Class number in non Galois quartic and non abelian Galois octic function fields over finite fields

Transcription

1 Class number in non Galois quartic and non abelian Galois octic function fields over finite fields Yves Aubry G. R. I. M. Université du Sud Toulon-Var La Garde Cedex France Abstract We consider a totally imaginary extension of a real extension of a rational function field over a finite field of odd characteristic. We rove that the relative ideal class number one roblem for such non Galois quartic fields is equivalent to the one for non abelian Galois octic imaginary functions fields. Then, we develo some results on characters which give a method to evaluate the ideal class number of such quartic function fields. 1 Introduction Let L be a totally imaginary extension of a function field K which is itself a real extension of a rational function field k F q (x). This means that the infinite lace of k totally slits in the extension K/k and that this laces have only one lace above each of them in the extension L/K. The determination of all such imaginary fields L with L/k Galois and cyclic and with ideal class number one has been done by S. Sémirat in [11]. The quartic bicyclic Galois case has been solved by X. Zhang in [13] in odd characteristic and by the author and Dominique Le Brigand in [3] in even characteristic. We are interested here in the determination of all such non Galois quartic fields L with ideal class number equal to one (which will be called the ideal class number one roblem), a roblem which has been solved by K. Uchida in [12] and by S. Louboutin in [8] in the number field case Mathematics Subject Classification. 11G20, 11R29, 11M06, 14G10, 14G15. Key words and hrases. Function fields, class number. 0 The author is very grateful to the Institut de Mathématiques de Luminy - C.N.R.S. - Marseille - France for its hositality during this work. 1

2 2 In section 2, we recall some definitions and some general results without any assumtion on the degrees of L/K and K/k. We give also an ambiguous classes formula for cyclic extensions. In section 3, we suose that L/K is quadratic. We give a result on the dyadic valuation of the relative ideal class number of L. Then, we give a formula for it in terms of a character χ, which is a articular case of the Galois situation studied in [10]. Section 4 is devoted to the quartic situation : L/K and K/k are suosed to be quadratic. Firstly, we show, as K. Uchida and S. Louboutin for number fields (see resectively [12] and [8]), that the relative ideal class number one roblem for such non Galois quartic fields is equivalent to the one for non abelian Galois octic imaginary fields. Secondly, if we suose furthermore that K has ideal class number one, we give a method to evaluate the character χ. Thirdly, we investigate the case where L has ideal class number one. 2 Preliminaries 2.1 The general setting Let K be an algebraic function field in one variable over a finite constant field F q with q elements (q odd) and let S K be a non emty finite set of laces (i.e. rime divisors) of K (called laces at infinity of K, the laces not in S K will be called finite laces). Let O K be the ring of elements of K whose oles are in S K. The ring O K is a Dedekind domain and we denote by Cl(O K ) its ideal class grou and by h OK its order, called the ideal class number of K. Let L be a finite extension of K contained in a searable closure of K with constant field F q and let S L be the set of laces of L which extend those in S K. By analogy with the number field case, the extension L/K is called real if the number of laces in S L is equal to S L [L : K] S K, i.e. every lace in S K slits comletely in L. Otherwise, the extension is called imaginary, and totally imaginary if S L S K (i.e. every lace in S K has only one lace above it in L). Let O L be the ring of elements of L whose oles are in S L. This ring is also the integral closure of O K in L. Let k F q (x) be a rational function field and S k { } be the lace at infinity of k corresonding to 1/x. Note that the ring of integers of k with resect to S k is O k F q [x] the olynomial ring with coefficients in F q. In this aer, all the functions fields will be suosed to have for exact constant field F q with q odd and will be contained in a searable closure of k F q (x). Now consider a real extension K/k of degree d and denote by S K {P 1,, P d } the set of laces of K above the infinite lace of k. Let us consider finally a totally imaginary extension L of K of degree n and denote by S L {P 1,, P d } the set of laces of L above the P i s. Our situation is the following one:

