The average sizes of two-torsion subgroups in quotients of class groups of cubic fields

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by  Zev Klagsbrun

THE AVERAGE SIZES OF TWO-TORSION SUBGROUPS IN
QUOTIENTS OF CLASS GROUPS OF CUBIC FIELDS
ZEV KLAGSBRUN
7 Abstract. We prove a generalization of a result of Bhargava regarding the average size
1 Cl(K)[2] as K varies among cubic fields. For a fixed set of rational primes S, we obtain
0 a formula for the average size of Cl(K)/hSi[2] as K varies among cubic fields with a fixed
2
signature,wherehSiisthesubgroupofCl(K)generatedbytheclassesofprimesofK above

n prime in S. We additionally obtain averagesizes for the relaxedSelmer groupSelS(K) and
2
a × × 2
for O /(O ) as K varies in the same families.
J K,S K,S

3
1

] 1. Introduction
T
N In addition to the Davenport-Heilbronn theorem, one of the few results proven concerning
. the distribution of class groups of number fields is a result of Bhargava in [1] and extended
h
t by Bhargava and Varma in [4] which states:
a
m
Theorem 1. When ordered by absolute discriminant,
[
(i) the average size of Cl(K)[2] as K ranges over totally real S -cubic fields is equal to 5/4,
3
2
(ii) the average size of Cl(K)[2] as K ranges over complex S -cubic fields is equal to 3/2,
v 3
8 and
3 (iii) the average size of Cl(K)+[2] as K ranges over totally real S -cubic fields is equal to 2.
3
8
2 Theorem1maybethought ofasananalogueoftheclassicalDavenport-Heilbronn theorem
0
regarding the average size of the 3-torsion subgroups of class groups of quadratic fields. In
.
1 [10], we generalized the Davenport-Heilbronn theorem to quotients of ideal class groups of
0
quadratic fields by the subgroup generated by the classes of primes lying above a fixed set
7
1 of rational primes S. The goal of this work is to do the same for Theorem 1.
:
v Explicitly: LetS beafinitesetofrationalprimes. ForeachcubicfieldK,defineCl(K) :=
S
Xi Cl(K)/hSKi, where SK is the set of primes of OK lying above the primes in S and hSKi is
the subgroup of Cl(K) generated by the ideal classes of the primes in S . Define Cl(K)+
r K S
a similarly using the narrow class group Cl(K)+ of K.
Theorem 2. When ordered by absolute discriminant,
(i) the average size of Cl(K) [2] as K ranges over totally real S -cubic fields is equal to
S 3
1 p2 +4
1+ 1+ ,
2|S|+2 4(p2 +p+1)
p∈S(cid:18) (cid:19)
Y
(ii) the average size of Cl(K) [2] as K ranges over complex S -cubic fields is equal to
S 3
1 p2 +4
1+ 1+ , and
2|S|+1 4(p2 +p+1)
p∈S(cid:18) (cid:19)
Y
(iii) the average size of Cl(K)+[2] as K ranges over totally real S -cubic fields is equal to
S 3
1 p2 +4
1+ 1+ .
2|S| 4(p2 +p+1)
p∈S(cid:18) (cid:19)
Y
1
2 ZEV KLAGSBRUN
Theorem 2 has the following immediate corollary.
Corollary 3. If S is non-empty, then
(i) a positive proportion of totally real S -cubic fields K have Cl(K)+[2] = 0,
3 S
(ii) a positive proportion of totally real S -cubic fields K have Cl(K)+ = Cl(K), and
3 S
(iii) a positive proportion of totally real S -cubic fields K have S -units of all possible sig-
3 K
natures.
In each case, the proportion of totally real S -cubic fields having the property claimed is at
3
1 p2 +4
least 1− 1+ .
