The parity problem for irreducible cubic forms

0.23 MB·English

by  H. A. Helfgott

Submittedexclusively to the London Mathematical Society
DOI:10.1112/S0000000000000000
THE PARITY PROBLEM FOR IRREDUCIBLE CUBIC FORMS
H. A. HELFGOTT
Abstract
5 Letf ∈Z[x,y]beanirreduciblehomogeneous polynomialofdegree3.Weshowthatf(x,y)hasanevennumberof
0 primefactorsasoftenasanoddnumberofprimefactors.
0
2
1. Introduction
n
a Let f Z[x,y] be a homogeneous, non-constant polynomial. Then, it is believed,
J ∈
1
2 lim µ(f(x,y))=0. (1.1)
1 N N2
→∞ N x,y N
− ≤X≤

] This conjecture can be traced to Chowla ([1], p. 96). It is closely related to the Bunyakovsky–
T
Schinzel conjecture on primes represented by irreducible polynomials.
N
The one-variableanalogueof(1.1)isclassicalfordegf =1andquite hopelessfordegf >1.We
h. know(1.1)itselfwhendegf 2.(ThemainideasoftheproofgobacktodelaVall´ee-Poussin([3],
t [4]); see [11], 3.3, for an exp≤osition.) The problem of proving (1.1) when degf 3 has remained
ma openuntil now§:sievingis forestalledby the parity problem([17]), whichChowla≥’sconjecture may
be said to embody in its pure form.
[
We prove (1.1) for f irreducible of degree 3. In a companion paper [12], we prove (1.1) for f

1 reducible of degree 3.
v In [12], we follow Chowla’s original formulation, using the Liouville function λ instead of µ in
7
(1.1). For degf =3, the two formulations are equivalent: see 5.
7 §
1
1 1.1. Background
0
5 Problems such as the one consideredin this paper were until recently consideredintractable. In
0 1998, Friedlander and Iwaniec ([5], [6]) proved that there are infinitely many primes of the form
/
h x2+y4.Oneofthemaindifficultiesinadaptingthisapproachtoapolynomialf otherthanx2+y4
t resides in the need to prove that f satisfies a certain bilinear condition. Another difficulty is that
a
m the frameworkin [6] seemingly breaks down when f represents no more than Oǫ(N2/3+ǫ) integers
from 1 to N for every ǫ>0.
:
v Heath-Brownand Moroz have proved([13], [14], [15]) that any irreducible, homogeneouscubic
Xi polynomialf satisfiesabilinearconditionakintothatdemandedby[6].Sincesuchanf represents
no more than O (N2/3+ǫ) integers from 1 to N, Heath-Brown had to abandon the setup in [6],
r ǫ
a which is based essentially on Vaughan’s identity, for one based on Buchstab’s identity.
Theframeworkin[6]isquiteflexible,andcaneasilybeadaptedtoshowthatµ(x2+y4)averages
to zero. Unfortunately, it is not clear that [13] could possibly adapted in this way; Buchstab’s
identityisinsomesenselessgeneralthanVaughan’s,ortoagreaterextentastatementonlyabout
primes.
Our strategy will be to extend the originalFriedlander-Iwaniec method to sequences containing
atleastN2/3(logN) A integersfrom1toN,whereA>0.Thisextensiongoesslightlybeyondthe
−
naturalreachofthemethod.Ofthetechnicalinnovationsrequired,thefollowingmaybeapplicable
in a wider context.
2000Mathematics Subject Classification 11N32(primary),11N35, 11N36(secondary).
2 h. a. helfgott
1.2. Anti-sieving
Thegeneralsituationisasfollows.Wearegiventhetaskofestimatingasum F .Assume
ab x ab
we know how to estimate ≤
P
F and F , (1.2)
ab ab
ab x ab x
X≤ X≤
a xα/y(x) a xαy(x)
≤ ≥
where 0 α 1 andlogy(x)=o((logx)1/2).In orderto eliminate the missing intervalxα/y(x)<
≤ ≤
a<xαy(x), we use a sieve λ with the primes larger than y(x)2 as our sieving set:
d
F = λ F λ F
ab d ab d ab
  −
aXb≤x aXb≤x Xd|a aXb≤x Xd|a
xα/y(x)<a<xαy(x) xα/y(x)<a<xαy(x)  xα/y(x)<a<xαy(x) d>(y(x))2
Notice the use of a sieve as a combinatorial identity, rather than as a means of approximation;cf.