3 3 L P 1 P d n O L K d O K k F q [x] P 1 P d Note that this data is equivalent to the data consisting of a degree n totally imaginary extension L/K of functions fields and a non emty finite set S K {P 1,, P d } of laces of K. Indeed, by Riemann-Roch theorem, we get the existence of a function x K such that K is a degree d extension of k(x) and such that the laces of K over are exactely P 1,, P d. Note also that if the imaginary extension L/k is Galois, K can also be described as the fixed field of the inertia grou of, that is the maximal real subfield of L (the maximal subfield of L in which slits totally). 2.2 Units, regulators and class numbers Let h F denotes the divisor class number of a function field F/F q, that is the number of rational oints over F q of the Jacobian of the smooth rojective algebraic curve associated to F. We have the following well-known relation due to F.K. Schmidt: δ OF h F r OF h OF, where h OF is the ideal class number of F with resect to the set of laces S F, δ OF is the gcd of the degrees of the laces of S F and r OF is the index of the grou of rincial divisors suorted on S F in the grou of zero degree divisors suorted on S F. Let R OF be the regulator of O F (which corresonds to the R q S K -regulator defined in [10]). We have the following relation which can be considered here as a definition of R OF : r OF δ O F R OF P S F deg P. The analogue of Dirichlet unit theorem states that the two unit grous O L and O K of L and K (modulo F q) are of rank S K 1 S L 1 d 1. The oint is that these two abelian grous have the same rank. We can estimate the index Q L/K of O K in O L, analogue to the Hasse index in the number field case (see [2] for instance for a roof): Proosition 1 Let L/K be totally imaginary. Then, we have: (i) Q L/K : [O L : O K ] divides [L : K] S K 1. (ii) R OL /R OK [L : K] S K 1 /Q L/K.

4 4 It is well-known that for any finite searable extension L/K of function fields, the divisor class number of K divides that of L (in fact, the olynomial P K (T ) on the numerator of the zeta function of K divides P L (T ), that of L, in Z[T ], see for examle [2] for a roof). For totally imaginary extensions L/K, the result holds also for ideal class numbers, as shown in the next roosition (see [9] for a roof with the additional assumtion that some finite lace of K is totally ramified in L or some infinite rime of K is inert in L). Proosition 2 If L/K is a totally imaginary extension, i.e. if every lace in S K has only one lace above it in L, then the ideal class number h OK of K divides that h OL of L. Define the relative ideal class number by h O L h OL /h OK. Proof. Let K O K denote the Hilbert class field of K with resect to O K i.e. K O K is the maximal unramified abelian extension of K in which every lace of S K slits comletely. The field LK O K is contained in the Hilbert class field L O L of L and thus [LK O K : L] divides [L O L : L] which is recisely the ideal class number of O L. The isomorhism given by the restriction Gal(LK O K /L) Gal(K O K /L K O K ) defined by σ σ K O K gives us [LK O K : L] [K O K : L K O K ]. Finally, we have L K O K K since first, the infinite laces of K slit in K O K and thus in L K O K, and secondly they are totally ramified or inert in L and thus in L K O K. Thus, [LK O K : L] [K O K : K] h OK divides [L O L : L] h OL. 2.3 The zeta function Let us introduce now the zeta function ζ OK (s) of the Dedekind domain O K by ζ OK (s) I 1 N(I) s where s C, the field of comlex numbers, where the sum ranges over the nonzero ideals I of O K, and where N(I) is the norm of the ideal I, that is, by definition the number of elements of the residue class ring O K /I. Unique factorization of ideals in O K imlies the following Euler roduct reresentation: ζ OK (s) (1 1 ) 1 N(P ) s P Sec(O K ) {0} where Sec(O K ) is the set of rime ideals of O K. In a same way, we define the zeta function ζ OL (s) of the Dedekind domain O L, and we have: ζ OL (s) (1 1 ) 1 (1 N(P) s 1 ) 1. N(P) s P Sec(O L ) {0} P Sec(O K ) {0} P P

