A Note on 3-Divisibility of Class Number of Quadratic Field?

Jianfeng XIE Kuok Fai CHAO

Abstract In this paper, the authors show that there exists infinitely many family of pairs of quadratic fields with D,n ∈Z whose class numbers are both divisible by 3.

Keywords Quadratic field, Class number, Hilbert class field

1 Introduction

The class number of algebraic number field is a classical topic being studied in a long history in number theory.Gauss proposed the following profound conjectures

Conjecture 1.1There are infinitely many real quadratic fields with class number one.

Conjecture 1.2There are only 9 imaginary quadratic fieldswith class number one, here D =?1,?2,?3,?7,?11,?19,?43,?67 and ?163.

Conjecture 1.2 has been verified by Baker and Stark respectively in 1967.But Conjecture 1.1 is still an open problem so far.It seems that the case of real quadratic field is quite different to that of imaginary quadratic field.Due to this reason partly, some scholars think about the divisibility of class numbers of quadratic fields.Komatsu [5] gives an infinite family of pairs of quadratic fieldswith m,D ∈Z whose class numbers are multiple of 3.In[2], Iizuka, Konomi and Nakano construct an infinite family of pairs of quadratic fieldsandwith D ∈Q,m,n ∈Z whose class numbers are both divisible by 3 or 5 or 7.Recently Iizuka [1] proposes the following conjecture and proves that this conjecture holds for imaginary quadratic fields when p=3,n=1.

Conjecture 1.3(see [1])For any prime number p and any positive integer n, there is an infinite family of n+1 successive real (or imaginary)quadratic fields

with D ∈Z whose class numbers are divisible by p.

Inspired by these results, one can consider such problem: For a given positive integer n,does there exist an infinite family of pairs of real (imaginary)quadratic fieldsandwith D ∈Z whose class numbers are both divisible by 3?

From now on we call the real (resp.imaginary)case for the case of pairs of real (resp.imaginary)quadratic fields for short.In this paper,we give the positive answer for the problem above, More concretely, we have

Theorem 1.1For arbitrary positive integern, there exist infinity many pairs of quadraticfieldswith someD ∈Zsuch that their class numbers can be divided by3for the real case and imaginary case, respectively.

We note that our main result is for any positive number.In particular, taking n=1 in the imaginary case, it is the case studied by Iizuka.

NotationThroughout this paper, Z,Q,Fpdenote the ring of rational integers, the field of rational numbers and the finite field of order p, respectively.For a prime number p and an integer a, vp(a)denotes the greatest exponent m such that pm|a, i.e., p-adic valuation.For a algebraic number field K, Kpdenotes its completion with respect to its nonzero prime ideal p.We denote the class group of K and the class number of K by ClKand h(K), respectively.

2 The Local-Global Principle for xm ?d

In this section, we would like discuss the reducibility of polynomial xm?d which will play a key role in our paper.At first, we recall a fact in algebraic number theory.

Theorem 2.1(see [3])LetL/Kbe a finite extension of number fields andOL,OKdenote their rings of integers respectively.Suppose thatpis a nonzero prime ideal ofOKandq1,··· ,qgare all the distinct prime ideals ofOLthat lie abovep.Takeα ∈Lsuch thatL=K(α), and letf(T)be an irreducible polynomial overKwithf(α)=0.Factorf(x)into the productof irreducible polynomials with coefficients inKp.Then, we haveg =h.By changing the order off1,··· ,fg, we obtain isomorphisms of fields overKp,

Next theorem is an immediate consequence of Chebotarev’s density theorem.Since we cannot find the proof in any literature, we give one for the sake of completeness.

Theorem 2.2LetL/Kbe a cyclic extension of number fields, then there are infinitely many places ofKwhich do not split inL.

ProofDenote G = Gal(L/K)and n = [L : K].Let q be a nonzero prime of OLand p=OK∩q.Indeed, there are only finitely many ramified nonzero prime ideals.Since we only focus on the existence of infinitely many places of K which do not split in L, we get rid of these

finitely many prime ideals.From now on, we can assume that p is unramified in L.Denote β={nonzero prime ideals of OKwhich is unramified in L}.Then we have a fact that p does not split in L if and only if Gq= G, where Gqis the decomposition group of q.Since G is cyclic, we choose a generator σ of G.Letis the Artin symbol.Thus for each p ∈S, p does not split in L.By density theorem,where δ(S)is the Dirichlet density of S.This means S consists of infinitely many elements.

