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Demjanenko matrix and recursion formula for relative class number over function fields
Journal of Number Theory 98 (2003) 55–66
http://www.elsevier.com/locate/jnt
Demjanenko matrix and recursion formula for relative class number over function fields Hwanyup Junga,*,1 and Jaehyun Ahnb a
Department of Mathematics, Korea University, Seoul 136-701, South Korea b Department of Mathematics, KAIST, Taejon 305-701, South Korea Received 5 June 2001; revised 28 March 2002 Communicated by D. Goss
Abstract In this paper, we define Demjanenko matrix in function field and express the relative ideal class numbers h ðOKPn Þ as the determinant of this matrix. We also define another matrix which give us a recursion formula for the relative divisor class numbers h ðKPn Þ: r 2002 Elsevier Science (USA). All rights reserved. Keywords: Demjanenko matrix; Relative class number; Recursion formula; Function field
1. Introduction Let A ¼ Fq ½T be the ring of polynomials over a finite field Fq with q-elements, and k ¼ Fq ðTÞ: For each MAA; one uses the Carlitz module to construct a field extension þ KM ; called the Mth cyclotomic function field and its maximal real subfield KM : Let þ OKM and OKMþ be the integral closure of A in KM and KM ; respectively. It is known that the divisor class number hðKM Þ of KM is divisible by the divisor class number þ þ þ hðKM Þ of KM : Write h ðKM Þ ¼ hðKM Þ=hðKM Þ; called the relative divisor class number of KM : It is also known that the ideal class number hðOKM Þ of OKM is divisible by the ideal class number hðOKMþ Þ of OKMþ : We write h ðOKM Þ ¼ hðOKM Þ=hðOKMþ Þ; called the relative ideal class number of KM :
*Corresponding author. E-mail addresses: [email protected] (H. Jung), [email protected] (J. Ahn). 1 Supported in part by BK21 Project at Korea University. 0022-314X/02/$ - see front matter r 2002 Elsevier Science (USA). All rights reserved. PII: S 0 0 2 2 - 3 1 4 X ( 0 2 ) 0 0 0 2 3 - 9
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H. Jung, J. Ahn / Journal of Number Theory 98 (2003) 55–66
Let P be an irreducible polynomial. In [4], Guo and Shu studied the behavior of class numbers h ðKPn Þ and hðKPþn Þ in the tower of cyclotomic function fields KP CKP2 C?CKPn C?: In [2], Bae and Kang defined some matrices which may be viewed as an analogue of Maillet matrix and Demjanenko matrix in classical case, and expressed h ðKPn Þ and hðKPþn Þ as the determinant of these matrices. In this paper, we define a matrix (we call it Demjanenko matrix) and express h ðOKPn Þ as the determinant of this matrix (Theorem 3.1). Our matrix is more natural analogue of the classical Demjanenko matrix [5] than Bae and Kang’s one because it only involves the arithmetic of ðA=Pn Þn and character does not appear. It is known [9, Lemma 3] that h ðKPn Þ ¼ ðq 1Þr h ðOKPn Þ;
ð1Þ
where r ¼ qðn1Þdeg P ðqdeg P 1Þ=ðq 1Þ 1: Thus our formula is also one for h ðKPn Þ: Let c be a prime factor of q 1 and c the highest power of c dividing q 1: From Eq. (1), we see that cr is a lower bound for the exponent of c dividing h ðKPn Þ: This lower bound cr is larger than that of Guo-Shu [4, Theorem 3.1] and equals to that of Bae–Kang [2, Theorem 3]. In [6], Ireland and Small asked whether cr is the highest exponent of c dividing h ðKPn Þ: We find an example which show that cr is not the highest one (Section 5, Example 3). Adopting the idea of Girstmair [3], we also define a matrix to get a recursion formula for the relative divisor class numbers h ðKPn Þ (Theorem 4.1). In the final section we give several examples of h ðKPn Þ by using the Demjanenko matrix and the recursion formula.
2. Basic facts and notations Let M be a polynomial in A: It is well known that the Galois group GM of KM over k is isomorphic to ðA=MÞn ; explicitly the isomorphism is given as j : ðA=MÞn -GM ;
A mod M/sA :
þ : aAFnq g is the Galois group of KM over KM : For Under this isomorphism, J ¼ fsaP any subset H of GM ; let sðHÞ ¼ sAH s: A character w of GM is called even if its restriction to J is trivial, and odd otherwise. Any character w of GM can be viewed as character of ðA=MÞn ; so the conductor Fw of w is defined as a divisor of M: For a polynomial NAA; we let MN (resp. Mþ N ) be the set of all the polynomials (resp. monic polynomials) in A with degree less than the degree of N and prime to N: Let
H. Jung, J. Ahn / Journal of Number Theory 98 (2003) 55–66
57
þ M N ¼ MN \MN : It is known [4] that
0 1 YB X C h ðKM Þ ¼ wðAÞA: @ w odd
AAMþ Fw
When M ¼ Pn ; a power of an irreducible polynomial P; then the formula can be written as Y
h ðKPn Þ ¼
w odd
0
X
@
1 wðAÞA:
ð2Þ
AAMþ Pn
% For each polynomial A prime to M; we let AAM M be the unique element such that % A A% mod M and sgnM ðAÞ denote the leading coefficient of A:
3. Demjanenko matrix For a monic polynomial MAA; let yM ¼
X
ZM ð0; tÞt1 ;
tAGM
where ZM ðs; tÞ is the partial zeta function associated to t: It is a Stickel-berger element of the extension KM =k: For tAGM ; let At AMM be the unique element such that sAt ¼ t; under the isomorphism j : ðA=MÞn -GM : Lemma 3.1. ZM ð0; tÞ is ðq 2Þ=ðq 1Þ if At is monic, and 1=ðq 1Þ otherwise. Proof. For each tAGM ; let Wt ¼ fAAA: A monic; ðA; MÞ ¼ 1; and ððAÞ; KM =kÞ ¼ tg: Then by definition, ZM ðs; tÞ ¼
X
qs deg ðAÞ :
AAWt
It is easy to check that ( Wt ¼
fQM þ At : Q ¼ 0 or Q is monicg fQM þ At : Q is monicg
if At is monic; if At is non-monic:
H. Jung, J. Ahn / Journal of Number Theory 98 (2003) 55–66
58
Thus
ZM ðs; tÞ ¼
8 s deg ðMÞ < qs deg ðAt Þ þ q 1s
if At is monic;
: qs deg ðMÞ 1q1s
if At is non-monic:
1q
By evaluating at s ¼ 0; we get the result.
