Math 594. Radical towers and roots of unity

Some time ago, there was a proposed quals problem which asked the students to prove that if f ∈ Q[T] is irreducible with odd degree > 1 and f splits over R with splitting field K ⊆ R having Gal(K/Q) solvable, then any radical tower F/Q containing K must contain a root of unity of order > 2 and hence F cannot be embedded into R. This turned out to be a bit too hard to be a quals problem. The crucial aspect of this problem is not that it takes place with ground field Q, or even that the fields have characteristic zero. Both of these conditions can be eliminated. What really matters is that 2 is a very small prime number (as the proof will explain); i.e., it is the oddness hypothesis that is the essential one. Here is a general theorem which we shall prove (after explaining how it relates back to the above proposed quals problem).

Theorem 0.1. Let E be a field, and let K/E be a finite Galois extension. Assume that [K : E] is not a power of 2 (in particular, [K : E] > 1). If F is a radical tower over E into which K admits an E-embedding, then F contains a root of unity of odd prime order. In particular, if E is a subfield of R, then F admits no E-embedding into R.

An important special case is to consider a separable irreducible polynomial f ∈ E[T] of odd degree > 1 and to take K/E to be a splitting field of f. Note that [K : E] is divisible by the odd integer deg f > 1 since f is irreducible, so [K : E] is not a power of 2. The hypothesis that K admits an E-embedding into a radical tower says that “f can be solved in radicals” over E, and in characteristic zero the existence of such a radical tower is equivalent to Gal(K/E) being solvable. The significance of the theorem is that when E ⊆ R, such a radical tower F/E must contain a root of unity of odd prime order and hence F cannot be found inside of R, even if we drop any desire for the radical tower to be Galois over E.

The case E = Q and cubic f is the classical “casus irreducibilis”, where an irreducible cubic over Q has 3 real roots but any attempt to describe them in terms of radicals seems (by experience, and in fact by necessity due to the above theorem!) to force the use of non-real complex numbers. This was most disturbing to the mathematicians who first worked with the cubic formula, at a time when complex numbers were still regarded with great suspicion (and so to seem to “need” the use of complex numbers to describe solutions to a cubic with three real roots was quite alarming).

Proof. Since any radical extension generated by a dth root (with d > 1) can be expressed as a successive extension of adjoining prime roots (as we run over a prime factorization of d), we can chop up the radical tower F/E into steps

E = E_{0} ⊆ E_{1} ⊆ ⋯ ⊆ E_{n} = F

where E_{i+1} = E_{i}(a_{i}) with a_{i}^{p_{i}} ∈ E_{i}, with p_{i} prime. We may certainly assume [E_{i+1} : E_{i}] > 1 for all i without loss of generality. For each i, I claim that either [E_{i+1} : E_{i}] = p_{i} or else E_{i+1} contains a non-trivial p_{i}th root of unity. For this claim, it suffices to show more generally that if p is prime and K′/K is an extension of fields with [K′ : K] > 1 and K′ = K(a) where a^{p} = b lies in K, then either [K′ : K] = p or else K′ contains a non-trivial pth root of unity. Consider the polynomial T^{p} − b in K[T]. If this is irreducible, then clearly K′ is K-isomorphic to K[T]/(T^{p} − b) and hence [K′ : K] = p. Thus, we may suppose T^{p} − b is reducible. In this case, [K′ : K] < p, so gcd([K′ : K], p) = 1, and applying the multiplicative norm map N_{K′/K} : K′ → K to the equation b = a^{p} gives b^{[K′:K]} = N_{K′/K}(a)^{p}. But [K′ : K] is relatively prime to p, so up + v[K′ : K] = 1 with u, v ∈ Z, and hence b = b^{up+v[K′:K]} = c^{p} with c = b^{u} · N_{K′/K}(a)^{v} ∈ K. In this case, we have a^{p} = b = c^{p}, so ζ = a/c is a pth root of unity. But a is not in K since K′ = K(a) with [K′ : K] > 1, so ζ ≠ 1 and hence ζ is a non-trivial pth root of unity in K′ (in particular, the characteristic is not equal to p).

Of course, when p_{i} = 2 then 1 < [E_{i+1} : E_{i}] ≤ 2 forces [E_{i+1} : E_{i}] = 2 = p_{i}. However, for odd p_{i} we have 1 < [E_{i+1} : E_{i}] ≤ p_{i} and the possibility 1 < [E_{i+1} : E_{i}] < p_{i} may happen. But in such cases, the preceding shows that E_{i+1} contains a root of unity of odd prime order p_{i}, so F contains such a root of unity too. This would provide what we want to prove, so we may assume that [E_{i+1} : E_{i}] = p_{i} for all i. Note that this step makes crucial use of a special property of the prime 2: there are no integers strictly between 1 and 2. This is the source of the special role of 2 in the proof.

Fix an E-embedding of K into F. Let K_{i} = K ∩ E_{i} for each i, so K/K_{i} is a Galois extension and [K_{i} : E] divides [E_{i} : E] (since E ⊆ K_{i} ⊆ E_{i}). Since K/K_{i} is finite Galois, the compositum L_{i} =^{def} KE_{i} inside of F is not only finite Galois over E_{i} but the following lemma ensures that [L_{i} : E_{i}] = [K : K_{i}].

Lemma 0.2. Let F/k be an extension of fields and let K/k be a subextension which is finite Galois. Let F_{0}/k be another intermediate extension with K ∩ F_{0} = k. Then the finite Galois extension KF_{0}/F_{0} has degree [KF_{0} : F_{0}] equal to [K : k], and the natural injection Gal(KF_{0}/F_{0}) ↪ Gal(K/k) is an isomorphism.

