Solvable Number Field Extensions of Bounded Root Discriminant

Let $K$ be a number field and $d_K$ the absolute value of the discrimant of $K/\mathbb{Q}$. We consider the root discriminant $d_L^{\frac{1}{[L:\mathbb{Q}]}}$ of extensions $L/K$. We show that for any $N>0$ and any positive integer n, the set of length n solvable extensions of $K$ with root discriminant less than $N$ is finite. The result is motivated by the study of class field towers.


Introduction
Let K be a number field with Hilbert class field H K . Set K = K (0) and define K (i) := H K (i−1) for all i ≥ 1. Let K (∞) = ∪ i K (i) . We say F has finite class field tower if K (∞) is a finite extension of Q, i.e., if the K (n) stabilize for large n. If K (∞) is infinite over Q, K has infinite class field tower. The main theorem on class field towers is that of Golod and Shafarevich, who proved in 1964 that number fields with infinite class field tower exist [7].
We define the root discriminant of K to be rd(K) := d Given a tower of number fields L/K/F , we have the following equality of ideals of F : where d L/F denotes the relative discriminant. (With this notation, d F denotes the absolute value of d F/Q ). It follows from (1) that if L is an extension of K, then rd(K) ≤ rd(L), with equality if and only if d L/K = 1, i.e., L/K is unramified at all finite primes. Thus if K has infinite class field tower, then all fields K (i) have the same root discriminant. In particular, the set Z N,K := {L : L/Q finite, L ⊇ K, and rd(L) ≤ N} is infinite for N ≥ rd(K). From discriminant bounds originally due to Odlyzko and others, there exist only finitely many number fields K with rd(K) ≤ Ω := 4πe γ ≈ 22, where γ is Euler's constant, and this number can be improved to 2Ω if we assume the Generalized Riemann Hypothesis [2].
It follows that a number field with infinite class field tower must have root discriminant larger than Ω. Martinet has constructed a number field with infinite class field tower and root discriminant ≈ 92.4. Using tamely ramified class field towers, Hajir and Maire gave an example of a number field K with Z 82.2,K infinite [2].
We will not be concerned with specific values of root discriminants but rather with the following question: fix a number field K and an arbitrary (large) real number N > 0. How can infinite subsets of Z N,K arise? For example, fix a positive integer n. If N were large enough, could the following set be infinite: {L : L/Q finite, L/K solvable length n, and rd(L) ≤ N}?
The answer to the question is no, and this is the main theorem of the paper. • Taking n = 1 gives finiteness for abelian extensions. The general solvable case follows by induction from the n=1 case, and the proof of the n=1 case occupies the bulk of the paper. We set Y N,K := Y 1,N,K .
• Rather than considering the root discriminant of extensions L of K, we could equivalently consider the quantity (N K/Q d L/K ) 1/[L:K] . This is evident by (1).
• Odlyzko mentions in [6] that Theorem 1 is known to be true in the case K = Q, but he does not provide a proof.
From here on, all field extensions are assumed to be finite unless otherwise stated.

Discriminants and Ramification Groups
Let L/K be a Galois extension of local fields, with K a finite extension of Q p . In [9], Serre gives the following formula for the relative different D L/K in terms of the ramification groups where v L denotes the normalized P-adic valuation of a fractional ideal of O L , P the unique prime ideal of O L . If L/K is now a Galois extension of global fields with P a prime of L lying above a prime p of K, then where G i are the ramification groups of P/p, g is the number of primes of L above p, and f is the residue degree of P/p. So for a Galois extension K/Q, we obtain Note that if we define the root discriminant of a finite extension K p of Q p to be , then p vp(rd(K)) = rd(K p ).

Proof of Theorem 1
Fix a number field K and a real number N > 0. Our first goal is to show that the set X N,K := {L : L/K abelian, L/Q Galois, and rd(L) ≤ N} is finite in the case when K/Q is Galois.
If E/F is a Galois extension of number fields ramified at a prime p of F with e = e E (p), then by (2), p (e−1)f g |d L/K . It follows that if L is a number field Galois over Q with rd(L) ≤ N, then L/Q can not ramify at any rational prime p with p > √ N . Let S be the union of the real places of K and the set of primes of K lying above the rational primes p with p ≤ √ N . Suppose that X N,K is infinite. Then there exists an increasing sequence of natural numbers n l such that [L n l : K] = n l and L n l ∈ X N,K . For a fixed positive integer m, the maximal abelian extesion of K of exponent m that is unramified outside S is finite (see, e.g., [10]). Thus we may assume that L n l /K is cyclic for each l. We deal with the following two cases separately: Case I: For every p ∈ S, lim sup e Ln l (p) < ∞, (the lim sup being indexed by l).