Then we can investigate the irreducibility of the polynomials in the form of xm?d with m ∈Z+,d ∈Q.At first we deal with a general case.

Theorem 2.3Suppose that the polynomialf(x)= xm?d ∈Q[x]is irreducible over the cyclotomic fieldQ(ζm),d ∈Q×.There are infinitely many prime placesqofQsuch thatf(x)is irreducible overQqand overFq.

ProofLet L be the splitting field of f(x)in C.Consider the finite extension L/Q(ζm).Since f(x)is irreducible over Q(ζm), this is a cyclic extension of degree m.By Theorem 2.2,there exists infinitely many prime places of Q(ζm)which do not split in L.We denote M the set of all such prime places.For any q ∈M, we factor f(x)into the product of irreducible polynomials with coefficients in Q(ζm)q, i.e.,.According to Theorem 2.1 and the fact that q does not split in L, we know that h=1, which means that f(x)is irreducible over Q(ζm)q.

Let q be the prime place of Q which lies under q.It is immediate to know that Qqis a subfield of Q(ζm)q.Since f(x)is irreducible over Q(ζm)q, then is also irreducible over Qq.

It may not be easy to determine whether a polynomial in the form of xm?d is irreducible over Q(ζm)or not.However, when m is odd, we can just consider this problem over Q.Let us recall a result in [7] which is needed in the proof of Lemma 2.2.

Lemma 2.1(see [7])Ifmis odd and the polynomialxm?d,d ∈Q×has no root inQ,then it has no root inQ(ζm).

Lemma 2.2Ifmis odd and the polynomialxm?d,d ∈Q×is irreducible overQ, then it is also irreducible overQ(ζm).

ProofLet α=∈R.Firstly we show that if 1 ≤i

It is clear that the splitting field of xm?d is Q(ζm)(α).Assume that xm?d is reducible over Q(ζm), then the degree of the minimal polynomial f(x)of α over Q(ζm)is less than m, namely,deg f(x)=h.So f(x)can be written in the form of1 ≤ij< m, 1 ≤j ≤h ?1.Hence the constant term ofThis means αh∈Q(ζm), which leads a contradiction.Therefore xm?d is irreducible over Q(ζm).

Combining Lemma 2.2 and Theorem 2.3, we get the following result.

Theorem 2.4Ifmis odd and the polynomialf(x)= xm?d,d ∈Q×is irreducible overQ, then there are infinitely many placesqofQsuch thatf(x)is irreducible overQqand then overFq.

Now we can get an equivalent statement,which can be seen as a kind of local-global principle.

Theorem 2.5Assume thatmis odd andf(x)=xm?d, d ∈Q×.Thenf(x)is reducible overQif and only iff(x)is reducible overFqexcept finitely many placesqofQ.

Remark 2.1There do exist polynomials f(x)=xm?d with m being even such that f(x)is irreducible over Q but is reducible over every Fq, where q runs through all prime places of Q.For example, x4+1 and x10?5 are the cases.

3 A Construction of Quadratic Fields

In this section, we discuss how we can construct a pair of fieldswhose class numbers can be divided by 3.Recall that Hilbert class field H of a number field K is the maximal unramified abelian extension of K, and there is a canonical isomorphic Gal(H/K)ClK.It is clear that the class number of K can be divided by 3 if and only if there exists a cyclic unramified cubic extension of K.It is a hint for a construction.Kishi and Miyake [4] give the following characterization of all quadratic fields which admits a cyclic unramified cubic extension.

Theorem 3.1(see [4])Choose(u,w)∈Z×Z, and letg(Z)=Z3?uwZ ?u2.If

then the normal closure ofQ(θ), whereθis a root ofg(Z), is a cyclic, cubic, unramified extension of;in particular, thenK =has class number divisible by3.Conversely,every quadratic number fieldKwith class number divisible by3and every unramified, cyclic,cubic extension ofKis given by a suitable choice of integersuandw.

Remark 3.1The condition (1)in (3.1)is critical.The reason why we develop the story of irreducibility of xm?d in Section 2 is to serve for the condition (3)in (3.1).

