Cyclotomic fields, part 2

In our previous article on cyclotomic fields we were talking about why the Galois group G of ℚ(μ_n)/ℚ is isomorphic to (ℤ/nℤ)^×, where n∈ℤ and μ_n is the group of n^th roots of unity, the roots of xⁿ-1=0 in some extension of ℚ. (Check here for a list of previous articles on algebraic number theory.)

So far we've shown that G is isomorphic to a subgroup of (ℤ/nℤ)^×. We still need to show it is actually isomorphic to the whole group, or equivalently that |G|, the order of G, which is equal to the degree of the field extension, [ℚ(μ_n):ℚ], actually equals |(ℤ/nℤ)^×|, which is φ(n), the number of positive integers less than and relatively prime to n.

The group μ_n is cyclic. Any generator of the group is, by definition, a primitive n^th root of unity. We let ζ be an arbitrary but fixed such generator. Then ℚ(μ_n)=ℚ(ζ). Let f(x) be the minimal polynomial of ζ. f(x)∈ℚ[x] is irreducible over ℚ. All other elements of μ_n are of the form ζ^a for some a∈ℤ, where a is well-defined modulo n. Further, ζ^a is a primitive n^th root of unity if and only if a is relatively prime to n, i. e. the greatest common divisor (a,n)=1. The degree of f(x) equals [ℚ(μ_n):ℚ] and |G|. At this point, all we know about these numbers is that they divide φ(n) (since |G| does).

The group homomorphism j:G→(ℤ/nℤ)^× was defined by the relation σ(ζ)=ζ^j(σ), for σ∈G. We showed that j(σ) is injective and independent of the choice of ζ. Hence G is isomorphic to a subgroup of (ℤ/nℤ)^×, and therefore the degree of f(x) (and [ℚ(μ_n):ℚ] and |G|) is ≤φ(n). G is isomorphic to a proper subgroup of (ℤ/nℤ)^×, i. e. not the whole group, if j is not surjective and the degree of f(x) is strictly less than φ(n).

As we noted last time, the problem here is that we don't yet know that all φ(n) primitive n^th roots of unity are zeroes of f(x), so that |G| and the degree of f(x) equal φ(n), and hence G(ℚ(μ_n)/ℚ) ≅ (ℤ/nℤ)^×. Stated another way, we don't yet know that the field homomorphism on ℚ(μ_n) induced by mapping ζ to ζ^a is actually an automorphism of the field, hence an element of the Galois group. It could fail to be if, say, the minimal polynomial of ζ^a is different from that of ζ, which could happen if the degree of f(x) is less than φ(n) because not all primitive n^th roots of unity are zeroes of f(x). In order to rule out this possibility, we will show that the degree of f(x) is ≥φ(n).

There are various ways to prove the isomorphism, and even a number of ways to prove that f(x) has φ(n) distinct roots, so its degree is ≥φ(n). Many of these proofs use machinery (such as discriminants, factorization and ramification of primes, etc.) that we haven't extensively discussed yet, so I'll avoid using such things in the proof. However, we'll get to these topics eventually, and also show a way to construct the automorphisms σ∈G explicitly – after finishing the proof that G(ℚ(μ_n)/ℚ) ≅ (ℤ/nℤ)^×.

So let's get started. The roots of the minimal polynomial f(x)∈ℚ[x] are all conjugates σ(ζ) for σ∈G, so f(x)=Π_σ∈G(x-σ(ζ)). Hence f(x) is monic (leading coefficient 1). The coefficients of f(x) are symmetric functions of all conjugates of ζ, so the coefficients are all left fixed by all σ∈G. f(x) divides xⁿ-1, so all its roots – the conjugates of ζ – are algebraic integers. So the coefficients are also algebraic integers (sums of products of powers of algebraic integers) – members of the ring of integers O_ℚ(ζ). They are also in the base field, since they're left fixed by G. A basic fact is that O_ℚ(ζ)∩ℚ = ℤ – any algebraic integer that lies in the base field is necessarily an integer of the base field. Hence f(x)∈ℤ[x].

Suppose that for any a∈ℤ relatively prime to n, i. e. (a,n)=1, ζ^a is also a root of f(x): f(ζ^a)=0. Since these ζ^a with 1≤a<n are distinct primitive n^th roots of unity if ζ is, and there are φ(n) of them, the degree of f(x), and hence |G|, is ≥ φ(n). But we already showed |G|≤φ(n), hence |G|=φ(n). Since G is isomorphic to a subgroup of (ℤ/nℤ)^×, we must actually have an isomorphism: G ≅ (ℤ/nℤ)^×.