[12], 2.1. The first term on the right is typically at most
§
(logy(x))2
O X,
logx
(cid:18) (cid:19)
where X = F . The second term on the right equals
1 a x a
≤ ≤
P
 
λ F ,
 d ab
aXb≤x  Xd|b 
a<xα/y(x)d>max((y(x))2,xα/(ay(x))) 
 d<xαy(x)/a 
 
which is akin to the first sum in (1.2), and can often be treated by the same methods.
1.3. Acknowledgements
Thanks are due to the author’s thesis adviser, H. Iwaniec.
2. Notation
2.1. Ideals
Given a number field K/Q, we write O for the ring of integers of K and I for the semigroup
K K
of non-zero ideals of O . For a I , define τ(a) to be the number of divisors of a, ω(a) to be
K K
∈
the number of prime divisors of a and µ(a) to be ( 1)ω(a) if a is not divided by the square of any
−
element of I (set µ(a)=0 otherwise). Define rad(a)= p.
K pa
For a ∈ IK and S a finite set of prime ideals of OQK,|we define rS(a) = p S pvp(a) and
r\S(a)= p∈/S pvp(a). Q ∈
Q
2.2. Lattices and convex sets
A lattice is a subgroupof Z2 of finite index; a lattice coset is a cosetof such a subgroup.By the
index [Z2 : L] of a lattice coset L we mean the index of the lattice of which it is a coset. For any
lattice cosets L , L with gcd([Zn : L ],[Zn : L ]) = 1, the intersection L L is a lattice coset
1 2 1 2 1 2
∩
with
[Zn :L L ]=[Zn :L ][Zn :L ]. (2.1)
1 2 1 2
∩
For L Z2 a lattice coset and S R2 a convex set contained in a square of side N,
⊂ ⊂
Area(S)
#(S L)= +O(N), (2.2)
∩ [Z2 :L]
the parity problem for irreducible cubic forms 3
where the implied constant is absolute.
2.3. Shorthand
We adopt the following convention from [6]: given a property P, we define
f(x) if P(x) holds,
f(P(x))=
(0 otherwise.
For example, f(u x v) equals f(x) if u x v, and 0 otherwise. We will abuse notation by
≤ ≤ ≤ ≤
writing a>u (or a<u, a u, etc.) when we mean Na>u (or Na<u, etc.). Thus f(u a v)
≤ ≤ ≤
equals f(a) if u Na v, and 0 otherwise.
≤ ≤
We will write f(x) g(x) or f(x) = O(g(x)) to mean that f(x) cg(x) for some positive
≪ | | ≤
constant c. By p we shall always mean a prime ideal, and by p a rational prime.
3. Postulates
Let K/Q be a number field. Let a sequence a of non-negative reals be given. We are
{ a}a∈IK
requiredtoshowthattheaverageofa µ(a)overalla I iszero.Weneedonlycertainproperties
a K
∈
of a to show as much; it will be convenientto list them as postulates for further reference,
{ a}a∈IK
andtoprovethembeforethebeginningofthe mainargument.Weakerpostulatesmighthavebeen
used to the detriment of clarity.
As is customary, we define
A(n)= a , A (n)= a
a d a
aX∈IK aX∈IK
Na n da,Na n
≤ | ≤
for any n > 0, d I . When bounding a µ(a), we may fix n, and require the
postulates only for t∈hatKparticular value of n. a∈IK:Na≤n a
P
3.1. Statements
We will work with an approximation
A (n)=g(d)A(n)+r ,
d d
forsomeg :I [0,1).AssumethatthereisafixedD Z+ andaD Z+ withgcd(D ,D )=
K 0 1 0 1
→ ∈ ∈
1, D (logn)ρ, ρ>0, such that the following postulates hold:
1
≪
(a) For every prime ideal p I with Np prime and p∤D D ,
K 0 1
∈
(Np) α
g(pα)= −
1+1/Np
forallα 1.Forp I prime,p∤D ,withNpnon-prime,wehaveg(pα)=0forallα 1.
K 0
≥ ∈ ≥
For p I prime, pD , with Np prime, either g(pα) = 0 or g(pα) = (Np) α holds for all
K 1 −
∈ |
α 1.
