Math 594. Galois group of cyclotomic fields over Q

1. Preparatory remarks

Fix n ≥ 1 an integer. Let K_{n}/Q be a splitting field of X^{n} − 1, so the group of nth roots of unity in K has order n (as Q has characteristic not dividing n) and is cyclic (as is any finite subgroup of the multiplicative group of a field, by an old homework). As was discussed in class, we have a natural injection of groups

Gal(K_{n}/Q) ↪ (Z/n)^{×},

and we proved in lecture that in the case when n is a prime power, this is an isomorphism. This was done by using Eisenstein’s criterion to prove that the polynomial Φ_{p^{e}}(X) = (X^{p^{e}} − 1)/(X^{p^{e−1}} − 1) = Φ_{p}(X^{p^{e−1}}) for e ≥ 1 is irreducible over Q. Concretely, the roots of Φ_{p^{e}} (in a splitting field over Q) are clearly the full set of primitive p^{e}th roots of unity. Hence, the statement that the extension K_{p^{e}}/Q is “as big as is possible” really says that all primitive p^{e}th roots of unity are Galois conjugates of each other over Q (i.e., have the same minimal polynomial over Q). This is a very special property of Q. For example, over R it is clear for odd n that primitive nth roots of unity break up into many Gal(C/R)-conjugate pairs. Likewise, over F_{p} (as well as many other interesting fields) one cannot say that “all primitive nth roots of unity are created equal”: they might have different minimal polynomials over the ground field. For example, if n|(p − 1) then F_{p}^{×} contains a full set of nth roots unity, so these are most definitely not “created equal”. As one instance, F_{7} contains 6 distinct 6th roots of unity; the primitive cube roots of unity in F_{7} are 2 and 4 (i.e., X^{2} + X + 1 = (X − 2)(X − 4) is a factorization of Φ_{3} in F_{7}[X]).

Back in the setup over Q, to handle n which may not be a prime power we cannot hope to use Eisenstein’s criterion (we lack explicit polynomials with which to work), so we shall require another method, due originally to Gauss, which proceeds by a totally different approach (and in particular proves the irreducibility of Φ_{p^{e}} in Q[X] by methods unrelated to Eisenstein’s criterion). The key to Gauss’ idea is to exploit the magical properties of the pth power map in F_{p}[X] for an auxiliary prime p. Since the number of primitive nth roots of unity in K_{n} is exactly |(Z/n)^{×}| (as shown on HW 11) and a Galois conjugate of a primitive nth root of unity is again a primitive nth root of unity (why?), to say these are all Galois conjugates over Q is to say that one of them (and hence all of them) has minimal polynomial over Q of degree |(Z/n)^{×}|. But since any such primitive nth root of unity ζ in K_{n} actually generates K_{n} over Q, we see that [K_{n} : Q] is equal to the degree of the minimal polynomial of ζ over Q. Since [K_{n} : Q] is also equal to the size of Gal(K_{n}/Q), we conclude that to say Gal(K_{n}/Q) ↪ (Z/n)^{×} is an isomorphism is equivalent to saying that all primitive nth roots of unity in K_{n} are Galois conjugate over Q (i.e., have the same minimal polynomial). This is what allows one to say that “all primitive nth roots of unity are created equal over Q”; there is no algebraic way to distinguish them from each other once it is seen that Gal(K_{n}/Q) acts transitively on this set (i.e., they’re all roots of the same minimal polynomial over Q). The examples in the preceding paragraph show that this fails miserably over other ground fields (such as R and finite fields).

As an important special case of the role of transitive Galois group actions to eliminate the possibility of algebraic distinguishability, when n = 4, one cannot say in any intrinsic sense that there is a “prefered” solution to T^{2} + 1 = 0 in C (view C as simply an abstract choice of splitting field of T^{2} + 1 over R, which we will later prove to be algebraically closed; it can be shown that there is only one absolute value on C extending the usual one on R, and this is how we give C a topological structure of normed field). Since the topology of C determines the subfield R as the closure of Q, this subfield is intrinsic to C with its topology, and Gal(C/R) permutes the two solutions to T^{2} + 1 = 0. No canonical mathematical construction should depend on a choice of solution to this equation (though constructions may use such a choice if the end result is independent of the choice, such as in the statement of the residue formula in complex analysis).

2. Main result

With the motivation and background set up, it is time to prove the main result. We begin with some notation. Fix n ≥ 1 and K_{n}/Q a splitting field of X^{n} − 1. Define

Φ_{n}(X) = ∏(X − ζ) ∈ K_{n}[X],

where ζ runs over all primitive nth roots of unity in K_{n} (i.e., all generators of the intrinsic order n cyclic group of solutions to T^{n} − 1 = 0 in K_{n}). It is clear from the intrinsic nature of primitive nth roots of unity that the action of Gal(K_{n}/Q) permutes these around. Hence, even without knowing if Gal(K_{n}/Q) is “big”, it is clear that the monic polynomial Φ_{n}(X) is invariant under the action of Gal(K_{n}/Q) (which just shuffles its linear factors under the natural action on K_{n}[X]). Hence, by Galois theory the coefficients of Φ_{n} must lie in Q! Its degree is clearly |(Z/n)^{×}|. The main aim is therefore to prove:

Theorem 2.1. (Gauss) The polynomial Φ_{n} ∈ Q[X] is irreducible.

