Every nonzero integral ideal can be expressed as the product of finite prime ideals in Dedekind domain. For each integral ideal A, it is essential to measure the multiplicity of its prime ideal factors. We define λ(A):=logN(A)logγ(A) to be the index of composition of A, where γ(A)=∏P|AN(P) and N(A) is the norm of ideal A. In this paper, we obtain an Ω-result for the mean value of the index of composition of integral ideal.
Citation: Jing Huang, Wenguang Zhai, Deyu Zhang. Ω-result for the index of composition of an integral ideal[J]. AIMS Mathematics, 2021, 6(5): 4979-4988. doi: 10.3934/math.2021292
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Every nonzero integral ideal can be expressed as the product of finite prime ideals in Dedekind domain. For each integral ideal A, it is essential to measure the multiplicity of its prime ideal factors. We define λ(A):=logN(A)logγ(A) to be the index of composition of A, where γ(A)=∏P|AN(P) and N(A) is the norm of ideal A. In this paper, we obtain an Ω-result for the mean value of the index of composition of integral ideal.
Let K be an algebraic number filed of degree d and OK be the ring of integers of K. For each integral ideal A∈OK, A=Pe11⋯Pegg, where the Pi(i=1,...,g) are prime ideals of OK, this expression is unique up to the order of the factors. Motivated by [17], we define λ(A):=logN(A)logγ(A) be the index of composition of A, where N(A) is the norm of ideal A and γ(A)=∏P|AN(P). We write λ(A)=γ(A)=1 if A=OK. The index of composition of an integral ideal measures the multiplicity of its prime factors.
Before stating our main results, we introduce some notations. The Dedekind zeta-function for the field K is defined by
ζK(s)=∑A≠01Ns(A)=∞∑n=1anns, ℜs>1, |
where an is the number of integral ideals of K with norm n. ζK(s) can be analytically continued to the whole complex plane, s=1 is a simple pole with residue
ρK=2r1(2π)r2RKωK√|d(K)|, |
where r1,r2 denote the number of real and complex places respectively, RK is the regular of K, d(K) is the discriminant of K and ωK is the order of the group of units. We know that an≤(τ(n))d, where τ(n) is the number of divisors of n. Let
C(x):=∑n≤xan=ρKx+Δ(x). | (1.1) |
Δ(x)=O(xθd+ϵ), |
where
θd:={131416,if d=2,4396,if d=3,1−2d+8d(5d+2),if d=4,5,6,1−2d+32d2,if d≥7.1−3d+6,if d≥10. |
In [15], Zhang and Zhai obtained a series of results about the mean value of λ±k(A). The results imply that the average order of λ(A) is ρK. Also they found that the mean value of λ−1(A) has a close connection with the zero free region of ζK(s) and got that
∑N(A)≤xλ−1(A)=ρKx+C1∫x21logzdz+OK(xϑd+ε), |
where
ϑd:={12,if d=2,31−2d+8d(5d+2),if d=4,5,6,1−2d+32d2,if d≥7. |
If K is a quadratic or cubic number field, Zhang and Zhai [16] proved the asymptotic formula
∑N(A)≤xλ−1(A)=ρKx+C1∫x21logzdz+C2∫x2z−12logzdz+OK(R(x)), | (1.2) |
where
R(x)=x12exp(−clog13x(loglogx)−13), |
and C1,C2 are computable constants, c>0 is a positive constant. Assuming the Riemann Hypothesis for ζK(s) is true, Zhang and Zhai [16] used the estimation of exponential sum and convolution method to get
R(x)={x5/12+ε,if d=2,x73/156+ε,if d=3. |
It is natural to consider how well the main term of ∑N(A)≤xλ−1(A) approximates it, that is, what can be said about Ω- result (for example, see [6,7,18]). In this paper, we shall get the following result.
Theorem 1.1. Let K be a quadratic or cubic number field. Then we have
∑N(A)≤xλ−1(A)=ρKx+C1∫x21logzdz+C2∫x2z−12logzdz+C3∫x2z−23logzdz+R0(x), |
where Ci(i=1,2,3) are computable constants and the error term R0(x) satisfies
∫Y1R0(x)dx=Ω(Y54logY). | (1.3) |
Remark. In order to prove Theorem 1.1, we will follow the line in Pintz [12] and Nowak [10], using Mellin transform and constructing an auxiliary function g(s) with some properties. Because λ(A) is not multiplicative, it is not easy to get the generating series of λ−1(A). Instead we shall study the mean value of ∑N(A)≤xlogγ(A). In fact, as a function of z, γz(A) is regular for |z|≤ε, then we can differentiate it and set z=0 to get the mean value and generating series for logγ(A). Theorem 1.1 can follows from the lower bound for the error term of ∑N(A)≤xlogγ(A).