2|S| 4(p2 +p+1)
p∈S(cid:18) (cid:19)
Y
The core idea behind Theorem 1 is how to use the geometry of numbers to count S -
4
quartic fields. The application to class groups as it originally appears in [1] arises from
a bijection established by Heilbronn and slightly refined by Bhargava between the set of
non-trivial two-torsion elements in the class group Cl(K) of an S -cubic field and the set of
3
nowhere overramified isomorphism classes of S -quartic fields L with resolvent field K (see
4
Definitions 2.1 and 2.2) that have a real place.
We establish a similar bijection between the set of non-trivial two-torsion elements in
Cl(K) and the set of isomorphism classes of S -quartic fields L with resolvent field K
S 4
satisfying certain local conditions. We then prove Theorem 2 by applying recent results of
Bhargava for counting the number of such fields having bounded discriminant [2].
The local product appearing in Theorem 2 is a consequence of the fact that the decompo-
sition type of any prime in S will vary with K. By assuming a fixed decomposition type for
each prime in S, we get a more natural answer.
Theorem 4. Let S be a set of rational primes. For each p ∈ S, fix a rank 3 Q -algebra R
p p
and set r = (r −1), where r is the number of irreducible components of R . When
p∈S p p p
ordered by absolute discriminant,
P
(i) the average size of Cl(K) [2] as K ranges over totally real S -cubic fields with K⊗Q ≃
S 3 p
R for all p ∈ S is equal to 1+2−(r+2),
p
(ii) the average size of Cl(K) [2] as K ranges over complexS -cubic fields with K⊗Q ≃ R
S 3 p p
for all p ∈ S is equal to 1+2−(r+1), and
(iii) the average size of Cl(K)+[2] as K ranges over totally real S -cubic fields with K⊗Q ≃
S 3 p
R for all p ∈ S is equal to 1+2−r
p
Remark 1.1. In Theorem 4, r is a the number of primes above p in K. If the primes above
p
p take classes uniformly at random in Cl(K) subject only to the relation arising from the
factorization of pO and all of the primes in S behave independently, then the subgroup
K
hS i ≤ Cl(K) may be throught of as a group generated by r elements chosen uniformly at
K
random from Cl(K). It is therefore natural to expect that for any finite abelian 2-group
H, the probability Prob(Cl(K)/hSi[2∞] ≃ H) is equal to what Cohen and Lenstra dub the
u-probability of H with u = r +r , where r is equal to 2, 1, and 0 in cases (i), (ii), and
∞ ∞
(iii) of the theorem respectively. The average sizes appearing Theorem 4 are precisely the
u-averages for these values of u [6].
In addition to Theorem 2, we prove similar distribution results for a pair of objects closely
connected to Cl(K) [2]. A standard result (see Section 8.3.2 of [5], for example) shows that
S
TWO-TORSION IN QUOTIENTS OF CLASS GROUPS OF CUBIC FIELDS 3
Cl(K) [2] sits in the short exact sequence
S
(1) 0 → O× /(O× )2 → SelS(K) → Cl(K) [2] → 0,
K,S K,S 2 S
where O× is the S units of O and SelS(K) is the 2-Selmer group of K relaxed at S ,
K,S K K 2 K
defined as
SelS(K) := α ∈ K×/(K×)2 : val (α) ≡ 0 (mod 2) for all p 6∈ S .
2 p K
We are able to compu(cid:8)te average sizes for both O× /(O× )2 and SelS(K). (cid:9)
K,S K,S 2
Theorem 5. When ordered by absolute discriminant,
(i) the average size of SelS(K) as K ranges over totally real S -cubic fields is equal to
2 3
1
2|S| +2|S|+3 2− and
p2 +p+1
p∈S(cid:18) (cid:19)
Y
(ii) the average size of SelS(K) as K ranges over complex S -cubic fields is equal to
2 3
1
2|S| +2|S|+2 2− .
p2 +p+1
p∈S(cid:18) (cid:19)
Y
Theorem 6. When ordered by absolute discriminant,
(i) the average size of O× /(O× )2 as K ranges over totally real S -cubic fields is equal
K,S K,S 3
1
to 2|S|+3 2− and
p2 +p+1
p∈S(cid:18) (cid:19)
Y
(ii) the average size of O× /(O× )2 as K ranges over complex S -cubic fields is equal to
K,S K,S 3
1
2|S|+2 2− .
p2 +p+1
p∈S(cid:18) (cid:19)
Y
Remark 1.2. All of the results we present are stated in terms of S -cubic fields. However,
3
the results remain correct and can be proven with modified versions of the current proofs
even if we remove the S assumption.