Proof. By construction, Φ_{n} ∈ Q[X] is monic, and over the extension field K_{n} we see that Φ_{n} divides X^{n} −1. Since formation of gcd commutes with extension of the ground field, the divisibility Φ_{n}|(X^{n} − 1) in Q[X] must hold because it is true in K_{n}[X] (i.e., Φ_{n} serves as a gcd of Φ_{n} and X^{n} − 1). By Gauss’ Lemma, since X^{n} − 1 ∈ Q[X] has integral coefficients, any monic factorization in Q[X] is necessarily in Z[X]. That is, if we write X^{n} − 1 = Φ_{n}h with h ∈ Q[X], then since h is visibly monic (as X^{n} − 1 and Φ_{n} are monic) it follows that both Φ_{n} and h must lie in Z[X].

Now suppose that Φ_{n} is not irreducible in Q[X], so there is a factorization Φ_{n} = fg in Q[X] with monic f and g of positive degree. We may also suppose f is irreducible. By Gauss’ Lemma applied to the monic factorization fg = Φ_{n} with Φ_{n} ∈ Z[X], we must have f, g ∈ Z[X]. We seek to derive a contradiction. In K_{n}[X] we have the monic factorization Φ_{n} = ∏(X − ζ) where the product runs over all primitive nth roots of unity in K_{n}. Since f and g both have positive degree, there must exist distinct primitive nth roots of unity ζ and ζ′ in K_{n} such that X − ζ is a factor of f and X − ζ′ is a factor of g; that is, f(ζ) = 0 and g(ζ′) = 0 in K_{n}.

We can write ζ′ = ζ^{r} for some unique r ∈ (Z/n)^{×} since ζ and ζ′ are both primitive nth roots of unity. Since ζ ≠ ζ′, we must have r ≠ 1. Choose a positive integer representing this residue class r, and again denote it by r, so r > 1 and gcd(r, n) = 1. Consider the prime factorization r = ∏ p_{j} with primes p_{j} not necessarily pairwise distinct. To go from ζ to ζ′ = ζ^{r} we successively raise to exponents p_{1}, then p_{2}, etc. Since f(ζ) = 0 and g(ζ′) = 0, so f(ζ′) ≠ 0 and g(ζ) ≠ 0 (as the factorization Φ_{n} = fg and separability of Φ_{n} forces f and g to have no common roots), there must exist a least j for which ζ^{p_{1}···p_{j−1}} is a root of f and its p_{j}th power is a root of g. In other words, we can find a primitive nth root of unity ζ_{0} and a prime p not dividing n such that f(ζ_{0}) = 0 and g(ζ_{0}^{p}) = 0. From this fact we shall deduce a contradiction.

Since f is irreducible over Q, it must be the minimal polynomial of ζ_{0}. But g(ζ_{0}^{p}) = 0, so g(X^{p}) ∈ Q[X] has ζ_{0} as a root. Thus, f|g(X^{p}) is Q[X]. We can therefore write g(X^{p}) = fq in Q[X], with q necessarily monic (as g(X^{p}) and f are monic). Since g(X^{p}) has coefficients in Z, Gauss’ Lemma once again ensures that q ∈ Z[X]. Thus, the identity g(X^{p}) = fq takes place in Z[X]. Now reduce mod p! In F_{p}[X], we get

\overline{f}\overline{q} = \overline{g}(X^{p}) = \overline{g}(X)^{p},

the final equality using the fact that a^{p} = a for all a ∈ F_{p}. The monicity of f and g, coupled with the condition that both have positive degree, ensures that \overline{f}, \overline{g} ∈ F_{p}[X] have positive degree (though may well be reducible). From the divisibility relation \overline{f}|\overline{g}^{p} we conclude that \overline{f} and \overline{g} must have a non-trivial irreducible factor in common. Hence, the product \overline{f}\overline{g} has a non-trivial irreducible factor appearing with multiplicity more than 1. But in Q[X] we have fg = Φ_{n}|(X^{n} −1), so by Gauss’ Lemma we even get fg|(X^{n} −1) in Z[X], so we may reduce mod p to get \overline{f}\overline{g}|(X^{n} −1) in F_{p}[X]. It follows that X^{n} −1 ∈ F_{p}[X] has a non-trivial square factor and hence is not separable. But this is absurd, since p doesn’t divide n and hence the derivative test ensures that X^{n} − 1 ∈ F_{p}[X] is separable! Contradiction. ■

Note that although the Φ_{n}’s were constructed abstractly, they’re easy to compute explicitly in Z[X]. In fact, since each nth root of unity is a primitive dth root of unity for a unique d|n, we obviously have X^{n} − 1 = ∏_{d|n} Φ_{d}(X). This allows a computer to recursively compute Φ_{n}(X) by long division using X^{n} − 1 and Φ_{d}(X)’s for proper divisors d of n; the abstract theory is needed to prove that when Φ_{d} is known for all d|n with d < n, then ∏_{d|n,d<n} Φ_{d} actually divides X^{n} − 1 in Q[X]; by thinking about the root structure in a splitting field, such a divisibility claim is obvious (without thinking about root structure in a big extension, such issues become quite tedious to handle).