&
By Lemma 3.1, we can write the Stickelberger element yM as follows: yM
X q 2 X 1 1 ¼ s þ s1 q1 A q1 A þ AAM AAMM M X 1 ¼ s1 sðGM Þ: A q1 þ
ð3Þ
AAMM
From now on, we assume M ¼ Pn ; P is monic irreducible polynomial and nX1: Definition 3.1. Let S 0 be the GM -submodule of Q½GM generated by sðGM Þ=ðq 1Þ and yM : Let S ¼ S 0 -Z½GM ; called the Stickelberger ideal of KM : By Tate [8, Chapter IV, 1.8 Proposition], we have corKM =KPr ðyPr Þ ¼ sðGalðKM =KPr ÞÞyM ; of the for 1prpn: Here corKM =KPr is the correstriction map. Thus our definition P Stickelberger ideal is coincided with Yin’s one [9, Definition 1]. Let Z ¼ AAMþ s1 A : M From Eq. (2), we have S 0 ¼ Z½GM Z þ ZsðGM Þ=ðq 1Þ; and so S ¼ Z½GM Z þ ZsðGM Þ: Let A ¼ fxAZ½GM : sðJÞxAZsðGM Þg: In [1], we have shown the following; Proposition 3.1. ðA : SÞ ¼ h ðOKM Þ: For AAMM ; let A1 be the unique element of MM such that AA1 1 mod M: Lemma 3.2. Y1 ¼ fZg,fðsa 1ÞsA1 : 1aaAFnq ; AAMþ M g is a Z-basis of A:
H. Jung, J. Ahn / Journal of Number Theory 98 (2003) 55–66
Proof. Write x ¼
P
aAFnq
P
AAMþ M
X
sðJÞx ¼
na;A sa sA1 AA: Then X na;A sA1 sðJÞ
aAFnq AAMþ M
0
X
¼
@
AAMþ M
So
P
aAFnq
X
1 na;A AsA1 sðJÞAsðGM ÞZ:
aAFnq
na;A ¼ mAZ for all AAMþ M : Thus 0 1 X X na;A sa sA1 ¼ @m na;A AsA1 þ aAFnq
59
aAFnq ;aa1
¼ msA1 þ
X
na;A sa sA1
aAFnq ;aa1
X
na;A ðsa 1ÞsA1 :
aAFnq ;aa1
Therefore, x¼
0
X
X
@msA1 þ
AAMþ M
¼ mZ þ
X
1 na;A ðsa 1ÞsA1 A
aAFnq ;aa1
X
na;A ðsa 1ÞsA1 :
aAFnq ;aa1 AAMþ M
Now we show the Z-independence of Y1 : Assume X X mZ þ na;A ðsa 1ÞsA1 ¼ 0: aAFnq ;aa1 AAMþ M
Then X
0 @m þ
AAMþ M
1
X
na;A AsA1 þ
aAFnq ;aa1
X
X
ðna;A Þsa sA1 ¼ 0:
aAFnq ;aa1 AAMþ M
P n þ Thus m þ aAFnq ;aa1 na;A ¼ 0 and na;A ¼ 0 for all aAFq ; aa1; AAMM : Thus m is also zero. & Lemma 3.3. Y2 ¼ fsðGM Þg,fsA Z: AAM M g is a Z-basis of S: Proof. For AAMþ M; sðJÞsA Z ¼ sA sðGM Þ ¼ sðGM Þ: P So sA Z ¼ aAFnq ;aa1 saA Z þ sðGM Þ: Thus Y2 generates S: Since the cardinality of Y2 is just the rank of S; Y2 is a Z-basis of S: &
H. Jung, J. Ahn / Journal of Number Theory 98 (2003) 55–66
60
For A; BAM M ; we let /ABS ¼ 1 if sgnM ðABÞ ¼ 1 and /ABS ¼ 0 otherwise. Define DM ¼ ð/ABSÞA;B ; where A; B run through M M: Theorem 3.1. jdet DM j ¼ h ðOKM Þ:
Proof. By Proposition 3.1, Lemmas 3.2 and 3.3, we see that ðA : SÞ is the absolute value of the determinant of the transition matrix T of Y2 with respect to Y1 : Since X
X
X
ðsaA1 s1 A Þ¼
aAFnq ;aa1 AAMþ M
X
saA1 ðq 2Þ
aAFnq ;aa1 AAMþ M
X
sA1
AAMþ M
¼ sðGM Þ ðq 1ÞZ; we have X
sðGM Þ ¼ ðq 1ÞZ þ
X
ðsa 1ÞsA1 :
ð4Þ
aAFnq ;aa1 AAMþ M
We introduce some notations. For A; BAMM ; let aAB ¼ sgnM ððABÞ1 Þ1 and ðABÞ0 ¼ sgnM ððABÞ1 ÞAB: Thus AB ¼ aAB ðABÞ0 : For AAM M; sA Z ¼
X BAMþ M
¼
X
X
sAB1 ¼
saAB1 sðAB1 Þ0
BAMþ M
ðsaAB1 1ÞsðAB1 Þ0 þ
BAMþ M
X
sðAB1 Þ0 :
BAMþ M
þ 1 It is easy to see that fðAB1 Þ0 : BAMþ M g ¼ fB : BAMM g for any AAMM : Thus we have
sA Z ¼ Z þ
X
ðsaAB1 1ÞsðAB1 Þ0 :
ð5Þ
BAMþ M þ Let C 1 ¼ ðAB1 Þ: If ðAB1 Þ1 AMþ M ; then CAMM and saAB1 1 ¼ 0: Assume 1 1 1 1 ðAB Þ AMM : Then CAMM and C AB mod M; equivalently AC
B mod M: Thus for A; CAM M ; ðA; CÞ-entry of the transition matrix T is 1 if ðACÞ
H. Jung, J. Ahn / Journal of Number Theory 98 (2003) 55–66
61
is monic and 0 otherwise. Therefore the transition matrix T is given as follows:
P Since sðGÞ aAFnq ;aa1 sa Z ¼ Z; by elementary row operations, we can convert the transition matrix T to
So we get the result.
