Abelian subgroups of any order in class groups of global ... · Journal of Number Theory 106 (2004) 26–49 Abelian subgroups of any order in class groups of global function ﬁelds

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Journal of Number Theory 106 (2004) 26–49

Abelian subgroups of any order in classgroups of global function fields

Allison M. Pacelli�

Department of Mathematics, Williams College, Bronfman Science Center, Room 200,

Williamstown, MA 01267, USA

Received 14 December 2002; revised 27 August 2003

Communicated by D. Goss

Abstract

Let F be a finite field with q elements, and T a transcendental element over F: In this

paper, we construct infinitely many real function fields of any fixed degree over FðTÞ with

ideal class numbers divisible by any given positive integer greater than 1. For imaginary

function fields, we obtain a stronger result which shows that for any relatively prime integers

m and n with m; n41 and relatively prime to the characteristic of F; there are infinitely

many imaginary fields of fixed degree m such that the class group contains a subgroup

isomorphic to ðZ=nZÞm�1:r 2004 Elsevier Inc. All rights reserved.

Keywords: Class group; Class number; Function field

1. Introduction

Determining the structure of the class group is a difficult problem in both numberfields and function fields that has been studied since Gauss first considered classgroups of quadratic fields. Although class groups can only be computed in specialcases at present, there are many interesting questions that one can ask: for example,how many class groups have class numbers divisible by a given integer n? Nagell [6]answered a special case of this question in 1922 by showing the existence of infinitelymany imaginary quadratic number fields with class number divisible by a given

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0022-314X/$ - see front matter r 2004 Elsevier Inc. All rights reserved.

doi:10.1016/j.jnt.2003.12.003

integer n: Yamamoto [10] proved the same result for real quadratic number fields in1969. Cohen and Lenstra [3] gave a heuristic argument in 1983 predicting, as x-N;a positive fraction of quadratic number fields with discriminant ox having classnumber divisible by a given integer n: Their argument has been generalized tonumber fields of any degree and to function fields, but has not been proven yet in anyof these cases.

In 1983, Azuhata and Ichimura [1] proved a result not only about the divisibilityof the class number but also about the structure of the class group in the numberfield situation. They constructed infinitely many number fields K of degree m

over Q such that the class group of K contains an abelian subgroup isomorphic

to ðZ=nZÞr2 in the special case that r241: As usual, r2 denotes half thenumber of complex embeddings of K into C: Unfortunately their result saysnothing about the class group of a totally real number field. In 1985, however, ShinNakano [7] was able to extend this result to the totally real case, constructing asubgroup isomorphic to Z=nZ in the class group of a totally real number field andincreasing the rank of the subgroup from r2 to r2 þ 1 in class groups of non-totallyreal fields. The smaller rank of the subgroup for totally real fields is a consequence ofthe extra units present.

The results above describe the class groups of number fields. Less is known aboutthe analogous question of class groups of global function fields. In 1992, Friesen [4]constructed infinitely many real quadratic function fields with class number divisibleby a given integer n: Murty and Cardon [2] proved a stronger result about quadraticfunction fields in 2001, giving a lower bound on the number of such fields with classnumber divisible by a given integer n43: Their results have been generalized to cyclicfunction fields of prime degree l over the rational function field in a forthcomingpaper [8]. In 1999, Ichimura [5] gave a partial function field analogue to his andAzuhata’s results in [1] for the special case that m is prime. In this case, the size of thesubgroup constructed depends on the factorization of X m � 1 in F½X �: In this paper,we give more general function field analogues of the results of Azuhata, Ichimura,and Nakano, above, which imply the existence of infinitely many function fields ofany fixed degree with ideal class numbers divisible by any given positive integergreater than 1.

Let q be a power of an odd prime, and let F be the field with q elements. Letk be the rational function field, and fix a transcendental element T of k so thatk ¼ FðTÞ: If K is an extension of k; then denote by OK the integral closure of F½T � inK : We write ClK to denote the ideal class group of OK : The main results are asfollows:

Theorem 1. For any relatively prime integers m; nAZ; not divisible by the

characteristic of FðTÞ; with m; n41; there exist infinitely many function fields K of

degree m over k ¼ FðTÞ such that

(1) the prime at infinity is totally ramified in K ; and

(2) ClK contains an abelian subgroup isomorphic to ðZ=nZÞm�1:

ARTICLE IN PRESSA.M. Pacelli / Journal of Number Theory 106 (2004) 26–49 27

Theorem 2. For any integers m; nAZ; not divisible by the characteristic of FðTÞ; with

m; n41; there exist infinitely many function fields K of degree m over k ¼ FðTÞ such

that

(1) the prime at infinity splits completely in K ; and

(2) ClK contains an abelian subgroup isomorphic to Z=nZ:

For the proofs of both theorems, we construct a polynomial

f ðX Þ ¼Ym�1

i¼0

ðX � BiÞ þ Dn;

where B0;y;Bm�1 and D are certain polynomials in F½T �: The method by which wechoose the Bi’s and D will be different for each case in order to force the behavior ofthe prime at infinity. If y is a root of f ðX Þ; then K ¼ kðyÞ satisfies the conditions ofthe theorems.

2. Imaginary case

In the function field situation, we no longer have the quantities r1 and r2; so wecannot refer to a function field as real or imaginary depending on whether or not r2 is0. The analogue is to consider the prime at infinity in k: If the infinite prime is totallyramified in K ; then we say that K is imaginary, and if the infinite prime splitscompletely, we say that K is totally real. In this section, we prove Theorem 1, whichdeals with the imaginary case.

Let L be the set of all prime divisors of n; and define n0 ¼Q

lAL l: Let m0 be the

least common multiple of the orders of all roots of unity contained in any functionfield of degree m: Let E and W denote, respectively, the group of units and the group

of roots of unity in the field K: For an element r in F½T �; let jrj ¼ qdegðrÞ: Givenpolynomials B0;y;Bm�1;DAF½T �; define

f ðX Þ ¼Ym�1

i¼0

ðX � BiÞ þ Dn;

and let y be a root. Set K ¼ kðyÞ: The next two lemmas and proposition show thatwith an appropriate choice of B0;y;Bm�1; and D; the field K satisfies the conditionsof Theorem 1.

