Valuations of $p$-adic regulators of cyclic cubic fields

0.2 MB·

by  Tommy Hofmann

Valuations of p-adic regulators of cyclic cubic fields
Tommy Hofmanna,∗, Yinan Zhangb
aFachbereich Mathematik, Technische Universita¨t Kaiserslautern, Postfach 3049, 67653
Kaiserslautern, Germany
bSchool of Mathematics and Statistics, The Universityof Sydney
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2 Abstract

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We compute the p-adic regulator of cyclic cubic extensions of Q with discriminant up
a
J to 1016 for 3 < p < 100, and observe the distribution of the p-adic valuation of the
regulators. We find that for almost all primes, the observation matches the model that
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the entries in the regulator matrix are random elements with respect to the obvious

] restrictions. Basedonthis randommatrix model, aconjecture onthe distributionofthe
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valuations of p-adic regulators of cyclic cubic fields is stated.
N
. Keywords: p-adic regulator, distribution of p-adic regulators, cubic fields
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2000 MSC: 11Y40, 11K41, 11R16, 11R27
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a
m
[

1. Introduction
1
v
0 The class group of a number field K is an important invariant of the field, providing
4 informationonthe multiplicativestructureofthe numberfield. Anotherinvariantofthe
3 field, the regulator, provides information on the unit group structure, and is intimately
0
linked to the the class group by the class number formula. Various algorithms for class
0
group computation have emerged over the years, culminating in Buchmann’s subexpo-
.
1 nential algorithm which can be used compute class groups of arbitrary number fields
0
[3]. Further improvements by Cohen, Diaz y Diaz and Olivier [7] allowed computation
7
of both the class group and regulator in the same algorithm. Despite this and further
1
: improvements to the algorithm in the last 25 years [1, 2], it is still impractical to com-
v
pute the classgroupandregulatorofnumberfields withlargediscriminants. The ability
i
X to compute class groups and regulators efficiently for arbitrary number fields has, and
r remains, a key focus in computational number theory.
a
With the development of these algorithms and the availability of computational re-
sources, one has seen a rise of what is nowadays called arithmetic statistics. Instead of
considering invariants of single arithmetic objects, one investigates the statistics of in-
variantsof families ofarithmetic objects. This has ledto variousdeep conjecturesabout
invariantsofnumberfields,mostnotablyMalle’sconjectureonthedistributionofGalois
∗Correspondingauthor
Email addresses: [email protected] (TommyHofmann),[email protected]
(YinanZhang)
Preprint submitted to Elsevier January 3, 2017
groups [14], the Cohen–Lenstra–Martinet conjecture on the distribution of class groups
[8,9]andMalle’srefinementthereof[16]. We extendthis listofconjecturesby providing
a heuristic on the distribution of p-adic regulators of cyclic cubic fields.
The p-adic regulator R (K) of a number field K was introduced by Leopoldt in his
p
investigation of p-adic L-functions, and he conjectured that it is non vanishing. Unlike
its classical counterpart, the p-adic regulator is only well defined if the number field is
totally real abelian or CM. As a result, very little information is known about the p-
adic regulator, in contrast to the classical regulator. Previous research on computing
p-adicregulatorshasbeenpredominantlyfocusedonnumericalverificationofLeopoldt’s
conjecture rather than computing their exact value.
Limited work on the valuation of p-adic regulator was carried out by Miki [18], who
attempted to provide an upper bound on v (R (K)). This did not turn out to be useful
p p
inpractice,astheformulacontainedtermswhosevaluesarenotexplicit. Asimplelower
bound on the valuation was also briefly mentioned by Hakkarainen in his PhD thesis
[12], but overall there has been little focus on this area, due to mainly difficulties with
computing R (K). In particular, Panayi, who was one of the first to compute R (K)
p p
explicitly in his PhD thesis [19], noted that there were significant practical difficulties
with computing in p-adic fields.