&
4. Recursion formula In this section, we assume M ¼ Pn ; where P is a monic irreducible polynomial and ˜ ˜ nX2: For AAA\PA; let AAM M=P be the unique element such that A A mod M=P: n n ˜ ˜ Let MM ¼ fAAMM : AAMM=P ; AaAg: For each A; BAMM ; we define cA;B ¼ ð/ABS /AB sgnM=P ðBÞ1 SÞ ˜ /AB˜ sgnM=P ðBÞ ˜ 1 SÞ ð/ABS and CM ¼ ðcA;B ÞA;BAMnM : Let Zn ¼ ðNP sðHÞÞyM ¼ ðNP sðHÞÞZ; where H ¼ GalðKM =KM=P Þ and NP ¼ qdeg P the norm of P: Lemma 4.1. For a character w of GM ; let Fw be the conductor of w: Then wðZn Þa0 if and only if Fw ¼ M and w is odd. Thus q2 1 FðMÞ 1 dimQ Q½GM Zn ¼ : q1 NP Here FðMÞ ¼ jðA=MÞn j ¼ qðn1Þdeg P ðqdeg P 1Þ:
H. Jung, J. Ahn / Journal of Number Theory 98 (2003) 55–66
62
Proof. If w is even, wðsðJÞZn Þ ¼ wðsðJÞðNP sðHÞÞZÞ ¼ wððNP sðHÞÞsðGM ÞÞ ¼ 0; and wðsðJÞZn Þ ¼ ðq 1ÞwðZn Þ: So wðZn Þ ¼ 0: Assume Fw is a proper divisor of M: Then w is trivial on H; so wðNp sðHÞÞ ¼ 0: Thus wðZn Þ ¼ 0: Suppose Fw ¼ M and w is odd. Then wðsðHÞÞ ¼ 0; so wðZn Þ ¼ NP wðyM Þa0 [9, Lemma 6]. & For AAMM ; let ZnA ¼ sA Zn : Lemma 4.2. fZnA : AAMnM g and fZnA1 : AAMnM g are Z-bases of Z½GM Zn : In particular, they are Q-bases of Q½GM Zn : f ¼ aA˜ and Proof. For AAMM and aAFnq ; aA X X ZnaA ¼ sa ZnA ¼ sðJÞsA Zn ¼ 0: aAFnq
aAFnq
P ˜ Thus Z½GM Zn ¼ A ZZnA ; where A runs P through Mn M with A non-monic. Since ˜ ¼ Bg ¼ sB H; we have fsA : AAMM ; A Z ¼ 0 for any BAMM=P : So we ˜ A AAMM ;A¼B P n get ZnB ¼ AAMn ;A¼B Z : By using these facts, we follow the proof of [3, Lemma 3] ˜ A M to get our result. & Let b ¼
P
AAMþ M
sA AQ½GM and b ¼ bZn ¼
P
AAMþ M
ZnA ¼
P
AAMM
/ASZnA :
Lemma 4.3. For each AAMM ; X
sA1 b ¼
cA;B ZnB :
BAMnM
Proof. For AAMM ; we have X X /BSsA1 ZnB ¼ /ABSZnB sA1 b ¼ BAMM
¼
X
BAMM
˜ BAMM ;BAM ;BaB˜ M=P
¼
/ABSZnB þ
n
BAMM
þ
X
X
/ABSZnB þ
þ ˜ BAMM ;BAM ;BaB˜ M=P
X
/ABSZnB
˜ BAMM ;BAM ;B¼B˜ M=P
X
þ
X
/ABSZnB þ
X
þ ˜ BAMM ;BAM ;B¼B˜ M=P
/ABSZnB
BAM M=P
þ ˜ BAMM ;BAM ;BaB˜ M=P
/ABSZnB þ
X BAMþ M=P
/ABSZnB
/ABSZnB
H. Jung, J. Ahn / Journal of Number Theory 98 (2003) 55–66
63
and X
X
X
BAM M=P
˜ CAMM ;C¼B
/ABSZnB ¼
BAM M=P
X
¼
/ABSZnC
n
˜ n; /ABSZ B
n
BAMM
X
/ABSZnB ¼
þ ˜ BAMM ;BAM ;BaB˜ M=P
0
X
/ABS@
þ ˜ BAMM ;BAM ;BaB˜ M=P
¼
X
1
X
ZnaB A n
1aaAFq
˜ 1 SZn : /AB sgnM=P ðBÞ B
n
BAMM
Finally, X
/ABSZnB ¼
BAMþ M=P
0
X
/ABS@
BAMþ M=P
¼
X
1
X
ZnaB A n
1aaAFq
/AB sgnM ðBÞ1 SZnB
BAM M=P
¼
X
0 /AB sgnM ðBÞ1 S@
BAM M=P
¼
X
X
1 ZnC A
˜ CAMnM ;C¼B 1
˜ SZnB : /AB˜ sgnM=P ðBÞ
n
BAMM
So we get the result.
&
Theorem 4.1. h ðKPn Þ ¼ jdet CPn j h ðKPn1 Þ: Proof. From the equation (2), we have h ðKPn Þ ¼ h ðKPn1 Þ
Y
wðbÞ:
Fw ¼M;odd
Let lb be the Q-linear map on Q½GM Zn defined by the multiplication by b: By Sinnott [7, Lemma 1.2(b)], we have Y wðbÞ ¼ jdet lb j: Fw ¼M;odd
H. Jung, J. Ahn / Journal of Number Theory 98 (2003) 55–66
64
Let g be the Q-linear map on Q½GM Zn defined by gðZnA Þ ¼ ZnA1 : Then by Lemma 4.3, jdet ðgÞj ¼ 1: By Lemma 4.3, X
ðlb 3gÞðZnA Þ ¼
cA;B ZnB
n
BAMM
for each AAMnM : Hence jdet lb j ¼ jdet ðlb 3gÞj ¼ jdet CPn j: So we get the result.