Lemma 1. Suppose there exist monic, irreducible polynomials p1;y; pm�1 with

jpij 1 ðmod m0n0) and polynomials B1;y;Bm�1; and D in F½T � such that

(2.1) f ð0Þ 0 ðmod p1?pm�1),(2.2) ð f 0ð0Þ; p1?pm�1Þ ¼ 1; and

(2.3) ðBi

piÞla1; ðBi

pjÞl ¼ 1 for iaj; 1pi; jpm � 1; and for each lAL:

ARTICLE IN PRESSA.M. Pacelli / Journal of Number Theory 106 (2004) 26–4928

For each lAL; the subgroup of K�=WK�l generated by the classes of y� B1;y� B2;y; y� Bm�1 is an elementary abelian group of rank m � 1:

Proof. Let p denote one of the primes p1;y; pm�1; and set P ¼ ðpÞ: Let k̂P be thecompletion of k at P: Notice that

K#kk̂PDYt

j¼1

Lj;

where ½Lj : k̂P� ¼ ejfj; and ej; fj denote, respectively, the ramification index and

relative degree of the prime in Lj lying over P̂Ak̂P: Since K ¼ kðyÞ; and f is the

minimal polynomial for y over k; we have KDk½x�=ð f ðX ÞÞ: If f1ðXÞ; f2ðXÞ;y; ftðXÞare the irreducible factors of f over k̂P; then

K#kk̂PD k½x�=ð f ðXÞÞ#kk̂P

D k̂½x�=ð f1ðXÞf2ðXÞ?ftðX ÞÞ

DMt

j¼1

k̂P½x�=ð fjðXÞÞ

DMt

j¼1

k̂PðyjÞ;

where each yj is a root of fjðX Þ in k̂P:

We claim that one of the irreducible factors of f over k̂P is linear. By condition(2.1), f ð0Þ 0 ðmod p), so we have

f ðXÞ XgðXÞ ðmod pÞ

for some gðXÞAk̂P½X �: Then

f 0ðX Þ Xg0ðX Þ þ gðX Þ ðmod pÞ:

By condition (2.2),

gð0Þ f 0ð0Þc0 ðmod pÞ:

By Hensel’s Lemma, then

f ðXÞ ðX � *yÞhðXÞAk̂p½X �;

where X � *y X ðmod p), and hðX Þ gðXÞ ðmod p). In particular, this implies that*yApF½T �P; where F½T �P is the completion of F½T � at P:


Let l denote the embedding from K ¼ kðyÞ to k̂P which takes y to *y: Given suchan embedding, there is a corresponding prime ideal pl in OK lying over P:

pl ¼ faAOK j ordPðlðaÞÞ40g:

Notice that yApl since lðyÞ ¼ *yApF½T �P: Furthermore, if el and fl denote the

ramification index and relative degree, respectively, of pl over P; then

elfl ¼ degðX � *yÞ ¼ 1:

Thus el ¼ fl ¼ 1; and so, for each prime pi; 1pipm � 1; there exists some primeideal piCOK over P ¼ ðpiÞCF½T � of relative degree 1 and such that y 0 ðmod pi).

Fix lAL: To prove the lemma, suppose that

ðy� B1Þb1?ðy� Bm�1Þbm�1AWK�l ;

for some b1;y; bm�1AZ: Since jpij 1 ðmod m0n0), then jpi j�1l

¼ m0z for some

integer z: It follows that any root of unity in K is an lth power residue mod pi

because if zAW ; then

zpi

� �l

zjpi j�1

l zm0z ¼ 1 ðmod piÞ:

Thus

y� B1

pi

� �b1

l

?y� Bm�1

pi

� �bm�1

l

¼ 1:

Since y� Bi �Bi ðmod pi),

�1

pi

� �b1þ?þbm�1

l

B1

pi

� �b1

l

?Bm�1

pi

� �bm�1

l

¼ 1:

By condition (2.3) then,

�1

pi

� �b1þ?þbm�1

l

Bi

pi

� �bi

l

¼ 1:

If l is odd, then clearly ð�1piÞl ¼ 1: If l is even, then n0 is also even, so jpij 1 ðmod 4Þ;

say jpij � 1 ¼ 4z for some integer z: In this case, it is also true that ð�1piÞl ¼ 1

since ð�1piÞl ð�1Þ

jpi j�1l ð�1Þ

4zl 1 ðmod pi). Thus

Bi

pi

� �bi

l

¼ 1:

But again by condition (2.3), this implies ljbi:


Repeating the argument above for each prime pi; 1pipm � 1; we see that ljbi foreach i; 1pipm � 1: It follows that the m � 1 elements y� B1; y� B2;y; y� Bm�1

are linearly independent over WK�l : &

The following standard lemma will be used to construct the desired subgroup ofClK :

Lemma 2. Suppose G is a finite abelian group of exponent n; and dimZ=lZ Gn=lXr for

all l dividing n: Then G contains a subgroup isomorphic to Z=nZ"?"Z=nZ of rank

r:

Proposition 1. Suppose that the polynomials B0;y;Bm�1 and D further satisfy the

following two conditions:

(2.4) y� B0; y� B1;y; y� Bm�1 are pairwise relatively prime,(2.5) the prime at infinity is totally ramified in K :

Then ClK contains an abelian subgroup isomorphic to ðZ=nZÞm�1:

Proof. Let ClK ½n� denote the set of all elements of the class group of OK whose orders

divide n: By the previous lemma, we need only show that ClK ½n�n=l has dimension atleast m � 1 over Z=lZ for all l in L: Consider the following sequence:

ð1Þ-ClKn

l

h i!i

ClK ½n�!h

K�=EK�l ;

where i is just inclusion, and the map h is defined as follows. Let %a denote the class ofthe ideal a in ClK ½n�: If %aAClK ½n�; then an ¼ ðaÞ for some aAOK : Set hðaÞ ¼ ½a�l ;where ½a�l denotes the class of a in K�=EK�l : One can show that h is a well-defined

homomorphism, and the above sequence is exact.Since the prime at infinity is totally ramified, we know that the rank of the unit

group in K is 0; therefore the only units in K are the roots of unity. Replacing E byW in the sequence above yields the following exact sequence:

ð1Þ-ClKn

l

h i!i

ClK ½n�!h

K�=WK�l : ð1Þ

Recall that

Ym�1

i¼0

ðy� BiÞ ¼ �Dn;

since y is a root of f ðX Þ: By condition (2.4) then, each ideal ðy� BiÞ is an nth power.