To ease notation, write L l for L n l . Write S = {p j }. Let f l = f(L l /K) be the conductor of L l /K. As l gets arbitrarily large, there exists j such that the power a j of p j dividing f l gets sufficiently large. This is because L l /K can only be ramified at the primes in S, so its conductor is divisible by only these primes, which means L l is contained in the ray class field of K modulo the product of the infinite real places of K and a power m l of the product of the finite primes ramifying in L l /K. As [L l : K] increases, the minimal such m l increases, which by definition of the ray class field implies a j increases for some j.
Set p = p j and a = a j . Let P l be a prime of L l lying above p. WriteL l for L l P l , and to keep notation consistent writeK for K p . Define the group of n-units U n ofK to be the group of units of OK that are congruent to 1 (mod p n ). The p-contribution to f l is the conductor of the abelian extension of local fieldsL l /K, which we denote byf l . Let θ be the local reciprocity mapK * → G(L l /K). Thenf l = p b l , where b l is by definition the smallest integer n satisfying θ(U n ) = 1. As θ maps U n onto the n th upper ramification group l G n of l G := G(L l /K), we see that l G b l −1 is non-trivial (in fact, l G b l −1 is the last non-trivial ramification group in the sense that l G b l −1+ǫ = 1 for any ǫ > 0). For a discussion of the upper ramification groups, see [9].
By the previous two paragraphs, l → ∞ implies a → ∞, which implies b l − 1 → ∞. As b l − 1 → ∞, the largest integral index c l for which the c l th lower ramification group l G c l of l G is non-trival tends to ∞. Now, let p be the rational prime over which p lies. We have l G ≤ G(L l /Q p ) := l Γ and l G n ≤ l Γ n for any n. (It follows from the definition of the ramification groups that l Γ e(n+1)−1 = l G n , where e = e(p/p)). Therefore, as l increases, the largest n for which l Γ n is non-trivial increases as well. From Section 2 we have that the exponent of p in rd(L l ) is The condition of Case I means that l Γ 0 can be bounded independently of l. So as l → ∞, (3) does as well, which in turn implies that rd(L l ), and therefore rd(L l ), tends to ∞, a contradiction.
Case II: : There exists p ∈ S such that lim sup e Ln l (p) = ∞.
Let m = m 0 m ∞ be a modulus of K, where m 0 is a product of finite primes and m ∞ is a product of r real places of K. We have the following exact sequence from class field theory where O * K are the units of the ring of integers Let p be the rational prime lying below p. Because L l /Q is assumed to be Galois, the assumption of Case II implies that lim sup e Ln l (p ′ ) = ∞ for every prime p ′ of K with p ′ ∩Z = (p). For any rational prime q lying below a prime of S, defineq = q∈S,q|q q. For any modulus m, let R m denote the ray class field of K modulo m. Let n be the modulus q∈S q (note that whether a prime of K is contained in S depends only on the rational prime over which it lies). L l is contained in R n s(l) for some positive integer s(l). Our intermediate goal q∈T Rqs(l) Because lim sup e L l (p) = ∞, to prove our intermediate goal, it suffices to show that L l /L l ∩ Rps(l) is ( * ), or equivalently that Rps(l)L l /Rps(l) is ( * ). For this, it suffices to show that R n s(l) /Rps(l) is ( * ). There is no ramification above p in the extension of the composite field q∈T Rqs(l) over Rps(l), where T consists of the rational primes ≤ √ N and the infinite real place of Q, i.e., the set of rational primes below the primes of S. Thus it suffices to prove the following lemma: Proof. By induction on the number of primes of T , it suffices to show that for any positive integers a, b: where h K is the class number of K, q is a finite prime of T distinct from p, and the products are over primes of K (if q were the infinite place of Q, one finds that the right-hand side of the inequality would be 2 r 1 h K , r 1 being the number of real places of K). Define maps Looking at (4) and using the fact that (O/pq) This completes the proof of our intermediate goal.
With this in hand, we now complete the proof that X N,K is finite under the assumption of Case II.
Set E l = L l ∩ Rps(l), and let P l be a prime of of E l lying above p. Since K/Q is assumed to be Galois, the Galois closure of E l /Q is an abelian extension of K ramified only above p, and it is contained in L l because L l /Q is Galois. But E l is the maximal such field; therefore E l /Q is Galois. In proving our intermediate goal, we showed that lim sup e E l (p) = ∞. Let E l = E l P l andK = K p , and consider the extension of local fieldsÊ l /K with Galois group l G. From the structure of (O/p s(l) ) * and the exact sequence (4) with m =p s(l) , it follows that there exists C > 0 such that the prime-to-p part of | l G| is less than C for all l. Let F l be the fixed field of the inertia subgroup of p of G(E l /K), so that E l /F l is totally ramified above p. Let Q l be the prime of F l below P l and setF l = F l Q l so thatÊ l /F l is a totally ramified extension of local fields. Set l H = G(Ê l /F l ). If l J denotes the p-Sylow subgroup of l H, then lim sup l→∞ | l J| = ∞. Recall that we are assuming that L l /K is cyclic, so all groups in sight are cyclic. LetM l be the fixed field of l J, with maximal ideal r l .
