Imaginary Quadratic Fields With Isomorphic Abelian Galois Groups

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1 Imaginary Quadratic Fields With Isomorphic Abelian Galois Groups Universiteit Leiden, Université Bordeaux 1 July 12, UCSD - X -

2 a Question Let K be a number field and G K = Gal(K/K) the absolute Galois group. Question: Does G K determine K? meaning, if G K1 = GK2 then K 1 = K2? Answer: Yes! Neukirch, Ikeda, Iwasawa & Uchida (around )! α Aut(Q) : α[k 1 ] = K 2 inducing G K1 = GK2 α Q α K 1 K 2

3 more Questions 1/2 Let K be a number field and G solv K = Gal(Ksolv /K) the maximal prosolvable quotient of G K. Question: Does G solv K determine K? meaning, if G solv K 1 = G solv K 2 then K 1 = K2? Answer: Yes! Neukirch, Ikeda, Iwasawa & Uchida (around )

4 more Questions 2/2 Let K be a number field and A K = G K /[G K, G K ] the maximal abelian quotient of G K. Question: Does A K determine K? meaning, if A K1 = AK2 then K 1 = K2? Answer: No!!! First examples found by Onabe (1976) Uses Kubota s (1957) description of the character group of G K in terms of Ulm invariants A K is not explicitly given but characterized by the number of Ulm invariants

5 summary K ab is not explicit, but A K is explicit (class field theory) Theorem Many imaginary quadratic fields K have the same minimal absolute abelian Galois group A K = M = Ẑ 2 n 1 Z/nZ here Ẑ = lim Z/nZ, the profinite completion of Z 2291 out of 2348, meaning, more than 97.5% of K of prime class number < 100 have this minimal group Conjecture: there are infinitely many such K

6 modules over Ẑ The topological group A K is a module over Ẑ: The exponentiation of elements of this group with ordinary integers extends to exponentiation with elements of Ẑ For any σ A K, we have lim n σn! = id A K Describe infinite abelian Galois groups as modules over Ẑ

7 example 1/2 Toy example: K = Q By Kronecker-Weber theorem Q ab = n=1 Q(ζ n) is the maximal cyclotomic extension of Q This yields the well-known isomorphism A Q = lim (Z/nZ) = Ẑ Ẑ = p Z p, (Chinese Remainder Theorem) Z p = Z/(p 1)Z (1 + pz p ) = Z/(p 1)Z Z p, (p 2) T Q = p (Z/(p 1)Z) = n 1 Z/nZ This is the countable product of finite cyclic groups having infinitely many cyclic components of prime power order for every prime

8 example 2/2 Taking the product over all p, we obtain A Q = Ẑ n 1 Z/nZ = Ẑ T Q T Q is the closure of the torsion subgroup of A Q It is not a torsion group itself A Q /T Q is a free Ẑ-module of rank 1 The subfield of Q ab left invariant by the subgroup T Q A Q = Gal(Q ab /Q) is the unique Ẑ-extension of Q

9 using cft K ab unknown for arbitrary K Class field theory: A K for arbitrary K A K = [( p K p )/K ]/(conn. comp. of unit element) From now on we take K to be imaginary quadratic. Then, A K = ( p< K p )/K

10 inertial part A K K ab U K =Ô /µ K H is the Hilbert class field Cl K is the class group of K µ K is the group of roots of unity in K (of order 6) Ô = p O p unit group of the profinite completion of the ring of integers O = O K H Cl K A K = ( p< K p )/K K U K = ( p O p)/o = Ô /µ K

11 the Ô & T K K ab K p µ p local roots of unity A K U K =Ô /µ K Ô = p O p p µ p = T K T K is the closure of the torsion subgroup of Ô Lemma (1) H Ô = Ẑ[K:Q] T K Cl K Ô = Ẑ2 T K K

12 the T K & T K /µ K Lemma (2) Let w K be the number of roots of unity in K. Then we have a non-canonical isomorphism of profinite groups T K = Z/nw K Z n 1 If w K is squarefree, then T K = TQ Lemma (3) We have a non-canonical isomorphism T K /µ K = Z/nw K Z n 1

13 structure of the Inertial part Theorem { U K = Ô /µ K = Ẑ 2 T K /µ K = Ẑ2 n=1 Z/nZ, K Q(i) Ẑ 2 n=1 Z/4nZ, K = Q(i) (isomorphisms of profinite groups) Corollary All imaginary quadratic K Q(i) of class number 1 have A K = Ẑ 2 Z/nZ n=1 This implies Onabe s observation: there are 8 such K!