Set Dni ¼ ðy� BiÞ for 1pipm � 1: It follows that hðDiÞ ¼ ½y� Bi�l : As a vector

space over Z=lZ; then, ImðhÞ has dimension at least m � 1 by Lemma 1. Since the


sequence in Eq. (1) is exact,

ImðhÞDClK ½n�=ClKn

l

h iDClK ½n�n=l ;

and so, ClK ½n�n=l has dimension at least m � 1 over Z=lZ:To prove Theorem 1, we will show that it is possible to choose irreducible

polynomials p1;y; pm�1 and polynomials B0;y;Bm�1; and DAF½T � so thatconditions (2.1)–(2.5) are satisfied, and f ðX Þ is irreducible. Finally, note that theexistence of infinitely many such fields K is a consequence of the existence of onesuch field because of the finiteness of the class number. Let p be prime. If K1 is a field

of degree m whose class group contains a subgroup isomorphic to ðZ=pZÞm�1; and h1

is the class number of K1; then let f1 be any integer such that p f14h1: Next, let K2 bea field of degree m whose class group contains a subgroup isomorphic to

ðZ=p f1ZÞm�1: Since p f14h1; we know that K1 and K2 are distinct. Moreover, the

class group of K2 also contains a subgroup isomorphic to ðZ=pZÞm�1: If h2 is the

class number of K2; then next choose an integer f2 with p f24h2; and a field K3 of

degree m whose class group contains a subgroup isomorphic to ðZ=p f2ZÞm�1: Again,K3 is distinct from K1 and K2; and the class group of each field contains a subgroup

isomorphic to ðZ=pZÞm�1: Continuing in this manner, we create infinitely manyfunction fields of degree m with class group containing a subgroup isomorphic to

ðZ=pZÞm�1: A similar argument shows that the existence of one field whose class

group contains a subgroup isomorphic to ðZ=nZÞm�1 for a general n implies theexistence of infinitely many such fields.

2.1. Choosing polynomials

Choose distinct irreducible polynomials pi; s in F½T �; 1pipm � 1; such that

jpij 1 ðmod m0n0Þ; 1pipm � 1;

and

jsj 1 ðmod mÞ:

Note that there are infinitely many such primes pi and s: Because m and n arerelatively prime to the characteristic of F; the primes whose norms are congruent to 1modulo an integer m are exactly those primes which split completely in kðzmÞ; wherezm is a primitive mth root of unity.

Since jpij 1 ðmod m0n0), we have l j ðjpij � 1Þ for all lAL: Let gi; 1pipm � 1; bea primitive root mod pi that satisfies the congruence

g2i þ ðm � 2Þgi þ 1c0 ðmod piÞ: ð2Þ


This is possible since jpij � 143: Since m j ðjsj � 1Þ; we also have that

X m � 1 Ym�1

i¼0

ðX � CiÞ ðmod sÞ;

where the Ci’s are distinct mod s for 1pipm � 1: Choose Bi; 1pipm � 1 such that

Bi 1 ðmod pjÞ if iaj;

gi ðmod piÞ;Ci ðmod sÞ:

8><>:

Choose B0 such that

B0 g�1

i ðmod piÞ for 1pipm � 1;

C0 ðmod sÞ:

�

Next choose an irreducible polynomial D such that the following properties hold.

D 1 ðmod sÞ;ð�1Þmþ1 ðmod piÞ for 1pipm � 1;

(

ðdeg ðDÞ;mÞ ¼ 1;

deg ðDnÞ4m maxfdegðBiÞgm�1i¼0 ;

ðBi � Bj ;DÞ ¼ 1 for 0pi; jpm � 1; iaj:

ð3Þ

Infinitely many D satisfying the first two properties exist by a strong version ofDirichlet’s Theorem for function fields [9, p. 40] which asserts that in any arithmeticprogression, there exist irreducible polynomials of each large degree. We need onlydiscard finitely many not satisfying the second two properties above.

2.2. Verification of divisibility conditions

Lemma 3. With polynomials B0;y;Bm�1 and D in F½T � chosen as above, conditions

(2.1)–(2.3) in Lemma 1 are satisfied.

Proof. Condition (2.1) is satisfied because

Dn þ ð�1ÞmB0B1?Bm�1 ð�1Þðmþ1Þn þ ð�1Þm

g�1i gi ðmod piÞ

ð�1Þmþ1 þ ð�1Þm ðmod piÞ

0 ðmod piÞ:


For condition (2.2), observe that

f 0ð0Þ ¼Xm�1

k¼0

Ym�1

j¼0jak

ð�BjÞ

ð�1Þm�1Ym�1

j¼1

Bj þXm�1

k¼1kai

Ym�1

j¼0jak

Bj þYm�1

j¼0jak

Bj

2664

3775 ðmod piÞ

ð�1Þm�1½gi þ m � 2 þ g�1i � ðmod piÞ

c 0 ðmod piÞ by Eq: ð2Þ:

Finally, condition (2.3) follows immediately from the choice of Bi for1pipm � 1: &

Lemma 4. y� B0; y� B1;y; y� Bm�1 are pairwise relatively prime, that is, condition

(2.4) in Proposition 1 is satisfied.

Proof. We chose D so that ðBi � Bj;DÞ ¼ 1 for iaj; 0pi; jpm � 1: If there exists a

prime ideal p of OK such that p divides both y� Bi and y� Bj for some i and j; then

pjðBi � BjÞ: SinceQm�1

h¼0 ðy� BhÞ ¼ �Dn; we also have that p j D: Thus, p j ðBi �Bj ;DÞ; a contradiction. &

Lemma 5. With the conditions on B0;y;Bm�1; and D given above, f ðXÞ is irreducible.

Proof. We show that f ðXÞ is an Eisenstein polynomial with respect to s: Notice first

that sjjðDn þ ð�1ÞmB0B1?Bm�1Þ; the constant term of f ðXÞ:

Dn þ ð�1ÞmB0B1?Bm�1 1 þ ð�1Þm

Ym�1

i¼0

Ci ðmod sÞ

1 � 1 ðmod sÞ

0 ðmod sÞ:

If s2 j ðDn þ ð�1ÞmB0B1?Bm�1Þ; then replace B0 by B0 þ p1?pm�1s: All the desired

properties of B1;y;Bm�1 and D above still hold since B0 B0 þ p1?pm�1s modulopi and s: But now we also have that

Dn þ ð�1ÞmðB0 þ p1?pm�1sÞB1?Bm�1

Dn þ ð�1ÞmB0?Bm�1 þ ð�1Þm

p1?pm�1sB1?Bm�1 ðmod s2Þ


0 þ ð�1Þmp1?pm�1sB1?Bm�1 ðmod s2Þ

c0 ðmod s2Þ:

Because f ðXÞ is monic, we need only show that the remaining coefficients of f aredivisible by s: Since Bi Ci ðmod s) for 0pipm � 1; then

Ym�1

i¼0

ðX � BiÞ X m � 1 ðmod sÞ:

So all coefficients ofQm�1

i¼0 ðX � BiÞ; excluding the leading and constant terms, are

divisible by s: Since these are exactly the coefficients of f ðXÞ under consideration,this completes the proof. &

2.3. The infinite prime

Lemma 6. The prime at infinity is totally ramified in K; that is, condition (2.5) in

Proposition 1 is satisfied.

