# Lower Bound for the Class Number ofℚn2+4

Authors: Hasan Sankari; Ahmad Issa
Journal: International Journal of Mathematics and Mathematical Sciences
 (2020)

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## Abstract

In this paper, we give an explicit lower bound for the class number of real quadratic fieldℚd, where d=n2+4 is a square-free integer, using ωn which is the number of odd prime divisors of n.


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## Body

## 1. Introduction

Letd be a positive square-free integer and let hd and Ck denote the class number and the class group of a real quadratic field k=ℚd, respectively.The class number problem of quadratic fields is one of the most intriguing unsolved problems in Algebraic Number Theory and has for a long time inspired the study of lower bounds ofhd.Many fruitful research studies have been conducted in this direction. Hasse [1] and Yokoi [2, 3] studied lower bounds for class numbers of certain real quadratic fields. Mollin [4, 5] generalized their results for certain real quadratic and biquadratic fields.In this work, we give a lower bound forhn2+4, and also we find a necessary and sufficient condition for k=ℚn2+4 to have class number ωn+1.


## 2. Notation and Preliminaries

Letk be a real quadratic field and ζks be its Dedekind zeta function. Siegel [6] developed a method of computing ζk1−2n, where n is a positive integer. By specializing Siegel’s formula for a real quadratic field, we obtain the following result.Theorem 1 (Zagier [7]).
Letk be a real quadratic field with discriminant D. Then(1)ζk−1=160∑t<Dt2≡Dmod4σ1D−t24,where σ1r denotes the sum of divisors of r.However, there is another method, according to Lang, of computing special values ofζks if k is a real quadratic field.Letk=ℚd be a real quadratic field of discriminant D and H an ideal class of k. Let I be any integral ideal belonging to H−1 with an integral basis {r1, r2}. We put(2)δI=r1r2′−r1′r2,where r1′ and r2′ are the conjugates of r1 and r2 respectively.Letε be the fundamental unit of k. Then, {εr1, εr2} is also integral basis of I, and thus we can find a matrix M=abcd with integer entries satisfying(3)εr1r2=M.r1r2.Now, we can state Lang’s formula.Theorem 2 (Lang [8]).
By keeping the abovementioned notation, we have(4)ζk−1,H=sgnδIr2r2′360NIc3a+d3−6a+dNε−240c3sgncS3a,c+180ac3sgncS2a,c−240c3sgncS3d,c+180dc3sgncS2d,c,where NI denotes the norm of an ideal I, Nε is the norm of ε, and Si−,− denotes the generalized Dedekind sum as defined in [9].To use Lang’s formula, we need to determine the values ofa,b,c,d, and generalized Dedekind sum.Lemma 1 (see ref [10]).
The entries ofM are given by(5)a=trr1r2′εδI,b=trr1r1′ε′δI,c=trr2r2′εδI,d=trr1r2′ε′δI.
Moreover, detM=Nε and bc≠0.Kim [11] obtained the following expressions for generalized Dedekind sum. These expressions are also needed to compute the values of zeta functions for ideal classes of the respective real quadratic fields.Lemma 2 (Kim [11]).
Letm be a positive integer. Then, we have(i)
S3±1,m=±−m4+5m2−4/120m3(ii)
S2±1,m=m4+10m2−6/180m3