≥
(b) For a,b I with gcd(Na,Nb)=1,
K
∈
g(ab)=g(a)g(b).
(c) For any two distinct prime ideals p ,p ∤ D lying above the same rational prime, we have
1 2 0
g(p p a)=0 for all a I .
1 2 K
∈
(d) For any C ,C 0, there is a κ >0 such that
1 2
≥
(τ(a))C1ra ≪C1,C2 A(n)(logn)−C2.
Na n2a/X∈3I(lKogn)−κ
≤
4 h. a. helfgott
(e) For any C ,C 0, there is a κ >0 such that
1 2
≥
(τ(d))C1Ad2(n)≪C1,C2 A(n)(logn)−C2.
Nd>dX∈(lIoKgn)κ
(f) For any C ,C 0, there is a κ >0 such that
1 2
≥
(τ(a))C1aa ≪C1,C2 A(n)(logn)−C2.
Na naX∈(loIKgn)−κ
≤
(g) Let b,c:I R be any functions with b(a) (τ(a))C1, c(a) (τ(a))C2. Let
K
→ ≪ ≪
d(a)= c(a/d)µ(d>ℓ), (3.1)
da
X|
gcd(d,D)=1
where D D D and ℓ nǫ for some ǫ>0.Then, for anyC 0,there is a κ >0 suchthat
0 1 3
| ≫ ≥
b(a)d(b)aab ≪C1,C2,C3 A(n)(logn)−C3
a,b
X
ab n
≤
v Nb<2v
≤
for all v [n1/2(logn)κ,n3/2(logn) κ]. Both κ and the implied constant are independent
−
∈
of b, c, D and ℓ.
In brief – (a) is a statement on g(pα), (b) and (c) establish what one may call the quasi-
multiplicativityofg,(d)statesthatthe residuesaresmallenough,(e)statesthatfewidealsawith
a =0 have large square factors,and (f) is a weak bound on growth.All of these are postulates of
a
6
a classical kind (“type I”) whereas (g) is a bilinear condition (and thus of “type II”).
3.2. Verification
Let K/Q be a cubic extension of Q. Let ω ,ω O be Q–linearly independent. Let ̟ >0 be
1 2 K
an arbitraryconstant.Given a lattice L Z2, we∈write b for the minimal idealof O containing
L K
⊂
the image of L under the map (x,y) (xω +yω ).
1 2
7→
Wemustshowthat,foranysufficientlylargeN,anyconvexsubsetS [ N,N]2withArea(S)>
⊂ −
N2(logN) ̟, and any lattice coset L of index [Z2 :L] (logN)̟, the sequence
−
≤
a = 1
a
(x,yX)∈S∩L
gcd(x,y)=1
(xω1+yω2)=abL
satisfies the postulates with n = max N(xω +yω ) const N3. The constants depend
(x,y) S 1 2
only on K, ω , ω and ̟. ∈ | | ∼ ·
1 2
3.2.1. Linear postulates. By [14], Lemma 2.2, there is a D Z+ such that, for any prime
0
∈
ideal p I , if p ∤ D and pxω +yω for some coprime x,y Z, then Np is prime. Define
K 0 1 2
∈ | ∈
D =[Z2 :L].
1
Givend I ,letL = x,y L:db xω +yω .Foranyd I whosenormisaprimepower
K d L 1 2 K
∈ { ∈ | } ∈
pα, define
[Z2 :L ] 1 [Z2 :pZ2 L ] 1
g(d)= d − − ∩ d − .
[Z2 :L] 1 [Z2 :pZ2 L] 1
− − ∩ −
(If L = , we set g(d) = 0.) Then postulate (a) follows easily. We may take postulate (b) as a
d
∅
definition, andthen postulate (c) alsofollows.Postulate(f) is aroutinaryconsequenceof(2.2) (cf.
[5], p. 1047).
Postulate (d) is in essence the same as Lemma 2.2 in [15], Lemma 3.2 in [14], or, ultimately,
Lemma 5.1 in [13]. The contribution of a with square factors can be bounded easily (use [10],
the parity problem for irreducible cubic forms 5
Lemma A.5, for factors (logN)C), and the validity for all convex sets S can be obtained by
≫
partitioning them into squares of side N(logN) C.