Recent developmentin a p-adic class number algorithmby Fieker andZhang [11] for
totallyrealabelianfieldsmadeitpossibletocomputethep-adicregulatorforthesefields
in a relatively efficient manner. This allowed us to compute the p-adic regulator for a
large number of cyclic cubic extensions of Q, and the experimental data allowed us to
conjecture and provide heuristics on the distribution of the values of v (R (K)).
p p
Let K be the set of all cyclic cubic extensions of Q inside a fixed algebraic closure
of Q. Note that such extensions are necessarily totally real. For a prime p let Kun and
p
Kram respectively denote the set of all fields in K which are unramified and ramified at
p
p respectively. Note that Kram =∅ and K=Kun in case p≡2mod3. For D >0 we set
p p
K(D) = {K ∈ K | |d(K)| ≤ D}, where d(K) is the discriminant of K. Moreover we set
Kun(D)=Kun∩K(D) andKram(D)=Kram∩K(D). Basedonheuristics andnumerical
p p p p
data, we make the following conjecture:
Conjecture 1. For primes p>3 the following hold:
(i) If p≡2mod3, then v (R (K))∈2Z for all K ∈K and for i≥0 we have
p p
#{K ∈K(D)|v (R (K))=2i} 1 1
p p
lim = 1− .
D→∞ #K(D) p2i−2 p2
(cid:18) (cid:19)
(ii) If p≡1mod3, then for i≥0 we have
#{K ∈Kpun(D)|vp(Rp(K))=i} i−1 1 2
lim = 1− .
D→∞ #Kpun(D) pi−2 (cid:18) p(cid:19)
and
#{K ∈Kpram(D)|vp(Rp(K))=i} i 1 2
lim = 1− .
D→∞ #Kpram(D) pi−1 (cid:18) p(cid:19)
2
2. Definition and algorithm
LetK beanumberfieldofdegreenwithunitgroupU . Wefirstdefinethe(classical)
K
regulatorR(K)anditsp-adicanalogueR (K),asthedefinitionofR (K)canbeambigu-
p p
ous. The field K has r embeddings into R, denoted as τ ,...,τ , and 2r embeddings
1 1 r1 2
intoC,denotedasconjugatepairsτ ,τ ,...,τ ,τ ,withr +2r =n. Let
r1+1 r1+1 r1+r2 r1+r2 1 2
u ,...,u be a system of fundamental units of U , that is, these elements form a
1 r1+r2−1 K
basis of the torsion-free quotient of U . Consider the submatrix formed by deleting one
K
column of the matrix
(δ log|τ (u )|) ∈Mat (R),
i i j i,j r1+r2−1
where δ = 1 for 1 ≤ i ≤ r and 2 if r +1 ≤ i ≤ r +r . As each row sums to zero,
i 1 1 1 2
the determinantofsuchasubmatrixisindependentofthecolumndeleted. Theabsolute
value of this determinant, denoted by R(K), is also independent of the choice of the
fundamental units and is known as the regulator of the number field K.
2.1. The p-adic regulator
The p-adic regulator can be defined similarly, with changes to the embeddings. By
fixing an embedding from C into C, any embedding of K into C can be considered as
p p
eitherrealorcomplex,dependingontheimageofK inthecompositeembeddingintoC.
For totally real or CM fields, whether an embedding from K to C is real or complex is
p
independent ofthe choiceofembedding fromC toC,butambiguitieswillariseinother
p
numberfields. Ascycliccubicfieldsaretotallyreal,thep-adicregulatorisawell-defined
object and does not depend on any choice.
Wedenotebyσ ,...,σ therealembeddingsandbyσ ,σ ,...,σ ,σ
1 r1 r1+1 r1+1 r1+r2 r1+r2
the complex embeddings from K to C . Then, the p-adic regulator R (K) is defined as
p p
the determinant of the submatrix obtained by removing one column of the matrix
(δ log σ (u )) ∈Mat (C ),
i p i j i,j r1+r2−1 p
where log is the p-adic logarithm, and u ,...,u and δ as defined previously.
p 1 r1+r2−1 i
As for the ordinary regulator, this value is up to units independent of the choice of the
fundamental units and the column deleted.
Alternativelyinsteadofdeletingacolumninthematrix,onecanaddarowof1’stoit.
Again, due to each row summing to zero, the determinant is unaffected. This definition
was introduced by Iwasawa [13] and was subsequently implemented in the algorithm by
Fieker and Zhang [11], part of which is used here with some modifications. This does
have the disadvantage of calculating the determinant of a matrix one dimension higher
than necessary,but it turned out that this is outweighed by leaving the structure of the
original matrix intact.