5 5 2.4 Ambiguous classes in cyclic extensions If we suose that the extension L/K is cyclic, then we can rove the following ambigous classes formula in the same way as the one in lemma of [6] for number fields. Lemma 3 Let L/K be a cyclic extension of Galois grou G, C L Cl(O L ) be the ideal class grou of O L, CL G be the ambiguous ideal class grou (the subgrou of C L of elements fixed under G) and e(l/k) P S K e(p ) P S K e(p )f(p ) where e(p ) and f(p ) stand for the ramification index e(p P ) and residual degree f(p P ) of any lace P in L over P. Then, CL G h OK e(l/k) [L : K][OK : N L/KL OK ] where N L/K denotes the norm of L over K. Proof. The roof given in [6] for number fields also holds in the function field case. The oint is the use of corollary of [5] which gives that: H 0 (G, O L ) H 1 (G, O L ) 1 [L : K] P S K G P where G P denotes the decomosition grou of P in the cyclic extension L/K, which has order e(p )f(p ). This gives the contribution of the laces at infinity in the definition of e(l/k). 3 The relative ideal class number for L/K quadratic In this section, we suose that the imaginary extension L/K is quadratic. 3.1 On the dyadic valuation of the relative ideal class number Proosition 4 Let L/K be an imaginary quadratic extension and K/k be a real extension of the rational function field k F q (x). Let t L/K be the number of finite laces (i.e. not in S K ) ramified in L/K. Then, 2 t L/K 1 divides h O L. Proof. By the ambiguous class formula (lemma 3), we have: C Gal(L/K) L h OK 2 t L/K+ S K 2[O K : N L/KL O K ], since e(l/k) P S K e(p ) P S K e(p )f(p ) 2 t L/K 2 S K. Moreover, we clearly we have O 2 K N L/K L O K

6 6 since [L : K] 2. Thus, [OK : N L/KL OK ] divides [O K : O 2 K ]. But [O K : O 2 K ] [F q : F 2 q ].2 rk(o K ) 2 S K since the rank rk(ok ) of the finitely generated grou O K is equal to S K 1 by Dirichlet s theorem and since the non zero squares in a finite field of odd characteristic have index 2 in its multilicative grou. Thus 2 tl/k 1 h OK divides the order of C Gal(L/K) L which divides itself the ideal class number h OL by Lagrange theorem on grou order, hence the result. 3.2 Relative ideal class number and L-functions The following investigation is actually a articular case of Galois extensions dealt with in [10] but we will write the things exlicitly because of the simlicity of the urose. As the rime ideals P of O K are inert, ramified or slits in L/K, we see immediately that the norm N( ) of P is equal to N(P ) 2, N(P ) or N(P ). Thus, we obtain: where L OK (s, χ) ζ OL (s) ζ OK (s)l OK (s, χ) (3) P Sec(O K ) {0} (1 χ(p ) ) 1 N(P ) s with χ(p ) 1, 0, 1 according as P is inert, ramified or slits in L/K. To obtain a relation between the zeta function ζ OK (s) of O K and its class number h OK, one can define the zeta function of the function field K by: ζ K (s) (1 1 ) 1 N(P ) s P where P ranges over all the laces (i.e. rime divisors) of K. Then, we have ζ K (s) ζ OK (s) P S K A residue calculus gives us (see [9]): (1 1 N(P ) s ) 1 ZK (q s ) P K (q s ) (1 q s )(1 q 1 s ) (4) and thus we obtain ζ OK (s) h O K R OK q 1 (ln q) S K 1 s S K 1 + O(s S K ), h OL R OL h OK R OK L OK (0, χ). (5) Proosition 5 Let L/K be an imaginary quadratic extension and K/k be a real extension of degree d of the rational function field k F q (x). Then, we have: h O L Q L/K 2 1 d L OK (0, χ)