Fix l for the moment and put r = r l . In this paragraph and the next, we summarize a few results from [8] and [9] and apply them to the situation at hand. For non-negative integers n, let U n be the n-units ofM l , as defined in the proof of Case I. For any real v ≥ 0, define U v = U n , where n − 1 < v ≤ n. For all v ≥ 0, the local reciprocity map θ :M l * → l J maps U v onto the upper ramification group l J v . It follows that any jumps in the filtration J v must occur at integral values of v (i.e., if J v+ǫ = J v for any ǫ > 0, then v is an integer), and that for v ≥ 1, the quotient l J v / l J v+1 of upper ramification groups is isomorphic to a quotient of U v /U v+1 . For integral n ≥ 1, we have isomorphisms U n /U n+1 ∼ = r n /r n+1 ∼ = F + M l , the last group being the additive group of the residue field ofM l , which is an elementary abelian p-group. Since l J is cyclic, we find that for v ≥ 1, the quotients l J v / l J v+1 are either trivial or are cyclic groups of order p.
For integral values of u, the function φ satisfying l J u = l J φ(u) is given by By (5), φ(n + 1) − φ(n) ≤ 1, from which it follows that the l J n / l J n+1 are also either trivial or cyclic of order p. Thus each subgroup of l J occurs as l J n for some n. Because jumps in the upper numbering occur at integral values, if n is a non-negative integer with l J n = l J n+1 , then φ(n) ∈ Z. Using this fact along with (5) and the fact that every subgroup of l J occurs as a ramification group, we derive what is essentially the example on p.76 of [9]: where | l J| = p t and l J(i) denotes the unique subgroup of l J of order p t−i . Less formally, the subgroup of order p t−i of l J occurs as at least p i lower ramification groups of l J.

Remark 2.
• That jumps in the upper ramification groups of l J v occur only at integral values of v also follows from the Hasse-Arf Theorem, which says that if L/K is an abelian extension of local fields with perfect residue fields, then jumps in the upper ramification groups G v of L/K occur only at integral values of v. The proof does not require class field theory [9].
• One can more directly obtain that l J n / l J n+1 is isomorphic to a subgroup of U n /U n+1 by considering the map l J n / l J n+1 ֒→ U n /U n+1 induced by σ → σ(β)/β where β is a uniformizer ofÊ l [9].
We use the information just obtained about the ramification groups of l J to obtain bounds for rd(L l ), which will complete the proof of Case II. Since E l ⊆ L l , it suffices to show that lim sup l→∞ rd(E l ) = ∞. By Section 2, it then suffices to show that lim sup l→∞ rd(Ê l ) = ∞.
The nth ramification group ofÊ l /M l is a subgroup of the nth ramification group ofÊ l /Q p . By Section 2, it suffices to show that lim sup where l Γ n are the ramification groups ofÊ l /Q p . Let l γ n denote the ramification groups of E l /M l . It follows from Lemma 2 that where | l J| = p t l . Note that lim sup l→∞ t l = ∞. We also have The quantity e(K/Q p ) is fixed and e(F l /K) = 1. By the definition of l J, and thus we obtain the second inequality in: which, using (7), gives the desired result (6), completing Case II . This completes the proof that X N,K is finite when K/Q is Galois. We use this to show that Y N,K is finite.
We initially maintain the assumption that K/Q is Galois. Suppose L ∈ Y N,K . We will show there is a constant A := A N ;K , independent of L, such that rd(L) < A N ;K , whereL denotes the Galois closure of L/Q. Suppose we have shown this to be true. The fieldL is a compositum of abelian extensions of K, thus abelian over K. The finiteness of X A N;K ,K implies that there is some large number field F A , which can be chosen independently of L, such thatL, and thus L, is contained in F A . So, upon showing the existence of A N ;K , we will have shown that Y N,K is finite under the assumption that K/Q is Galois.
Existence of A N ;K : For any σ : L ֒→ C, σ maps a prime P of L ramifying in L/K to a prime P ′ of σ(L) ramifying in σ(L)/K, where P and P ′ lie over the same rational prime p. It follows that the rational primes ramifying in L/Q are the same as those ramifying inL/Q, and using the fact that the discriminant is the norm of the different, one checks that rd(L) = rd(σ(L)).