To achieve our goals, a natural idea is to find integer pairs (u1,w1)and (u2,w2)such that both of them satisfy all the conditions in (3.1), and

In order to find such integer pairs(u1,w1)and(u2,w2),we consider the integer pairs(x1,y1)and (x2,y2)such that

If so, let u1=,w1=y1andfor some k ∈Z{0}.Then we have

From (3.3)–(3.4), we get

Thus we are back to the situation (3.2).Clearly we only need to find integer pairs (x1,y1)and (x2,y2)satisfying equation (3.3).Furthermore, we mention that

3.1 First step

Now we start to construct some solutions for equation(3.3).Assume that there exist integer pairs (x1,y1)and (x2,y2)satisfying equation (3.3)and let

It follows that

Put

One can check that 4(k3?c2)=nc2.Then we can simplify the right-hand side of (3.6)

In order to ensure this equality holds, we can put

here p is a prime number and l is a positive integer.We will explain why these two integers are introduced.Solving this equation by regarding x1and x2as variables, we get

It is clear that (3.3)holds if we set (x1,y1)and (x2,y2)as in (3.5)and (3.8).We note that we can ensure x1,x2are integers by choosing proper integer t.This will be discussed in Theorem 3.2.

where

It is clear that the leading coefficient of f(t)is 4 ?27α2.Since α ∈Q, 4 ?27α2≠ 0.It is clear that there exist infinitely many integers t0such that f(t0)> 0 (resp.f(t0)< 0)when 4 ?27α2>0 (resp.4 ?27α2<0).Based on this fact, we always can choose proper t to ensure that both quadratic fieldsare real or imaginary.For the imaginary case, we should require 4 ?27α2< 0.It is easy to be achieved by choosingNow we consider the real case.We realize that 4 ?27α2> 0 if and only ifIt is equivalent toThe following lemma asserts that the condition 4 ?27α2> 0 can be achieved as well.This is the reason why we introduce the integer l above.

Lemma 3.1Letn,k,pbe positive integers such that.Then there exists integerlsuch that

ProofLetNote that 0 ≤d<1.We have

3.2 Second step

Set (x1,y1)and (x2,y2)as in (3.5)and (3.8).Let u1=,w1= y1,Here we mention that w1=w2=t2.Consider the polynomials

respectively.We will show that (u1,w)and (u2,w)satisfy all the conditions in Theorem 3.1 under a suitable choice of t.If so, let

then

Thus we get two quadratic fieldswhose class numbers are divisible by 3.

Now the aim is to ensure the polynomials F(Z)and G(Z)are irreducible by choosing suitable t.Recall

and let

We require that p is a prime number and is coprime to 6nkl.Then it is easy to check thatwhich means f1(Z)and f2(Z)are irreducible over Q.By Theorem 2.4, there exist two primes q1and q2(not necessary to be different)such that f1(Z)is irreducible over Fq1and f2(Z)is irreducible over Fq2, respectively.Thus we get the following lemma.

Lemma 3.2Ift ≡0(mod q1q2), thenF1(Z)andF2(Z)are both irreducible overQ.

ProofSince w =t2in the setting above and t ≡0(mod q1q2),it is clear that w ≡0(mod q1).From(3.8), we can see that(mod q1), then F1(Z)≡f1(Z)(mod q1), which means F1(Z)is irreducible over Fq1.Hence we know that F1(Z)is irreducible over Q.The irreducibility of F2(Z)follows in a similar way.

Remark 3.2In the proof of Lemma 3.2 above, we abuse the notation of congruence and hence consider the congruence of a rational numbermodulus prime p.Indeed if we restrict n not having any p factor,can be viewed as the reciprocal (or inverse)of n modulus p and it still makes sense under this restriction.

For convenience, recall that

and denote g =18k2(p ?1)(3nk2(p ?1)2+l2p2)|(3nk2(p ?1)2?l2p2)|.By Theorem 2.3, there exist infinitely many primes q such that ?3 is quadratic non-residue in Fq.Choose such a prime number q0.Applying Theorem 3.1, we have

Theorem 3.2Choose a prime numberpsuch thatp ≡1(mod 18nlk)and let

hereq0is prime to integersg,p,q2,q2.Then pairs(u1,w1)and(u2,w2)satisfy the conditions of Theorem3.1, soboth admit an unramified cyclic cubic extension.In particular, their class numbers are both divisible by3.