7 7 Proof. The equality follows from (5) and roosition 1. Proosition 6 Let L/K be an imaginary quadratic extension and K/k be a real extension of the rational function field k F q (x). Then, L OK (s, χ) is a olynomial in q s of degree 2(g L g K ) + j, with j {P i S K P i inert in L/K} and where g L and g K are the genus of the functions fields L and K. Proof. Equations (3) and (4) imlies: L OK (s, χ) ζ O L (s) ζ OK (s) ζ L(s) ζ K (s) ( ) P S K N(P ) ( s 1 1 S L ) 1 P S K (1 1 N(P ) s ) 1 S L (1 1 N( ) s ζ L (s) ζ K (s) L (s) where L (s) ) 1. The divisibility of the numerators of the zeta function N( ) s of the functions fields L and K shown in [2] imlies that (ζ L /ζ K )(s) is a olynomial in q s of degree 2(g L g K ). The extension K/k being totally real, the infinite laces of K have degree 1 and for P S K we get: 1 1 N(P ) 1 q s. The result follows from the s fact that N(P i ) N(P i ) 2 if and only if P i is inert in L/K. Now, we extend χ multilicatively to the nonzero ideals of O K. If I is a nonzero ideal of O K, define the degree of I by N(I) q deg I. For any integer i, consider as in [4], the sum S i (χ) χ(i) deg Ii where the sum ranges over all nonzero ideal I of O K of degree i. Remark that we have S 0 (χ) 1. Consider the sum i0 S i (χ) with defined in roosition 6. This sum is finite since there exist only finitely many ideals in O K of fixed degree. We have: Proosition 7 Let be as in roosition 6. We have L OK (0, χ) S i (χ). i0

8 8 Proof. In the following, the sums range over nonzero ideals I of O K. L OK (s, χ) I P Sec(O K ) {0} χ(i) N(I) s I i0 deg Ii i0 χ(i) q is (q s ) i S i (χ) where the last equality holds by roosition 6. (1 χ(p ) ) 1 N(P ) s χ(i) q s deg I i0 i0 1 q is deg Ii (q s ) i S i (χ), χ(i) 4 The quartic case 4.1 The relation with the Galois closure Lemma 8 let L/K be an imaginary quadratic extension and K/k be a real quadratic extension. If L/k is non Galois then the Galois closure of L is a dihedral octic function field (i.e. [L : k] 8 and Gal(L/k) D 4 the dihedral grou of order 8). Proof. The Galois closure N of L is just the comositum of L and its conjugate L by the non trivial element of the Galois grou Gal(L/K). The function field N has degree 8 over k, is not abelian and has more than one subfield of degree 4 over k, which excludes the quaternionic case. Thus, we are in the following situation, where all the extensions are quadratic: L L K N N + E k

9 9 where L is the conjugate of L. Lemma 9 Let L/K be imaginary quadratic, K/k be real quadratic, L/k be non Galois and N be the Galois closure of L. Then, we have the following relation between their zeta functions: ζ N (s)/ζ N +(s) (ζ L (s)/ζ K (s)) 2. Proof. Since the extension N/K is abelian, we can show as in [3] (see also chater 14 of [10]) that we have the factorization: which imlies that: ζ N (s)/ζ K (s) (ζ L (s)/ζ K (s))(ζ L (s)/ζ K (s))(ζ N +(s)/ζ K (s)) ζ N (s)/ζ N +(s) (ζ L (s)/ζ K (s))(ζ L (s)/ζ K (s)). But the functions fields L and L, with L the conjugate of L under Gal(K/k), have the same zeta function. Then, the result follows. We are now in osition to give a relation between the relative ideal class numbers of N and L: Proosition 10 Let L/K be imaginary quadratic, K/k be real quadratic, L/k be non Galois and N be the Galois closure of L. Then we have: h O N Q N/N + (h O 2 L ) 2. Proof. We have seen that for any function field K, we have: ζ OK (s) h O K R OK (ln q) SK 1 s SK 1 + O(s S K ). q 1 Considering the function Λ OK (s) ζ OK (s)/s S K 1, we get by lemma 9 ( ) ( ) 2(0) Λ ON /Λ ON + (0) Λ OL /Λ OK since S N S N + S L S K 0. Thus, we obtain: R ON R ON + h O N Thus, roosition 1 and S N + 2 S K give us: ( ROL R OK h O L ) 2. h O N Q ( N/N + h ) O L 2. 2 Q L/K

