REAL QUADRATIC EXTENSIONS OF THE RATIONAL FUNCTION FIELD IN CHARACTERISTIC TWO

REAL QUADRATIC EXTENSIONS OF THE RATIONAL FUNCTION FIELD IN CHARACTERISTIC TWO

by

Dominique Le Brigand

Abstract. — We consider real quadratic extensions of the rational field over a finite field of characteristic two. After recalling the equation of such extensions, we present a geometric approach of the continued fraction expansion algorithm to compute the regulator. Finally, we study the ideal class number one problem and give numerous examples for which the ideal class number equals one.

Résumé (Extensions quadratiques réelles du corps rationnel en caractéristique2)

Nous ´etudions les extensions quadratiques r´eelles du corps rationnel sur un corps fini de caract´eristique 2. On rappelle la forme g´en´erale de telles extensions puis on donne une approche g´eom´etrique de l’algorithme des fractions continues qui permet de calculer le r´egulateur. Enfin on s’int´eresse aux extensions quadratiques r´eelles dont le nombre de classes d’id´eaux de l’anneau des entiers est ´egal `a un et on donne un grand nombre d’exemples pour lesquels cette situation est r´ealis´ee.

1. Introduction

We consider a separable quadratic extensionK of the rational fieldk =Fq, such that the full constant field of the function field K/Fq is Fq. We denote by Ox the integral closure of Fq[x] in K and by hx the ideal class-number of Ox. It is easy to prove that there is only a finite number of imaginary quadratic extensions such that hx= constant. For real quadratic extensions and when the constant fieldFq is fixed, it is not known whether this result is false or not. The Gauss conjecture for function fields pretends that there is an infinite number of real quadratic extensions such thathx= 1. The main motivation for this paper was to examine the validity of the Gauss conjecture in the characteristic 2 case. Unfortunately, we have no answer.

This paper is organized as follows. In Section 2, we recall basic results about quadratic extensions. In Section 3, we focus on real quadratic extensions in characteristic 2 and

2000 Mathematics Subject Classification. — 11R58, 11A55.

Key words and phrases. — Real quadratic extension, continued fraction expansion algorithm, regulator, ideal class number.

give some geometric approach of the continued fraction expansion (CFE) algorithm.

In Section 4, we study the ideal class number one problem in characteristic 2 and give examples. In particular, we give all the real quadratic extensions of a particular form such thathx= 1.

2. Quadratic extensions

Let q = p^e, and let x be transcendental over Fq, k = Fq(x), finally let K/k be a (separable) quadratic extension. We always assume that Fq is the full constant field of the hyperelliptic function field K/Fq and that the genus of K isg >1. The places of the rational function field k = Fq(x) are ∞, the pole of x, and the other places, calledfinite places ofk/Fq, are in one to one correspondence with the monic irreducible polynomials of Fq[x]. We denote by (P) the place corresponding to the monic irreducible polynomialP ∈Fq[x]. The degree of the place (P) is equal to the degree, DegP, of the polynomial P. If℘is a place ofK/Fq which is above a finite place (P) ofk(we denote this by℘|(P)), we say that℘is afinite place ofK. We say that a finite place ℘ of K, ℘|(P), is inert (resp. split, resp. ramified) if (P) is inert (resp. split, resp. ramified) in the extensionK/k. We denote by suppD the support of a divisor D of K/Fq, by degD its degree. The principal divisor of a u ∈ K^∗ is denoted by div(u) and div(u) = div0(u)−div∞(u), with div0(u) (resp. div∞(u)) the zero divisor (resp. the pole divisor) of u. We denote by h the divisor class number of K/Fq,i.e.the order of the jacobian over Fq, Jac(K/Fq), considered as the group of classes of zero degree divisors modulo principal ones. The class inJac(K/Fq) of a zero-degree divisorRis denoted by [R]. LetOxbe the integral closure ofFq[x] inK.

ThenOxis the ring ofSx-integers,Sxbeing the set of places ofKabove the infinite place∞of the rational fieldk. Oxis a Dedekind domain and ak[x]-module of rank 2.