Remark 3.3We should ensure g,p,q0,q1,q2are all coprime mutually.We claim that this can be done.Firstly, once n is given, so k = n+4 and then we can get l as in Lemma 3.1.According to Dirichlet’s Prime Number theorem, we can choose a prime number p such that

? p ≡1(mod 18lnk)and

ProofAccording to the congruence conditions(3.9),one can check that x1,x2are integers,and k2| x2.At first we show x1,x2are integers.Indeed, since p ?18lk(p ?1), it suffices to show that

Due to t ≡?1(mod p), we have p|t3+1, then

Since 18l|(p?1),we have 18lk(p?1)|3nk2(p?1)2(t3+1).Because t ≡1(mod 18k(p?1)),we have 18k(p ?1)|t3?1 and then 18lk(p ?1)|l2p2(t3?1).Now it is clear that

18lk(p ?1)|3nk2(p ?1)2(t3+1)±l2p2(t3?1).

Now we turn to check that k2| x2.Since a = 2k, it suffices to show that 9lp(p ?1)k2|3nk2(p ?1)2(t3+1)and 9lp(p ?1)k2|l2p2(t3?1).We realize that p|3nk2(p ?1)2(t3+1)and p|l2p2(t3?1)have been proved already.Since 9l|(p?1),we get 9l(p?1)k2|3nk2(p?1)2(t3+1).Because t ≡1(mod 9l(p ?1)k2), we have 9l(p ?1)k2|l2p2(t3?1).

Moreover, we have

We clarify that the conditions(3.1)in Theorem 3.1 are satisfied in the settings of u1,u2,w1and w2above.We note that (t,a)= 1.Then we have (w1,u1)= 1 and (w2,u2)= 1 immediately.By the assumption t ≡0(mod q0q1q2), it implies that t ≡0(mod q1q2).By Lemma 3.2, we know that F1(Z)and F2(Z)are both irreducible.It is clear that conditions(2)and(3)in (3.1)are fulfilled.Then we check that the condition (I)in (3.1)is satisfied.We recall t ≡1(mod g),and it implies (3,t)=1.Then 3w1and 3w2follow by w1=w2=t2.

It remains to show that condition(1)in Theorem 3.1 is satisfied as well,namely,4u1w3?and 4u2w3?are not squares in Z.By the choice of q0, 4u1w3?≡?m2(mod q0), here m can be any integer.This means 4u1w3?is not a square.Similarly we can show that 4u2w3?is not a square as well.By Theorem 3.1,admit an unramified cyclic cubic extension, respectively.

4 The Proof of Theorem 1.1

To show that our construction can generate infinitely many pairswith their class numbers being divisible by 3, we recall a celebrated result on integral points by Siegel [6].Let MQbe the set of all standard absolute values on Q.

Theorem 4.1(see [6])LetSbe a finite set such that{∞}?S ?MQandf(x)∈Q[x]be a polynomial of degreed ≥3with distinct roots inC.Then

?{(x,y)∈RS×RS|y2=f(x)}<∞,

whereRSis the ring ofS-integers ofQ, i.e.,RS={x ∈Q|vp(x)≥0for allp ∈MQS}.

Lemma 4.1Suppose thatf(x)∈Q[x]is a polynomial of degreed ≥3with distinct roots inC, andT ?Zconsists of infinitely many integers and put, thenEcontains infinitely quadratic fields.

ProofWe assume that f(x)∈Z[x].Otherwise we choose an integer d such that d2f(x)∈Z[x] and consider the polynomial d2f(x)instead sinceBecause T is a countable set, we can denote this ordered set by T = {ti| i ∈I} with a countable set I.Assume E is a finite set, then there exist finitely many primes p1,p2,··· ,pN(N ∈Z+)such that for any i ∈I, we have

Furthermore, there exists a integerfor a specific i such that there are infinitely many i′∈I,

Let S ={∞}.Then RS=Z.Consider the set

It follows that there exist infinitely many pairs of(ti′,ai′)in A.But by Siegel’s theorem 4.1, A is a finite set, which leads to a contradiction.Hence E is a set with infinite many elements.

Proof of Theorem 1.1For the real case, we choose a proper l in Lemma 3.1.Set

are real quadratic fields.

For the imaginary case,choose an integerWhen t>0, we have<0.This means the quadratic fields

are imaginary.

Let

Chinese Remainder theorem implies that the set T consists of infinite elements.Let f(t)=?and

Since f(t)has no repeated roots which is in the form ofwith α,β ∈Q×, Lemma 4.1 implies that E contains infinitely many quadratic fields.Moreover, let Dt= f(t), t ∈T, Theorem 3.2 implies thatThen we complete our proof.

AcknowledgementWe thank the anonymous referees for the valuable comments on our manuscript.