10 10 Now, we show that, under the hyothesis of the roosition, we have Q L/K 1. Suose on the contrary that Q L/K 2 (by roosition 1, it is 1 or 2) and let us show that this imlies that L/k is Galois. If Q L/K 2, we can write L K( ε K ) with ε K a fundamental unit of O K. Consider the constant field extensions K K Fq F q 2 for K k, K, L and N. The extension L/ k is Galois since L K( ε K ), K/ k is real and the norm N K/ k(ε K ) is a square in K (see lemma 16 (i)). Moreover, the extension Ñ/k is Galois of Galois grou D 4 Z/2Z and thus Ñ/ k is Galois of Galois grou the dihedral grou D 4. Hence L is the fixed field of Ñ by a normal subgrou H of order 2 of D 4. Thus: L L N Ñ H {1} Ñ {1} Z/2Z Ñ H Z/2Z. Since H is a normal subgrou of D 4 this imlies that H Z/2Z is a normal subgrou of D 4 Z/2Z and we obtain that L/k is Galois, which is in contradiction with our hyothesis. Finally, the Hasse index Q L/K 1 and the roosition is roved. The Hasse index Q N/N + is equal to 1 or 2 according to roosition 1. Thus, remarking that relative class numbers are integers, we have the following corollary. Corollary 11 Let L/K be imaginary quadratic, K/k be real quadratic, L/k be non Galois and N be the Galois closure of L. Then we have: h O N 1 h O L 1. In other words, the relative ideal class number one roblem for imaginary quartic non Galois function field having a real subquadratic field is equivalent to the one in the octic Galois dihedral case. In fact, it remains to consider the general octic Galois case since the quaternionic case can be set aside by the following roosition. Proosition 12 Let N/N + be imaginary, N + /k be real with N/k Galois with Galois grou Q 8 the quaternionic grou of order 8. Then, h O N is even. Proof. By roosition 4, it suffices to show that the number t N/N + of finite ramified laces in N/N + is at least 2. But since the extension N/k is quaternionic, this imlies that the extension N + /k is biquadratic (i.e. Gal(N + /k) Z/2Z Z/2Z). Combined with the fact that N + /k is real, this imlies that there is at least two finite laces ramified in N + /k. But they are also ramified in N/N + since the subgrou of order 2 of Q 8 is contained in all the non trivial ones and thus any inertia grou contains this grou. Hence, any lace ramified in N + /k is ramified in N/N +.

11 A formula for quartic extensions with rincial real quadratic subfield We are interested now in the case where L/k has degree 4 with h OL 1. Thus, we suose that L/K is imaginary quadratic and that K/k is real quadratic and we assume furthermore that O K is a rincial domain, i.e. that h OK 1 (since h OL 1 imlies that h OK 1 by roosition 2). Now, let us study the symbol χ(p ) reviously defined (recall that L has odd characteristic). Suose that K k( m) with m m K F q [X] a square-free olynomial. Since we assume that the lace slits in K/k, then m is necessarily a olynomial of even degree with leading coefficient a square in F q. Suose that L K( M) with M M L O K square-free in O K. Let P πo K be a rincial rime ideal of O K generated by π. Let F q [X] be such that P, that is such that P F q [X] F q [X]. Definition. If Q is an element of O K, we define the symbol [ Q P ] to be 0 if Q P, 1 if Q is congruent to a square modulo P, and 1 otherwise. Lemma 13 We have χ(p ) [ M P ]. Proof. The result follows from the fact that M is square-free in O K. Lemma 14 If n F q [X] and if is not inert, we have: [ n P ] (n ) where ( n ) is the quadratic character on F q[x] defined to be 0 if n, 1 if n is congruent to a square modulo, and 1 otherwise. Proof. We have [ n N(P ) 1 P ] n 2 mod P and ( n ) n N() 1 2 mod. But N(P ) N() q deg if is not inert. Let Gal(K/k) {Id, σ}. We set σ(π) π, and Tr will denote the trace of K/k. The following theorem is an analogue of the one that holds in the number field case (see [7]). Theorem 15 (i) If slits in K/k then [ M P ] [ M T r(π)t r(m π) (π)] ( ) and [ M (π) ][ M ( π) ] ( M M ). (ii) If is inert in K/k then [ M P ] ( M M ). (iii) If is ramified in K/k then [ M 2T r(m) P ] ( ).