So all we have to show is f(ζ^a)=0 for 1≤a<n and (a,n)=1. The first thing to note is that it suffices to prove this just for primes p with (p,n)=1. For suppose we had that. For general a with (a,n)=1, let p be a prime that divides a. Then (p,n)=1. Consider ζ^a/p. Since (a/p,n)=1, ζ^a/p is a primitive n^th root of unity with one fewer prime divisor in the exponent than ζ^a. So by induction on the number of prime divisors of the exponent f(ζ^a/p)=0. But if the result is true for prime powers of primitive n^th roots of unity that satisfy f(x)=0, then f(ζ^a)=0 since ζ^a=(ζ^a/p)^p. Alternatively, you can recall that (according to a theorem of Dirichlet), there are infinitely many primes p in the arithmetic progression a+nk for (a,n)=1 and k∈ℤ. Since ζⁿ=1, ζ^a=ζ^p for all such p.

So let p be prime and (p,n)=1. Then note that f(x)^p-f(x^p)∈pℤ[x] for any f(x)∈ℤ[x]. This can be proved by induction on the degree of f(x). Suppose the highest degree term of f(x) is Ax^m, with p∤A. Then (Ax^m)^p-Ax^mp∈pℤ[x] because A^p≡A (mod p), because (ℤ/pℤ)^× is cyclic of order p-1 (Fermat's theorem). So if h(x)=f(x)-Axⁿ, then we just have to show h(x)^p-h(x^p)∈pℤ[x]. But that can be assumed true by induction, unless the degree of h(x) is 1. In the latter case, if h(x)=Ax+B, we need (Ax+B)^p-(Ax^p+B)∈pℤ[x]. But all the coefficients in (Ax+B)^p except the first and last contain binomial coefficients divisible by p, and the remaining terms are handled with Fermat's theorem as before.

Finally, then, suppose the opposite of what we want to show, namely that there is a prime p with p∤n and f(ζ^p)≠0. By what we just showed, f(ζ^p) is divisible by p in O_ℚ(ζ). We have f(x)=Π_i∈I(x-ζⁱ) for I={i∈ℤ: 1≤i<n and (i,n)=1}. So f(ζ^p) divides a product of nonzero factors ζ^p-ζⁱ. By a lemma we'll prove in a moment, if J={(i,j): i,j∈ℤ, 0≤i,j<n, i≠j}, Π_(i,j)∈J(ζⁱ-ζ^j) = (-1)^n-1nⁿ. Hence f(ζ^p) divides nⁿ and p|n, contrary to assumption. This contradiction means f(ζ^p)=0, as required. We've now shown f(x) has at least φ(n) roots, hence G(ℚ(μ_n)/ℚ) ≅ (ℤ/nℤ)^×.

Now for the last lemma: We have xⁿ-1=Π_0≤i<n(x-ζⁱ). Equating the constant terms gives (-1)^n-1=Π_0≤i<nζⁱ. And by taking derivatives of both sides, nx^n-1=Σ_0≤i<nΠ_{0≤j<n, j≠i}(x-ζ^j). Substituting x=ζ^k, nζ^k(n-1)=Π_{0≤j<n, j≠k}(ζ^k-ζ^j). Taking products of this for 0≤k<n gives, with the set J as above, Π_(i,j)∈J(ζⁱ-ζ^j) = nⁿ(Π_0≤k<nζ^k)^n-1. But the last product on the right side was evaluated above, so finally we are left with (-1)^n-1nⁿ on the right (since n-1 has the same even/odd parity as its square).

Well, that was a bit of work, wasn't it? But nothing too esoteric, apart from a little Galois theory and some classic number theoretical facts. (Thanks to [1, pp 96-8] for the bulk of the proof.)

Actually, it is possible to do this without the lemma, using the theorem on primes in an arithmetic progression. Suppose f(x) is any polynomial in ℤ[x] such that f(ζ)=0 when ζ is a primitive n^th root of unity. Then for any a∈ℤ with (a,n)=1, since f(x)^p-f(x^p)∈pℤ[x] for any prime p∈ℤ, we have 0=f(ζ)^p≡f(ζ^p) mod pO_ℚ(ζ). But there are infinitely many primes p≡a mod n, and for such p, ζ^p=ζ^a. Consequently, f(ζ^a) is a member of an infinite number of distinct prime ideals, which is possible only if f(ζ^a)=0. Hence f(x) has degree ≥φ(n), which is the crucial fact we found before.