The group of fractionary ideals modulo principal ones is finite and its orderhxis the ideal class-number of Ox. The ring Ox is principal if and only if hx = 1. In this paper, we will say that hx is the ideal class-number of Ox or the ideal class-number of the quadratic extensionK/k. We recall that

– if cardSx = 1, K/k is an imaginary quadratic extension: if Sx ={P∞}, with degP∞= 1,K/k is ramified and ifSx={℘∞}, with deg℘∞= 2,K/k is inert;

– if cardSx= 2,K/k is a real quadratic extension and we setSx={∞1,∞2}.

This situation was studied by Artin [1] in his thesis, when p = charFq > 2. The two class numbers h andhx are linked by Schmidt’s formula (cf.[29]) hxrx =hδx, where rx is the regulator of the extension K/k and δx = gcd{deg℘, ℘|∞}. If the extensionK/kis an imaginary quadratic extension,rx= 1 andhx=h(resp.hx= 2h) if∞ramifies (resp. is inert) inK. If the extensionK/kis real quadratic,rxis the order of the subgroup of the jacobian of K/Fq generated by the classC∞ = [∞2− ∞1].

Moreover, we have hx = 1 if and only if Jac(K/Fq) is a cyclic group generated

by C∞. Finally, notice that the study of the jacobian of a hyperelliptic function field is of theoretical interest in cryptography in relation with the discrete logarithm problem. Many papers deal with that subject (see for instance [25] and [33] for odd characteristic and [26] forp= 2).

2.1. Affine model of a quadratic extension. — In characteristic p = 2, the equation defining a real extensionK/kis less well known than in the odd characteristic case. For sake of completeness we recall both situations.

Theorem 1. — Letq=p^eand letK/Fq be a hyperelliptic function field of genusg>1, such that the full constant field ofK/Fq isFq. Letx∈K be transcendental over Fq, k=Fq(x), such thatK/k is separable and quadratic. We denote byλxthe number of finite places of kwhich ramify in K.

(1) Casep >2. ThenK=k(y), withF(x, y) =y²−f(x) = 0, wheref ∈Fq[x]and f =aP1· · ·Pr∈Fq[x], thePi’s being pairwise distinct monic irreducible polynomials and a ∈ F^∗_q. Moreover the finite places of k which ramify in K are the (Pi)’s, so λx=r. Setm= Degf.

(a) If the quadratic extensionK/kis imaginary and∞ramifies inK,y may be chosen such thata= 1,m= 2g+ 1.

(b) If the quadratic extensionK/kis imaginary and∞is inert inK,y may be chosen such thata is a non-square,m= 2g+ 2.

(c) If the quadratic extensionK/k is real,y may be chosen such that a= 1, m= 2g+ 2.

(2) Case p= 2. Then K = k(y), with F(x, y) =y²+B(x)y+C(x) = 0, where B, C∈(Fq[x])^∗ are such thatB is monic and all irreducible factors ofB (if any) are simple factors ofC, i.e.

B = Yr i=1

B_iⁿⁱ and C=aN Yr i=1

Bi,

theBi’s are pairwise distinct monic irreducible polynomials,N ∈Fq[x]^∗ is monic and prime toB,a∈F^∗_q. Moreover the finite places ofkwhich ramify inKare the (Bi)’s, soλx=r. Setm= DegC.

(a) If the quadratic extensionK/kis imaginary and∞ramifies inK,y may be chosen such thatm= 2g+ 1,DegB6g,a= 1.

(b) If the quadratic extensionK/kis imaginary and∞is inert inK,y may be chosen such thatm= 2g+ 2,DegB=g+ 1,trace_Fq/F2(a) = 1.

(c) If the quadratic extensionK/kis real,ymay be chosen such thatDegB= g+ 1, andm <2g+ 2.

Reciprocally, any separable quadratic extension K of the rational function fieldk = Fq(x)is of the preceding form according to the behaviour of the infinite place of k in the extensionK/k.

Remark 2. — We give some comments about this theorem for the characteristic 2 case (compare with [8]). First of all, everything goes back to Hasse (see also [35] for instance), since settingv =y/B, one obtains an equation in Hasse normal form (see [14]):

(1) G(v, s) =v²+v+ aN

Qr

i=1 B_i²ⁿⁱ⁻¹ = 0.