Proof. Consider the Newton polygon for f ðX Þ with respect to the prime at N: Inthis case, it is not hard to see that the Newton Polygon consists of the single linesegment from ð0;�degðB0?Bm�1 þ DnÞÞ to ðm; 0Þ with slope equal todegðB0?Bm�1 þ DnÞ=m: Thus f ðXÞ has m roots yi each with

ordNðyiÞ ¼�1

mdegðB0?Bm�1 þ DnÞ ¼ �n degðDÞ

m;

by Eq. (3). Since m is relatively prime to both n and degðDÞ; N must be totallyramified in K : &

3. Real case

We now address the analogue of the totally real case for number fields, that is, thesituation in which the infinite prime splits completely. As in [7], we increase the rank

of the subgroup of K�=WK�l by one in order to obtain a copy of Z=nZ in ClK : Inorder to obtain this extra rank, we need one more linearly independent element over

WK�l : This requires choosing a new polynomial AAF½T � such that f ðAÞ is an nthpower. We also choose B0;y;Bm�1 and D differently than in the last section; we useparametric representations as in [7].


Lemma 7. Suppose there exist monic, irreducible polynomials p1;y; pm�1 and s with

jsj 1 mod ðm0n0Þ and polynomials B1;y;Bm�1;A;C; and D in F½T � such that for

each lAL

(3.1) f ð0Þ 0 ðmod p1?pm�1sÞ;(3.2) ð f 0ð0Þ; p1?pm�1sÞ ¼ 1;(3.3) Bi

pi

� �la1; Bi

pj

� �l¼ 1 for iaj; 1pipm � 1; 1pjpm � 1; and

(3.4) Bi

s

� �l¼ 1; A

s

� �la1 for 1pipm � 1:

Then for each lAL; the subgroup of K�=WK�l generated by the classes of

y� B1; y� B2;y; y� Bm�1; and y� A is an elementary abelian group of rank m:

Proof. The proof is similar to that of Lemma 1 for the imaginary case. Since we alsohave that f ð0Þ 0 and f 0ð0Þc0 ðmod s) by conditions (3.1) and (3.2) above, theargument in Lemma 1 also proves that there exists a prime s of K over s of relativedegree 1 which contains y:

Fix lAL: To prove the lemma, suppose that

ðy� B1Þb1?ðy� Bm�1Þbm�1ðy� AÞaAWK�l ;

where bi; aAZ: Since jsj 1 ðmod m0n0), any root of unity in K is an lth powerresidue mod s; so

y� B1

s

� �b1

l

?y� Bm�1

s

� �bm�1

l

y� A

s

� �a

l

¼ 1:

Since y� Bi �Bi ðmod s),

�1

s

� �b1þ?þbm�1þa

l

B1

s

� �b1

l

?Bm�1

s

� �bm�1

l

A

s

� �a

l

¼ 1:

By condition (3.4) then,

�1

s

� �b1þ?þbm�1þa

l

A

s

� �a

l

¼ 1:

If l is odd then clearly ð�1sÞl ¼ 1:If l is even, then n0 is also even, so jsj 1 ðmod 4). In

this case, it is also true that ð�1sÞl ¼ 1: Thus

A

s

� �a

l

¼ 1:

But again by condition (3.4), this implies l j a:The proof is now reduced to showing that bi 0 ðmod l) for 1pipm � 1 which

follows just as in Lemma 1. &


Proposition 2. Suppose that the primes p1;y; pm�1 and s and polynomials

B0;y;Bm�1;A;C; and D further satisfy the following four conditions:

(3.5) f ðAÞ ¼ Cn;(3.6) ð f 0ðAÞ;CÞ ¼ 1;(3.7) ð f 0ðBiÞ;DÞ ¼ 1 for i ¼ 1;y;m � 1; and

(3.8) the prime at infinity splits completely in K :

Then the ideal class group of OK contains an abelian subgroup isomorphic to Z=nZ:

Proof. As before, let ClK ½n� denote the set of all elements of the class group of OK

whose orders divide n: By Lemma 2, we need only show that ClK ½n�n=l has dimensionat least 1 over Z=lZ for all l in L: The following sequence is still exact:

ð1Þ-ClKn

l

h i!i

ClK ½n�!h

K�=EK�l : ð4Þ

Because the infinite prime splits completely, however, we cannot just replace E by

W in the sequence as before. Instead, let SCK�=WK�l be the subgroup generated

by the classes of y� A; y� B1;y; y� Bm�1; and let S0CK�=EK�l be the image of S

under the natural reduction map from K�=WK�l to K�=EK�l . The followingsequence is also exact:

ð1Þ-S-EK�l=WK�l-S-S0-ð1Þ:

As a result,

dimZ=lZ ðSÞ ¼ dimZ=lZ ðE=WÞ þ dimZ=lZ ðS0Þ:

By Lemma 7, dimZ=lZ ðSÞXm; and by condition (3.8), we know that the rank of the

unit group in K is m � 1: Thus

m � 1 þ dimZ=lZ ðS0ÞXm;

and so, dimZ=lZ ðS0ÞX1: We claim that S0 is contained in ImðhÞ; from which the

proposition follows. If S0CImðhÞ; then ImðhÞ has dimension at least 1 over Z=lZ:Since the sequence in Eq. (4) is exact,

ImðhÞDClK ½n�=ClKn

l

h iDClK ½n�n=l ;

and so, ClK ½n�n=l has dimension at least 1 over Z=lZ for all primes l dividing n:Applying Lemma 2 to ClK ½n� completes the proof.