† −
∼
Thenon-trivialpartofpostulate(e)residesinbounding thecontributionoftermswithdprime,
Nd n(logn) C. Use the bound onthe number of points per fibre in,e.g., [10], Proposition4.13,
−
toge≥ther with the crudest bound for τ. (The techniques in [8] would be enough: postulate (e) is
simply stating that few values of a homogeneous cubic have large square factors.)
3.2.2. Bilinear postulate. While our general framework resembles that of Friedlander and
Iwaniec ([6]), the bilinear condition (g) is that of Heath-Brown ([13], [14], [15]). As (g) is an
postulate of the less familiar kind, it is worthwhile to specify the minor changes we must make to
the statement and the proof of Proposition6.1(ii) in [14].
First, note that any function defined as in 3.1 can be extended to ideal numbers so as to fulfil
condition(6.1)in[14],i.e.,soastoaveragetozeroatleastasfastasexp( c√logx)whenrestricted
−
to particular ideal classes and lattices of index (logx)C. This is simply Siegel-Walfisz; here the
≪
condition ℓ nǫ in (g) is crucial.
Whilethe≫conditionsonb(a)areleftunspecifiedinProposition6.1(ii)of[14],itisenoughtohave
b(a) (τ(a))C. The term b(a) disappears by Cauchy’s inequality before the second equation of p.
279 o≪f [14]. While a lacunarity condition on b(a) (not fulfilled here) is implicitly used to eliminate
smallcommonfactorsatthe beginningofthe proofof[14],Proposition6.1(see also[13], 11),we
may remove the condition gcd(x,y) = 1 by carrying the argument in [14], pp. 278–284,§for each
lattice coset L defined by (x,y) L,d gcd(x,y), where d (logN)C, and then sieving out such
d
lattice cosets. A riddle (vd. Prop∈osition| 3.2 and Corollary≪3.3 in [10]) enables us to sieve with a
totalerrortermofrelativesizeO((logN) C+O(1))whilehandlingonlytheL ’swithd (logN)C;
− d
cf. [10], Propositions 3.11 and 3.12. ≪
4. Proof
Lemma 4.1 (A variant of Vaughan’s identity). Let y, u, w be positive numbers satisfying y
u w. Let K/Q be a number field; let Q be a finite set of prime ideals thereof, and let h:I ≤C
K
≤ →
be an arbitrary function. Then, for any a I ,
K
∈
4 7
h(a)= β (a) β (a),
j j
−
j=1 j=5
X X
where
β (a)=h(a u)+ h(b)µ(c u),
1
≤ ≤
X∗
β (a)= h(u<b w)µ(c>u), β (a)= h(b>w)µ(u<c w),
2 3
≤ ≤
X∗ X∗ (4.1)
β (a)= h(b>w)µ(c>w), β (a)= h(b u)µ(c y),
4 5
≤ ≤
X∗ X∗
β (a)= h(b y)µ(y <c u), β (a)= h(y <b u)µ(y <c u),
6 7
≤ ≤ ≤ ≤
X∗ X∗
and stands for .
bca,rQ(b)=rQ(a)
∗ |
P P
Proof. For any a I ,
K
∈
β (a)+β (a)+β (a)= h(b>u)µ(c>u), β (a)+β (a)+β (a)= h(b u)µ(c u).
2 3 4 5 6 7
≤ ≤
X∗ X∗
†While[13]takesasitsobjectasquarewithacorneroftheform(x,y),x=y,itsargumentsworkforanysquare
whosecorners(x,y)satisfyN(logN)−C ≪x,y≪N(logN)C andwhosesides,asin[13],are≫N(logN)−C.
6 h. a. helfgott
Thus
h(a)=h(a u)+h(a>u)=h(a u)+ h(b>u)µ(c)=h(a u)+ h(b)µ(c u)
≤ ≤ ≤ ≤
X∗ X∗
7 4
h(b u)µ(c u)+ h(b>u)µ(c>u)=β (a) β (a)+ β (a).