&
Remark 4.1. Assume q ¼ 3: We claim that cA;B 0 mod 2: If B˜ is monic, we have cA;B ¼ 0 trivially. Suppose B˜ is non-monic. Then cA;B ¼ ð/ABS / ABSÞ ˜ / ABSÞ: ˜ ˜ have an opposite sign, ð/ABS Since AB and AB (resp. AB˜ and AB) ˜ ˜ /ABS / ABS 1 mod 2 (resp. /ABS / ABS 1 mod 2). So cAB
0 mod 2; and we prove the claim. Let c0A;B ¼ cA;B =2 and C0Pn ¼ ðc0A;B ÞA;B : Thus we have h ðKPn Þ ¼ 2rn jdet C0Pn j h ðKPn1 Þ;
ð6Þ
where rn ¼ rank CPn ¼ 3ðn2Þdeg P ð3deg P 1Þ2 =2: By substituting (1) into (6), we have the following recursion formula for the relative ideal class numbers: h ðOKPn Þ ¼ jdet C0Pn j h ðOKPn1 Þ: It is interesting to find a recursion formula for the relative ideal class numbers h ðOKPn Þ in general.
5. Examples We give several examples which are obtained by using MAPLE. Example 5.1. Let q ¼ 3 and P ¼ T: Then M P2 ¼ f2; 2T þ 1; 2T þ 2g; and the Demjanenko matrix DP2 is 2
1
6 41 1
1 1 0
1
3
7 05 0
and h ðOKP2 Þ ¼ jdet DP2 j ¼ 1; so h ðKP2 Þ ¼ 22 : The M P3 is f2; 2T þ 1; 2T þ 2; 2T 2 þ 1; 2T 2 þ 2; T þ 2T 2 þ 1; T þ 2T 2 þ 2; 2T þ 2T 2 þ 1; 2T þ 2T 2 þ 2g; the
H. Jung, J. Ahn / Journal of Number Theory 98 (2003) 55–66
Demjanenko matrix DP3 is 2
1 61 6 6 61 6 6 61 6 61 6 6 61 6 61 6 6 41 1
1 1 1 1
1 1 0 0
1 1 1 1
1 1
1 1 0 1
1 1 1 0
1 1 1 1
0 1
0 1 1 1 1 1
0 0 1 0 1 0
0 0 0 1 1 1
1 1 0
1 0 0 0
1 1 0 0
1 0 0 1
0 1
65
3 1 07 7 7 07 7 7 07 7 07 7 7 07 7 17 7 7 15 0
and h ðOKP3 Þ ¼ 7; so h ðKP3 Þ ¼ 28 7: The MnP3 is fT 2 þ 2; T 2 þ 2T þ 1; T 2 þ 2T þ 2; 2T 2 þ 2; 2T 2 þ 2T þ 1; 2T 2 þ 2T þ 2g; the matrix CP3 is given by 3 2 2 0 2 2 2 2 7 6 6 2 2 2 2 2 0 7 7 6 6 2 2 0 2 0 2 7 7 6 7 6 6 0 2 0 2 2 2 7 7 6 6 2 0 0 0 2 2 7 5 4 0 0 2 2 2 0 and jdet CP3 j ¼ 26 7: Thus we see that h ðKP3 Þ=h ðKP2 Þ ¼ jdet CP3 j: Example 5.2. Let q ¼ 3 and M ¼ P ¼ T 3 þ T 2 1: Then M M ¼ f2; 2 T; 1 þ 2 T; 2 þ 2 T; 2 T 2 ; 1 þ 2 T 2 ; 2 þ 2 T 2 ; T þ 2 T 2 ; 1 þ T þ 2 T 2 ; 2 þ T þ 2 T 2 ; 2 T þ 2 T 2 ; 1 þ 2 T þ 2 T 2 ; 2 þ 2 T þ 2 T 2 g and the Demjanenko matrix DM is given by 3 2 1 1 1 1 1 1 1 1 1 1 1 1 1 61 1 1 1 0 0 0 1 1 1 1 0 17 7 6 7 6 61 1 1 1 1 1 1 1 0 0 0 0 07 7 6 61 1 1 1 1 0 1 0 0 0 1 1 17 7 6 7 6 61 0 1 1 1 1 0 0 1 1 1 0 17 7 6 7 6 61 0 1 0 1 0 1 1 0 0 0 1 07 7 6 61 0 1 1 0 1 1 1 0 1 1 0 07 7 6 7 6 61 1 1 0 0 1 1 1 0 0 1 0 17 7 6 61 1 0 0 1 0 0 0 0 1 0 1 07 7 6 7 6 61 1 0 0 1 0 1 0 1 0 1 1 07 7 6 61 1 0 1 1 0 1 1 0 1 1 0 17 7 6 7 6 41 0 0 1 0 1 0 0 1 1 0 1 15 1 1 0 1 1 0 0 1 0 0 1 1 0 and h ðOKM Þ ¼ jdet DM j ¼ 5 79:
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H. Jung, J. Ahn / Journal of Number Theory 98 (2003) 55–66
Example 5.3. Let q ¼ 3 and deg P ¼ 1: For np5; one can find h ðKPn Þ in [4, Table 1]. When n ¼ 6; using the matrix CPn ; we calculate h ðKP6 Þ ¼ 2252 52 717 135 174 1917 314 377 432 613 67 733 793 1093 1272 151 181 241 379 397 433 5232 1423 1531 1747 3511 5779 6823 7057 7687 12619 25183 31177 156151 192889 196687 208009 347707:
References [1] S. Bae, H. Jung, J. Ahn, Cyclotomic units and Stickelberger ideals of global function fields, Trans. Amer. Math. Soc., to appear. [2] S. Bae, P. Kang, Class numbers of cyclotomic function field, Acta. Arith. 102 (2002) 251–259. [3] K. Girstmair, A recursion formula for the relative class number of the pn th cyclotomic field, Abh. Math. Sem. Univ. Hamburg. 61 (1991) 131–138. [4] L. Guo, L. Shu, Class numbers of cyclotomic function field, Trans. Amer. Math. Soc. 351 (1999) 4445– 4467. [5] F. Hazama, Demjanenko matrix, class numbers, and Hodge group, J. Number Theory. 34 (1990) 174– 177. [6] K. Ireland, R. Small, Class numbers of cyclotomic function fields, Math. Comput. 46 (1986) 337–340. [7] W. Sinnott, On the Stickelberger ideal and the circular units of an abelian field, Invent. Math. 62 (1980/81) 181–234. [8] J. Tate, Les Conjecture de Stark sur les functions L d’Artin en s ¼ 0; Birkha¨user, Boston. [9] L. Yin, Stickelberger ideals and relative class numbers in function fields, J. Number Theory. 81 (2000) 162–169.