It remains to show that S0CImðhÞ: Since f ðAÞ ¼ Cn by condition (3.5), there existsa polynomial gðXÞAF½T �½X � with

f ðXÞ ¼ ðX � AÞgðXÞ þ Cn:


Then

ðy� AÞgðyÞ ¼ �Cn:

Notice that the ideals ðy� AÞ and gðyÞ are relatively prime. If some prime p dividedboth, then

gðAÞ gðyÞ 0 ðmod pÞ;

and C 0 ðmod pÞ: This would imply that f 0ðAÞ ¼ gðAÞ 0 ðmod pÞ; contradictingcondition (3.6) that f 0ðAÞ and C are relatively prime. Thus the ideal ðy� AÞ is an nthpower; set Dn

m ¼ ðy� AÞ for some ideal Dm:Replacing C with D above, the same argument shows that ðy� BiÞ is also an nth

power for 1pipm � 1 since f 0ðBiÞ is relatively prime to D by condition (3.7). Set

Dni ¼ ðy� BiÞ for 1pipm � 1: We have hðDiÞ ¼ ½y� Bi�l for 1pipm � 1; and

hðDmÞ ¼ ½y� A�l ; where ½��l denotes the class of an element in K�=EK�l : This

completes the proof. &

To prove Theorem 2, we will show that it is possible to choose primes p1;y; pm�1

and s; and polynomials B0;y;Bm�1;A;C; and DAF½T � so that conditions (3.1)–(3.8)are satisfied and f ðXÞ is irreducible. As in the imaginary case, the existence ofinfinitely many such fields K is a consequence of the existence of one such fieldbecause of the finiteness of the class number.

3.1. Choosing polynomials

Let N ¼ 2mn; and define polynomials

F0ðXÞ ¼ X n � 1;

FiðX Þ ¼ X 2in þ 1 for 1pipm � 2;

and

Fm�1ðX Þ ¼ ðX n þ 1ÞðX 2m�1n þ 1Þ:

Note that F0ðX Þ?Fm�1ðXÞ ¼ X N � 1: Set mi ¼ degðFiÞ; and observe that

m0o?omm�1;

andPm�1

i¼0 mi ¼ N: Denote by FiðX ;Y Þ the homogeneous polynomial defined by

FiðXÞ:

FiðX ;Y Þ ¼ Ymi Fi

X

Y

� �AF½T �½X ;Y �:


Lemma 8. There exist infinitely many primes sAF½T � with jsj 1 ðmod m0N) and for

which there exists some wAF½T � with

w

s

� �la1; and

ðw þ 1Þm � 1

s

� �N

¼ 1

for all primes l dividing n:

Proof. For each natural number v; let zv be a fixed primitive v-th root of unity. SetL ¼ kðzm0N ; zmÞ: Choose any primes p1; p2 of L with ðp2;mn0Þ ¼ 1: Let p1 and p2 be

uniformizing parameters for p1 and p2; respectively. Let bAL� be such that

b p1 ðmod p21Þ;

�1 þ zm þ p2 ðmod p22Þ:

(

Because b ¼ ðb� p1Þ þ p1; and b� p1 and p1 have different ords, then

ordp1ðbÞ ¼ minfordp1

ðb� p1Þ; ordp1ðp1Þg ¼ 1:

So b is also a uniformizing parameter for p1: Since ðp2;mÞ ¼ 1 and

ðbþ 1Þm � 1 mzm�1m p2 ðmod p2

2), we also have that

ordp2ððbþ 1Þm � 1Þ ¼ ordp2

ðmÞ þ ðm � 1Þordp2ðzmÞ þ ordp2

ðp2Þ ¼ 1:

Thus ðbþ 1Þm � 1 is a uniformizing parameter for p2: It follows that X n0 � b is

Eisenstein with respect to p1; and X N � ðbþ 1Þm þ 1 is Eisenstein with respect to p2:

If E1 ¼ Lðffiffiffibn0

pÞ and E2 ¼ Lð

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðbþ 1Þm � 1N

pÞ; then p2 is totally ramified in E2: In E1;

however, p2 is unramified. Recall that sinceffiffiffibn0

pgenerates E1 over L; the different

DE1=L contains n0ðffiffiffibn0

pÞn0�1: If p2 was ramified in E1; that would imply that

n0ðffiffiffibn0

pÞn0�1AP for some P in E1 lying over p2: But then n0b would be contained in

P-L ¼ p2; a contradiction since b �1 þ zmc0 ðmod p2) and ðp2; n0Þ ¼ 1:Thus E1-E2 ¼ L; and so, GalðE1E2=LÞDGalðE1=LÞ � GalðE2=LÞ: Let s be a

generator of the cyclic group GalðE1=LÞ; and let t be the identity element ofGalðE2=LÞ: By the Chebotarev Density Theorem, there exist infinitely many primes s

of L; unramified in E1E2; for which there exists S over s with st ¼ ðS;E1E2=LÞ; thatis, with s inert in E1 and split completely in E2:

Let s ¼ s-F½T �; and let w b ðmod s). Because s is unramified in E1 ¼ Lðffiffiffibn0

pÞ;

w

s

� �l¼ b

s

� �l

a1;

and since s splits completely in E2 ¼ Lðffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðbþ 1Þm � 1N

pÞ; we also have

ðw þ 1Þm � 1

s

� �N

¼ ðbþ 1Þm � 1

s

� �N

¼ 1:


Finally, notice that since s splits completely in L; s must also split completely inFðTÞðzm0NÞ; so we must have jsj 1 ðmod m0N). &

By the previous lemma, we can choose wAF½T � and a prime s satisfying

jsj 1 ðmod m0NÞ;w

s

� �la1;

and

ðw þ 1Þm � 1

s

� �N

¼ 1:

Notice that m0n0 j m0N; so s satisfies the hypothesis of Lemma 7 thatjsj 1 ðmod m0n0).

Since ðwþ1Þm�1s

� �N¼ 1; there exists some zAF½T � with

zN ðw þ 1Þm � 1 ðmod sÞ: ð5Þ

Because jsj 1 ðmod 4Þ; there also exists some yAF½T � with

yN �1 ðmod sÞ: ð6Þ

Choose distinct monic, irreducible polynomials p1;y; pm�1as such that

jpij 1 ðmod m0n0Þ:

Let F denote the Euler phi function for polynomials. Choose primitive rootsgi ðmod pi) such that FðpiÞ44 and

g2i � ðm � 2Þgi � 1c0 ðmod piÞ: ð7Þ

Choose AAF½T � with

A g�1i ðmod piÞ; for 1pipm � 1;

w ðmod sÞ:

�ð8Þ

Choose irreducible polynomials h1;y; hm�1 with

hj

Fjð1Þg�1

i þ 1ðmod piÞ; jai;

Fjð1Þg�1

i þ gi

ðmod piÞ; j ¼ i;

Fjðz; yÞw þ 1

ðmod sÞ;

8>>>>>>><>>>>>>>:

ð9Þ

jhjj 1 ðmod NÞ:


