, let's establish some general conventions and terminology. The terminology will be used to keep the upcoming discussion succinct, so unfortunately you'll need to memorize the details or else plan to refer back here if you want to follow beyond this point. As a reward, we'll be able to state some fairly simple rules for when primes of ℤ do or do not split in extension fields of ℚ.

Let's set out the terminology to describe the general possibilities for how prime ideals split in arbitrary extension fields. We continue to suppose E⊇F is a finite extension, of degree n=[E:F], not necessarily Galois. Let P be a prime ideal of OF. P⋅OE is an ideal of OE, so it factors in a unique way as a product of powers of distinct ideals that are prime in OE:
P⋅OE = ∏1≤j≤g Qjej
We say that each Qj is a prime lying above P in E. Since every Qj divides P, each one contains P: Qj⊇P. In fact, P=Qj∩OF for each j.

If we start with a prime ideal P of the integers of an extension of ℚ, then P∩ℤ is a prime ideal (p) of ℤ for some rational prime p, and P lies above (p). Given the factorization of P shown above, each Qj also lies above (p). By general commutative ring theory, since Qj is a prime of OE, the quotient ring OE/Qj is a finite field Fq where q is some power of p. (In Dedekind domains, any prime ideal is a maximal ideal, and for any commutative ring R with identity, the quotient of R by a maximal ideal is a field.) Likewise, since P is prime in OF, OF/P is a finite field Fq′ where q′ is also a power of p (and divides q). Each field OE/Qj is a finite extension of the field OF/P, and we denote its degree by fj. Finally, it turns out that P⋅OE is a vector space of degree n=[E:F] over the field OF/P.

These facts allow one to make the following definitions. The exponent ej is called the ramification index of Qj (relative to the given extension). The degree fj is the inertial degree of Qj. And g is called the decomposition number. All of these numbers are specific to how the prime P of OF splits (or decomposes) in OE. These numbers are the key data one wants to have for each prime P for any given extension E⊇F. A prime P is said to be ramified unless all exponents ej=1, in which case it is unramified. Primes that ramify tend to make life more complicated, but fortunately there are only finitely many for any particular extension. All these numbers are related as follows:
n = [E:F] = ∑1≤j≤g ejfj
If P is unramified and all inertial degrees are 1, then we must have g=n, and one says that P splits completely. This is a fairly unusual circumstance. Although only finitely many primes are ramified, most have some inertia.

If E⊇F is a Galois extension, things are much simpler. In that case, all ej have the same value, say e, and all fk have the same value, say f. (This is because the Galois group G(E/F) acts on the various Qj and permutes them among themselves. Since each σ∈G(E/F) is an automorphism, these ideals are all essentially alike.) Hence [E:F]=efg. So there are exactly g primes lying above P in E, and each has the same ramification index and inertial degree.

In the next installment we'll, look at more examples involving quadratic extensions of ℤ, and state the rules for how primes split. The rules will turn out to be based on the classical law of "quadratic reciprocity".

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5 Comments:

Blogger Chuck said...

Are you planning to continue your algebraic number theory series?

2/16/2009 11:40:00 AM  
Blogger Charles Daney said...

Are you planning to continue your algebraic number theory series?

Yes. Thanks for asking.

However, I have the feeling that it's becoming more technical than most people are interested in. So I haven't made it a high priority.

But knowing at least one person might be interested is enough to stimulate me to put up another installment. Might take a couple of weeks, but I'll try to squeeze it in.

2/16/2009 04:33:00 PM  
Blogger Adam said...

Make that two people!

3/02/2009 10:24:00 PM  
Blogger Charles Daney said...

OK, OK, the next installment is now up - here.

3/08/2009 10:25:00 PM  
Blogger Adam said...

That's awesome!

6/05/2009 08:22:00 PM  

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