3
We have nonetheless chosen to maintain the S -assumption throughout the paper for
3
the purposes of clarity. Instead, we present the following argument showing that the total
contributiontotheaveragesizeofCl(K)[2]coming fromcyclic cubicfieldsasK variesamong
all totally real cubic fields is zero.
It is well-known that the total number of cyclic cubic fields of discriminant less than X
grows like O(X1/2) [7]. By a result of Wong [11], the size of Cl(K)[2] in any cyclic cubic
field K is bounded by O(|Disc(K)|3/8+ǫ) for any ǫ > 0 (see also [3] for a better bound).
As a result, the combined number of elements in Cl(K)[2] from all cyclic cubic fields with
discriminant less than X is bounded by O(X7/8+ǫ). Since the number of totally real cubic
fields of discriminant less than X grows like O(X), the total contribution to the average size
of Cl(K)[2] coming from cyclic cubic fields must be zero.
Notation
We will use the following notation throughout this paper:
• S will be a set of rational primes.
• K will be a cubic field.
• O will be the ring of integers of K.
K
4 ZEV KLAGSBRUN
• S will denote the set of primes of O lying above primes in S.
K K
• O× will denote the S -units of K.
K,S K
• SelS(K) will be the 2-Selmer group of K relaxed at the primes in S .
2 K
• Cl(K) will be the ideal class group of O .
K
• Cl(K) will be the quotient Cl(K)/hS i, where hS i is the subgroup of Cl(K) gen-
S K K
erated by primes in S .
K
Acknowledgement
I would like to thank Frank Thorne for suggesting the use of the Artin relation in the
proof of Lemma 2.7 and for pointing out the result of Wong mentioned in Remark 1.2. I
would also like to thank Alyson Deines for a number of useful suggestions.
2. Class Fields Theory
Bhargava’s results about Cl(K)[2] in [1] rely on a correspondence of Heilbronn detailed in
[9]. We briefly describe this correspondence before establishing a similar correspondence for
Cl(K) [2] in Proposition 2.6.
S
Definition 2.1. Given an S -quartic field L, the cubic resolvent field Res(L) is the unqiue
4
(up to isomorphism) cubic subfield of the Galois closure N of L/Q.
While the resolvent field of L is unique, non-isomorphic L may share the same resolvent
field.
Following Section 3.1 in [1], we make the following definition.
Definition 2.2. Let p be a rational prime. A quartic field L is called overramified at p if
L⊗Q is either irreducible and ramified or the direct sum of two ramified fields. The field
p
L is called nowhere overramified if L is not overramified at p for any prime p.
Remark 2.3. Note that unlike Bhargava’s definition in [1], our definition of nowhere over-
ramified does not include any restriction on the ramification of L at infinity.
The fields L and Res(L) have the same discriminant precisely when L is nowhere over-
ramified.
Proposition 2.4. Let K be an S -cubic field.
3
(A) The following are in bijective correspondence.
(i) The set of index two subgroups of Cl(K).
(ii) The set of unramified quadratic extensions F of K.
(iii) The set of isomorphism classes of nowhere overramified S -quartic fields L having
4
a real place with Res(L) = K.
(B) The following are in bijective correspondence.
(i) The set of index two subgroups of Cl(K)+.
(ii) The set of quadratic extensions F of K that are unramified at all finite places.
(iii) The set of isomorphism classes of nowhere overramified S -quartic fields L with
4
Res(L) = K.
