Monday, December 13, 2010

Splitting of prime ideals in algebraic number fields

Our series of articles on algebraic number theory is back again. Maybe this time it won't be so sporadic. Stranger things have happened. The previous installment, of which this is a direct continuation, is here. All previous installments are listed here.

When we left off, we were talking about how to determine the way a prime ideal factors in the ring of integers of a quadratic extension of ℚ. Such a field is of the form ℚ(√d) for some square-free d∈ℤ. We were using very simple elementary reasoning with congruences, and we found a fairly simple rule, namely:

If p∈ℤ is an odd prime (i. e., not 2), and K=ℚ(√d) is a quadratic extension of ℚ (where d is not divisible by a square) then
  1. p splits completely in K if and only if p∤d and d is a square modulo p.
  2. p is prime (i. e. inert) in K if and only if d is not a square modulo p.
  3. p is ramified in K if and only if p|d.
The prime 2 behaves a little more weirdly, but the result is that 2 ramifies if and only if d≡2 or 3 (mod 4); 2 is inert if and only if d≡5 (mod 8); 2 splits if and only if d≡1 (mod 8).

One limitation was that our simple reasoning made it necessary to assume that OK, the ring of integers of K, was a PID (principal ideal domain).

Let's review what we were trying to do. We were investigating the factorization of a prime ideal (p)=pOℚ(√d) in Oℚ(√d). If Oℚ(√d) is a PID, then there is a simple approach to investigate how p splits. If p splits then (p)=P1⋅P2, where Pi=(αi), i=1,2. Any quadratic extension is Galois, and the Galois group permutes the prime ideal factors of (p). The factors are conjugate, so if α1=a+b√d we can assume α21*=a-b√d. Hence (p)=(α1)⋅(α1*)= (α1α1*)= (a2-db2).

Taking norms (to eliminate possible units ε∈Oℚ(√d)) reduces the problem to a Diophantine equation of the form ±p=a2-db2. With the problem thus reduced, a necessary condition for (p) to split (or ramify) is that the equation can be solved for a,b∈ℤ. A sufficient condition to show that (p) is inert, i. e. doesn't split or ramify, is to show that the equation can't be solved.

Let's look at how that might work. For example, let d=3. Looking at the equations modulo 3, we have ±p≡a2 (mod 3). That is, either p or -p is a square modulo 3. Say p=5. The only nonzero square mod 3 is 1, and 5≢1 (mod 3). However -5≡1 (mod 3), so could we have -5=a2-3b2? Suppose there were some a,b∈ℤ such that -5=a2-3b2. Then instead of looking at the equation modulo 3, we could look at it modulo 5, and find that then a2≡3b2 (mod 5). If 5 divides either a or b, it divides both, and so 25 divides a2-3b2, which is impossible since 25∤5. Therefore 5∤b. ℤ/(5) is a field, so b must have an inverse c such that cb≡1 (mod 5). Therefore, (ac)2 ≡ 3(bc)2 ≡ 3 (mod 5), and so 3 is a square mod 5. But that can't be, since only 1 and 4 are squares modulo 5. The contradiction implies -5=a2-3b2 has no solution for a,b∈Z.

All that does show 5 doesn't split or ramify in ℚ(√3), hence it must be intert, but this approach is messy and still requires knowing that the integers of ℚ(√3) form a PID. We need to find a better way. Fortunately, there is one. But first let's observe that this elementary discussion shows there is a fairly complicated interrelationship among:
  1. Factorization of (prime) ideals in extension fields,
  2. Whether a given ring of integers is a PID,
  3. Whether an integer prime can be represented as the norm of an integer in an extension field,
  4. Whether an integer can be represented by an expression of the form a2+db2 for a,b∈Z (in the case of quadratic extensions),
  5. Whether, for primes p,q∈Z, p is a square modulo q and/or q is a square modulo p.
The problem of representing an integer by an expression like a2+db2 is a question of solving a Diophantine equation, and more specifically is of the type known as representing a number by the value of a quadratic form. This question was studied extensively by Gauss, who proved a remarkable and very important result, known as the law of quadratic reciprocity, which relates p being a square modulo q to q being a square modulo p, for primes p,q.

We will take up quadratic reciprocity soon (and eventually much more general "reciprocity laws"), but right now, let's attack head on the issue of determining how a prime of a base field splits in the ring of integers of an extension field. We will use abstract algebra instead of simple arithmetic to deal with this question. For simplicity, we'll assume here that the base field is ℚ, even though many results can be stated, and are often valid, for more arbitrary base fields.

Chinese Remainder Theorem

The first piece of abstract algebra we'll need is the Chinese Remainder Theorem (CRT). Although it's been known since antiquity to hold for the ring ℤ, generalizations are actually true for any commutative ring.

Let R be a commutative ring, and suppose you have a collection of ideals Ij, for j in some index set, j∈J. Suppose that the ideals are relatively prime in pairs. In general that means that Ii+Ij=R if i≠j, and further, the product of ideals, Ii⋅Ij, is Ii∩Ij when i≠j. If R is Dedekind, then each ideal has a unique factorization into prime ideals, and they are relatively prime if Ii and Ij have no prime ideal factors in common when i≠j. Let I be the product of all Ij for j∈J, which is also the intersection of all Ij for j∈J, since the ideals are coprime in pairs.

The direct product of rings Ri for 1≤i≤k is defined to be the set of all ordered k-tuples (r1, ... ,rk), for ri∈Ri, with ring structure given by element-wise addition and multiplication. The direct product is written as R1×...×Rk, or &Pi1≤i≤kRi.