12 12 Proof. (i) First, let us remark that if slits in K/k then Tr(π) (π). Indeed, Tr(π) (π) iff π (π) iff ( π) (π) iff does not slit. Now, Tr(M π) M π + Mπ M(π + π) + π( M M) M Tr(π) mod π. Thus, Tr(π) Tr(M π) M(Tr(π)) 2 mod π and [ Tr(π) Tr(M π) ] [ M(Tr(π)) 2 ]. (π) (π) But Tr(π) Tr(M π) F q [X] thus by lemma 14 we get: and [ Tr(π) Tr(M π) ] ( Tr(π) Tr(M π) ) (π) [ M(Tr(π)) 2 ] ( Tr(π) Tr(M π) ). (π) Since Tr(π) (π) we obtain [ M P ] [ M (π) that: [ M ][ M (π) ( π)] Tr(π) Tr(M π) ] ( ). Furthermore, we can show easily [ M M ] (ii) If is inert, we have O K /O K O K /πo K which is an extension of degree 2 of F q [X]/() F q deg F N(). We have in this case M M qdeg mod O K. Furthermore, we can easily show that, if n F q [X], then Then, we have ( M M ( n ) n (qdeg 1)/2 mod() N() 1 ) (M M) 2 (MM qdeg ) qdeg 1 2 M N2 () 1 2 [ M (π) ] mod O K. (iii) By the roerty of the different, we know that (π) divides the different D OK /F q[x] and that D OK /F q[x] is the greater common divisor of all ideal (f α(α)) where α is an integral generator of K over k and f is the irreducible olynomial for α over k (see [5]). Since f M (X) X 2 Tr(M)X + N(M), we have f M (M) M M and thus D OK /F q[x] divides M M. Hence, M M (π) and 2 Tr(M) 2M +2 M 4M +2( M M) 4M mod(π). Thus, [ M (π) ] [ 4M (π) ] [ 2 Tr(M) (π) ] ( 2 Tr(M) () ), by lemma 14. This theorem rovides us with a method for calculating χ(p ) for any nonzero rincial rime ideal P of O K, hence for calculating S i (χ), hence L OK (0, χ) by roosition 7, hence h O L by roosition 5.

13 Princial Non Galois quartic extension The ideal class number one roblem for bicyclic quartic Galois extension is treated in [13] and [3] according as the characteristic of k is odd or even. The case of cyclic quartic Galois extension L/k is derived from [11] which solved the rime ower cyclic case. We now investigate the non Galois quartic case. Consider, as in the revious section, a quartic function field extension L/k with L/K imaginary quadratic and K/k real quadratic. We set also K k( m) with m F q [X] a square-free olynomial, Gal(K/k) {Id, σ} and L K( M) with M O K square-free in O K. Lemma 16 (i) The extension L/k is Galois if and only if the norm N K/k (M) : σ(m)m is a square in K. Moreover, N K/k (M) is a square in K if and only if N K/k (M) is a square in k or N K/k (M)/m is a square in k. (ii) If h OL 1 then at least one of the infinite laces P 1 or P 2 of K is ramified in L. Proof. (i) It is easily seen that L/k is Galois if and only if σ(m)/m is a square in K which is equivalent to N K/k (M) is a square in K. Moreover, if N K/k (M) (a(x) + b(x) m) 2 a(x) 2 + b(x) 2 m + 2a(x)b(x) m and since it lies in k, we get the equivalence. (ii) As remarked in [11], the constant field extension (F q δ OL.L)/L has degree δ OL and is contained in the Hilbert class field of L, thus δ OL divides h OL. Since δ OL gcd(deg P 1, deg P 2 ) where P 1 and P 2 are the infinite laces of L, we obtain that at least one of these degrees is equal to one (L/K is not suosed to be Galois) and thus the corresonding lace is ramified. Proosition 17 With the notations above, if we suose that h OL 1 and that the genus g K of K is non zero then the cardinality of the finite base field F q is less than or equal to 5. Proof. Since the function field L is an extension of the function field K, it follows that the numerator olynomial P K (T ) of the zeta function of K divides that of L (see [2] for a roof). This means that the relative divisor class number h L P L(1) P K (1) can be written as a roduct h L 2(g L g K ) i1 (1 ω i ) with ω i comlex numbers of modulus q (Riemann Hyothesis). Thus, we have the following lower bound: h L ( q 1) 2(g L g K ) ( q 1) 2(g K 1)+deg Diff L/K where deg Diff L/K is the degree of the different of L/K (the last equality comes from Riemann-Hurwitz theorem). Moreover, using the Schmidt relation, we obtain the following lower bound for the ideal class number of L: h OL h OK Q L/K 2 2 deg P i ( q 1) 2(g K 1)+deg Diff L/K. (6) i1