It is important to note that we can replace the fundamental units with a set of
independent units which generate a p-maximal subgroup of U . More precisely, let
K
u′,...,u′ be multiplicatively independent units and assume that the index of
1 r1+r2−1
hu′,...,u′ ,µ i in U is coprime to p (here µ is the set of roots of unity of
1 r+1+r2−1 K K K
K). Then one can show that the determinant of the submatrix obtained by removing
one column of the matrix
(δ log σ (u′)) ∈Mat (C ),
i p i j i,j r1+r2−1 p
3
is equal to the p-adic regulator modulo p-adic units.
By applying saturationtechniques ofBiasse andFieker [2]on a tentative unit group,
wecanobtainap-maximalsubgroupofU . Theunitsarerepresentedasapowerproduct
K
of some small elements α and some large exponents e ∈Z in the form of
j i,j
r
u = αei,j ,
i j
j=1
Y
instead of in terms of some fixed basis of the field. This is due to the fact that u
i
may require up to O( |d(K)|/n) digits, and the power product representationnot only
permits more efficient storage of the result, but also speeds up the actual computation
p
of log σ (u ). The different p-adic embeddings are computed using standard techniques
p j i
for p-adic factorisation or root finding. Alternatively once we have one fixed p-adic
embedding we can make use of Q-automorphisms of the field to find the remaining
embeddings.
2.2. Tables of number fields
In order to test our conjecture, we were in need of a large number of cyclic cubic
extensions of Q. One possibility is to use the tables of cyclic cubic fields of discriminant
<1016 compiledbyMalle,whichwereusedin[15]andbasedonthe algorithmdescribed
in [5]. Instead, we took the opportunity to validate the data by computing the same
table using a different method. More precisely, we used the algorithms based on global
class field theory as provided by Fieker in [10] (see also [6]). Since the base field is
just Q, theory tells us that the cyclic cubic extensions K|Q are in bijection with pairs
(f,U), where f ∈Z is an integer and U ⊆(Z/fZ)× is a subgroup of index 3, modulo
>0
an appropriate equivalence relation. In fact f can be taken to be the conductor of K,
which is the smallestinteger suchthat K ⊆Q(ζ ). Hence in order to list all cyclic cubic
f
fields of bounded conductor we enumerate all pairs (f,U) with bounded f. Since the
discriminant d of a cyclic cubic field K of conductor f satisfies d = f2, in this way
K k
wecanenumerateallcycliccubic fieldswithboundeddiscriminant. Toobtainadefining
equation ofK|Q giventhe pair (f,U) requiresmore work andinvolvesthe computing of
discretelogarithmsin(Z/fZ)× (seethereferencesformoredetails). Sincetheconductor
f mustsatisfy#(Z/fZ)× ≡0mod3andtheprimedivisorsoff areexactlytheramified
primes of K|Q, this shows that if p is a ramified prime in K, then p≡1mod3.
Note that we have found the same number of cyclic cubic fields of discriminant less
than 1016, namely 15852618.
3. p-adic regulators of cyclic cubic fields
Ourconjectureisbasedonseveralobservationsmadefromtheresultsofthecomputa-
tionandstatisticsofrandommatrices. Inthissection,foreachprimepwewillconstruct
sets Msplit,Mram,Minert ⊆ GL (Q ) with the following property: for each cyclic cubic
p p p 3 p
field K in which p is split, there exists A ∈ Msplit such that R (K) = det(A), and
p p
similarly for the other two cases.
We first note that there is a lower bound on the p-adic valuation of the regulator,
depending on whether the field is ramified or not at p.
4
Lemma 2. For a field cyclic cubic field K we have
1, if p is ramified in K,
v (R (K))≥
p p
(2, otherwise.