So this is well known. Observe thatK/kis anArtin-Schreier extension. The condition B monic is not a restriction, since otherwise change y in y⁰ = y/b, if b 6= 1 is the leading coefficient ofB. If the quadratic extension K/k is real, it is unnecessary to consider the case DegB = g+ 1, m = 2g+ 2 and the leading coefficient a of C is such that a=c+c², with c∈ F^∗_q (i.e.trace_Fq/F2(a) = 0), since otherwise changey in y⁰ =y+cx^g+1 and then DegB =g+ 1, andm <2g+ 2. Finally, the condition:

“all irreducible factors ofBare simple factors ofC” is quoted in [4] (for instance) and used in [20] to obtain the characterization of imaginary quadratic extensions.

Definition 3. — IfK/k is a quadratic extension, we callnormal affine model of K/k a plane affine curve C with equation F(x, y) = 0 satisfying the conditions of the preceding Theorem and say thatF is a normal equation ofK/k.

2.2. Hyperelliptic involution. — Consider a quadratic extension K/k and let C={F(x, y) = 0}be an affine normal model ofK/k. Thehyperelliptic involution σ is thek-automorphism ofK such that

σ(y) =

( −y ifp >2 y+B(x) ifp= 2.

Foru∈K, we seteu=σ(u). Thenorm of uis defined by N(u) =uu.e

The hyperelliptic involution acts on the finite places ℘ of K/Fq and ℘e= ℘^σ is the conjugated place of ℘. Considering σ as an Fq(x)-automorphism of K = FqK, it acts on the affine points of C: if P = (a, b) ∈ Fq

2 is such that F(a, b) = 0, then P^σ = (a,−b) (resp.P^σ = (a, b+B(a))) if p > 2 (resp.p = 2) is an affine point of C. We set Pe =P^σ. Since an affine normal modelC is a smooth affine curve in any characteristic, we identify the finite (degree one) places of K = KFq with the (smooth) affine pointsP = (a, b) of a normal affine model C. Given any finite place (a, b) ofK, there is a unique finite place℘ofK, such that its conorm in the constant field extensionK/Fq ofK/Fq is

Conorm_K/K(℘) = X

τ∈Gal(Fq/Fq)

(a, b)^τ.

2.3. Representation of elements in the jacobian of a hyperelliptic function field

2.3.1. Representation with reduced divisors

Definition 4. — Let K/k be a quadratic extension. An effective divisor A of the hyperelliptic function fieldK/Fq is calledquasi-reducedif its support does not contain a pole of x, nor conorms (with respect to K/k) of places of k/Fq. A quasi-reduced divisorAofK/Fq is called reduced if degA6g. We consider thatA= 0 is reduced.

We denote byD⁺_redthe set of reduced divisors.

Note that ifAis quasi-reduced, then its support suppAdoes not contain any inert finite place℘of K. Moreover, if a ramified finite place is in the support of A, then its valuation equals one and if a split finite place℘is in the support onA, then℘eis not in the support ofA. In [27], the following representation of the elements of the jacobian ofK/k is given (in the ramified case it goes back to [1] or [6] forp6= 2 and [16] for p= 2). Observe that the authors of [27] assume thatp6= 2. But the results are also true forp= 2 considering an appropriate affine model.

Proposition 5. — Let K/k be a quadratic extension and let g be the genus of the hy- perelliptic function fieldK/Fq.

(1) If K/k is ramified, then

Jac(K/Fq) ={[A−(degA)P∞], A∈ D⁺_red}.

(2) If K/k is real, then

Jac(K/Fq) ={[A−(degA)∞2+n(∞1− ∞2)], A∈ D⁺_redand06n6g−degA}.

Proof. — see [27].

Corollary 6. — Let K/k be a real quadratic extension. The regulator ofK/k is such that rx>g+ 1, whereg is the genus of the hyperelliptic function fieldK/Fq. Proof. — This a trivial consequence of the previous proposition, since

rx= inf{n∈N^∗, n(∞1− ∞2) is a principal divisor}

andn(∞1− ∞2) is not principal for all 06n6g.