We can now define the cyclotomic polynomial Φ_n(x)=Π_{0<i<n, (i,n)=1}(x-ζⁱ), for any primitive n^th root of unity ζ. From the foregoing, we know a lot about Φ_n(x): its roots are precisely all the primitive n^th roots of unity (in ℂ), its degree is φ(n), it is irreducible (over ℚ), its coefficients are in ℤ, and it is the minimal polynomial of ζ. The notation Φ_n(x) is on account of its relation to the Euler function φ(n).

We also have this factorization of xⁿ-1 in ℤ[x]: xⁿ-1 = Π_d|nΦ_d(x). This holds, since the roots of each Φ_d(x) are precisely the roots of unity in the cyclic group μ_n that have exact order d for each d that divides n. (Each root has one and only one exact order d satisfying d|n.) This relation is occasionally useful, and it yields interesting facts such as Σ_d|nφ(d) = n (by taking degrees of polynomials on both sides).

It turns out that the irreducibility of Φ_n(x) is relatively easy to prove for certain n, namely those that are powers of a single prime. So let p be prime and q=p^r for an integer r≥1. Let f(x)=Φ_q(x). The roots of f(x) are primitive q^th roots of unity, namely ζ∈μ_q such that ζ has order q. There are φ(q) of these and φ(q)=q-q/p=q(1-1/p)=(p-1)p^r-1 (because every p^th element of the set {0,1,...,q-1} is divisible by p). So clearly f(x)=(x^q-1)/(x^q/p-1). Let g(x)=(x^p-1)/(x-1) and h(x)=g(x+1)=((x+1)^p-1)/x=x^p-1+Σ_0<j<p(^p _j)x^j-1, where (^p _j) is a binomial coefficient, which is divisible by p if 0<j<p. Finally, consider the polynomial h(x^q/p)=g(x^q/p+1).

Suppose f(x) splits in ℚ[x]. Then since f(x)=g(x^q/p), the latter splits, and consequently g(x^q/p+1)=h(x^q/p) does too. But h(x^q/p) is what's known as an Eisenstein polynomial, because the leading coefficient is not divisible by p, the constant term is p (not divisible by p²), and all other nonzero coefficients (the binomial coefficients) are divisible by p. However, Eisenstein polynomials are irreducible over ℚ. This contradiction means f(x) must be irreducible over ℚ. QED.

The fact that Φ_q(x) is irreducible if q=p^r, and hence G(ℚ(μ_q)/ℚ)≅(ℤ/qℤ)^×, can be used as the basis for yet another proof of this isomorphism for arbitrary n, by considering prime power divisors q of n, the corresponding extensions ℚ(μ_q)/ℚ, and their Galois groups in building up the full extension ℚ(μ_n)/ℚ and its Galois group. But we won't go into that now.

In the next installment, we'll discuss many more fun facts about cyclotomic fields.

References:

[1] Goldstein, Larry Joel - Analytic Number Theory

Roots of unity and cyclotomic fields

In preparation for many good things that are to come, we need to have a talk about another important class of field extensions of ℚ – the cyclotomic extensions. (Check here for a list of previous articles on algebraic number theory.)

A cyclotomic field in general is a field that is an extension of some base field formed by adjoining all the roots of the polynomial f(x) = xⁿ-1=0 for some specific positive n∈ℤ to the base field. Usually, though not always, this will mean roots that lie in some large field in which f(x) splits completely and that contains ℚ as the base field, such as ℂ, the complex numbers. f(x) is known as the n^th cyclotomic polynomial. Mostly the same theory applies if the base field is a finite algebraic extension of ℚ, but we'll use ℚ as the base field for simplicity.

Since f(1)=0, x-1 is one factor of f(x), and f(x)/(x-1) = x^n-1 + … + x + 1 ∈ ℤ[x], with all coefficients equal to 1. If n is even, -1 is also a root of f(x). However, all other roots of f(x) in ℂ are complex numbers that are not in ℝ. Some of these roots, known as the "primitive" n^th roots of unity – denoted by ζ_n (or just ζ if the context is clear) – have the property that all other roots are a power ζ^k for some integer k, 1≤k<n. So the smallest subfield of ℂ that contains ℚ and all roots of f(x) is ℚ(&zeta), known as the n^th cyclotomic field.