14 diagrams 1/2 K ab T U K L Ẑ 2 K H Cl K The invariant field L of the closure T of the torsion subgroup of U K is an extension of H with group Ẑ2 A K

15 diagrams 2/2 class number 1: Ẑ 2 T K ab L H = K U K =A K The invariant field L of the closure T of the torsion subgroup of U K is an extension of K with group Ẑ2

16 Question: A K = M for hk > 1? K Q(i) imaginary quadratic 1 U K A K Cl K 1 (1) U K = Ẑ 2 T Does not depend on K It is isomorphic to the minimal Galois group M = Ẑ2 Z/nZ n=1 Question: Can A K be isomorphic to M for h K > 1? If (1) splits, then A K is isomorphic to M

17 the main sequence does not split Theorem For every imaginary quadratic field K, the sequence 1 U K A K ψ Cl K 1 is totally non-split, i.e., there is no non-trivial subgroup C Cl K for which the associated subextension is split 1 U K ψ 1 [C] C 1

18 ... still A K = M K ab T M L Ẑ 2 H Cl K K A K We will show: even though 1 M A K Cl K 1 is non-split, we may still have A K = M

19 modding out T 1 U K A K Cl K 1 M = Ẑ2 T is totally non-split, but the quotient sequence 1 U K /T A K /T Cl K 1 Ẑ 2 can be split or non-split. If it is totally non-split, we have A K /T = Ẑ2 and A K = Ẑ 2 T = M!

20 Galois diagram K ab T M T 0 L H Ẑ 2 L 0 Ẑ 2 Cl K L 0 H =Ẑ2 K A K Translation under Galois theory: A K /T = Gal(L/K) L 0 = maximal Ẑ-extension of K the sequence 1 U K /T A K /T Cl K 1 is totally non-split if and only if L = L 0 A K = M

21 Galois diagram: L = L 0 K ab T M L 0 = L Ẑ 2 =Ẑ2 K H Cl K A K Totally non-split case: H L 0 Question: How can we decide numerically whether this is the case? Answer:... do not use fields! Class field theory suffices!

22 the Lemma for computing Lemma 1 U K /T A K /T Cl K 1 Ô / µ K p /K p p Suppose Cl K = [a] is cyclic of prime order p. Then the above sequence is split iff every α K generating a p is locally everywhere a p th power up to roots of unity This means: for all primes p we have α = ζ p x p p µ p with ζ p µ p and x p K p This condition is trivially satisfied at p p It is a local computation at p p

23 the Algorithm Algorithm: to decide whether A K = M Input: K imaginary quadratic Checks: for each prime p h K whether φ p : Cl K [p] ( p p O p/µ p )/(p th powers) a α, (if a p = αo) is injective Output: Yes, if for every prime p h K the map φ p is injective Involves class group computation local computation at p p and becomes linear algebra over F p Remark: For p = 2 we have a theorem, so no computation is necessary

24 Onabe s extended list Prime h K = p < 100. A K = M for more than 97.5% of all K p #K : hk = p non-split split Q( 35), Q( 51), Q( 91), Q( 115), Q( 123), Q( 187), Q( 235), Q( 267), Q( 403), Q( 427) Q( 643), Q( 331), Q( 107) Q( 1723), Q( 1123), Q( 1051), Q( 739), Q( 443), Q( 347) Q( 5107), Q( 2707), Q( 1163), Q( 859) Q( 9403), Q( 5179), Q( 2027), Q( 10987), Q( 13267) Q( 11923), Q( 2963), Q( 1667) Q( 25243), Q( 16699), Q( 8539), Q( 383) Q( 17683), Q( 17539), Q( 17299), Q( 4327) Q( 21163), Q( 9587), Q( 2411) Q( ), Q( 74827), Q( 47563) Q( 46867), Q( 12923), Q( 9203) Q( 28283), Q( 20011) Q( 96763), Q( 21487), Q( 14887) Q( 42683) Q( ), Q( ) Q( ) Q( ) Q( ), Q( ) Q( 64303) Q( ), Q( ), Q( 48779) Q( )

25 a Conjecture Numerical observation: for h K = p, we have A K = M for a fraction 1 1/p of fields Conjecture There are infinitely many imaginary quadratic fields K for which the absolute abelian Galois group is isomorphic to M = Ẑ2 n 1 Z/nZ The numerical evidence may be strong, but we do not even have a theorem that there are infinitely many prime numbers that occur as the class number of an imaginary quadratic field And even if we had, we have no theorem telling us what the distribution between split and non-split will be

26 checked already We checked numerically: For p h K, splitting at p occurs with frequency 1/p Splitting at different primes p, q h K is independent No influence of the splitting over the three kinds of local behavior in K of the prime p Splitting at p, where p 2 h K and Cl K = Cp C p C m, p m, occurs with frequency 1/p 2

27 future work Future work: Cohen-Lenstra should yield asymptotic number of K with A K = M Extend the theory to Real quadratic fields! Go further... to any number field!!

28 229 - X Thank You for Your Attention!

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