Proof. For both (A) and (B), the correspondence (i) ↔ (ii) is class field theory. The corre-
spondence between (ii) and (iii) in (B) is detailed in [9] and the correspondence between (ii)
and (iii) in (A) comes from restricting this correspondence to the sets in (A). While we do
TWO-TORSION IN QUOTIENTS OF CLASS GROUPS OF CUBIC FIELDS 5
not include a proof of the equivalence between (ii) and (iii) here, we will however describe
the maps yielding the correspondence.
We begin with the map (iii) → (ii). Let N be the Galois closure of L/Q. By assumption,
Gal(N/Q) ≃ S . The group S contains three distinct D subgroups, all of which are
4 4 4
conjugate. The resolvent field K = Res(L) may be taken to be the fixed field of any of these
D subgroups. The group D contains two distinct Klein four subgroups, only one of which
4 4
contains a transposition when D is viewed as a subgroup of Gal(N/Q) ≃ S . Letting V be
4 4
that subgroup, the extension F/K is given by the fixed field of V.
We next describe the map (ii) → (iii). While the field F/Q is not Galois, the lack of
ramification (at finite primes) in F/K forces its Galois closure N to be an S -extension of
4
Q. The group S contains four distinct S subgroups, all of which are conjugate. The field
4 3
L may be taken to be the fixed field of any of these S subgroups. (cid:3)
3
Remark 2.5. In both Proposition 2.4 and Proposition 2.6 which follows, if K is taken to
be a complex cubic field, then Cl(K) = Cl(K)+ and parts (A) and (B) are equivalent.
Proposition 2.6. Let K be an S -cubic field.
3
(A) The following are in bijective correspondence.
(i) The set of index two subgroups of Cl(K) .
S
(ii) The set of unramified quadratic extensions F of K in which all primes in S split
K
completely.
(iii) The set of isomorphism classes of nowhere overramified S -quartic fields L such
4
that Res(L) = K, L has a real place, and L⊗Q has a component equal to Q for
p p
all p ∈ S.
(B) The following are in bijective correspondence.
(i) The set of index two subgroups of Cl(K)+.
S
(ii) The set of quadratic extensions F of K that are unramified at all finite places and
in which all primes in S split completely.
K
(iii) The set of isomorphism classes of nowhere overramified S -quartic fields L such
4
that Res(L) = K and L⊗Q has a component equal to Q for all p ∈ S.
p p
Proof. For both (A) and (B), the correspondence (i) ↔ (ii) is class field theory. The equiva-
lences (ii) ↔ (iii) in (A) and (B) follow from the similar equivalences in Proposition 2.4 and
Lemma 2.7 appearing below. (cid:3)
Lemma 2.7. Let K be an S -cubic field field and F/K a quadratic extension unramified
3
at all finite places. If L is the S -quartic field corresponding to F/K under Heilbronn’s
4
correspondence as in Proposition 2.4, then L⊗Q has a component equal to Q if and only
p p
if all primes of K above p split in F/K.
Proof. Let N be the Galois closure of L/Q. If L⊗Q has a component equal to Q , then
p p
Gal(N /Q ) ≤ S for all primes P | p of N. Letting p be a prime of K above p, we have
P p 3
Gal(N /K ) ≤ Gal(N/K). Since Gal(N /K ) must simultaneously embed into a copy of S
P p P p 3
and a copy of D inside of S , we find that Gal(N /K ) must be a subgroup of the Klein
4 4 P p
four group Gal(N/F). As a result, p splits in F/K.
For the opposite direction, we rely on the Artin relation of zeta functions (see [9], for
example)
ζ(s)ζ (s)
F
(2) ζ (s) = .
L
ζ (s)
K
6 ZEV KLAGSBRUN
(p) (p) (p)
If all primes of K above p split in F/K, then we have ζ (s) = ζ (s), where ζ (s) denotes
F K ∗
the part of the Euler product for ζ (s) coming from primes above p. As a result, if all primes
∗
of K above p split in F/K, then (2) yields
(3) ζ(p)(s) = (1−p−s)−1ζ(p)(s).