http://www.elsevier.com/locate/jnt
Demjanenko matrix and recursion formula for relative class number over function fields Hwanyup Junga,*,1 and Jaehyun Ahnb a
Department of Mathematics, Korea University, Seoul 136-701, South Korea b Department of Mathematics, KAIST, Taejon 305-701, South Korea Received 5 June 2001; revised 28 March 2002 Communicated by D. Goss
Abstract In this paper, we define Demjanenko matrix in function field and express the relative ideal class numbers h ðOKPn Þ as the determinant of this matrix. We also define another matrix which give us a recursion formula for the relative divisor class numbers h ðKPn Þ: r 2002 Elsevier Science (USA). All rights reserved. Keywords: Demjanenko matrix; Relative class number; Recursion formula; Function field
1. Introduction Let A ¼ Fq ½T be the ring of polynomials over a finite field Fq with q-elements, and k ¼ Fq ðTÞ: For each MAA; one uses the Carlitz module to construct a field extension þ KM ; called the Mth cyclotomic function field and its maximal real subfield KM : Let þ OKM and OKMþ be the integral closure of A in KM and KM ; respectively. It is known that the divisor class number hðKM Þ of KM is divisible by the divisor class number þ þ þ hðKM Þ of KM : Write h ðKM Þ ¼ hðKM Þ=hðKM Þ; called the relative divisor class number of KM : It is also known that the ideal class number hðOKM Þ of OKM is divisible by the ideal class number hðOKMþ Þ of OKMþ : We write h ðOKM Þ ¼ hðOKM Þ=hðOKMþ Þ; called the relative ideal class number of KM :
*Corresponding author. E-mail addresses: [email protected] (H. Jung), [email protected] (J. Ahn). 1 Supported in part by BK21 Project at Korea University. 0022-314X/02/$ - see front matter r 2002 Elsevier Science (USA). All rights reserved. PII: S 0 0 2 2 - 3 1 4 X ( 0 2 ) 0 0 0 2 3 - 9
56
H. Jung, J. Ahn / Journal of Number Theory 98 (2003) 55–66
Let P be an irreducible polynomial. In [4], Guo and Shu studied the behavior of class numbers h ðKPn Þ and hðKPþn Þ in the tower of cyclotomic function fields KP CKP2 C?CKPn C?: In [2], Bae and Kang defined some matrices which may be viewed as an analogue of Maillet matrix and Demjanenko matrix in classical case, and expressed h ðKPn Þ and hðKPþn Þ as the determinant of these matrices. In this paper, we define a matrix (we call it Demjanenko matrix) and express h ðOKPn Þ as the determinant of this matrix (Theorem 3.1). Our matrix is more natural analogue of the classical Demjanenko matrix [5] than Bae and Kang’s one because it only involves the arithmetic of ðA=Pn Þn and character does not appear. It is known [9, Lemma 3] that h ðKPn Þ ¼ ðq 1Þr h ðOKPn Þ;
ð1Þ
where r ¼ qðn1Þdeg P ðqdeg P 1Þ=ðq 1Þ 1: Thus our formula is also one for h ðKPn Þ: Let c be a prime factor of q 1 and c the highest power of c dividing q 1: From Eq. (1), we see that cr is a lower bound for the exponent of c dividing h ðKPn Þ: This lower bound cr is larger than that of Guo-Shu [4, Theorem 3.1] and equals to that of Bae–Kang [2, Theorem 3]. In [6], Ireland and Small asked whether cr is the highest exponent of c dividing h ðKPn Þ: We find an example which show that cr is not the highest one (Section 5, Example 3). Adopting the idea of Girstmair [3], we also define a matrix to get a recursion formula for the relative divisor class numbers h ðKPn Þ (Theorem 4.1). In the final section we give several examples of h ðKPn Þ by using the Demjanenko matrix and the recursion formula.
2. Basic facts and notations Let M be a polynomial in A: It is well known that the Galois group GM of KM over k is isomorphic to ðA=MÞn ; explicitly the isomorphism is given as j : ðA=MÞn -GM ;
A mod M/sA :
þ : aAFnq g is the Galois group of KM over KM : For Under this isomorphism, J ¼ fsaP any subset H of GM ; let sðHÞ ¼ sAH s: A character w of GM is called even if its restriction to J is trivial, and odd otherwise. Any character w of GM can be viewed as character of ðA=MÞn ; so the conductor Fw of w is defined as a divisor of M: For a polynomial NAA; we let MN (resp. Mþ N ) be the set of all the polynomials (resp. monic polynomials) in A with degree less than the degree of N and prime to N: Let
H. Jung, J. Ahn / Journal of Number Theory 98 (2003) 55–66
57
þ M N ¼ MN \MN : It is known [4] that
0 1 YB X C h ðKM Þ ¼ wðAÞA: @ w odd
AAMþ Fw
When M ¼ Pn ; a power of an irreducible polynomial P; then the formula can be written as Y
h ðKPn Þ ¼
w odd
0
X
@
1 wðAÞA:
ð2Þ
AAMþ Pn
% For each polynomial A prime to M; we let AAM M be the unique element such that % A A% mod M and sgnM ðAÞ denote the leading coefficient of A:
3. Demjanenko matrix For a monic polynomial MAA; let yM ¼
X
ZM ð0; tÞt1 ;
tAGM
where ZM ðs; tÞ is the partial zeta function associated to t: It is a Stickel-berger element of the extension KM =k: For tAGM ; let At AMM be the unique element such that sAt ¼ t; under the isomorphism j : ðA=MÞn -GM : Lemma 3.1. ZM ð0; tÞ is ðq 2Þ=ðq 1Þ if At is monic, and 1=ðq 1Þ otherwise. Proof. For each tAGM ; let Wt ¼ fAAA: A monic; ðA; MÞ ¼ 1; and ððAÞ; KM =kÞ ¼ tg: Then by definition, ZM ðs; tÞ ¼
X
qs deg ðAÞ :
AAWt
It is easy to check that ( Wt ¼
fQM þ At : Q ¼ 0 or Q is monicg fQM þ At : Q is monicg
if At is monic; if At is non-monic:
H. Jung, J. Ahn / Journal of Number Theory 98 (2003) 55–66
58
Thus
ZM ðs; tÞ ¼
8 s deg ðMÞ < qs deg ðAt Þ þ q 1s
if At is monic;
: qs deg ðMÞ 1q1s
if At is non-monic:
1q
By evaluating at s ¼ 0; we get the result.