2.3.2. Representation with reduced ideals. — LetK/kbe a ramified or real quadratic extension given by a normal equationF(x, y) = 0. Then an integral basis of Ox is (1, y) and we write thisOx= [1, y]. We recall the following definitions.

Definition 7. — An ideal A of Ox is called an integral ideal. Two integral ideals A and B are said to beequivalent if there exist non-zeroα, β ∈ Ox such that (α)A= (β)B. An integral idealAisprincipal if there existsα∈Ksuch thatA= (α)Ox. The conjugate of an integral ideal Ais the integral idealAe such thatAe ={α, αe ∈A}. An

integral idealAisambiguous ifA=A. Thee polynomial normofAis the polynomial NA∈Fq[x] such thatAeA= (NA)Ox. Thedegree ofAis degA= DegNA.

Definition 8. — LetK/k be a ramified or real quadratic extension. LetAbe a non- zero integral ideal and consider its factorization in prime ideals A= Q

i∈I℘^e_iⁱ. We say that Aisquasi-reduced (resp.reduced) if the corresponding effective divisorA= P

i∈Iei℘i is quasi-reduced (resp. reduced). We consider that{0}is a reduced ideal.

For a ramified or real quadratic extension K/k, Proposition 5 can be written in terms of reduced ideals (see [27]). Let us recall the result for a real quadratic exten- sion.

2.3.3. Case of a real quadratic extension. — IfK/kis a real quadratic extension, an integral idealAhas aFq[x]-basis such that: A= (S)[Q, y+P], whereS, Q, P ∈Fq[x]

andQdivides N(y+P). If p >2, this representation goes back to [1] (see also [27]

and, ifp= 2, [37, Th. 11].

Definition 9. — Aprimitive ideal is an integral ideal such thatA= [Q, y+P], with Q|N(y+P) (i.e.S= 1).

It can be proved that a primitive idealAis quasi-reduced and it is reduced if and only if degA6g. We will see this further in casep= 2.

Corollary 10. — LetK/kbe a real quadratic extension. There is a canonical bijection between Jac(K/Fq)and the following set

A={(a, n), a reduced ideal and06n6g−dega}.

Proof. — This is a straightforward consequence of Proposition 5 and of the definitions.

We will not explain the group law onA(see [27] or [37] ifp= 2).

Corollary 11. — LetK/k be a real quadratic extension.

(1) There is a canonical bijection between the cyclic subgroup of Jac(K/Fq)gen- erated by the class[∞2− ∞1]and the following set

{(a, n), a principal and reduced such that 06n6g−dega}.

(2) The ideal class number hx ofK/k equals1 if and only if all reduced ideals are principal.

Proof. — Clear.

Note that if the order h of the jacobian Jac(K/Fq) is a prime number, then of coursehx= 1.

2.4. Regular differentials of a hyperelliptic function field. — Let K/Fq be a hyperelliptic function field. In [2, p. 261] and forp > 2, it is said that any quasi- reduced divisorDof degree>gis non-special. A similar result holds for characteristic p= 2. Let us recall both cases.

Lemma 12. — Let K/Fq be a hyperelliptic function field and letxandy be such that K =Fq(x, y), with F(x, y) = 0for F a normal equation of K/k =Fq(x). Then a Fq-basis of the vector space of regular differentials ΩK(0) is (ω0, xω0, . . . , x^g−1ω0), where

ω0=

( dx/y ifp6= 2 dx/B(x) ifp= 2.

Proof. — By [34, Proposition VI.2.4], a basis of ΩK(0) is (ω0, xω0, . . . , x^g−1ω0), where

div(ω0) = (g−1) div∞(x).