It is possible to express all the roots of f(x) in the form e^2πi/n, where e^z is the complex-valued exponential function, which can be defined in various ways. The most straightforward way is in terms of an infinite series, e^z = Σ_0≤n<∞zⁿ/n!. The exponential function e^z can also be defined as the solution of the differential equation dF(z)/dz = F(z) with initial value F(1)=e, the base of the natural logarithms. So there is the rather unusual circumstance that the roots of an algebraic equation can be expressed as special values of a transcendental function. Mathematicians long hoped that other important examples like this could be found (a problem sometimes referred to as "Kronecker's Jugendtraum", a special case of Hilbert's twelfth problem), but that hope has mostly not been fulfilled.

The most well-known nontrivial root of unity is the fourth root, i=&radic(-1), which satisfies x⁴-1 = (x²+1)(x²-1) = 0.

All complex roots of unity have absolute value 1, i. e. |ζ|=1, since |ζ| is a positive real number such that |ζ|ⁿ=1. The set of all complex numbers with |z|=1 is simply the unit circle in the complex plane, since if z=x+iy, then |z|² = x²+y² = 1. (Note that the linguistic root of words like "circle", "cyclic", and "cyclotomic" is the Greek κύκλος (kuklos).) Since e^iθ = sin(θ) + i⋅cos(θ) for any θ, with θ=2πk/n the real and imaginary parts of a general n^th root of unity ζ=e^2πi⋅k/n are just Re(ζ)=sin(2πk/n) and Im(ζ)=i⋅cos(2πk/n).

There are many reasons why cyclotomic fields are important, and we'll eventually discuss a number of them. One simple reason is that roots of algebraic equations can sometimes be expressed in terms of real-valued roots (such as cube roots, d^1/3 for some d), and roots of unity. See, for example, this article, where we discussed the Galois group of the splitting field of f(x)=x³-2.

The set of all complex n^th roots of unity forms a group under multiplication, denoted by μ_n. This group is cyclic, of order n, generated by any primitive n^th roots of unity. (Any finite subgroup of the multiplicative group of a field is cyclic.) As such, it is isomorphic to the additive group ℤ_n = ℤ/nℤ, the group of integers modulo n. Because of this, many of the group properties of μ_n are just restatements of number theoretic properties of ℤ_n. For instance, each element of order n in μ_n is a generator of the whole group – one of the primitive n^th roots of unity. Since μ_n⊆ℚ(ζ_n), adjoining all of μ_n gives the same extension ℚ(ζ_n) = ℚ(μ_n).

Now, ℤ/nℤ is a ring, and its elements that are not divisors of zero are invertible, i. e they are units of the ring. They form a group under ring multiplication, which in this case is written as (ℤ/nℤ)^× (sometimes U_n for short). An integer m is invertible in ℤ/nℤ if and only if it is prime to n, i. e. (m,n)=1 (because of the Euclidean algorithm). The number of such distinct integers modulo n is a function of n, written φ(n). This number is important enough to have its own symbol, because it was studied by Euler as fundamental to the arithmetic of ℤ/nℤ. Thus φ(n) is also the order of the group (ℤ/nℤ)^×.

Let ζ=e^(2πi)m/n, for 0≤m<n, be an element of μ_n. The correspondence m↔e^(2πi)m/n establishes a group isomorphism between the additive cyclic group ℤ/nℤ and the multiplicative group μ_n. Modulo n, m generates ℤ/nℤ additively if and only if (m,n)=1, which is if and only if the corresponding ζ generates μ_n. So the number of generators of μ_n – which is the number of primitive n^th roots of unity – is the same as the order of (ℤ/nℤ)^×, i. e. φ(n).

One has to be careful, because the multiplicative structure of μ_n parallels the additive structure of ℤ/nℤ, not the multiplicative structure of (ℤ/nℤ)^×. (Because if ζ^M and ζ^N are typical elements of μ_n then ζ^M×ζ^N=ζ^M+N.) Hence even though there are φ(n) generators of μ_n, these generators do not form a group by themselves (a product of generators isn't in general a generator), so the set of them isn't isomorphic to (ℤ/nℤ)^×, even though the latter also has φ(n) elements. Give this a little thought if it seems confusing.