It is easy to see that (1.3) implies the following Ω-result.
Theorem 1.2. Let K be a quadratic or cubic number field. Then we have
R0(x)=Ω(x1/4logx). |
Notation. Throughout the paper ε always denotes a fixed but sufficiently small positive constant. We write f(x)≪g(x), or f(x)=O(g(x)), to mean that |f(x)|≤Cg(x). ∑n∼N denote that the sum over N<n≤2N. f(x)=Ω(g(x)) means that there exists a suitable constant C>0 such that |f(x)|>Cg(x) holds for a sequence x=xn such that limn→∞xn=∞.
In this section, suppose ε>0 is a small positive constant, z is a complex number such that |z|≤ε. Let s=σ+it be a complex number with ℜ(s−z)>1. Define
G(s,z):=∑Aγz(A)N−s(A). |
Lemma 2.1. For |z≤ε, we have
G(s,z)=ζK(s−z)ζK(2s−z)ζK(3s−z)ζK(4s−z)ζK(4s−3z)ζK(2s−2z)ζK(3s−2z)ζK(4s−2z)G1(s,z), |
where G1(s,z) can be expanded into a Dirichlet series of s, which is absolutely convergent for σ>15+ε.
Proof. By Euler product representation, we have
G(s,z)=∏P(1+Nz(P)Ns(P)+Nz(P)N2s(P)+Nz(P)N3s(P)+⋯)=ζK(s−z)G∗(s,z), |
where ζK(s) is the Dedekind ζ-function, and
G∗(s,z):=∏P(1−Nz(P)Ns(P))∏P(1+Nz(P)Ns(P)+Nz(P)N2s(P)+Nz(P)N3s(P)+⋯)=ζK(2s−z)G∗1(s,z), |
where
G∗1(s,z)=∏P(1−N2z(P)N2s(P)−N2z(P)N4s(P)+N3z(P)N4s(P)−⋯). |
Arguing similarly, we can get
G(s,z)=ζK(s−z)ζK(2s−z)ζK(3s−z)ζK(4s−z)ζK(4s−3z)ζK(2s−2z)ζK(3s−2z)ζK(4s−2z)G1(s,z), |
where
G1(s,z)=∏P(1+Nz(P)N5s(P)−N2z(P)N5s(P)+N3z(P)N5s(P)−⋯). |
By the similar method as before, we know that G1(s,z) can be written as the product of Dedekind ζ-functions. If we note that |z|≤ε, then G1(s,z)can be expanded to a Dirichlet series, which is absolutely convergent for ℜs>1/5+ε.
To obtain the mean value of ∑N(A)≤xlogγ(A), we need the following Lemma.
Lemma 2.2. If σ>1, then
ζK(s)≪log(|t|+2). | (2.1) |
Let k≥0 be an integer. Uniformly for 12≤σ≤1, we have
ζ(k)K(s)≪(|t|+2)d3(1−σ)logk+1(|t|+2), | (2.2) |
where ζ(k)K(s) is the k-th derivative of ζK(s).
Proof. The order of ζK(s) can be found in [16,Lemma 2.3]. For the cases k≥1 of (2.2), we use the Cauchy derivative formula to get
ζ(k)K(s)=k!2πi∫|z−s|=RζK(z)(z−s)k+1dz. |
Let z=s+Reiθ and R=1/log(|t|+2). Then the above formula can be written as
ζ(k)K(s)=k!2π∫2π0ζK(s+Reiθ)(Reiθ)kdθ≪logk(|t|+2)|ζK(s+Reiθ)|. |
Therefore Lemma 2.2 follows from the order of ζK(s) (k=0 in (2.2)).