## 3. Main Results

Letn be a positive integer and let d=n2+4 is a square-free integer. Clearly, d≡5mod8 and n is odd. In this case, the fundamental unit of k is ε=n+d/2 and Nε=−1. If pn, then p splits in k=ℚd as(6)p=p,p+2+d2p,p+2−d2.By [12], Theorem 2.4, we know that(7)ζk−1,A=n3+11n360,where A will always denote the principal ideal class in k.In this section, we will prove our main results. As a start, we record the following proposition.Proposition 1.
Letk=ℚd, where d=n2+4 is a square-free integer. Let p be an odd prime divisor of n, and let C be the ideal class containing p,p+2+d/2 or p,p+2−d/2. Then(8)ζk−1,C=n3+np4+10p2360p2.Proof.
Let us assumeI≔p,p+2+d/2∈C−1. Then, r1=p+2+d/2,r2=p is an integral basis for I, and thus δI=pd. By Lemma 1, we get(9)abcd=n+p2+1d−p+224ppn−p2−1.
Sincepn, n±p/2±1≡±1modp. Hence, by Lemma 2, we obtain(10)240c3sgncS3a,c=240×p3S31,p=2−p4+5p2−4,240c3sgncS3d,c=240×p3S3−1,p=−2−p4+5p2−4,180ac3sgncS2a,c=180n+p2+1p3S21,p=p4+10p2−6n+p+22,180dc3sgncS2d,c=180n−p2−1p3S2−1,p=p4+10p2−6n−p−22.
By Theorem2, we get (11)ζk−1,C=n3+np4+10p2360p2.We use Proposition1 in order to prove the following theorem, which gives a lower bound for hn2+4.Theorem 3.
Letn be a positive integer, ωn≥3 and let d=n2+4 is a square-free integer. Then(12)hd≥ωn+1.Proof.
By [12], Proposition 2.3, it follows that (2) remains prime in k=ℚd.
We assume thatA denotes the principal ideal class in k, and then ζk−1,A is given by (7). If pin1≤i≤ωn, then pi splits in k as in (6), so let Ci1≤i≤ωn be the ideal class in k such that pi,pi+2+d/2∈Ci; then, by Proposition 1, we obtain(13)ζk−1,Ci=n3+npi4+10pi2360pi2.
By computingζk−1,Ci=ζk−1,A, we arrive at n=pi. This contradicts the fact that ωn≥3. Similarly, ζk−1,Ci=ζk−1,Cj for i≠j, and then n=pipj. This again contradicts the fact that ωn≥3.
We noticeC1,…,Cωn are a distinct nonprincipal ideal class in k. This completes the proof.Now, we give a necessary and sufficient condition fork=ℚn2+4 to have class number ωn+1.Theorem 4.
By keeping the abovementioned notation, we havehd=ωn+1 if and only if (14)∑m=0n−1/2σ11+mn−m2=n3+11n12+∑i=1ωnn3+npi4+10pi212pi2.Proof.
Now, by Theorem1,(15)ζk−1=160∑t<n2+4t2≡n2+4mod4σ1n2+4−t24,=160∑t≤ntis oddσ11+n2−t24,=130∑1≤t≤ntis oddσ11+n−tn+t4.
If we replacet by n−r, where r is even, then (16)ζk−1=130∑0≤r≤n−1ris evenσ11+r2n−r4.
Now, we replacer by 2m, and then we have(17)ζk−1=130∑m=0n−1/2σ11+mn−m2.
Necessary: lethd=ωn+1. Then, the class group of k is Ck=A,C1,…,Cωn. Now, by definition, we have(18)ζk−1=∑J∈Ckζk−1,J,ζk−1=ζk−1,A+∑i=1ωnζk−1,Ci.
This implies(19)ζk−1=n3+11n360+∑i=1ωnn3+npi4+10pi2360pi2.
Finally, by (17), we obtain(20)∑m=0n−1/2σ11+mn−m2=n3+11n12+∑i=1ωnn3+npi4+10pi212pi2.
Sufficiency: let(21)∑m=0n−1/2σ11+mn−m2=n3+11n12+∑i=1ωnn3+npi4+10pi212pi2.
Hence, by (17), we find(22)ζk−1=n3+11n360+∑i=1ωnn3+npi4+10pi2360pi2.
By Theorem3, we get hd≥ωn+1.
Supposehd>ωn+1. Then, there exist at least ωn+2 ideal classes in k.
Since for any ideal classE, ζk−1,E>0; thus,(23)ζk−1>ζk−1,A+∑i=1ωnζk−1,Ci.
It is a contradiction.



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