1 j j
− ≤ ≤ −
j=5 j=2
X∗ X∗ X X
Set h = µ in Lemma 4.1. Let z = e(loglogx)(logloglogx)ǫ/2 for some ǫ > 0. Let y = x1/3z 2,
−
u = x1/3z, w = x1/2z 1, Q = p : pD D . Since we wish to estimate a µ(a), we will
− { | 0 1} a x a
evaluate a β (a) for j =1,2,...,7. The cases j =1, j =5 and j =6 are≤easy.
a≤x a j P
LemmaP4.2. Let a be a sequence satisfying the postulates for n=x. Then, for any C >0,
a
{ }
a β (a) A(x)(logx) C.
a 1 −
≪
a x
X≤
Proof. By postulate (f), we have a µ(a u) A(x)(logx) C and
a x a ≤ ≪ −
≤
aaµ(b)µ(c u)= aaµ(PrQ(a))µ(r Q(a) u) τ(a)aa A(x)(logx)−C.
≤ \ ≤ ≪ ≪
aX≤xX∗ aX≤x a≤uXD03D13
Lemma 4.3. Let a be a sequence satisfying the postulates for n=x. Then, for any C >0,
a
{ }
a β (a) A(x)(logx) C.
a 5 C −
≪
a x
X≤
Proof. By exclusion–inclusion and postulate (d),
a β (a)= µ(e) µ(b) µ(c) a
a 5 ebcd
aX≤x e:rQX(e)=e bX≤u Xc≤y d≤xX/N(ebc)
r\Q(b)=b r\Q(c)=c r\Q(d)=d
= µ(e) µ(f) µ(b) µ(c) a
efbcd
e:rQX(e)=e f:rXQ(f)=f bX≤u Xc≤y d≤x/XN(efbc)
r\Q(b)=b r\Q(c)=c
A(x) µ(e) µ(f) µ(b)g(b) µ(c)g(c)
≤
e:rQX(e)=e f:rXQ(f)=f bX≤u Xc≤y
r\Q(b)=b r\Q(c)=c
gcd(Nb,Nc)=1
+O(A(x)(logx) C).
−
By the standard zero-free region for ζ (s), the innermost sum is e c√logx, c>0.
K c ≪ −
Lemma 4.4. Let a be a sequence satisfying the postulatePs for n=x. Then, for any C >0,
a
{ }
aaβ6(a) C A(x)(logx)−C.
≪
a x
X≤
Proof. Same as Lemma 4.3.
The following two lemmas are direct consequences of the bilinear postulate.
Lemma 4.5. Let a be a sequence satisfying the postulates for n=x. Then, for any C >0,
a
{ }
a β (a) A(x)(logx) C.
a 2 C −
≪
a x
X≤
the parity problem for irreducible cubic forms 7
Proof. Apply postulate (g) with D =D D , ℓ=u,
0 1
1 if gcd(a,D D )=1,
b(a)=µ(u<a w), c(a)= 0 1
≤ (0 otherwise,
and v ranging fromxw 1(logx) C to xu 1/2. Use postulate (g) to bound the remaining terms.
− − −
Lemma 4.6. Let a be a sequence satisfying the postulates for n=x. Then, for any C >0,
a
{ }
a β (a) A(x)(logx) C.
a 3 C −
≪
a x
X≤
Proof. Apply postulate (g) with D =1, ℓ=w,
µ(u<a w) if gcd(a,D D )=1, 1 if gcd(a,D D )=1,
b(a)= ≤ 0 1 c(a)= 0 1
(0 otherwise, (0 otherwise,
and v ranging fromxw 1(logx) C to xu 1/2. Use postulate (g) to bound the remaining terms.
− − −
It remains to consider the sums a β (a) for j = 4,7. We will recur to anti-sieving and a
a a j
certain kind of cancellation.
P
Lemma 4.7. Let a be a sequence satisfying the postulates for n=x. Then
a
{ }
(loglogx)4(logloglogx)ǫ
a β (a) A(x),
a 7
≪ logx
a x
X≤
where ǫ is as in the definition of z.
Proof. Let λ be a Rosser-Iwaniec sieve [2] with sieved set A = b I : y < Nb u ,
d K
multiplicities θ{(b)}= g(b)µ2(b), sieving set P = p : uy 1 < Np wu{1 ∈and upper cut w≤u }1.
− − −
{ ≤ }
(Brun’spuresievewoulddoalmostaswell.)Bydefinition,λ =0if1<Nd uy 1orNd>wu 1.
d − −
≤
Since λ =1, we have the identity
1
1= λ λ
d d
−
Xd|b uy−1<Xd≤wu−1
db
|
for every b I . Hence β (a)=β (a) β (a), where
K 7 8 9
∈ −
β (a)= λ µ(y <b u)µ(y <c u)
8 d
≤ ≤
bca db
X| X|
pa/b p/Q
| ⇒ ∈
β (a)= λ µ(y <b u)µ(y <c u).