Moreover, the group (ℤ/nℤ)^× is not necessarily cyclic. It is cyclic if n is 1, 2, 4, p^e, or 2p^e for odd prime p, but not otherwise. Confusingly, if the group does happen to be cyclic then integers modulo n that generate the whole group are called "primitive roots" for the integer n. If (ℤ/nℤ)^× happens to be cyclic, then only those m∈(ℤ/nℤ)^× having order φ(n) are "primitive roots" that generate the group, while all m∈(ℤ/nℤ)^× have the property that if ζ∈μ_n has order n and generates the latter group, then so does ζ^m, as we showed above. Got that straight, now? This needs to be understood when working in detail with roots of unity.

Another reason for the importance of cyclotomic fields is that the Galois group of the extension [ℚ(ζ_n):ℚ] is especially easy to describe. Indeed, it is isomorphic to the group of order φ(n) we've just discussed: (ℤ/nℤ)^×. There's a little work in proving this isomorphism, but let's first note what it implies. Let G=G(ℚ(ζ)/ℚ) be the Galois group. It is an abelian group of order φ(n) since it's isomorphic to (ℤ/nℤ)^×. Further, any subgroup of G′ of G is abelian and by Galois theory determines an abelian extension (i. e., an extension that is Galois with an abelian Galois group) of ℚ as the fixed field of G′. Conversely, it can be shown (not easily) that every abelian extension of ℚ is contained in some cyclotomic field. (This is the Kronecker-Weber theorem.)

Half of the proof of the isomorphism is easy. Pick one generator ζ of μ_n, i. e. a primitive n^th root of unity. We'll see that it doesn't matter which of the φ(n) possibilities we use. Suppose σ∈G is an automorphism in the Galois group. Since σ is an automorphism and ζ generates the field extension, all we need to know is how σ acts on ζ. Since σ is an automorphism, σ(ζ) has the same order as ζ, so it's also a primitive n^th root of unity. Therefore &sigma(ζ) = ζ^m for some m, 1≤m<n. As we saw above, m is uniquely determined and has to be a unit of ℤ/nℤ, with (m,n)=1, in order for ζ^m to be, like ζ, a generator of the cyclic multiplicative group μ_n. Hence m∈(ℤ/nℤ)^×. Call this map from G to (ℤ/nℤ)^× j, so that σ(ζ)=ζ^j(σ). To see that it's a group homomorphism, suppose σ₁,σ₂∈G, with j(σ₁)=r, j(σ₂)=s. Then σ₂(σ₁(ζ)) = σ₂(ζ^r) = (ζ^s)^r = ζ^sr, hence j(σ₂σ₁) = j(σ₂)j(σ₁). j is clearly injective since j(σ)=1 means σ(ζ)=ζ, so σ is the identity element of G. Finally, to see that j doesn't depend on the choice of primitive n^th root of unity, suppose ζ^m with m∈(ℤ/nℤ)^× is another one. Then σ(ζ^m) = σ(ζ)^m = (ζ^j(σ))^m = (ζ^m)^j(σ).

Thus G is isomorphic to a subgroup of (ℤ/nℤ)^×. That's enough to show G is abelian, so the extension ℚ(ζ)/ℚ is abelian. To complete the proof of an isomorphism G≅(ℤ/nℤ)^× we would need to show that the injective homomorphism j is also surjective, i. e. every m∈(ℤ/nℤ)^× determines some σ∈G such that m=j(σ). We can certainly define a function from ℚ(ζ) to ℚ(ζ) by σ(ζ)=ζ^m for a generator ζ of the field ℚ(ζ). One might naively think that's enough, but the problem is that one has to show that σ is a field automorphism of ℚ(ζ).

The map σ defined that way certainly permutes the n^th roots of unity in μ_n, the roots of the polynomial f(x)=xⁿ-1. However, not all permutations of elements of μ_n, of which there are n!, yield automorphisms of ℚ(ζ). The problem here is that if z(x) is the minimal polynomial of some ζ, i. e. the irreducible polynomial of smallest degree in ℤ[x] such that z(ζ)=0, then by Galois theory the order |G| of the Galois group G is the degree of the field extension, which is the degree of z(x). Since G is isomorphic to a subgroup of the group (ℤ/nℤ)^×, and the latter has order φ(n), all we know is that |G| divides φ(n). It could be that other primitive n^th roots of unity have minimal polynomials in ℤ[x] that are not the same as z(x), though they have the same degree |G|. For σ to be an automorphism, σ(ζ) needs to have the same minimal polynomial as ζ, and we don't know that immediately from the relation σ(ζ)=ζ^m.

