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 nth 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) = xn-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” nth roots of unity – denoted by ζn (or just ζ if the context is clear) – have the property that all other roots are a power ζk for some integer k, 1≤k<n. So the smallest subfield of ℂ that contains ℚ and all roots of f(x) is ℚ(ζ), known as the nth cyclotomic field.
It is possible to express all the roots of f(x) in the form e2πi/n, where ez is the complex-valued exponential function, which can be defined in various ways. The most straightforward way is in terms of an infinite series, ez = Σ0≤n<∞zn/n!. The exponential function ez 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 eiθ = sin(θ) + i⋅cos(θ) for any θ, with θ=2πk/n the real and imaginary parts of a general nth root of unity ζ=e2πi⋅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, d1/3 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 nth roots of unity forms a group under multiplication, denoted by μn. This group is cyclic, of order n, generated by any primitive nth roots of unity. (Any finite subgroup of the multiplicative group of a field is cyclic.) As such, it is isomorphic to the additive group ℤn = ℤ/nℤ, the group of integers modulo n. Because of this, many of the group properties of μn are just restatements of number theoretic properties of ℤn. For instance, each element of order n in μn is a generator of the whole group – one of the primitive nth roots of unity. Since μn⊆ℚ(ζn), adjoining all of μn gives the same extension ℚ(ζn) = ℚ(μn).
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 Un 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 μn. The correspondence m↔e(2πi)m/n establishes a group isomorphism between the additive cyclic group ℤ/nℤ and the multiplicative group μn. Modulo n, m generates ℤ/nℤ additively if and only if (m,n)=1, which is if and only if the corresponding ζ generates μn. So the number of generators of μn – which is the number of primitive nth 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 μn parallels the additive structure of ℤ/nℤ, not the multiplicative structure of (ℤ/nℤ)×. (Because if ζM and ζN are typical elements of μn then ζM×ζN=ζM+N.) Hence even though there are φ(n) generators of μn, 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, pe, or 2pe 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∈μn has order n and generates the latter group, then so does ζm, 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(ζn):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 μn, i. e. a primitive nth 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 nth root of unity. Therefore &sigma(ζ) = ζm 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 ζm to be, like ζ, a generator of the cyclic multiplicative group μn. Hence m∈(ℤ/nℤ)×. Call this map from G to (ℤ/nℤ)× j, so that σ(ζ)=ζj(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(ζr) = (ζs)r = ζsr, 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 nth root of unity, suppose ζm with m∈(ℤ/nℤ)× is another one. Then σ(ζm) = σ(ζ)m = (ζj(σ))m = (ζm)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)=ζm 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 nth roots of unity in μn, the roots of the polynomial f(x)=xn-1. However, not all permutations of elements of μn, 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 nth 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)=ζm.
We will defer discussion of the rest of the proof that G(ℚ(μn)/ℚ)≅(ℤ/nℤ)× for the next installment, since some new and important concepts will be introduced.