9 d
≤ ≤
Xbc|a uy−1<Xd≤wu−1
pa/b p/Q db
| ⇒ ∈ |
We begin by bounding a β (a). Changing the order of summation, we obtain
a x a 9
≤
a β (a)= P µ(c) λ µ(de) a
a 9 d cdef
aX≤x pyc<Xcp≤/uQ uy−1<Xd≤wu−1 y/Nd<Xe≤u/Nd f≤pxfX/Np(/cdQe)
| ⇒ ∈ |⇒ ∈
= λ µ(d)µ(g/d) µ(e) a .
d gef
u<Xg≤w Xd|g y/Nd<Xe≤u/Nd f≤xX/N(ge)
d>uy−1,d Ng/u gcd(e,d)=1 pf p/Q
≥ |⇒ ∈
d wu−1,d<Ng/y
≤
pg/d p/Q
| ⇒ ∈
Since d has no factors of norm less than z3 when λ =0, we may removethe condition gcd(e,d)
d
6
with an error of at most O(A(x)(logx) C) by means of postulates (a)–(d). We can make the
−
8 h. a. helfgott
intervalsofsummationofdandeindependentfromeachotherbyslicing[uy 1,wu 1]intointervals
− −
of the form [K,K(1+(logx) C)). There are at most O((logx)C+1) such intervals, and the error
−
incurred during the slicing is at most O(A(x)(logx) C+O(1)). Hence
−
(cid:12) (cid:12)
a β (a) A(x)(logx) C+O(1)+(logx)C+1 max (cid:12) b (g)d (h)a (cid:12),
a 9 − (cid:12) K K gh(cid:12)
≪ uy−1<K wu−1(cid:12) (cid:12)
aX≤x ≤ (cid:12)(cid:12) gXhg,hx (cid:12)(cid:12)
(cid:12) ≤ (cid:12)
(cid:12)u<g w (cid:12)
(cid:12) ≤ (cid:12)
where (cid:12) (cid:12)
(cid:12) (cid:12)
b (g)= λ µ(d)µ(g/d), d (h)= µ(e). (4.2)
K d K
dg eh
X| X|
K d<(1+(logx)−C)K y/K<e u/K
≤ ≤
y<g/d u ph/e p/Q
≤ | ⇒ ∈
pg/d p/Q
| ⇒ ∈
We apply postulate (g) with D =1, b=b , c(a)=1 when gcd(a,D D )=1, c(a)=0 otherwise,
K 0 1
and ℓ=y/K or ℓ=u/K (in succession). We obtain a bound a β (a) A(x)(logx) C′ for
a x a 9 ≪ −
C arbitrarily large. ≤
′
P
We must now bound
a β (a)= λ µ(b) µ(y <c u)a . (4.3)
a 8 d be
≤
aX≤x y<Xb≤uXd|b e≤Xx/Nb Xc|e
pe p/Q
| ⇒ ∈
We must find cancellation in the innermost sum and lower bc below (logx) κx2/3. For e>1,
−
µ(y <c u)= µ(c) µ(c y) µ(c>u)
≤ − ≤ −
ce ce ce ce
X| X| X| X| (4.4)
= µ(c y) µ(rad(e)) µ(c<rad(e)/u).
− ≤ −
ce ce
X| X|
Wewillfirstboundthecontributionfrom µ(c y).Aboundforµ(rad(e)) µ(c<rad(e)/u)
ce ≤ ce
will be obtained later in a similar fashion. T|he terms with e=1 may be ignored|by postulate (f).