L K
Observe that if L⊗Q contains a ramified component, then K⊗Q must contain a ramified
p p
component of the same degree. As a result, (3) holds even if we restrict to the Euler factors
coming from unramified primes. As a result, the Euler product for ζ (s) contains a factor
L
of (1−p−s)−1 coming from an unramified prime above p, and the L⊗Q has a component
p
equal to Q . (cid:3)
p
We therefore get the following corollary:
Corollary 2.8. If K is an S -cubic field, then
3
nowhere overramified S -quartic fields L (up to iso.)
4
(i) |Cl(K) [2]| = 1+ such that Res(L) = K, L has a real place, and and
S
(cid:12) (cid:12)
(cid:12) L⊗Q has a component equal to Q for all p ∈ S (cid:12)
 p p 
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12)(cid:12) nowhere overramified S4-quartic fields L (cid:12)(cid:12)
(ii) Cl(K)+[2] = 1+ (up to iso.) such that Res(L) = K and L⊗Q .
S (cid:12) p(cid:12)
(cid:12) has a component equal to Q for all p ∈ S (cid:12)
 p 
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
Proof. Since Cl(K) is a(cid:12)finite abelian group, the number of index two subgr(cid:12)oups of Cl(K)
S   S
(cid:12) (cid:12)
is the same as the number of non-trivial two-torsion elements of Cl(K) . By Proposition
S
2.6, this is equal to the number of quartic fields L such that Res(L) = K, L has a real place
and L⊗Q has Q component for all p ∈ S. The result for Cl(K)+ follows similarly. (cid:3)
p p S
3. Counting Fields
In order to prove Theorems 2 and 4, we will need to be able to count the number of quartic
fields of bounded discriminant satisfying a given set of local conditions.
For a set Σ of Q -algebras, define µ (Σ ) as
p p p p
p−1 1 1
µ (Σ ) := · · ,
p p
p Disc (R) |Aut(R)|
p
RX∈Σp
where Disc (R) is the p-part of the discriminant of R.
p
For each p ∈ S, let Σ be a set of non-overramified rank 4 Q and set Σ = (Σ ) . For
p p p p∈S
each i ∈ {0,2,4}, define N(i)(X,Σ) to be the number of nowhere overramified S -quartic
4
fields L (up to isomorphism) such that |Disc(L)| < X, L has i real places, and L⊗Q ∈ Σ
p p
for all p ∈ S.
We then have the following specialization of a theorem of Bhargava.
Theorem 3.1 (Theorem 1.3 in [2]). For each i ∈ {0,2,4},
1 µ (Σ )
(4) N(i)(X,Σ) = p p ·X +o(X)
2n ζ(3) µ
i p
p∈S
Y
8 if i = 0
where µ = 1− 1 and n = 4 if i = 2 .
p p3 i 
24 if i = 4


TWO-TORSION IN QUOTIENTS OF CLASS GROUPS OF CUBIC FIELDS 7
Proof. This will follow from Theorem 1.3 in [2]. For each prime p, define Σˆ to be the set of
p
all rank 4 non-overramified Q -algebras. An easy computation shows that µ (Σ ) = 1− 1 .
p p p p3
We then set Σˆ = (Σˆ ) .
p p
For each prime p ∈ S, let Σ be as above and for p 6∈ S, set Σ = Σˆ . Applying Theorem
p p p
1.3 in [2], we then get
1 µ (Σ ) µ (Σ )
N(i)(X,Σ) = µ (Σˆ ) p p = N(i)(X,Σˆ) p p
ni p p p p∈S µp(Σˆp) p∈S µp(Σˆp)
Y Y Y
However, N(i)(X,Σˆ) is simply the number of nowhere totally ramified S -quartic fields L
4
(up to isomorphism) such that |Disc(L)| < X and L has i real places. By Lemma 27 in [1],
this is known to be 1 ·X +o(X), so the result follows. (cid:3)
2niζ(3)
We may similarly count the number of cubic fields of bounded discriminant satisfying a
set of local conditions. For a set of local conditions Σ = (Σ ) where each Σ is a set of
p p∈S p
rank 3 Q -algebras and each i ∈ {1,3}, define M(i)(X,Σ) to be the number of S -cubic fields
p 3
K (up to isomorphism) such that |Disc(K)| < X, K has i real places, and K ⊗Q ∈ Σ for
p p
all p ∈ S.
