Chapter 3 Application to unit groups
3.3 The main theorem
3.4.3 Complex D 2 p extensions
Assume that K/Q is a Galois extension with G= Gal(K/Q)∼= D2p. Then
K/Q satisfies the conditions of Theorem 3.3.1 at p. Let σ, τ ∈ D2p be of order p
and 2 respectively and label the intermediate subfields as in the following diagram:
K
L
L
σp−1F
0Q
2 p p. . .
2In Section 2.7.1 we described the representation theory ofD2poverZp, which
we now briefly recall. The groups D2p are one of the few infinite families of groups
for which all integral lattices have been classified. Write 1, , ρ for the three irre- ducible Qp[D2p]-representations, where 1, denote the trivial and non-trivial one
dimensional representations, and ρ the (p−1)-dimensional irreducible. There are six indecomposable Zp[D2p]-lattices:
1, , A, A0,(A, ),(A0,1), (3.13)
where1, extend to the corresponding Qp[D2p]-representations,A, A0 are two non-
isomorphicZp[D2p]-lattices withinρ, and (A, ),(A0,1) are non-trivial extensions of
,1by A, A0 respectively (these are explicitly constructed in [Lee64]).
If K has no real embeddings, then rkO×K = 2 and OK× ⊗Qp ∼= ρ. Thus,
O×K/µK⊗Zpcould potentially lie in one of two isomorphism classes ofZp[G]-lattices,
namely,
Referring to the matrix given on p65 for ` = p, we find that these two cases are distinguished by the D2p/Cp-isomorphism class of H1(P,OK×/µK), which is a one
dimensional Fp-vector space with non-trivial or trivial C2-action for A and A0 re-
spectively. This together with Corollary 3.3.2 and Proposition 3.2.4 gives the first part of:
Lemma 3.4.5.Let K/Q be a D2p-extension, for p odd, which has no real embed-
dings. Then
i) if K does not contain any pth roots of unity, then
dimFpker(Cl(F
0
)→Cl(K)) + dimFpker(W →Cl(K)/c(Cl(F
0
))) = 1,
where W is as in Notation 3.2.1. Moreover, OK×/µK ⊗Zp ∼= A0 if and only
if ker(W → Cl(K)/c(Cl(F0))) is one dimensional and acted on trivially by
D2p/Cp.
ii) if K does contain the pth roots of unity (i.e. p = 3 and F0 = Q(ζ3), where ζ3
denotes a primitive third root of unity), then
dimFpker(W →Cl(K)/c(Cl(F 0 ))) + dimFp((NK/Q(ζ3)O × K)/±1) = 2, with dimFpker(W →Cl(K)/c(Cl(F 0)))τ=1 + dimFp((NK/Q(ζ3)O × K)/±1)τ= −1 ! = 1 if O×K/µK⊗Zp ∼=A 2 if O×K/µK⊗Zp ∼=A0 .
Proof. For the last part ifi), we know thatO×K/µK⊗Zp∼=A0 when the one dimen-
sional space ker(Cl(F0)→Cl(K))⊕ker(W →Cl(K)/c(Cl(F0))) is fixed byD2p/Cp.
But,τ acts by multiplication by−1 on Cl(F0) (τ acts by the identity on primes of
F0 lying over ramified or inert primes and swaps the two primes lying over a split prime. In each case,τ takes it to its inverse in the class group) and so non-trivially on the F3-vector space ker(Cl(F0) → Cl(K)) whenever it is non-zero. As a result,
in theA0 case it must be ker(W →Cl(K)/c(Cl(F0))) which is non-trivial.
In the case ofii), again the representation theory ofD2p and Theorem 3.3.1
ensure thatH1(P,O×K/µK) is of rank one. Therefore, the right hand side of (3.3) is
also of rank one. We may further simplify (3.3) in our case.