Proof. For notational ease, put K = E ∩ F . Using (1) multiple times, we have It follows from the definition of ramification groups that the natural restriction map G(EF Q /E P ) → G(F q /K p ) takes G (Q/P),i (injectively) into G (q/p),⌈ i+1 e −1⌉ , where ⌈n⌉ denotes the least integer greater than or equal to n and e = e(Q/q). From this we find that One checks that We can rewrite b as Using [EF : F ] = ef EF/F (q)g EF/F (q) and α ≤ eγ, we find that a ≥ b, as desired. This complete the proof of the lemma.
Thus rd(L) ≤ rd(L) [K:Q] , and we may take Thus, as long as K/Q is Galois, we have shown that Y N,K is finite. We now show that the assumption K/Q Galois is unnecessary. Recall the following theorem from algebraic number theory (see, e.g., [5] or [9]). Theorem 2. Let E/F be a finite Galois extension of degree n and let p be a prime of F lying below a prime P of E with ramification index e = e(P/p). Then LetK denote the Galois closure of K/Q and suppose L/K is abelian with rd(L) ≤ N. We show there is a constant C N,K such that rd(LK) ≤ C N,K rd(L).
Suppose we know this to be true. SinceK/Q is Galois, there exists a large number field We show such a C N,K exists. It follows from (1) that The primes ramifying inK/K are fixed with K, so there exists a finite set of rational primes R, independent of L, such that every prime of L ramifying in LK/L lies above a prime in R. Let q be a prime of L ramifying in LK/L lying above a rational prime q.
Let e = e LK (q) (note LK/L is Galois). By Theorem 2, the exponent of q in d LK/L is at most We have the same bound for every prime q above q in L. Let e q = e L/Q (q), f q = f L/Q (q), respectively (do not confuse e with e q ). We obtain: gives the result.
When n = 1, this completes the proof of Theorem 1 in its entirety. We obtain the Theorem for general n by induction: Let L ∈ Y n,N,K . Let G (i) denote the ith derived subgroup of G, so n is the smallest integer with G n = 1. The quotient G/G n−1 corresponds to an intermediate field L 0 of L/K with G/G n−1 ∼ = G(L 0 /K). L 0 ⊆ L implies rd(L 0 ) ≤ rd(L), so L 0 ∈ Y n−1,N,K . By induction there are only a finite number of possibilities for L 0 . For each such L 0 , G(L/L 0 ) = G n−1 is abelian, so by Theorem 1 for n = 1, L can only be one of finitely many fields.
Remark 3. The proof of the Theorem 1 does not lend itself to an effective bound of the size of Y n,N,K in terms of n, N and K. Corollary 1. Fix N > 0, a positive integer n, and a number field K. Then #{L : L/K is solvable, G(L/K) ⊆ GL n (F) and rd(L) ≤ N } is finite, where F is a finite field that is allowed to vary with L.
Proof. It is a theorem of Zassenhaus [11] that the length of any solvable subgroup of GL n (F) has a finite bound depending on n, independent of F (in fact F need not even be finite). The result now follows from Theorem 1.

Further Questions
As mentioned in the introduction, a natural extension of Theorem 1 is to consider the size of various subsets of Z N,K . Kedlaya [3], generalizing work by Yamamoto [12], has shown that for any n, there exist infinitely many real quadratic number fields K admitting an unramified degree n extension L with G(L/K) isomorphic to the alternating group A n , whereL denotes the Galois closure of L/K. It is unknown, however, whether or not a fixed (say real quadratic) number field may admit an unramified degree n extension with Galois closure having Galois group A n for infinitely many n.
Fix a number field K. Suppose that Z N,K is infinite. One can ask whether Z N,K must contain an infinite class field tower. Maire [4] has proven the existence of infinite unramified extensions of number fields with class number one. Let K be such a number field with infinite degree unramified extension L. Maire constructs L by constructing an infinite class field tower of a finite unramified extension of K. This leaves open the question of whether a general infinite unramified extension L/K must contain an infinite class field tower-i.e., whether G(L/K) must have a pro-solvable subquotient. One can also ask the same question but replace the condition L/K unramfied with the condition rd(L) < N (meaning rd(M) < N for every field M between L and K with M/K finite). The answers to such questions are likely beyond the scope of class field theory.
Since there are only finitely many number fields of bounded root discriminant of any degree, finding infinite subsets of Z N,K amounts to finding L ∈ Z N,K with arbitrarily large degree. One may alternatively consider how many number fields L of degree n over K with rd(L) < N exist, with n fixed and N varying. Significant work in this direction has been done by Ellenberg and Venkatesh [1] and others.