Proof. For a prime ideal p of K lying abovep let ν be the number ofp-powerrootsof
p
unity in the completion K . By [4, Appendix, Lemma 5] we know that
p
1 1
vp(Rp(K))≥vp(d(K)2)−vp(3)−1+ (f(p|p)+vp(νp))≥vp(d(K)2)−1+ f(p|p),
p|XpOK p|XpOK
where f(p|p) is the inertia degree of p over p. As v (d(K)) = 2 if p is ramified and
p
v (d(K))=0 otherwise, the claim follows.
p
To analyse the p-adic regulators,we will make use of the Galois module structure of
the unit group: we denote by U the unit group of O and for a subgroup U ⊆U we
K K K
setU∗ =U/{−1,1}. LetG=hσibetheGaloisgroupofK overQ. By[17]weknowthat
for everyprime p6=3 there exists a unit ε ∈U suchthat [U∗ :hGε i∗]is notdivisible
p K K p
by p. It follows that hε ,σ(ε )i is a subgroup of U of full rank and [U∗ : hε ,σ(ε )i∗]
p p K K p p
is prime to p. In particular the p-adic regulator can be computed using ε and σ(ε ).
p p
3.1. The split case
For a prime p>3 let us define
1 1 1
Msplit = a b −b−a a,b∈pZ ⊆Mat (Z ).
p −b−a a b  (cid:12) p 3×3 p
 (cid:12) 
(cid:12)
Proposition 3. Let Kbea cyclic cubic fieldan(cid:12)dp>3 aprime which splits in K. Then
 (cid:12) 
there exists A∈Msplit such that R (K)=det(A).
p p
Proof. We set ε = ε . Let µ ∈ Z[X] be the minimal polynomial of ε. Since p splits
p
we know that µ factors over Z as µ = (X − τ (ε))(X − τ (ε))(X − τ (ε)), where
p 1 2 3
τ ,τ ,τ : K →Q are the distinct p-adic embeddings. As f is also the minimal polyno-
1 2 3 p
mialofσ(ε),wemayassumethatτ (ε)=τ (σ(ε)),τ (ε)=τ (σ(ε))andτ (ε)=τ (σ(ε)).
1 2 2 3 3 1
In particular, the p-adic regulator can be computed as
1 1 1
R (K)=det log (τ (ε)) log (τ (ε)) log (τ (ε)) .
p  p 1 p 2 p 3 
log (τ (ε)) log (τ (ε)) log (τ (ε))
p 3 p 1 p 2
 
Now as τ (ε) ∈ 1 + pZ we have log (τ (ε)) ∈ pZ (for all n ≥ 1 the restriction
i p p i p
of log induces an isomorphism 1 + pnZ → pnZ ). Finally ε being a unit implies
p p p
3 log (τ (ε))=0.
i=1 p i
P The set Msplit is isomorphic to Z2 via
p p
1 1 1
Z2 −→Msplit, (a,b)7−→ pa pb −pb−pa .
p p  
−pb−pa pa pb
 
Using this isomorphism, we equip Msplit with its unique Haar measure µ such that
p
µ(Msplit)=1 and consider the random variable Msplit →R , A7→v (det(A)).
p p ≥0 p
5
3.2. The ramified case
Let p be a prime with p ≡ 1mod3. As the group F×/(F×)3 has order 3 and Q
p p p
containsthirdrootsofunity,thereareexactlythreecubicramifiedextensionsK ,K ,K
1 2 3
of Q , which are cyclic and totally ramified. We denote their valuations rings by O ,O
p 1 2
and O . Let Gal(K |Q )=hσ i. For i=1,2,3 we set
3 i p i
1 1 1
Mram = a σ (a) σ2(a) a∈O , a6=0, Tr (a)=0 ⊆GL (Q )
p,i σ (a) σ2i(a) ia  (cid:12) i Ki|Qp  3 p
 i i (cid:12) 
(cid:12)
  (cid:12)
and Mram =Mram ∪Mram ∪Mram ⊆(cid:12) GL (Q ). Since for a ∈ K ,a 6= 0, we have
p p,1 p,2 p,3 3 p i
Tr (a) = 0 if and only if a 6∈ Q this is in fact a disjoint union Mram = Mram ∪˙
Ki|Qp p p p,1
Mram∪˙ Mram.
p,2 p,3
Proposition 4. Let K be a cyclic cubic field and p>3 a prime which is ramified in K.
Then there exists A∈Mram such that R (K)=det(A).
p p
Proof. We set ε = ε . Let p be the prime ideal of K lying above p and K be the
p p
p-adiccompletion. As ε∈K ⊆K andK |Q iscyclic withGaloisgrouphτi, the p-adic
p p p
embeddings of ε are just ε,τ(ε),τ2(ε). Thus
1 1 1
R (K)= log (ε) log (τ(ε)) log (τ2(ε)) .
p  p p p 
log (τ(ε)) log (τ2(ε)) log (ε)
p p p
 
As τ commutes with logp and ε is a unit, the claim follows using the fact that Kp ∼=Ki
for some i=1,2,3.