Theorem 3.2 (Theorem 1.3 in [2]). For each i ∈ {1,3},
1 µ (Σ )
(5) M(i)(X,Σ) = p p ·X +o(X)
2m ζ(3) µ
i p
p∈S
Y
where m = 2, m = 6, and µ is as in Theorem 3.1.
1 3 p
The proof of Theorem 3.2 is extremely similar to that of Theorem 3.1 so we omit it.
4. Dealing With Local Conditions
In general, if K is an S -cubic field and L an S -quartic field such that the resolvent
3 4
Res(L) = K, then K ⊗Q does not determine L⊗Q . However, if we further assume that
p p
L⊗Q has a Q component, then this is no longer the case.
p p
Lemma 4.1. Let K be an S -cubic field and L an S -quartic field such that the resolvent
3 4
Res(L) = K. If p is a prime such that L⊗Q has a component equal to Q , then L⊗Q ≃
p p p
(K ⊗Q )⊕Q .
p p
Proof. Let N be the Galois closure of L/Q. Since L⊗Q has a component equal to Q , we
p p
have Gal(N /Q ) ≤ S for any prime P of N above P.
P p 3
Let K be the Galois closure of K/Q contained in N/Q and let p be any prime above p in
K. The Galois group of N/K is the unique order four normal subgroup V of S . Therefore,
4
if P is eany prime of N above p, we have Gal(N /K ) ≤ V. However, the subgroups V and
P p
Se of S intersect trivially. Aes a result, we have Gal(N /K ) = 0 and p splits completely in
3 4 P p
N/K˜. e
We therefore see that N ⊗ Q = (K ⊗ Q )4. Taking Geal(N/L) invariants, we get that
p p
L⊗Q = (K ⊗Q )⊕Q . (cid:3)
p p p
e
Lemma 4.1 motivates the following definition. Given a rank 3 Q -algebra R , we define a
p p
rank 4 Q -algebra R as R = R ⊕Q . We then get the following corollary.
p p p p p
f f
8 ZEV KLAGSBRUN
Proposition 4.2. Let Σ = (Σ ) where each Σ is a set of rank 3 Q -algebras. Set
p p∈S p p
µ(Σ) = µ (Σ )/µ (Σ ), where Σ = {R : R ∈ Σ }. Then
p∈S p p p p p p p p
(1) the average size of Cl(K) [2] as K ranges over totally real S -cubic fields with K ⊗
Q S 3
e Q ∈ Σ ffor all p ∈ S is equafl to 1+f1µ(Σ)
p p 4
(2) the average size of Cl(K) [2] as K ranges over complex S -cubic fields with K⊗Q ∈
S 3 p
Σ for all p ∈ S is equal to, and 1+ 1µ(Σ)
p 2 e
(3) the average size of Cl(K)+[2] as K ranges over totally real S -cubic fields with K ⊗
S 3
Q ∈ Σ for all p ∈ S is equal to 1+µ(Σ).
p p e
Proof. By combining Corollary 2.8 with Lemma 4.1, we have
e
|Cl(K) [2]| = M(3)(X,Σ)+N(4)(X,Σ).
S
K cubic,
X
K⊗Qp≃Rp forallp∈S e
0<Disc(K)<X
Part (i)thenfollowsfromTheorems 3.1and3.2. Part(ii) followsfromanessentially identical
calculation.
For part (iii), again by Corollary 2.8 combined with Lemma 4.1, we have
|Cl(K)+[2]| = M(3)(X,Σ)+N(4)(X,Σ)+N(0)(X,Σ).