14 14 But now, if we suose that h OL 1 then by roosition 2, we obtain that h OK 1 and by lemma 16, (ii), we obtain that deg Diff L/K 0 (which imlies that deg Diff L/K 2 since it is even by the Riemann-Hurwitz theorem). Thus, the inequality becomes : with a 2 which gives the bound on q. ( q 1) a 2 Remarks. For q 5, the inequality (6) imlies that g L 2, g K 1, deg Diff L/K 2, deg P 1 deg P 2 1 (the two laces that ramify in L/K are P 1 and P 2 ) and Q L/K 1. We have just finitely many cases to consider. Recall that the 2-rank rk 2 (Cl(O F )) dim F2 Cl(O F )/ Cl(O F ) 2 of the ideal class grou Cl(O F ) of a quadratic function field F k( m) in odd characteristic is given by (see [1]): rk 2 (Cl(O F )) n 1 µ F if m has an irreductible factor of odd degree and rk 2 (Cl(O F )) n µ F otherwise, where n is the number of monic irreducible olynomial factors of m and µ F is equal to 0 or 1 according as F/k is imaginary or real. Thus, for a real quadratic function field extension F/k, we have that the ideal class number h OF is odd if and only if m is an irreducible olynomial of even degree (with leading coefficient a square in F q) or m is the roduct of two monic irreducible olynomials of odd degree. Since we have found that the genus g K must be equal to 1 for q 5 and since deg m 2 g K 2 for a real quadratic function field k( m), we obtain that the olynomial m must have degree 4. Then, the results of subsection 3.2 combined with those of subsection 4.2 rovide us a method for calculating class numbers in the remain cases given by 4.3. For q 3, unfortunately, the inequality (6) doesn t give us any bound on the genus. Acknowledgments. The author would like to thank Stéhane Louboutin, Marc Perret and Dominique Le Brigand for many helful discussions. References [1] E. Artin, Quadratische Körer in Gebiete der höheren Kongruenzen, I, II, Math. Zeit., 19, , (1924). [2] Y. Aubry, Class number in totally imaginary extensions of totally real function fields, Third International Conference on Finite fields and Alications, Lecture Note Series of the London Mathematical Society, Cambridge University Press (1996), [3] Y. Aubry, D. Le Brigand, Imaginary bicyclic biquadratic functions fields in characteristic two, J. Number Theory 77, (1999).

15 15 [4] S. Galovich, M. Rosen, The class number of cyclotomic function fields, J. Number Theory 13 (1981), [5] S. Lang, Algebraic number theory, Addison-Wesley Series in Math., (1970). [6] S. Lang, Cyclotomic fields I and II, Graduate Texts in Math. 121, Sringer-Verlag (1990). [7] S. Louboutin, L-functions and class numbers of imaginary quadratic fields and of quadratic extensions of an imaginary quadratic field, Math. Com. 59, (1992), [8] S. Louboutin, R. Okazaki, Determination of all non-normal quartic CM-fields and of all non-abelian normal octic CM-fields with class number one, Acta Arith. 67, no. 1, (1994), [9] M. Rosen, The Hilbert class field in function fields, Exo. Math. 5 (1987), [10] M. Rosen, Number theory in function fields, Graduate Texts in Maths 210, Singer- Verlag (2002). [11] S. Sémirat, Class number one roblem for imaginary function fields: the cyclic rime ower case, J. Number Theory 84, (2000). [12] K. Uchida, Relative class numbers of normal CM-fields, Tôhoku Math. Journ. 25, (1973). [13] X. Zhang, Ambiguous classes and 2-rank of class grou of quadratic function field, J. China Univ. Sci. Technol. 17, , (1987).