P P
Suppose e has a prime divisor p l, where l >0 is fixed. Then the set of all square-freedivisors
≤
of e can be partitioned into pairs o,op . Evidently, µ(o)= µ(op). We have either o y, op y
{ } − ≤ ≤
or o>y, op>y, unless o lies in the range y/l<c y. Hence
≤
µ(c y) 1. (4.5)
(cid:12) ≤ (cid:12)≤
(cid:12)ce (cid:12) ce
(cid:12)X| (cid:12) X|
(cid:12) (cid:12) y/l<c y
(cid:12) (cid:12) ≤
Define l =22j for j 0. Note that(cid:12)x1/2 <l (cid:12) x. Let
j log log x
≥ ⌊ 2 2 ⌋ ≤
L = e:2e,(pe p / Q) ,
0
{ | | ⇒ ∈ }
L = e I :( p l s.t. pe) ( p l , p∤e) (pe p / Q) .
j K j j 1
{ ∈ ∃ ≤ | ∧ ∀ ≤ − ∧ | ⇒ ∈ }
Then, by (4.5),
µ(c y) a a a . (4.6)
be be bcj
(cid:12) ≤ (cid:12) ≤ ≤
e≤Xx/Nb(cid:12)(cid:12)Xc|e (cid:12)(cid:12) e≤Xx/Nb Xc|e y/lXj<c≤y j≤xX/N(bc)
e∈Lj (cid:12)(cid:12) (cid:12)(cid:12) e∈Lj y/lj<c≤y p|c⇒p>lj−1,p∈/Q
Bypostulates(a)–(cid:12)(d),itfollow(cid:12)sthatthecontributionofthesums µ(c y)to(4.3)isatmost
ce ≤
|
P
⌊log2log2x⌋
A(x) λd g(bc)+O(A(x)(logx)−C) (4.7)
y<Xb≤u Xd|b Xj=0 y/lXj<c≤y
µ(b)=±1 p|c⇒p>lj−1,p∈/Q
the parity problem for irreducible cubic forms 9
in absolute value. Now g(bc) g(b)/Nc, and
≤
1 y 1 logl
j
(logy log ) 1 1.
Nc ≪ − l · − Np ≪ logl ≪
y/lXj<c≤y j p≤Ylj−1(cid:18) (cid:19) j−1
p|c⇒p>lj−1,p∈/Q
Thus, the quadruple sum in (4.7) is at most O( λ g(b)) loglogx. The main
result on the Rosser-Iwaniec sieve ([2], Lemma 3),y<grba≤nut,eµd(b)p=o±st1uladt|ebs (da)–(c),·gives:
P P
loguy 1
λ g(b) µ(b)g(b) (logu logy) −
d ≪ | | ≪ − logwu 1
y<b u db y<b u −
µ(bX)=≤ 1X| uy−1<p<Xw≤u−1 p∤b (4.8)
± ⇒
(logz)2 (loglogx)2(logloglogx)ǫ
= .
≪ logx logx
It remains to bound the contribution from the terms µ(rad(e)) µ(c<rad(e)/u), viz.,
ce
|
P
λ µ(c<rad(e)/u) a . (4.9)
d be
(cid:12) (cid:12)
y<Xb≤u Xd|b e≤Xx/Nb (cid:12)(cid:12)Xc|e (cid:12)(cid:12)
µ(b)= 1 pe p/Q(cid:12) (cid:12)
± | ⇒ ∈ (cid:12) (cid:12)
Bypostulates(e)and(f),thereisaC >0sucht(cid:12)hatthetermswithN(cid:12) e x/(logx)CNbcontribute
≤
less than A(x)/logx to the total. Proceeding as in (4.5)–(4.7), we obtain that (4.9) is at most
⌊log2log2x⌋⌊log2log2x⌋
A(x) λ g(bc)+O(A(x)(logx) C).
d −
y<Xb≤u Xd|b Xj=0 Xk=0 x/(lju2k+1·NbX)≤c<x/(u2k·Nb)
µ(b)=±1 p|c⇒p>lj−1,p∈Q
Again as before, we obtain that this is at most O( λ g(b))A(x)(loglogx)2.
y<b u,µ(b)= 1 db d
By (4.8), we are done. ≤ ± |
P P
Lemma 4.8. Let a be a sequence satisfying the postulates for n=x. Then
a
{ }
(loglogx)4(logloglogx)ǫ
a β (a) A(x),
a 4
≪ logx
a x
X≤
where ǫ is as in the definition of z.