S
K cubic,
X
K⊗Qp≃Rp forallp∈S e e
0<Disc(K)<X
The result then follows from Theorems 3.1 and 3.2. (cid:3)
The quotients µ (Σ )/µ (Σ ) have a nice formula when either Σ = {R } or Σ contains
p p p p p p p
all rank 3 Q -algebras (up to isomorphism).
p
f
Lemma 4.3. If Σ = {R }, then µ (Σ )/µ (Σ ) = 2−(rp−1), where R is the number of
p p p p p p p
irreducible components of R .
p
f
Proof. Since R = R ⊕ Q , we have Disc (R ) = Disc (R ), so µ ({R })/µ ({R }) =
p p p p p p p p p p p
|Aut(R )|/|Aut(R )| = n!/(n+1)! = 1/(n+1), where n is the number of components of R
p p p
equal to Q . Efxamination shows that this is eqfual to 2−(rp−1). f (cid:3)
p
e
Lemma 4.4. If Σ contains all rank 3 Q -algebras, then µ (Σ )/µ (Σ ) = 1 1+ p2+4 .
p p p p p p 2 4(p2+p+1)
(cid:16) (cid:17)
Proof. By Lemma 4.3, we have µ ({R })/µ ({R }) = 2−(r−1) for each R with r irreducible
p p p p f p
components. Letting Σ denote the set of all rank 3 Q -algebras with r irreducible compo-
p,r p
nents, we get f
3
µ (Σ ) = µ (Σ ) = 2−(r−1)µ (Σ ).
p p p p,r p p,r
RXp∈Σp Xr=1
Calculating µ (Σ ) for eafch r ∈ {1,2,3},gwe find
p p,r
p3 −p2 +3p−3 p2 +p−2 p−1
µ (Σ ) = ,µ (Σ ) = , and µ (Σ ) = .
p p,1 3p3 p p,2 2p2 p p,3 6p
As a result, we get µ (Σ ) = p3−p2+4p−8, so
p p 8p3
p3 −p2 +4p−8 p3 5p2 +4p+8 1 p2 +4
µ (Σ )/µ (Σ ) = f · = = 1+ .
p p p p 8p3 p3 −1 8(p2 +p+1) 2 4(p2 +p+1)
(cid:18) (cid:19)
f
TWO-TORSION IN QUOTIENTS OF CLASS GROUPS OF CUBIC FIELDS 9
(cid:3)
We are now able to prove Theorems 2 and 4.
Proof of Theorem 2. For each p ∈ S, let Σ be the set of all rank 3 Q algebras. The result
p p
then follows from combining Proposition 4.2 with Lemma 4.4. (cid:3)
Proof of Theorem 4. Foreachp ∈ S,letΣ = {R }. ByLemma4.3,wehaveµ (Σ )/µ (Σ ) =
p p p p p p
2−(rp−1) for each p ∈ S. Letting r = (r −1), we have µ(Σ) = 2−r and the result follows
p∈S p
from Proposition 4.2. f (cid:3)
P
e
Corollary 3 will now follow from Theorem 2.
Proof of Corollary 3. Let λ be the proportion of totally real S -cubic fields K having
3
Cl(K)+[2] = 0. We then have Avg(Cl(K)+[2]) ≥ λ + 2 · (1 − λ) = 2 − λ. By Theo-
S S
rem 2, we have
1 p2 +4
Avg(Cl(K)+[2]) = 1+ 1+ ,
S 2|S| 4(p2 +p+1)
p∈S(cid:18) (cid:19)
Y
so it follows that
1 p2 +4
(6) λ ≥ 1− 1+ .
2|S| 4(p2 +p+1)
p∈S(cid:18) (cid:19)
Y
If S is non-empty, then the right-hand side of (6) is at least 5/14, proving (i).
To see (ii), observe that the kernel of the surjection Cl(K)+ → Cl(K) is a subgroup of
S S
Cl(K)+ having order at most two. As a result, if Cl(K)+ has odd order, then Cl(K)+ =
S S S
Cl(K) . The result then follows from (i).