Given all that, the CRT says the quotient ring R/I is isomorphic to the direct product of quotient rings &Pi1≤i≤k(R/Ii) via the ring homomorphism f(x)=(x+I1, ... ,x+Ik) for all x∈R.

The CRT is very straightforward, since f is obviously a surjective ring homomorphism, and the kernel is I, since it's the intersection of all Ii. (It's straightforward, at least, if you're used to concepts like "surjective" and "kernel".)

Now we'll apply the CRT in two different situations. First let R be the ring of integers OK of a finite extension K/ℚ, and Ii=Pi, 1≤i≤g, be the set of all distinct prime ideals of OK that divide (p)=pOK for some prime p∈ℤ. Then (p)=P1e1 ⋅⋅⋅ Pgeg, where ei are the ramification indices of each prime factor of (p). An application of CRT then shows that OK/(p) ≅ Π1≤i≤g(OK/Piei). Recall that for each i, OK/Pi is isomorphic to the finite field Fqi, where qi=pfi for some fi, known as the degree of inertia of Pi. (This field is the extension of degree fi of Fp=ℤ/pℤ.) Further, Σ1≤i≤geifi=[K:ℚ], the degree of the extension. Check here if you need to review these facts. Specifying how (p) splits in OK amounts to determination of the Pi and the numbers ei, fi, and g.

The second situation where we apply CRT involves the ring of polynomials in one variable over the finite field Fp=ℤ/pℤ, denoted by Fp[x]. Let f(x) be a monic irreducible polynomial with integer coefficients, i. e. an element of ℤ[x]. Let f(x) be f(x) with all coefficients reduced modulo p, an element of Fp[x]. f(x) will not, in general, be irreducible in Fp[x], so it will be a product of powers of irreducible factors: Π1≤i≤g(fi(x)ei), where fi(x)∈Fp[x]. Each quotient ring Fp[x]/(fi(x)) is a finite field that is an extension of Fp of some degree fi. In general, ei, fi, and g will be different, of course, from the same numbers in the preceding paragraph. But the CRT gives us an isomorphism Fp[x]/(f(x)) ≅ &Pi1≤i≤g(Fp[x]/(fi(x)ei)).

Now, here's the good news. For many field extensions K/ℚ, there exists an appropriate choice of f(x)∈ℤ[x] such that for most primes (depending on K and f(x)), the numbers ei, fi, and g will be the same for both applications of the CRT. Consequently, we will have OK/(p) ≅ Fp[x]/(f(x)), because for corresponding factors of the direct product of rings, OK/PieiFp[x]/(fi(x)ei). As it happens, most primes don't ramify for given choices of K and f(x), so that things are even simpler, since all ei=1, and all factors of the direct products are fields.

We can't go into all of the details now as to how to choose f(x) and what the limitations on this result are. However, here are the basics. Any finite algebraic extension of ℚ (and indeed of any base field that is a finite algebraic extension of ℚ) can be generated by a single algebraic number θ: K=ℚ(θ), called a "primitive element". In fact, &theta can be chosen to be an integer of K. Then the ring of integers of K, OK, is a finitely generated module over ℤ. (A module is like a vector space, except that all coefficients belong to a ring rather than a field.) The number of generators is the index [OK:ℤ[θ]]. (ℤ[θ] is just all polynomials in θ with coefficients in ℤ.) If p∈ℤ is any prime that does not divide [OK:ℤ[θ]], then the result of the preceding paragraph holds. If for some p and some choice of θ p does divide the index, then there may be another choice of θ for which p doesn't divide the index. Unfortunately, there are some fields (even of degree 3 over ℚ) where this isn't possible for some choices of p.

The situation is especially nice in the case of quadratic fields, K=ℚ(√d), square-free d∈ℤ. If d≢1 (mod 4); we can take θ=√d and f(x)=x2-d, since OK=ℤ[√d]. If d≡1 (mod 4), then the index [OK:ℤ[√d]]=2, and there's a possible problem only for p=2. However, we still have OK/(p) ≅ Fp[x]/(x2-d) for all p≠2. From that it's obvious that, except for p=2, (p) ramifies if p|d, (p) splits if d is a square modulo p, or else (p) is inert. That is exactly the conclusion we began with at the beginning of this article, on the basis of elementary considerations. Only now we need not assume that OK is a PID.

There are four important lessons to take away from this discussion.

First, there is a very close relationship between the arithmetic of algebraic number fields and the arithmetic of polynomials over a finite field. Not only do we have the isomorphism discussed above, but it turns out that a number of similar powerful theorems are true for both algebraic number fields and the field of quotients of polynomial rings over a finite field.

Second, a lot of the arithmetic of algebraic number fields can be analyzed in terms of what happens "locally" with the prime ideals of the ring of integers of the field.

Third, many of the results of algebraic number theory are fairly simple if the rings of integers are PIDs (or, equivalently, have unique factorization). Such results often remain true when the rings aren't PIDs, though they can be a lot harder to prove. Often the path to proving such results involves considering the degree to which a given ring of integers departs from being a PID.

Fourth, and perhaps most importantly, abstract algebra is a very powerful tool for understanding algebraic number fields – and it is much easier to work with and understand than trying to use "elementary" methods with explicit calculations involving polynomials and their roots.

We will see these lessons validated time and again as we get deeper into the subject.

So where do we go from here? There are a lot of directions we could take, so we'll probably jump around among a variety of topics.


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