We will defer discussion of the rest of the proof that G(ℚ(μ_n)/ℚ)≅(ℤ/nℤ)^× for the next installment, since some new and important concepts will be introduced.

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 O_K, 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)=pO_ℚ(√d) in O_ℚ(√d). If O_ℚ(√d) is a PID, then there is a simple approach to investigate how p splits. If p splits then (p)=P₁⋅P₂, where P_i=(α_i), 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 α₂=α₁*=a-b√d. Hence (p)=(α₁)⋅(α₁*)= (α₁α₁*)= (a²-db²).

Taking norms (to eliminate possible units ε∈O_ℚ(√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 I_j, for j in some index set, j∈J. Suppose that the ideals are relatively prime in pairs. In general that means that I_i+I_j=R if i≠j, and further, the product of ideals, I_i⋅I_j, is I_i∩I_j when i≠j. If R is Dedekind, then each ideal has a unique factorization into prime ideals, and they are relatively prime if I_i and I_j have no prime ideal factors in common when i≠j. Let I be the product of all I_j for j∈J, which is also the intersection of all I_j for j∈J, since the ideals are coprime in pairs.

The direct product of rings R_i for 1≤i≤k is defined to be the set of all ordered k-tuples (r₁, ... ,r_k), for r_i∈R_i, with ring structure given by element-wise addition and multiplication. The direct product is written as R₁×...×R_k, or &Pi_1≤i≤kR_i.

Given all that, the CRT says the quotient ring R/I is isomorphic to the direct product of quotient rings &Pi_1≤i≤k(R/I_i) via the ring homomorphism f(x)=(x+I₁, ... ,x+I_k) 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 I_i. (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 O_K of a finite extension K/ℚ, and I_i=P_i, 1≤i≤g, be the set of all distinct prime ideals of O_K that divide (p)=pO_K for some prime p∈ℤ. Then (p)=P₁^e₁ ⋅⋅⋅ P_g^e_g, where e_i are the ramification indices of each prime factor of (p). An application of CRT then shows that O_K/(p) ≅ Π_1≤i≤g(O_K/P_i^e_i). Recall that for each i, O_K/P_i is isomorphic to the finite field F_qi, where q_i=p^f_i for some f_i, known as the degree of inertia of P_i. (This field is the extension of degree f_i of F_p=ℤ/pℤ.) Further, Σ_1≤i≤ge_if_i=[K:ℚ], the degree of the extension. Check here if you need to review these facts. Specifying how (p) splits in O_K amounts to determination of the P_i and the numbers e_i, f_i, and g.

The second situation where we apply CRT involves the ring of polynomials in one variable over the finite field F_p=ℤ/pℤ, denoted by F_p[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 F_p[x]. f(x) will not, in general, be irreducible in F_p[x], so it will be a product of powers of irreducible factors: Π_1≤i≤g(f_i(x)^e_i), where f_i(x)∈F_p[x]. Each quotient ring F_p[x]/(f_i(x)) is a finite field that is an extension of F_p of some degree f_i. In general, e_i, f_i, and g will be different, of course, from the same numbers in the preceding paragraph. But the CRT gives us an isomorphism F_p[x]/(f(x)) ≅ &Pi_1≤i≤g(F_p[x]/(f_i(x)^e_i)).

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 e_i, f_i, and g will be the same for both applications of the CRT. Consequently, we will have O_K/(p) ≅ F_p[x]/(f(x)), because for corresponding factors of the direct product of rings, O_K/P_i^e_i ≅ F_p[x]/(f_i(x)^e_i). As it happens, most primes don't ramify for given choices of K and f(x), so that things are even simpler, since all e_i=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, O_K, 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 [O_K:ℤ[θ]]. (ℤ[θ] is just all polynomials in θ with coefficients in ℤ.) If p∈ℤ is any prime that does not divide [O_K: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 O_K=ℤ[√d]. If d≡1 (mod 4), then the index [O_K:ℤ[√d]]=2, and there's a possible problem only for p=2. However, we still have O_K/(p) ≅ F_p[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 O_K 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.