&
By Lemma 3.1, we can write the Stickelberger element yM as follows: yM
X q 2 X 1 1 ¼ s þ s1 q1 A q1 A þ AAM AAMM M X 1 ¼ s1 sðGM Þ: A q1 þ
ð3Þ
AAMM
From now on, we assume M ¼ Pn ; P is monic irreducible polynomial and nX1: Definition 3.1. Let S 0 be the GM -submodule of Q½GM generated by sðGM Þ=ðq 1Þ and yM : Let S ¼ S 0 -Z½GM ; called the Stickelberger ideal of KM : By Tate [8, Chapter IV, 1.8 Proposition], we have corKM =KPr ðyPr Þ ¼ sðGalðKM =KPr ÞÞyM ; of the for 1prpn: Here corKM =KPr is the correstriction map. Thus our definition P Stickelberger ideal is coincided with Yin’s one [9, Definition 1]. Let Z ¼ AAMþ s1 A : M From Eq. (2), we have S 0 ¼ Z½GM Z þ ZsðGM Þ=ðq 1Þ; and so S ¼ Z½GM Z þ ZsðGM Þ: Let A ¼ fxAZ½GM : sðJÞxAZsðGM Þg: In [1], we have shown the following; Proposition 3.1. ðA : SÞ ¼ h ðOKM Þ: For AAMM ; let A1 be the unique element of MM such that AA1 1 mod M: Lemma 3.2. Y1 ¼ fZg,fðsa 1ÞsA1 : 1aaAFnq ; AAMþ M g is a Z-basis of A:
H. Jung, J. Ahn / Journal of Number Theory 98 (2003) 55–66
Proof. Write x ¼
P
aAFnq
P
AAMþ M
X
sðJÞx ¼
na;A sa sA1 AA: Then X na;A sA1 sðJÞ
aAFnq AAMþ M
0
X
¼
@
AAMþ M
So
P
aAFnq
X
1 na;A AsA1 sðJÞAsðGM ÞZ:
aAFnq
na;A ¼ mAZ for all AAMþ M : Thus 0 1 X X na;A sa sA1 ¼ @m na;A AsA1 þ aAFnq
59
aAFnq ;aa1
¼ msA1 þ
X
na;A sa sA1
aAFnq ;aa1
X
na;A ðsa 1ÞsA1 :
aAFnq ;aa1
Therefore, x¼
0
X
X
@msA1 þ
AAMþ M
¼ mZ þ
X
1 na;A ðsa 1ÞsA1 A
aAFnq ;aa1
X
na;A ðsa 1ÞsA1 :
aAFnq ;aa1 AAMþ M
Now we show the Z-independence of Y1 : Assume X X mZ þ na;A ðsa 1ÞsA1 ¼ 0: aAFnq ;aa1 AAMþ M
Then X
0 @m þ
AAMþ M
1
X
na;A AsA1 þ
aAFnq ;aa1
X
X
ðna;A Þsa sA1 ¼ 0:
aAFnq ;aa1 AAMþ M
P n þ Thus m þ aAFnq ;aa1 na;A ¼ 0 and na;A ¼ 0 for all aAFq ; aa1; AAMM : Thus m is also zero. & Lemma 3.3. Y2 ¼ fsðGM Þg,fsA Z: AAM M g is a Z-basis of S: Proof. For AAMþ M; sðJÞsA Z ¼ sA sðGM Þ ¼ sðGM Þ: P So sA Z ¼ aAFnq ;aa1 saA Z þ sðGM Þ: Thus Y2 generates S: Since the cardinality of Y2 is just the rank of S; Y2 is a Z-basis of S: &
H. Jung, J. Ahn / Journal of Number Theory 98 (2003) 55–66
60
For A; BAM M ; we let /ABS ¼ 1 if sgnM ðABÞ ¼ 1 and /ABS ¼ 0 otherwise. Define DM ¼ ð/ABSÞA;B ; where A; B run through M M: Theorem 3.1. jdet DM j ¼ h ðOKM Þ:
Proof. By Proposition 3.1, Lemmas 3.2 and 3.3, we see that ðA : SÞ is the absolute value of the determinant of the transition matrix T of Y2 with respect to Y1 : Since X
X
X
ðsaA1 s1 A Þ¼
aAFnq ;aa1 AAMþ M
X
saA1 ðq 2Þ
aAFnq ;aa1 AAMþ M
X
sA1
AAMþ M
¼ sðGM Þ ðq 1ÞZ; we have X
sðGM Þ ¼ ðq 1ÞZ þ
X
ðsa 1ÞsA1 :
ð4Þ
aAFnq ;aa1 AAMþ M
We introduce some notations. For A; BAMM ; let aAB ¼ sgnM ððABÞ1 Þ1 and ðABÞ0 ¼ sgnM ððABÞ1 ÞAB: Thus AB ¼ aAB ðABÞ0 : For AAM M; sA Z ¼
X BAMþ M
¼
X
X
sAB1 ¼
saAB1 sðAB1 Þ0
BAMþ M
ðsaAB1 1ÞsðAB1 Þ0 þ
BAMþ M
X
sðAB1 Þ0 :
BAMþ M
þ 1 It is easy to see that fðAB1 Þ0 : BAMþ M g ¼ fB : BAMM g for any AAMM : Thus we have
sA Z ¼ Z þ
X
ðsaAB1 1ÞsðAB1 Þ0 :
ð5Þ
BAMþ M þ Let C 1 ¼ ðAB1 Þ: If ðAB1 Þ1 AMþ M ; then CAMM and saAB1 1 ¼ 0: Assume 1 1 1 1 ðAB Þ AMM : Then CAMM and C AB mod M; equivalently AC
B mod M: Thus for A; CAM M ; ðA; CÞ-entry of the transition matrix T is 1 if ðACÞ
H. Jung, J. Ahn / Journal of Number Theory 98 (2003) 55–66
61
is monic and 0 otherwise. Therefore the transition matrix T is given as follows:
P Since sðGÞ aAFnq ;aa1 sa Z ¼ Z; by elementary row operations, we can convert the transition matrix T to
So we get the result.