Note that the denominators above are relatively prime to the moduli because

FðpiÞ44 for each i with 1pipm � 1 and ws

� �la1: Also note that there are infinitely

many choices for each hi by the strong version of Dirichlet’s Theorem for function

fields mentioned above in Section 2.1. Let h0 ¼ ðh1?hm�1Þ�1: Let S;V;U be thesets of primes S;V ; and U ; respectively, which satisfy:

S V U 1 ðmod p1?pm�1Þ;S 1; V y; U z ðmod sÞ;Sc0; V 1; FiðUÞ 0 ðmod hiÞ; 1pipm � 1:

ð10Þ

By Dirichlet’s Theorem, S and V are infinite. To see that U is also infinite, we justneed to show that FiðUÞ 0 is solvable modulo hi: But this follows from the factthat jhij 1 ðmod N). Because all the Nth roots of unity are contained in

F½T �=hiF½T �; X N � 1 factors into linear factors mod hi; and therefore so does FiðX Þ:Now we are ready to define B0;y;Bm�1;D; and C: Take integers v0;y; vm�1;

multiples of n; with

v0ov1o?ovm�1:

Given any S0;y;Sm�1AS; VAV; and UAU; set

Bi ¼ A � h�1i Svi

i FiðU ;VÞ;

D ¼ V N=nQm�1

i¼0

Svi=ni ;

C ¼ UN=nQm�1

i¼0

Svi=ni :

ð11Þ

It is clear that B0;C; and D are in F½T �: To see that Bi is as well for 1pipm � 1; notethat by Eq. (10), hi j FiðU ;VÞ:

FiðU ;VÞ ¼ Vmi FiU

V

� � FiðUÞ 0 ðmod hiÞ: &

3.2. Verification of divisibility conditions

Lemma 9. For any S0;y;Sm�1AS;VAV; and UAU; conditions (3.1)–(3.5) in

Lemma 7 and Proposition 2 are satisfied.

Proof. For (3.5), notice that

f ðAÞ ¼Ym�1

i¼0

ðA � BiÞ þ Dn


¼Ym�1

i¼0

h�1i Svi

i FiðU ;VÞ þ Dn

¼Ym�1

i¼0

Svi

i

!ðUn � V nÞ þ Dn

¼Cn � Dn þ Dn

¼Cn:

For (3.1)–(3.4), first observe that the following congruences follow from Eqs. (5), (6),and (8)–(11) and the definition of h0:

B0 g�1

i ðmod piÞ;�1 ðmod sÞ;

�

Bj �gi ðmod piÞ; i ¼ j;

�1 ðmod piÞ; iaj;

�1 ðmod sÞ;

8><>: ð12Þ

Dn 1 ðmod piÞ;�1 ðmod sÞ:

�

It is now not hard to see that conditions (3.1)–(3.4) in Lemma 7 hold. For example,(3.1) is satisfied since

f ð0Þ ð�1ÞmB0B1?Bm�1 þ Dn

ð�1Þmðg�1i Þð�giÞð�1Þm�2 þ 1 �1 þ 1 0 ðmod piÞ;

ð�1Þmð�1Þm � 1 1 � 1 0 ðmod sÞ:

(

Condition (3.2) follows from Eqs. (7) and (12):

f 0ð0Þ ¼Xm�1

k¼0

Ym�1

j¼0jak

ð�BkÞ

ð�1Þm�1Ym�1

j¼1

Bj þYj¼0jak

Bj þXm�1

k¼1kai

Yj¼0jak

Bk

2664

3775 ðmod piÞ


ð�1Þm�1 ð�1Þm�2ð�giÞ þ ð�1Þm�2g�1

i þXm�1

k¼1kaj

ð�1Þm�2

2664

3775 ðmod piÞ

ð�1Þm�1ð�1Þm�2½�gi þ g�1i þ m � 2� ðmod piÞ

c 0 ðmod piÞ;

f 0ð0Þ ¼Xm�1

k¼0

Ym�1

j¼0jak

ð�BkÞ ð�1Þm�1Xm�1

k¼0

ð�1Þm�1 mc0 ðmod sÞ:

The first part of condition (3.3) follows from Eq. (12) since clearly �gi is not an lth

power. The second part holds because Eq. (12) implies that for iaj; Bi

pj

� �l¼ �1

pj

� �l¼

1: Finally, for condition (3.4), Bi

s

� �l¼ �1

s

� �l¼ 1 by Eq. (12). The fact that A

s

� �la1

follows from Eq. (8) and Lemma 8. &

We can choose S0;y;Sm�1;V and U more specifically so that conditions (3.6) and(3.7) are also satisfied. For ease of notation, let

si ¼ h�1i Svi

i ;

for 0pipm � 1: Define polynomials

GiðX ;Y Þ ¼Ym�1

j¼0jai

sjFjðX ;YÞYm�1

0pjokpm�1j;kai

skFkðX ;YÞ � sjFjðX ;YÞ� �

for 1pipm � 1;

H1ðXÞ ¼Y

0piojpm�1

½ðsjFjðX ; 0Þ � siFiðX ; 0Þ�

Y0pejpvjð0pjpm�1Þ|aICf1;y;m�1g

YiAI

½s1F1ðX ; 0Þ � s0F0ðX ; 0Þ�2 �Ym�1

j¼0

S2ej

j

" #;

H2ðXÞ ¼Xm�1

i¼0

Ym�1

j¼0jai

sjFjð0;X Þ:

Lemma 10. If for S0;y;Sm�1AS; VAV; and UAU; we further have that

GiðU ;VÞc0 ðmod SiÞ;


H1ðUÞc0 ðmod VÞ;

and H2ðVÞc0 ðmod UÞ; ð13Þ

then B0;y;Bm�1;C; and D parametrized by S0;y;Sm�1;V ; and U also satisfy

conditions (3.6) and (3.7) in Proposition 2, and f ðX Þ is irreducible.

Proof. If either condition (3.6) or condition (3.7) is not satisfied, then at least one ofthe following congruences must hold:

Xm�1

i¼0

Ym�1

j¼0jai

ðA � BjÞ 0 ðmod SkÞ or ðmod UÞ; ð14Þ

Bi � Bj 0 ðmod SkÞ or ðmod VÞ: ð15Þ

Each of these congruences, however, implies a contradiction to one of the conditionsin (13). For example, if (14) holds, then by Eq. (11)

Xm�1

i¼0

Ym�1

j¼0

sjFjðU ;VÞ 0 ðmod SkÞ or ðmod UÞ:

If the modulus is U ; this contradicts the assumption that H2ðVÞc0 ðmod U), and ifthe modulus is Sk; the expression reduces to

Ym�1

j¼0jak

sjFjðU ;VÞ;

contradicting GkðU ;VÞc0 ðmod Sk). If (15) holds, then by Eq. (11), we have

sjFjðU ;VÞ � siFiðU ;VÞ 0 ðmod SkÞ or ðmod VÞ:

If the modulus is V ; this contradicts the hypothesis that H1ðUÞc0 ðmod V ), and ifthe modulus is Sk; this contradicts the hypothesis that GkðU ;VÞc0 ðmod Sk).