3.3. The inert case
For a prime p > 3 denote by Q(3) the unique unramified cubic extension of Q
p p
and by Z(3) its ring of integers. This is a cyclic Galois extension with Galois group
p
Gal(Q(3)|Q)=hτi. We define
p
1 1 1
Minert = a τ(a) τ2(a) a∈Z(3), a6=0, Tr (a)=0 .
p τ(a) τ2(a) a  (cid:12) p Q(p3)|Qp 
 (cid:12) 
(cid:12)
  (cid:12)
Proposition 5. Let K be a cyclic cubic fie(cid:12)ld and p > 3 a prime which is inert in K.
Then there exists A∈Minert such that det(A)=R (K).
p p
Proof. Let p be the prime ideal of K lying above p. Then Kp ∼= Q(p3). Morevover,
the GaloisgroupGal(K |Q )is cyclicandthe restrictionGal(K |Q )→Gal(K|Q)is an
p p p p
isomorphism. The proof is now the same as in Proposition 4.
Let p>3 be a prime. Note that for each x, the set Mx is a subset of Mat (O) for
p 3×3
some ring of integers O of a finite extension of Q and therefore is naturally equipped
p
with canonical finite measure allowing us to speak of random variables on Mx. Based
p
on numerical observations, we make the following conjecture:
6
Conjecture 6. Let p>3 be a prime. For x∈{split, ram, inert} and i∈Z we have
≥0
#{K ∈Kx |v (R (K))=i and |d(K)|≤D}
p p p
lim =pr(X =i),
D→∞ #{K ∈Kpx ||d(K)|≤D}
where X: Mx →R , A7→v (det(A)).
p ≥0 p
4. Distribution of determinants of random matrices
For each x ∈ {split,ram,inert} we now compute the distribution of the random
variable Mx →R , A7→v (det(A)) and use this to make Conjecture 6 about the dis-
p ≥0 p
tributionsofthevaluationsofp-adicregulatorsmoreexplicitandaccessibletonumerical
investigations.
For a quadratic form g ∈ Z [X,Y] we denote by X the random variable X : Z2 →
p g g p
R , (a,b) 7→ v (g(a,b)). From the definition of equivalence of quadratic forms we
≥0 p
immediately obtain the following consequence for the associated random variables. If
h ∈ Z [X,Y] is another quadratic form, we write g ∼ h if g and h are equivalent, that
p
is, if there exists B ∈GL (Z ) such that g(X,Y)=h((X,Y)B).
2 p
Lemma 7. Let g,h∈Z [X,Y] be quadratic forms and α∈Z such that g ∼αh. Then
p p
X is in distribution equal to X +v (α).
g h p
Lemma 8. The random variable Msplit →R , A7→v (det(A)) is in distribution equal
p ≥0 p
to X +2, where f =X2+3Y2.
f
Proof. For an element
1 1 1
A= pa pb −pb−pa
 
−pb−pa pa pb
 
of Msplit we have det(A)=3p2(a2+b2+ab). Therefore Msplit →R , A7→v (det(A))
p p ≥0 p
is in distribution equal to X , where h = 3p2(X2+Y2+XY) ∈ Z [X,Y]. The claim
h p
follows from applying Lemma 7.
Lemma 9. Let p≡1mod3. The random variable Mram →R , A7→v (det(A)) is in
p ≥0 p
distribution equal to X +1, where f =X2+3Y2.
f
Proof. Since Mram is the disjoint union of the Mram it is sufficient to show it for
p p,i
each random variable Mram → R , A 7→ v (det(A)). Let K = K , O = O , σ = σ
p,i ≥0 p i i i
and denote by ζ ∈ O a third root of unity. As K|Q is totally ramified we can find a
p
primitive element α ∈ O such that O = Z [α] and α has minimal polynomial X3 −c
p
with c∈Z and v (c)=1. In particular we have a bijection
p p
1 1 1
ϕ: Z2 −→Mram∪{0}, (a,b)7−→ β σ(β) σ2(β) , β =aα+bα2.