&
4. Recursion formula In this section, we assume M ¼ Pn ; where P is a monic irreducible polynomial and ˜ ˜ nX2: For AAA\PA; let AAM M=P be the unique element such that A A mod M=P: n n ˜ ˜ Let MM ¼ fAAMM : AAMM=P ; AaAg: For each A; BAMM ; we define cA;B ¼ ð/ABS /AB sgnM=P ðBÞ1 SÞ ˜ /AB˜ sgnM=P ðBÞ ˜ 1 SÞ ð/ABS and CM ¼ ðcA;B ÞA;BAMnM : Let Zn ¼ ðNP sðHÞÞyM ¼ ðNP sðHÞÞZ; where H ¼ GalðKM =KM=P Þ and NP ¼ qdeg P the norm of P: Lemma 4.1. For a character w of GM ; let Fw be the conductor of w: Then wðZn Þa0 if and only if Fw ¼ M and w is odd. Thus q2 1 FðMÞ 1 dimQ Q½GM Zn ¼ : q1 NP Here FðMÞ ¼ jðA=MÞn j ¼ qðn1Þdeg P ðqdeg P 1Þ:
H. Jung, J. Ahn / Journal of Number Theory 98 (2003) 55–66
62
Proof. If w is even, wðsðJÞZn Þ ¼ wðsðJÞðNP sðHÞÞZÞ ¼ wððNP sðHÞÞsðGM ÞÞ ¼ 0; and wðsðJÞZn Þ ¼ ðq 1ÞwðZn Þ: So wðZn Þ ¼ 0: Assume Fw is a proper divisor of M: Then w is trivial on H; so wðNp sðHÞÞ ¼ 0: Thus wðZn Þ ¼ 0: Suppose Fw ¼ M and w is odd. Then wðsðHÞÞ ¼ 0; so wðZn Þ ¼ NP wðyM Þa0 [9, Lemma 6]. & For AAMM ; let ZnA ¼ sA Zn : Lemma 4.2. fZnA : AAMnM g and fZnA1 : AAMnM g are Z-bases of Z½GM Zn : In particular, they are Q-bases of Q½GM Zn : f ¼ aA˜ and Proof. For AAMM and aAFnq ; aA X X ZnaA ¼ sa ZnA ¼ sðJÞsA Zn ¼ 0: aAFnq
aAFnq
P ˜ Thus Z½GM Zn ¼ A ZZnA ; where A runs P through Mn M with A non-monic. Since ˜ ¼ Bg ¼ sB H; we have fsA : AAMM ; A Z ¼ 0 for any BAMM=P : So we ˜ A AAMM ;A¼B P n get ZnB ¼ AAMn ;A¼B Z : By using these facts, we follow the proof of [3, Lemma 3] ˜ A M to get our result. & Let b ¼
P
AAMþ M
sA AQ½GM and b ¼ bZn ¼
P
AAMþ M
ZnA ¼
P
AAMM
/ASZnA :
Lemma 4.3. For each AAMM ; X
sA1 b ¼
cA;B ZnB :
BAMnM
Proof. For AAMM ; we have X X /BSsA1 ZnB ¼ /ABSZnB sA1 b ¼ BAMM
¼
X
BAMM
˜ BAMM ;BAM ;BaB˜ M=P
¼
/ABSZnB þ
n
BAMM
þ
X
X
/ABSZnB þ
þ ˜ BAMM ;BAM ;BaB˜ M=P
X
/ABSZnB
˜ BAMM ;BAM ;B¼B˜ M=P
X
þ
X
/ABSZnB þ
X
þ ˜ BAMM ;BAM ;B¼B˜ M=P
/ABSZnB
BAM M=P
þ ˜ BAMM ;BAM ;BaB˜ M=P
/ABSZnB þ
X BAMþ M=P
/ABSZnB
/ABSZnB
H. Jung, J. Ahn / Journal of Number Theory 98 (2003) 55–66
63
and X
X
X
BAM M=P
˜ CAMM ;C¼B
/ABSZnB ¼
BAM M=P
X
¼
/ABSZnC
n
˜ n; /ABSZ B
n
BAMM
X
/ABSZnB ¼
þ ˜ BAMM ;BAM ;BaB˜ M=P
0
X
/ABS@
þ ˜ BAMM ;BAM ;BaB˜ M=P
¼
X
1
X
ZnaB A n
1aaAFq
˜ 1 SZn : /AB sgnM=P ðBÞ B
n
BAMM
Finally, X
/ABSZnB ¼
BAMþ M=P
0
X
/ABS@
BAMþ M=P
¼
X
1
X
ZnaB A n
1aaAFq
/AB sgnM ðBÞ1 SZnB
BAM M=P
¼
X
0 /AB sgnM ðBÞ1 S@
BAM M=P
¼
X
X
1 ZnC A
˜ CAMnM ;C¼B 1
˜ SZnB : /AB˜ sgnM=P ðBÞ
n
BAMM
So we get the result.
&
Theorem 4.1. h ðKPn Þ ¼ jdet CPn j h ðKPn1 Þ: Proof. From the equation (2), we have h ðKPn Þ ¼ h ðKPn1 Þ
Y
wðbÞ:
Fw ¼M;odd
Let lb be the Q-linear map on Q½GM Zn defined by the multiplication by b: By Sinnott [7, Lemma 1.2(b)], we have Y wðbÞ ¼ jdet lb j: Fw ¼M;odd
H. Jung, J. Ahn / Journal of Number Theory 98 (2003) 55–66
64
Let g be the Q-linear map on Q½GM Zn defined by gðZnA Þ ¼ ZnA1 : Then by Lemma 4.3, jdet ðgÞj ¼ 1: By Lemma 4.3, X
ðlb 3gÞðZnA Þ ¼
cA;B ZnB
n
BAMM
for each AAMnM : Hence jdet lb j ¼ jdet ðlb 3gÞj ¼ jdet CPn j: So we get the result.