The Galois group aspect is not relevant for our purposes.

Proof. Once the degree result is proven, the isomorphism on Galois groups is immediate by counting. To compute the degrees, let y be a primitive element for K/k, say with minimal polynomial f ∈ k[T]. Note that K/k is a splitting field of f over k. Since KF_{0}/F_{0} is generated by a root of f, it is necessary and sufficient to prove that f ∈ F_{0}[T] is irreducible. Suppose f = gh is a monic factorization of f over F_{0}; it suffices to show that this factorization is trivial. Since f splits over K and hence over F, when we consider the factorization f = gh in F [T] we see that g and h split over F and hence their coefficients may be expressed as Z-polynomials in the roots of f (recall that g and h were chosen to be monic). But the roots of f in F lie in K, so the coefficients of g and h in F lies in K. Hence, g and h as elements in F [T] lies in (K ∩ F_{0})[T] = k[T], so our factorization of f takes place in k[T]. But f is irreducible in k[T], so our factorization is indeed trivial. ■

Now consider the intersection L_{i} ∩ E_{i+1} inside of F for some 0 ≤ i < n. This is an intermediate field between E_{i+1} and E_{i} (since E_{i} ⊆ KE_{i} = L_{i}), so since [E_{i+1} : E_{i}] = p_{i} is prime we conclude that L_{i} ∩ E_{i+1} is equal to either E_{i+1} or E_{i}. In the first case, the finite Galois extension L_{i}/E_{i} contains E_{i+1}, so E_{i+1}/E_{i} is separable and L_{i} contains a Galois closure of E_{i+1} over E_{i}. But E_{i+1} is generated over E_{i} by a root of an irreducible polynomial T^{p_{i}} − b_{i} for some b_{i} in E_{i}, and hence the Galois closure contains a non-trivial p_{i}th root of unity. Thus, L_{i} would contain a non-trivial p_{i}th root of unity, so F would too. If p_{i} is odd, we’d be done. Thus, for the purposes of what we are trying to prove, there is no loss of generality in assuming that p_{i} = 2 whenever L_{i} ∩ E_{i+1} = E_{i+1}. In particular, when p_{i} is odd then L_{i} ∩ E_{i+1} = E_{i}. We will now proceed to deduce a contradiction from this situation.

When p_{i} is odd, since L_{i}∩E_{i+1} = E_{i} and L_{i}/E_{i} = KE_{i}/E_{i} is finite Galois, by Lemma 0.2 we conclude that L_{i}E_{i+1} is finite Galois over E_{i+1} with degree equal to [L_{i} : E_{i}]. But L_{i}E_{i+1} = KE_{i}E_{i+1} = KE_{i+1} = L_{i+1}, so we see that when p_{i} is odd then [L_{i+1} : E_{i+1}] = [L_{i} : E_{i}]. On the other hand, if p_{i} = 2 then either L_{i} ∩ E_{i+1} = E_{i}, in which case the same application of Lemma 0.2 shows [L_{i+1} : E_{i+1}] = [L_{i} : E_{i}], or else L_{i} ∩ E_{i+1} = E_{i+1}. But in this latter case we have E_{i+1} ⊆ L_{i} and [E_{i+1} : E_{i}] = p_{i} = 2, so in fact [L_{i} : E_{i+1}] = [L_{i} : E_{i}]/2. But such situations also must have L_{i+1} = L_{i} since L_{i+1} = L_{i}E_{i+1} and E_{i+1} ⊆ L_{i+1}, so in fact [L_{i+1} : E_{i+1}] = [L_{i} : E_{i}]/2. To summarize, for every i < n we see that [L_{i+1} : E_{i+1}] is equal to either [L_{i} : E_{i}] or [L_{i} : E_{i}]/2. In particular, the odd parts of [L_{i} : E_{i}] and [L_{i+1} : E_{i+1}] are the same for all i < n. But [L_{n} : E_{n}] = 1 since E_{n} = F, so it follows that [L_{i} : E_{i}] is a power of 2 for all i. We will now show that there is actually an i for which [L_{i} : E_{i}] has a non-trivial odd factor, and this contradiction will complete the proof.

Since [K : E] has a non-trivial odd factor by hypothesis and [K : E]|[F : E], we conclude that [F : E] = ∏ p_{i} has a non-trivial odd factor. Thus, some p_{i} must be odd! Hence, we may choose a least i_{0} (possibly i_{0} = 0) for which p_{i_{0}} is odd. We then have p_{i} = 2 for all i < i_{0}, so [E_{i_{0}} : E] = ∏_{i<i_{0}} p_{i} is a (possibly trivial) power of 2. Since [K_{i_{0}} : E]|[E_{i_{0}} : E], as K_{i} ⊆ E_{i} for any i, we conclude that [K_{i_{0}} : E] is also a power of 2. But [K : E] has an odd factor > 1 by hypothesis, so the quotient [K : K_{i_{0}}] = [K : E]/[K_{i_{0}} : E] by a power of 2 also has (the same) odd factor > 1. Since K_{i_{0}} = K ∩ E_{i_{0}}, by Lemma 0.2 we have [K : K_{i_{0}}] = [KE_{i_{0}} : E_{i_{0}}] = [L_{i_{0}} : E_{i_{0}}]. It follows that [L_{i_{0}} : E_{i_{0}}] has a non-trivial odd factor, and we saw above that this yields a contradiction. ■