AsF0 =Q(ζ3) is imaginary quadratic, rkOF×0 = 0, so the summand given by
coker(OF×0/µF0 →(O×
On the other hand, by (3.11), ˆH−1(P, µK) ∼= µK[p] is of rank one with
non-trivial action ofNG(P)/P. Therefore, after twisting, H1(P, µK) is of rank one
with trivial action. For the (µQ(ζ3) ∩NK/Q(ζ3)O×K)/NK/Q(ζ3)µK term, note that
NK/Q(ζ3)O
×
K ⊆µQ(ζ3)=O
×
Q(ζ3), whilstNK/Q(ζ3)µK ={±1}. Finally, as Cl(Q(ζ3)) =
0, ker(Cl(F0)→Cl(K)) is always trivial.
Remark 3.4.6.Our proof of Lemma 3.4.5 relies on the fact that we are able to describe the, relatively few, isomorphism classes of potentialD2p-lattices. It would
be interesting to ask for a proof of, say, the formula of i) without appealing to classification results within integral representation theory. Another consequence of there being relatively few possibilities is that we are also able to distinguish theA
and A0 cases instead using regulator constants (this follows from Section 2.7.1 or directly from [Bar12]).
Example 3.4.7.Recall from Remark 2.7.1 that if p ≤ 67 (or p ≤ 157 assuming the generalised Riemann hypothesis), then a Z[D2p]-lattice M is determined by
the isomorphism classes of M ⊗Z` for ` = 2, p as a Z`[G]-lattice. If K/Q is a D2p-extension and has no real embeddings, then OK×/µK ⊗Z2 as a Z2[G]-lattice
is independent of K. So, if p ≤ 67, by considering OK×/µK ⊗Zp, there are two
candidate isomorphism classes for OK×/µK as a Z[G]-lattice. We shall also denote
these byA, A0. Moreover, Lemma 3.4.5 in fact determines theZ[G]-lattice structure.
IfK/F is an arbitrary D2p-extension, withF not necessarily Q, then we may still
calculate the isomorphism class ofO×K/µK by applying Theorem 3.3.1 at bothpand
2.
Example 3.4.8.In the p = 3 case, we applied Lemma 3.4.5 to create a Magma function [Magma] which returns the isomorphism class of O×K/µK. Using a list of
complexS3-extensions obtained from the LMFDB [LMFDB], we calculated the total
number of such fields with each isomorphism class:
A A0 total 10480 8652 19132
≈54.8% ≈45.2%
The largest discriminant amongst these fields is ≈ −2.4×1015, but the list is not complete for all fields up to that discriminant (but did contain all complex S3-
only over extensions containingQ(ζ3), we find the relative proportions are
A A0 total 334 1009 1343
≈24.9% ≈75.1%
If we exclude fields containingQ(ζ3), the relative proportions are
A A0 total 10146 7643 17789
≈57.0% ≈43.0%
Such calculations are of interest in modern approaches to the study of Cohen– Lenstra heuristics. More specifically, it is expected that the distribution of possible Galois module structures of OK× should link to the distribution and average sizes of class groups. When ordered by discriminant, a zero proportion of complex S3-
extensions of Q contain Q(ζ3), so for asymptotic calculations we may discount all
such fields and only apply Lemma 3.4.5i).
Lemma 3.4.9.Amongst complex S3 extensions K/Q ordered by discriminant, a
zero proportion haveQ(ζ3) as their quadratic subfield.
Proof. The idea of the following proof was suggested by Alex Bartel. We first show the following claim:
Claim. AllS3-extensionsK/QcontainingQ(ζ3)are of the formQ(3
√
a, ζ3)for some
cubefreea∈Q×.
Proof of Claim. By Kummer theory, C3-extensions K of F0 := Q(ζ3) must be of
the formF0(√3b) forb∈F0×/(F0×)3. If K is Galois over
Q, then hbi ⊆F0×/(F0×)3
must be stable under the action of Gal(F0/Q). Letτ generate Gal(F0/Q). We now
split into two cases, the first where τ(b) ≡ b (mod (F0×)3) and the second where
τ(b)≡b2 (mod (F0×)3).