To see that f ðX Þ is irreducible, suppose for contradiction that there exists some

monic polynomial f̃ðX Þ with f̃ðXÞj f ðX Þ and 1pdegð f̃Þodegð f Þ: We have

f ðXÞ ¼Ym�1

i¼0

ðX � BiÞ þ Dn Ym�1

i¼0

ðX � BiÞ ðmod VÞ:

There must exist a nonempty set ICf0;y;m � 1g for which

f̃ðX Þ YiAI

ðX � BiÞ ðmod VÞ:


By condition (3.7) in Proposition 2, B0cBi ðmod V ), so B0 is not a multiple root of

f ðmod V ). We can assume, then, that 0eI ; replacing f̃ with f =f̃ if necessary. It

follows that V � f̃ðB0Þ: But f̃ðB0Þj f ðB0Þ; where

f ðB0Þ ¼ Dn ¼ V NYm�1

i¼0

Svi

i ;

so it must be that f̃ðB0Þ ¼ 7Qm�1

i¼0 Sei

i for some 0peipvi: Then

YiAI

ðB0 � BiÞ ¼ f̃ðB0Þ 7Ym�1

i¼0

Sei

i ðmod VÞ:

Squaring both sides and using Eq. (11) gives that

YiAI

siFiðU ; 0Þ � s0F0ðU ; 0Þð Þ2Ym�1

i¼0

S2ei

i ðmod VÞ;

which implies that H1ðUÞ 0 ðmod V ), a contradiction.It remains to show that we can, in fact, choose primes S0;y;Sm�1;V ; and U with

GiðU ;VÞc0 ðmod SiÞ;

H1ðUÞc0 ðmod VÞ;

and H2ðVÞc0 ðmod UÞ:

Let P ¼ p1?pm�1sh1?hm�1: Given S0AS with degðS0Þ4degðPÞ; choose anirreducible polynomial Si in S; 1pipm � 1; with degðSiÞ ¼ degðSi�1Þ þ 1: This ispossible by the strong version of Dirichlet’s Theorem for function fields. Set

UðS0Þ ¼ fUAU j jSm�1jojU joqmþ1jS0jg;

and

VðS0Þ ¼ fVAV j qmþ1jS0jojV joqmþ3jS0jg:

We claim that as jS0j-N; jVðS0Þj; jUðS0Þj-N: By the function field version of theTchebotarev Density Theorem,

#fVAVðS0Þ j degðVÞ ¼ x;V a ðmod PÞgB qx

FðPÞx;

for any fixed aAFq½T � with ða;PÞ ¼ 1: As jS0j increases, the interval containing jV j;which has length jS0jðqmþ3 � qmþ1Þ; increases as well. It follows that jVðS0Þj-N as

jS0j-N: A similar argument can be used for UðS0Þ: Since jSm�1j ¼ qm�1jS0j; the

interval containing U has length jS0jðqmþ1 � qm�1Þ which also goes to N withjS0j: &


Lemma 11. For sufficiently large S0AS; there exist VAVðS0Þ and UAUðS0Þ which

satisfy the conditions in Eq. (13).

Proof. Fix S0AS; and consider Si;Gi; ð1pipm � 1Þ;H1; and H2 to be fixedas well. For each VAVðS0Þ; the polynomial GiðX ;VÞ mod Si is not null,and its degree di is independent of the choice of S0 and V : Since each UAUðS0Þhas jSijojU joqmþ1jSij; it follows that if U1;U2AUðS0Þ with U1 U2 ðmod Si),then U1 � U2 ¼ rSi for some rAFq½T �: Then degðrÞ ¼ degðU1 � U2Þ � degðSiÞom þ1: Thus for a given U in UðS0Þ with GiðU ;VÞ 0 ðmod Si), there are at most qmþ1

solutions in UðS0Þ that are congruent to U modulo Si: Therefore, for each

VAVðS0Þ; the number of UAUðS0Þ with GiðU ;VÞ 0 ðmod Si) is at most qmþ1di:

Thus at most qmþ1dijVðS0Þj pairs ðV ;UÞAVðS0Þ �UðS0Þ satisfy GiðU ;VÞ 0 ðmod Si).

Next, notice that the leading term of H1ðX Þ is

Ym�1

i¼1

siXmi

Y|aICf1;y;m�1g

YiAI

ðsiXmiÞ2ð1þv0Þ?ð1þvm�1Þ:

For any VAVðS0Þ; then, the polynomial H1ðXÞ is not null and its degree d 01 is

independent of the choice of S0 and V : Each UAUðS0Þ has jU jojV j so UðS0Þ iscontained in one set of representatives modulo V : The congruenceH1ðXÞ 0 ðmod V ) then has at most d 0

1 solutions in UðS0Þ; and so, the number of

pairs ðV ;UÞAVðS0Þ �UðS0Þ with H1ðUÞ 0 ðmod V ) is at most d 01jVðS0Þj:

Finally, if d 02 ¼ degðH2Þ; then at most q4d 0

2jUðS0Þj pairs ðV ;UÞAVðS0Þ �UðS0Þsatisfy H2ðVÞ 0 ðmod U). This follows because each VAVðS0Þ has

jU jojV joq4jU j:If no pair ðV ;UÞAVðS0Þ �UðS0Þ satisfies all four congruence conditions, then we

must have

jVðS0Þ �UðS0Þjpðqmþ1d0 þ?þ qmþ1dm�1 þ d 01ÞjVðS0Þj þ q4d 0

2jUðS0Þj;

that is,

1pðqmþ1d0 þ?þ qmþ1dm�1 þ d 01ÞjUðS0Þj�1 þ q4d 0

2jVðS0Þj�1: ð16Þ

Since d0;y; dm�1; d 01 and d 0

2 are independent of the choice of S0; the right-hand side

of (16) goes to 0 as jS0j-N: For large enough S0 then, we can find V and U tosatisfy the congruences. &

3.3. The infinite prime

Lemma 12. The prime at infinity splits completely in K:


Proof. We claim that the Newton polygon for f ðX Þ with respect to the prime atinfinity consists of m distinct line segments. It then follows that the m roots of f ðXÞin %kN have distinct ordN: We will show that each of these roots is actually in kN;and therefore, the m roots must be distinct.