Proof. Let λ be a Rosser-Iwaniec sieve with sieved set A = b I : w < Nb < x/w ,
d K
multiplicities θ{(b)}= g(b)µ2(b), sieving set P = p : xw 2 < Np{ ∈wx 1y2 and upper cu}t
− −
{ ≤ }
wx 1y2. By definition, λ =0, if 1<Nd xw 2 or Nd>wx 1y2. Then β (a)=β (a) β (a)
− d − − 4 10 11
≤ −
for a I with a x, where
K
∈ ≤
β10(a)= λdµ(w <b xw−1)µ(w <c xw−1)
≤ ≤
bca db
X| X|
pa/b p/Q
| ⇒ ∈
β (a)= λ µ(w <b xw 1)µ(w <c xw 1).
11 d − −
≤ ≤
Xbc|a xw−2<Xd<wx−1y2
pa/b p/Q db
| ⇒ ∈ |
Much as for a β (a) in the proof of Lemma 4.8, we obtain a β (a) A(x)(logx) C from
a 9 a 11 −
≪
an application of postulate (g) with D =D D , ℓ=w or ℓ=xw 1 (in succession),
0 1 −
P P
b(a)=µ(max(xy 2,K) a<min(w,K(1+(logN) C)), c(a)= λ ,
− − d
≤
da
X|
w/K<d xw−1/K
≤
pa/d p/Q
| ⇒ ∈
10 h. a. helfgott
and K between xy 2 and w; cf. (4.2). It remains to bound the contribution of β , viz.,
− 10
aaβ10(a)= λdµ(b) µ(w <c xw−1)abe.
≤
aX≤x w<bX≤xw−1Xd|b e≤Xx/Nb Xc|e
pe p/Q
| ⇒ ∈
Since (cf. (4.4))
µ(w <c xw 1)= µ(c>w) µ(c>xw 1)
− −
≤ −
ce ce
X| X|
=µ(rad(e)) µ(c<rad(e)/w) µ(rad(e)) µ(c<rad(e)/xw 1),
−
−
ce ce
X| X|
we may proceed as in the proof of Lemma 4.7 and obtain
(loglogx)4(logloglogx)ǫ
a β (a) A(x).
a 11
≪ logx
a x
X≤
Proposition 4.9. Let a be a sequence satisfying the postulates for n = x. Then, for any
a
{ }
ǫ>0,
(loglogx)4(logloglogx)ǫ
a µ(a) A(x).
a ǫ
≪ logx
a x
X≤
Proof. Immediate from Lemmas 4.1 – 4.8.
5. Conclusion
The Liouville function λ:Z 0 1,1 is defined as follows:
−{ }7→{− }
λ(n)= ( 1)vp(n).
−
Yp|n
Main Theorem. Let f Z[x,y] be an irreducible homogeneous polynomial of degree 3. Let N
∈
be a positive number, S [ N,N]2 a convex set with Area(S) > N2(logN) A, and L a lattice
−
⊂ −
coset of index [Z2 :L] (logn)A. Let α=λ, α=µ, or α(n)=( 1)ω(n). Then, for every ǫ>0,
≪ −
(loglogN)4(logloglogN)ǫ
α(f(x,y)) 1,
f,A,ǫ
≪ logN ·
(x,yX)∈S∩L (x,yX)∈S∩L
gcd(x,y)=1 gcd(x,y)=1
(5.1)
(loglogN)4(logloglogN)ǫ
α(f(x,y)) 1,
f,A,ǫ
≪ logN ·
(x,yX)∈S∩L (x,yX)∈S∩L
(x,y)=(0,0)
6
Proof. By [14], Lemma 2.1, there are a cubic extension K/Q and Q–linearly independent
elementsω ,ω K suchthatf(x,y)= N(xω1+yω2),wheredistheminimalidealofO containing
1 2 ∈ Nd K
Zω +Zω . Proposition 4.9 implies that
1 2
N(xω +yω ) (loglogN)4(logloglogN)ǫ
1 2
µ 1
Nb ≪f,̟,ǫ logN ·
(x,yX)∈S∩L (cid:18) (cid:19) (x,yX)∈S∩L
gcd(x,y)=1 gcd(x,y)=1
b|xω1+yω2 b|xω1+yω2
for all non-zero ideals b d with Nb (logN)̟, ̟ > A. There are few pairs (x,y) such
that d2 f(x,y) for d large⊂; see postulate≪(e), or, ultimately, [8]. The rest is routine. Use, e.g.,
Proposi|tion 3.2 in [10] as in the proofs of Propositions 3.11–3.12 (ibid.). Once the first equation
in (5.1) is proved, the second one follows easily.