In the first case, the fact that Q×/(Q×)3 = (F0×/(F0×)3)Gal(F
0/
Q) (as |F0 :
Q|= 2) ensures that we may assume that b∈ Q×. As a result, K/Q must be the S3-extension ofQgiven by the splitting field of (x3−b).
We claim that the second case does not result in anyS3-extensions. Suppose
that K/Q as above is Galois with group G and fix an extension of τ to K, which
of unityζ3. Then Gmust act faithfully on the set
Σ :={β, ζ3β, ζ32β, τ(β), ζ3τ(β), ζ32τ(β)}.
Letσbe the element ofGof order 3 which sendsβ 7→ζ3β. We claim thatG=hσ, τi
acts on Σ in the following way:
β ζ3β ζ32β τ(β) ζ3τ(β) ζ32τ(β) σ τ σ τ σ τ σ σ σ
The action ofσ on the top row is by definition and the action ofτ is also given by assumption together with that fact thatτ(ζ3) =ζ32. The only non-obvious claim is
the action ofσonτ(β). Using the assumption thatτ(b)≡b2 (mod (F0×)3), we find thatτ(β) =β2λfor some λ∈Q(ζ3) (asF0 contains the third roots of unity). As a
result:
σ(τ(β)) =σ(β2)σ(λ) =ζ32β2λ
=ζ32τ(β)
Finally, note that this action on Σ is given byC6 acting on itself faithfully, and so
Gcannot be isomorphic to S3.
Without changing the extension, we may additionally assume that a is a cubefree integer and, given that−1 is a cube inQ(ζ3), that ais positive.
Claim. For a cubefree integera, the discriminant ofQ(3
√
a, ζ3) is divisible by (a0)4,
where a0 is the largest squarefree divisor ofawhich is coprime to 3.
Proof of Claim. The conductor-discriminant formula [Neu13, VII.11.9] gives that: ∆Q(√3a,ζ
3)=f(1)f()f(ρ)
2.
Here f(−) denotes the Artin conductor, 1 and are the trivial and non-trivial one dimensional complex representations of S3 = Gal(Q(3
√
mensional representation. We claim that, for every primep6= 3 dividinga, we have thatp2 dividesf(ρ). As the Artin conductor is a product
f(ρ) =Y
p
pfp(ρ)
of local Artin conductors, it suffices to calculate fp(ρ) forp dividing abut coprime
to 3. By definition, fp(ρ) = X i≥0 |Gi| |G0| codimρGi,
whereGi denotes theith ramification group ofGatp. As Q(ζ3) is unramified away
from 3, we find thatG0 is theC3-subgroup ofS3. But,Gi is a p-group for alli≥1,
and so must vanish. As a result
fp(ρ) = |G0|
|G0|
codimρG0 = 2.
This gives the desired result.
We now wish to bound the number of extensionsNX of the formQ(3
√
a, ζ3)
with discriminant at mostX. By the above claim this is less than or equal to the number of cubefree integersafor which a0 ≤X1/4 =:Y. By writing each such aas
3dbc withband c squarefree and not divisible by 3 we obtain the bound:
NX ≤3 X 1≤b≤Y X 1≤c≤Y /b 1 ≤3 X 1≤b≤Y Y b
Here the sums run only over integers and the coefficient 3 corresponds to the number of possible choices of exponent of 3 in the prime decomposition of a. By breaking the summation into the first term and 2≤b≤Y-terms we obtain the bound:
NX ≤3 Y + Z Y 1 Y bdb = 3Y(logY + 1).
In other words, the number ofS3-extensions ofQ containing Q(ζ3) of discriminant ≤X is O(X1/4(logX)1/4). In contrast, Bhargava and Wood have shown that the
for somec >0 [BW08, Thm. 2]. This concludes the proof.