p p,i  
σ(β) σ2(β) β
 
We may assume that σ(α) = ζα. Let A = ϕ(a,b). An easy calculations shows that
det(A) = det(ϕ(a,b)) = −9abc. As p ∤ 9 and v (c) = 1 we conclude that Mram →
p p,i
7
R , A7→v (det(A)) isindistributionequalto X +1,whereg =XY. As p≡1mod3
≥0 p g
we know that −3 is a quadratic residue modulo p, that is, there exists s ∈ Z such that
s2 ≡−3modp. Consider the matrix
1 1
B = .
s −s
(cid:18) (cid:19)
Now det(B) = −2s ∈ Z× and therefore g = XY ∼ (X +sY)(X −sY) = X2−s2Y2 =
p
X2+3Y2.
We nowturnto the inertcase. Unfortunately, inthis casewe onlyhavethe following
conjecture,whichbasicallystatesthatthematricesusedtocomputethep-adicregulator,
are uniformly distributed.
Conjecture 10. The random variable Minert →R , A7→v (det(A))is in distribution
p ≥0 p
equal to X +2, where f =X2+3Y2.
f
Remark 11. While we could not prove this conjecture, we have numerically tested it
for a large number of primes p. Moreover in Section 5 we have collected numerical evi-
dence for Conjecture 1, which is a consequence of Conjecture 10, therefore also justifying
Conjecture 10.
Let M = Mx for some x ∈ {split,ram,inert}. We now want to compute the proba-
p
bility thatthe valuationofthe determinantofa randomlychosenmatrixinM hasgiven
value. Since we have seen that M → R , A 7→ v (det(A)) is in distribution equal to
≥0 p
X +1 or X +2 respectively, where f = X2 +3Y2, it is sufficient to determine the
f f
probability pr(X =i), i∈Z . Now pr(X =i)=µ(M ), where
f ≥0 f i
M ={(a,b)∈Z ×Z |v (f(a,b))=i}
i p p p
and µ is the Haar measure on Z ×Z with µ(Z ×Z ) = 1. Note that since our form
p p p p
f has integer coefficients, for any commutative ring we have a unique evaluation map
R×R→R, (a,b)7→f(a,b) compatible with the unique morphism Z→R.
Lemma 12. For i≥0 we have
1
pr(X =i)= lim #{(x¯,y¯)∈(Z/pkZ)2 |v (f(x¯,y¯))=i}.
f k→∞p2k p
Proof. For k ≥1 let us denote by π : Z2 →(Z/pkZ)2 the canonical projectionand by
k p
µ the normalisedcountingmeasureon(Z/pkZ)2. As (Z )2 is the projectivelimit ofthe
k p
measure spaces ((Z/pkZ)2) , we have
k≥1
µ(M )= lim µ (π (M )).
i k k i
k→∞
Now π (M ) = {(x¯,y¯) ∈ (Z/pkZ)2 | v (f(x,y)) = i} and for k large enough we have
k i p
v (f(x¯,y¯))=i if and only if v (f(x,y))=i.
p p
Lemma 13. For k ≥i we have
#{(x,y)∈(Z/pkZ)2 |x2+3y2 ≡0modpi}=p2(k−i)#{(x,y)∈(Z/piZ)2 |x2+3y2 =0}.
8
ForacommutativeringRletusdenotebyX(R)theset{(x,y)∈R2 |x2+3y2 =0}.
Our aim is to compute the right hand side in Lemma 12 by calculating #X(R) for the
residue rings R=Z/pkZ. By the properties of f, this investigation naturally splits into
two cases.
4.1. The case p≡1mod3
Lemma 14. Let i≥0 and p≡1mod3. Then the following hold:
(i) We have #X(Z/piZ)=pi−1((p−1)i+p).
(ii) We have
2
i+1 1
µ(M )= 1− .
i pi p
(cid:18) (cid:19)
Proof. (i): As p is congruent 1 modulo 3, the Legendre symbol (−3) evaluates to 1,
p
thatis,−3isaquadraticresiduemodulop. Theprimepbeingodd,wecanfindtherefore
α∈Z/piZsuchthatα2 =−3inZ/piZ. Inparticularourbinaryquadraticformf factors
as X2+3Y2 =(X +αY)(X −αY). As
{(x,y)∈(Z/piZ)2 |x2+3y2 =0}−→{(u,v)∈(Z/piZ)2 |u·v =0},
(x,y)7−→(x+αy,x−αy),
is a bijection, it is sufficient to count the cardinality of the set on the right hand side.