Proposition 2.3. We have
∑N(A)≤xlogγ(A)=ρKxlogx+c1x+c2x12+c3x13+E0(x), |
where ci (1≤i≤3) are computable constants, and for Y→∞, the error term E0(x) satisfies
∫Y1|E0(x)|dx≫Y54. |
Proof. By Lemma 2.1, we have
G(s,z)=∑Aγz(A)N−s(A)=ζK(s−z)ζK(2s−z)ζK(3s−z)ζK(4s−z)ζK(4s−3z)ζK(2s−2z)ζK(3s−2z)ζK(4s−2z)G1(s,z). | (2.3) |
Firstly, we take the first partial derivative with respect to z form both sides of (2.3), and put z=0 to get
∂G(s,z)∂z|z=0=∑Aγz(A)logγ(A)Ns(A)|z=0=∑Alogγ(A)Ns(A)=ζK(s)ζK(4s)G1(s,0){ζ′K(2s)ζK(3s)+ζK(2s)ζ′K(3s)}ζK(2s)ζK(3s)+G2(s), |
where
G2(s)=ζK(s)ζK(4s)G′1(s,0)−ζ′K(s)ζK(4s)G1(s,0)−2ζK(s)ζ′K(4s)G1(s,0), |
and G1(s,0) is absolutely convergent for ℜs>15+ε. We can easily see that ∑Alogγ(A)Ns(A) has a pole of order 2 at s=1 and poles of order 1 at s=12,13, which prompts us to consider that ∑N(A)≤xlogγ(A) should have the following asymptotic formula
∑N(A)≤xlogγ(A)=ρKxlogx+c1x+c2x12+c3x13+E0(x), |
where E0(x)=O(x14+ε).
Following the idea of Pintz [12] and Nowak [10], we use the Mellin transform to get
H(s):=∫∞1E0(x)x−s−1dx=∫∞1(∑N(A)≤xlogγ(A)−ρKxlogx+c1x+c2x12+c3x13)×x−s−1dx | (2.4) |
for ℜs>1. Now we deal with the first term of (2.4). By the partial integration and (2.4)., we have
∫∞1∑N(A)≤xlogγ(A)xs+1dx=−1s(∑N(A)≤xlogγ(A)x−s|∞1−∫∞1x−sd(∑N(A)≤xlogγ(A)))=1s∑Alogγ(A)Ns+1(A)=ζK(s)ζK(4s)G1(s,0)(ζ′K(2s)ζK(3s)+ζ(2s)ζ′K(3s))sζK(2s)ζK(3s)+G2(s)s. |
Let K(s)=ζK(s)ζK(4s)G1(s,0)(ζ′K(2s)ζK(3s)+ζ(2s)ζ′K(3s)). After computing another four terms of (2.4), we can get
H(s)=K(s)sζK(2s)ζK(3s)+G2(s)s−ρK(s−1)2−c1s−1−c2s−12−c3s−13=F(s)s(s−1)2ζK(2s)ζK(3s)(2s−1)2(3s−1)2(4s−1)2, |
where
F(s)={K(s)+G2(s)ζK(2s)ζK(3s)}(s−1)2(2s−1)2(3s−1)2(4s−1)2−sζK(2s)ζK(3s)(2s−1)(3s−1)(4s−1)2M(s), |
here M(s)=(2s−1)(3s−1)(ρK+c1(s−1))+(s−1)2(2c2(3s−1)+3c3(2s−1)). It is easy to see that F(s) is an entire function for ℜs>15+ϵ. We choose z0=14+iβ0, according to the results of [1] (or [2,3,4,13,14]), we get 2z0 is a single zero of the Dedekind ζ-function and ζK(z0)ζK(3z0)G1(z0,0)≠0. In addition,
ζ′K(2z0)≠0, ζK(4z0)=ζK(1+i4β0)≠0. |
We write
g(s):=s(s−1)2ζK(2s)(2s−1)2ζK(3s)(3s−1)2(4s−1)2(s−z0)(s+2)13, |
which is regular in ℜs>−2, and
g(s)H(s)=F(s)(s−z0)(s+2)13 |
is regular in ℜs>15+ϵ apart from a simple pole at s=z0, since
F(z0)=ζK(z0)ζ′K(2z0)ζK(3z0)ζK(4z0)G1(z0,0)×(z0−1)2(2z0−1)2(3z0−1)2(4z0−1)2≠0. |
Using the order of ζK(s) and ζ′K(s) (Lemma 2.2), we know that the integrals
∫β+i∞β−i∞|g(s)|ds, ∫β+i∞β−i∞|g(s)H(s)|ds |
converge for β∈{15,2} as |t|→∞. Now, for η>0, we define a weight function
ω(η):=∫2+i∞2−i∞g(s)ηs+1ds |
which satisfies