&
Remark 4.1. Assume q ¼ 3: We claim that cA;B 0 mod 2: If B˜ is monic, we have cA;B ¼ 0 trivially. Suppose B˜ is non-monic. Then cA;B ¼ ð/ABS / ABSÞ ˜ / ABSÞ: ˜ ˜ have an opposite sign, ð/ABS Since AB and AB (resp. AB˜ and AB) ˜ ˜ /ABS / ABS 1 mod 2 (resp. /ABS / ABS 1 mod 2). So cAB
0 mod 2; and we prove the claim. Let c0A;B ¼ cA;B =2 and C0Pn ¼ ðc0A;B ÞA;B : Thus we have h ðKPn Þ ¼ 2rn jdet C0Pn j h ðKPn1 Þ;
ð6Þ
where rn ¼ rank CPn ¼ 3ðn2Þdeg P ð3deg P 1Þ2 =2: By substituting (1) into (6), we have the following recursion formula for the relative ideal class numbers: h ðOKPn Þ ¼ jdet C0Pn j h ðOKPn1 Þ: It is interesting to find a recursion formula for the relative ideal class numbers h ðOKPn Þ in general.
5. Examples We give several examples which are obtained by using MAPLE. Example 5.1. Let q ¼ 3 and P ¼ T: Then M P2 ¼ f2; 2T þ 1; 2T þ 2g; and the Demjanenko matrix DP2 is 2
1
6 41 1
1 1 0
1
3
7 05 0
and h ðOKP2 Þ ¼ jdet DP2 j ¼ 1; so h ðKP2 Þ ¼ 22 : The M P3 is f2; 2T þ 1; 2T þ 2; 2T 2 þ 1; 2T 2 þ 2; T þ 2T 2 þ 1; T þ 2T 2 þ 2; 2T þ 2T 2 þ 1; 2T þ 2T 2 þ 2g; the
H. Jung, J. Ahn / Journal of Number Theory 98 (2003) 55–66
Demjanenko matrix DP3 is 2
1 61 6 6 61 6 6 61 6 61 6 6 61 6 61 6 6 41 1
1 1 1 1
1 1 0 0
1 1 1 1
1 1
1 1 0 1
1 1 1 0
1 1 1 1
0 1
0 1 1 1 1 1
0 0 1 0 1 0
0 0 0 1 1 1
1 1 0
1 0 0 0
1 1 0 0
1 0 0 1
0 1
65
3 1 07 7 7 07 7 7 07 7 07 7 7 07 7 17 7 7 15 0
and h ðOKP3 Þ ¼ 7; so h ðKP3 Þ ¼ 28 7: The MnP3 is fT 2 þ 2; T 2 þ 2T þ 1; T 2 þ 2T þ 2; 2T 2 þ 2; 2T 2 þ 2T þ 1; 2T 2 þ 2T þ 2g; the matrix CP3 is given by 3 2 2 0 2 2 2 2 7 6 6 2 2 2 2 2 0 7 7 6 6 2 2 0 2 0 2 7 7 6 7 6 6 0 2 0 2 2 2 7 7 6 6 2 0 0 0 2 2 7 5 4 0 0 2 2 2 0 and jdet CP3 j ¼ 26 7: Thus we see that h ðKP3 Þ=h ðKP2 Þ ¼ jdet CP3 j: Example 5.2. Let q ¼ 3 and M ¼ P ¼ T 3 þ T 2 1: Then M M ¼ f2; 2 T; 1 þ 2 T; 2 þ 2 T; 2 T 2 ; 1 þ 2 T 2 ; 2 þ 2 T 2 ; T þ 2 T 2 ; 1 þ T þ 2 T 2 ; 2 þ T þ 2 T 2 ; 2 T þ 2 T 2 ; 1 þ 2 T þ 2 T 2 ; 2 þ 2 T þ 2 T 2 g and the Demjanenko matrix DM is given by 3 2 1 1 1 1 1 1 1 1 1 1 1 1 1 61 1 1 1 0 0 0 1 1 1 1 0 17 7 6 7 6 61 1 1 1 1 1 1 1 0 0 0 0 07 7 6 61 1 1 1 1 0 1 0 0 0 1 1 17 7 6 7 6 61 0 1 1 1 1 0 0 1 1 1 0 17 7 6 7 6 61 0 1 0 1 0 1 1 0 0 0 1 07 7 6 61 0 1 1 0 1 1 1 0 1 1 0 07 7 6 7 6 61 1 1 0 0 1 1 1 0 0 1 0 17 7 6 61 1 0 0 1 0 0 0 0 1 0 1 07 7 6 7 6 61 1 0 0 1 0 1 0 1 0 1 1 07 7 6 61 1 0 1 1 0 1 1 0 1 1 0 17 7 6 7 6 41 0 0 1 0 1 0 0 1 1 0 1 15 1 1 0 1 1 0 0 1 0 0 1 1 0 and h ðOKM Þ ¼ jdet DM j ¼ 5 79:
66
H. Jung, J. Ahn / Journal of Number Theory 98 (2003) 55–66
Example 5.3. Let q ¼ 3 and deg P ¼ 1: For np5; one can find h ðKPn Þ in [4, Table 1]. When n ¼ 6; using the matrix CPn ; we calculate h ðKP6 Þ ¼ 2252 52 717 135 174 1917 314 377 432 613 67 733 793 1093 1272 151 181 241 379 397 433 5232 1423 1531 1747 3511 5779 6823 7057 7687 12619 25183 31177 156151 192889 196687 208009 347707:
References [1] S. Bae, H. Jung, J. Ahn, Cyclotomic units and Stickelberger ideals of global function fields, Trans. Amer. Math. Soc., to appear. [2] S. Bae, P. Kang, Class numbers of cyclotomic function field, Acta. Arith. 102 (2002) 251–259. [3] K. Girstmair, A recursion formula for the relative class number of the pn th cyclotomic field, Abh. Math. Sem. Univ. Hamburg. 61 (1991) 131–138. [4] L. Guo, L. Shu, Class numbers of cyclotomic function field, Trans. Amer. Math. Soc. 351 (1999) 4445– 4467. [5] F. Hazama, Demjanenko matrix, class numbers, and Hodge group, J. Number Theory. 34 (1990) 174– 177. [6] K. Ireland, R. Small, Class numbers of cyclotomic function fields, Math. Comput. 46 (1986) 337–340. [7] W. Sinnott, On the Stickelberger ideal and the circular units of an abelian field, Invent. Math. 62 (1980/81) 181–234. [8] J. Tate, Les Conjecture de Stark sur les functions L d’Artin en s ¼ 0; Birkha¨user, Boston. [9] L. Yin, Stickelberger ideals and relative class numbers in function fields, J. Number Theory. 81 (2000) 162–169.