We have
i−1
˙
{(u,v)∈(Z/piZ)2 |u·v =0}=U ∪˙ (E ×E ),
l ≥i−l
l=1
[
where E resp. E denotes the set of elements with valuation equal to l resp. greater
l ≥i−l
or equal to i−l and U is the set {(u,v) ∈ (Z/piZ)2 | u = 0 or v = 0}. As #E =
k
(p−1)pi−k−1 we obtain
i−1 i−1
#{(u,v)∈(Z/piZ)2 |u·v =0}=#U + #E #E
l j
l=1 j=i−l
X X
i−1 i−1
=2pi−1+ (p−1)pi−l−1 (p−1)pi−j−1.
l=1 j=i−l
X X
Now
i−1 l−1 l−1 1 j pl−1
pi−j−1 = pl−j−1 =pl−1 = ,
p p−1
j=i−l j=0 j=0(cid:18) (cid:19)
X X X
and
i−1 i−1 l
1
(p−1) pi−l−1(pl−1)=(p−1)pi−1 i− =(p−1)pi−1i−pi+1.
p
l=1 l=0(cid:18) (cid:19) !
X X
Thus
#X(Z/piZ)=2pi−1+(p−1)pi−1i−pi+1=pi−1((p−1)i+p).
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(ii): First note that for k ≥i we have
{(x¯,y¯)∈(Z/pkZ)2 |v (f(x¯,y¯))=i}
p
={(x¯,y¯)∈(Z/pkZ)2 |f(x¯,y¯)≡0modpi and f(x¯,y¯)6≡0modpi+1}
={(x¯,y¯)∈(Z/pkZ)2 |f(x¯,y¯)≡0modpi}\{(x¯,y¯)∈(Z/pkZ)2 |f(x¯,y¯)6≡0modpi+1}
=p2(k−i)·#X(Z/piZ)−p2(k−i+1)·#X(Z/pi+1Z).
Thus we conclude that µ(M ) is equal to
i
2
1 i+1 1
lim p2k−i−1((p−1)i+p)−p2k−(i+1)−1((p−i)(i+1)+p) = 1− .
k→∞p2k pi p
(cid:16) (cid:17) (cid:18) (cid:19)
4.2. The case p≡2mod3
Lemma 15. Let i≥0 and p≡2mod3. Then the following hold:
(i) We have #X(Z/piZ)=p2i−2⌈2i⌉.
(ii) We have
0, if i is odd,
µ(M )=
i (p1i 1− p12 , if i is even.
(cid:16) (cid:17)
Proof. First note that by assumption, −3 is not a square modulo pi for all i ≥ 1. (i):
Assume that (x,y)∈X(Z/piZ). If y is a unit, then −3=(x/y)2 is a square modulo pi,
acontradiction. Nowassume2v (y)<j,thatis, j <⌈i⌉. Wewritex=ε·pj, y =ε′·pj′
p 2
with units ε,ε′. Denoting by s the minimum min(2j,2j′) we obtain i>s and moreover
ε2p2j−s ≡−3ε′2p2j′−s modpi−s.
Since s = 2j or s = 2j′ this implies that −3 is a square modulo pi−s, a contradiction.
This shows that all elements (x,y) ∈ X(Z/piZ) have to satisfy y2 = 0, x2 = 0 and we
necessarily have X(Z/piZ)=#{x∈Z/piZ|x2 =0}2. As
i−1
˙
{x∈Z/piZ|x2 =0}={0}∪˙ E
l
l=[⌈2i⌉
we obtain
i−1
#{x∈Z/piZ|x2 =0}=1+ #El =1+pi−⌈2i⌉−1=pi−⌈2i⌉.
l=X⌈2i⌉
(ii): As in the proof of Lemma 14 we obtain
1
µ(Mi)=kl→im∞p2k p2k−2ip2i−2⌈2i⌉−p2k−2i−2p2i+2−2⌈i+21⌉ =p−2⌈2i⌉−p−2⌈i+21⌉.
(cid:16) (cid:17)
We can now combine these results:
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