ω(η)={O(1),η≥1,0,if 0<η<1. | (2.5) |
Therefore,
V(Y):=1Y∫∞1E0(x)ω(Yxdx)=1Y∫∞1E0(x)(∫2+i∞2−i∞g(s)(Yx)s+1ds)dx=∫2+i∞2−i∞g(s)Ys(∫∞1E0(x)x−s−1dx)ds=∫2+i∞2−i∞g(s)H(s)Ysds. |
For Y large, we shift the line of integration to ℜs=15, then we have
V(Y)=2πiRess=z0(g(s)H(s)Ys)+∫15+i∞15−i∞g(s)H(s)Ysds=2πiα0Yz0+O(Y15), |
where
α0=F(z0)(z0+2)13=ζK(z0)ζ′K(2z0)ζK(3z0)ζK(4z0)G1(z0,0)(z0−1)2(2z0−1)2(3z0−1)2(4z0−1)2(z0+2)13. | (2.6) |
By (2.6), we can evident that
|V(Y)|≫|Yz0|=Y14 |
as Y→∞. On the other hand, by (2.5), we can obtain
|V(Y)|=|1Y∫Y1E0(x)ω(Yx)dx|≪1Y∫Y1E0(x)dx. |
Consequently, for Y→∞, we have
1Y∫Y1|E0(x)|dx≫Y14. |
Now we prove Theorem 1.1. From Proposition 2.3, we get
∑N(A)≤xlog(γ(A))=ρKxlogx+c1x+c2x12+c3x13+E0(x), | (3.1) |
with
∫Y1|E0(x)|dx≫Y54, |
where Y→∞. Using partial integration, we get
∑N(A)≤xλ−1(A)=∑2≤N(A)≤xlogγ(A)logN(A)=∫x2−1logtd(∑2≤N(A)≤tlogγ(A))=ρKx+C1∫x21logtdt+C2∫x2t−12logtdt+C3∫x2t−23logtdt+R0(x), |
where
R0(x)=∫x2dE0(t)logt=∫x2E′0(t)logtdt. | (3.2) |
Taking the derivative of the above formula we get E′0(x)=R′0(x)logx. Integrating both sides with respect to x, we have
E0(x)=∫x1R′0(t)logtdt=R0(x)logx−∫x1R0(t)tdt. | (3.3) |
And also
∫Y1|E0(x)|dx=∫Y1|R0(x)logx−∫x1R0(t)tdt|dx≤∫Y1|R0(x)logx|dx+∫Y1|∫x1R0(t)tdt|dx. | (3.4) |
We prove Theorem 1.1 by contradiction. Suppose that
∫Y1|R0(x)|dx≤ϵY54logY, | (3.5) |
where ϵ is a small constant. Thus we have
∫Y1|E0(x)|dx≤logY∫Y1|R0(x)|dx+∫Y1(∫x1R0(t)tdt)dx≤ϵY54+∫Y1(∫x1|R0(t)|tdt)dx. | (3.6) |
We use dyadic arguments to the inner integral to get
∫x1|R0(t)|tdt=logx2−1∑j=0∫2−jx2−j−1x|R0(t)|tdt≤logx∑j=012−j−1x∫2−jx2−j−1x|R0(t)|dt. | (3.7) |
By (3.5), we can obtain
∫x1|R0(t)|tdt≤logx∑j=0ϵ2−j−1x⋅(2−jx)54log(2−jx)≤ϵx14. | (3.8) |
Inserting (3.8) into (3.6), we have
∫Y1|E0(x)|dx≤ϵY54, |
which contradict with Proposition 2.3. Then we have
∫Y1|R0(x)|dx=Ω(Y54logY). |
For each integral ideal A, it is essential to measure the multiplicity of its prime ideal factors. In this paper, we define λ(A):=logN(A)logγ(A) to be the index of composition of A and consider how well the main term of ∑N(A)≤xλ−1(A) approximates it, that is, what can be said about Ω-results for the index of composition of integral ideal. The results imply that the average order of λ(A) is ρK.
The authors would like to express their gratitude to the referee for his or her careful reading and valuable suggestions. This work was supported by National Natural Science Foundation of China (Grant Nos. 11771256, 11971476).
The authors declare there is no conflict of interest.
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