First note that we can choose jS0j large enough so that

* ðv1 � v0Þ degðS0Þ42 degðh1Þ þ degðh2Þ þ?þ degðhm�1Þ;* ðviþ1 � viÞ degðS0Þ4degðhiþ1Þ � degðhiÞ for 1pipm � 2; and* v0 degðS0Þ þ m0 degðVÞ4degðAÞ:

Also recall that vioviþ1; m0omiþ1; and degðSiÞodegðSiþ1Þ for 0pipm � 2: Given

these conditions, we see that

degðB0Þ ¼ degðA � h�10 Sv0

0 F0ðU ;VÞÞ

¼ degðh1Þ þ?þ degðhm�1Þ þ v0 degðS0Þ þ m0 degðVÞ

o � degðh1Þ þ v1 degðS1Þ þ m1 degðVÞ

¼ degðA � h�11 Sv1

1 F1ðU ;VÞÞ

¼ degðB1Þ;

and

degðBiÞ ¼ degðA � h�1i Svi

i FiðU ;VÞÞ

¼ � degðhiÞ þ vi degðSiÞ þ mi degðVÞ

o � degðhiþ1Þ þ viþ1 degðSiþ1Þ þ miþ1 degðVÞ

¼ degðA � h�1iþ1S

viþ1

iþ1 Fiþ1ðU ;VÞÞ

¼ degðBiþ1Þ for 1pipm � 2:

Thus

degðB0ÞodegðB1Þo?odegðBm�1Þ: ð17Þ

Also notice that

degðB0Þ þ degðB1Þ þ?þ degðBm�1Þ ¼ n degðDÞ; ð18Þ


since

degðB0Þ þ?þ degðBm�1Þ ¼Xm�1

i¼0

degðA � h�1i Svi

i FiðU ;VÞÞ

¼Xm�1

i¼0

ðdegðh�1i Þ þ vi degðSiÞ þ mi degðVÞÞ

¼Xm�1

i¼0

vi degðSiÞ þ NdegðVÞ

¼ n degðDÞ:

Given these conditions on the degrees of the Bi’s, we see that the points to considerfor the Newton polygon are

ð0;�degðB0?Bm�1 þ DnÞÞ;

ði;�degðBiBiþ1?Bm�1ÞÞ; 1pipm � 1;

and ðm; 0Þ:

We must show that the line segments connecting these points do, in fact, constitutethe edges of the Newton polygon, and no two segments have the same slope. Let L0

denote the line segment between the points ð0;�degðB0?Bm�1 þ DnÞÞ andð1;�degðB1?Bm�1ÞÞ: From Eqs. (17) and (18), it follows that

degðB0Þon degðDÞ

m:

Subtracting n degðDÞ from both sides gives that

�degðB1Þ �?� degðBm�1Þo1

m� 1

� �n degðDÞ:

Because the slope of the secant line from ð0;�degðB0?Bm�1 þ DnÞÞ to ðm; 0Þ is1m

n degðDÞ; this implies that L0 lies strictly below that secant line and is therefore one

edge of the Newton polygon.Let Li; 1pipm � 2; denote the line segment connecting the points

ði;�degðBiBiþ1?Bm�1ÞÞ and ði þ 1;�degðBiþ1?Bm�1ÞÞ; let Lm�1 be the linesegment between ðm � 1;�degðBm�1ÞÞ and ðm; 0Þ: If mi denotes the slope of Li;for 0pipm � 1; then notice that mi ¼ degðBiÞ:

Next we show that each Li constitutes an edge of the Newton polygon. FromEqs. (17) and (18), observe that

degðBiÞodegðBiBiþ1?Bm�1Þ

m � i:


Subtracting degðBi?Bm�1Þ from each side yields:

�degðBiþ1?Bm�1Þo1

m � i� 1

� �degðBiBiþ1?Bm�1Þ:

But this is equivalent to stating that Li lies strictly below the secant line connecting

ði;�degðBiBiþ1?Bm�1ÞÞ and ðm; 0Þ

which has slope 1m�i

degðBi?Bm�1Þ: Thus Li is one edge of the Newton polygon for

f ðX Þ with respect to the infinite prime. Finally, by Eq. (17), the slopes of the linesegments Li are distinct which implies that the m roots are in kN: Otherwise, if a andb are roots with a; bekN; then there must exist some isomorphism s from kNðaÞ tokNðbÞ which leaves kN fixed. It follows that jbj ¼ jsðaÞj ¼ jaj; that is, degðbÞ ¼degðaÞ; a contradiction. This completes the proof. &

Acknowledgments

I thank my advisor Michael Rosen for providing many helpful suggestions as Iworked on my dissertation, of which this work is a major part. I also thank thereferee for his valuable comments.

References

[1] T. Azuhata, H. Ichimura, On the divisibility problem of the class numbers of algebraic number fields,

J. Fac. Sci. Univ. Tokyo 30 (1984) 579–585.

[2] D. Cardon, R. Murty, Exponents of class groups of quadratic function fields over finite fields, Canad.

Math. Bull. 44 (4) (2001) 398–407.

[3] H. Cohen, H.W. Lenstra Jr., Heuristics on class groups of number fields, Number Theory,

Noordwijkerhout 1983, Lecture Notes in Mathematics, Vol. 1068, Springer, Berlin, New York, 1984,

pp. 33–62.

[4] C. Friesen, Class number divisibility in real quadratic function fields, Canad. Math. Bull. 35 (1992)

361–370.

[5] H. Ichimura, On the class groups of pure function fields, Proc. Japan Acad. 64 (1988) 170–173

(corrigendum Proc. Japan Acad. 75 (1999) 22).

[6] T. Nagell, Uber die Klassenzahl imaginar quadratischer Zahlkorper, Abh. Math. Sem. Univ.

Hamburg 1 (1922) 140–150.

[7] S. Nakano, On ideal class groups of algebraic number fields, J. Reine Angew. Math. 358 (1985)

61–75.

[8] A. Pacelli, A lower bound on the number of cyclic function fields with class number divisible by n;

preprint.

[9] M. Rosen, Number Theory in Function Fields, Springer, Berlin, 2002.

[10] Y. Yamamoto, On unramified Galois extensions of quadratic number fields, Osaka J. Math. 7 (1970)

57–76.


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