S
Finally, we note that K has S -units of all signatures if and only if Cl(K)+ = Cl(K) .
K S S
Part (iii) of the corollary therefore follows from (ii). (cid:3)
5. Averages for O× /(O× )2 and SelS(K)
K,S K,S 2
We now wish to consider the average size of O× /(O× )2.
K,S K,S
Defineafunctionν onS -cubicfieldsbyν (K) = r (K),wherer (K)isthenumber
S 3 S p∈S p p
of irreducible components of K ⊗Q . By Dirichlet’s unit theorem, if K is a real cubic field,
p
P
then O× /(O× )2 has rank ν (K)+3 as an F -vector space. Similarly, if K is a complex
K,S K,S S 2
cubic field, then O× /(O× )2 has rank ν (K)+2.
K,S K,S S
We now proceed to prove Theorem 6.
Proof of Theorem 6. We only consider the totally real case, since the proof in the complex
case is nearly identical. We proceed by induction on the set S. For the base case, consider
the case where S contains a single prime p. By Dirichlet’s unit theorem, we have
3
(7) Avg(O× /(O× )2) = 8· 2r ·Prob(K ⊗Q has r irreducible components ).
K,S K,S p
n=1
X
By Theorem 3.2, we have
µ (Σ )
(8) Prob(K ⊗Q has r irreducible components ) = p p,r ,
p µ
p
10 ZEV KLAGSBRUN
where Σ is as in the proof of Lemma 4.4 and µ = 1− 1 . Evaluating (8), we get
p,r p p3
3
(9) 2r ·Prob(K ⊗Q has r irreducible components )
p
n=1
X
p3 p3 −p2 +3p−3 p2 +p−2 p−1
= 2· +4· +8·
p3 −1 3p3 2p2 6p
(cid:18) (cid:19)
2 1
= 4− = 2· 2− .
p2 +p+1 p2 +p+1
(cid:18) (cid:19)
Combining (7) with (9), we get
1
Avg(O× /(O× )2) = 2|S|+3 2− .
K,S K,S p2 +p+1
(cid:18) (cid:19)
For the inductive step, let S′ = S ∪{p} for some prime p 6∈ S. Applying Dirichlet’s unit
theorem once again, we have
(10) Avg(O× /(O× )2) =
K,S′ K,S′
3
2s+3 ·Prob(ν (K) = s)· 2r ·Prob(K ⊗Q has r irred. components |ν (K) = s)
S p S
!
s n=1
X X
By Theorem 3.2, the behavior of K ⊗ Qp is independent K ⊗ Qp′ for all p′ ∈ S, and
therefore independent of ν (K). As a result, we may switch the order of summation and get
S
(11) Avg(O× /(O× )2) =
K,S′ K,S′
3
= 2r ·Prob(K ⊗Q has r irred. components ) 2s+3 ·Prob(ν (K) = s)
p S
!
n=1 s
X X
3
= 2r ·Prob(K ⊗Q has r irred. components ) Avg(O× /(O× )2)
p K,S K,S
!
n=1
X
By the inductive hypothesis, we have
1
(12) Avg(O× /(O× )2) = 2|S|+3 2−
K,S K,S p2 +p+1
p∈S(cid:18) (cid:19)
Y
and by (9), we have
3
1
(13) 2r ·Prob(K ⊗Q has r irreducible components ) = 2· 2− .
p p2 +p+1
n=1 (cid:18) (cid:19)
X
The result follows from combining (11), (12), and (13). (cid:3)
6. Averages of Selmer Groups
In the setting of Theorem 4, we are given a fixed rank 3 Q -algebra R for each p ∈ S and
p p
we are able to compute the average size of of Cl(K) [2] as we range over S -cubic fields K
S 3
with a fixed signature such that K ⊗Q ≃ R for every p ∈ S.
p p
As seen in Theorem 4, the average size does not depend on the collection of R , only on
p
r = (r −1). As a result, we get the following:
p∈S p
P