(1) Case p6= 2. We have F(x, y) = y²−f(x) = y²−aP1(x). . . Pr(x). Denote bypi the unique place of Kabove (Pi). Moreover

div∞(x) =









2P∞ ifK/k ramifies

℘∞ ifK/k is inert

∞1+∞2 ifK/k is real It is easy to show that div∞(y) =Pr

i=1pidiv∞(y), where div∞(y) =









(2g+ 1)P∞ ifK/k ramifies (g+ 1)℘∞ ifK/k is inert (g+ 1)(∞1+∞2) ifK/k is real

Since the extension K/k is a Kummer extension, the different of K/k is (see [34, III.7.6. p. 113])

DiffK/k= Xr

i=1

pi+ηP∞

withη= 1 ifK/k is ramified (resp.η= 0 otherwise). Then div(dx) = DiffK/k−2 div∞(x) =

Xr i=1

pi+ηP∞−2 div∞(x) and the result follows.

(2) Case p = 2. This result is classical concerning Artin-Schreier extension (one can adapt [35, p.168] to the non-algebraically closed case). We haveF(x, y) =y²+ B(x)y+C(x) =y²+yQr

i=1B_iⁿⁱ+aNQr

i=1 Bi. In case DegB >1, let bi be the

unique place ofKabove the finite place (Bi) ofk. Then div(dx) =

(2Pr

i=1nibi−2 div∞(x) ifK/k is real or inert 2Pr

i=1nibi−(Pr

i=1nidegbi−g+ 1) div∞(x) ifK/k is ramified and

div(B(x)) = 2 Xr i=1

nibi− X^r

i=1

nidegbi

div∞(x).

The result follows.

Proposition 13. — Let K/k be a quadratic extension.

(1) If K/k is ramified or inert, any quasi-reduced divisorD of degree >g is non- special.

(2) AssumeK/k is real. LetD =A+r∞i,i= 1or 2 andr∈N, be an effective divisor of K/k of degree>g such that Ais quasi-reduced. Then D is non-special.

Proof. — Let us show that i(D) = dim_FqΩ(D) = 0. Assume ω ∈ Ω(D)^∗. Then, using Lemma 12, ω = (Pg−1

i=0λixⁱ)ω0 and div(ω)>D. SetT(x) =Pg−1

i=0 λixⁱ 6= 0, e= DegT < g. ThenT = div0(T(x)) is a conorm with respect toK/kof degree 2e.

Moreover

div(ω) =T−ediv∞(x) + div(ω0) =T+ (g−1−e) div∞(x).

(1) If K/k is ramified or inert, div(ω) > D ⇐⇒ T > D. Since D is a quasi- reduced divisor andT is a conorm,T >D impliese>degD and we obtain degD6 g−1 which is not true.

(2) IfK/k is real, div(ω)>D=A+r∞i ⇐⇒ T+ (g−1−e−r)∞i>A, which is equivalent toT>Aandr6g−1−e. As before, we havee>degAand we obtain degD= degA+r6g−1 which is not true.

2.5. Ideal class number for quadratic extensions. — The classification of all imaginary quadratic extensions which have a given ideal class number is the analogue of the ideal class number problem for imaginary quadratic number fields. The ideal class number one problem (hx = 1) for imaginary quadratic extensions has been settled by R.E. MacRae [21]. He proved that there is only one imaginary quadratic field ifp >2 (as predicted by Artin [1]) and three ifp= 2. The analogue for function fields of the famousGauss Conjecture for number fields is the following. For a fixed finite field Fq, is there infinitely many real quadratic extensions K/Fq(x) such that the integral closure of Fq[x] inK is a principal domain? In [7], S. Chowla “presents a case where (this) conjecture ... is proved in a parallel case in function field theory, under the assumption of a very plausible conjecture in number theory”. Without the assumptionqfixed, there is a positive answer to the question. It has been proved first by M.L. Madan in [22] in the odd characteristic case. Other papers deal with similar results (see [30], [13], [15], [19], [9],...). But, as far as we know, the precise analogue

for function fields of the Gauss conjecture remains unproved. The ideal class number hxof a quadratic extensionK/kis always even ifK/kis inert, since thenhx= 2h. If K/kis ramified,hxandhhave the same parity, sinceh=hx. We recall the following result concerning the parity of the ideal class number of a quadratic extension.

Proposition 14. — Let K/k be a real quadratic extension, k=Fq(x). The ideal class numberhx ofK/k is odd if and only if

– casep >2: K=k(y)with y²=f(x), wheref ∈Fq[x]is such that f is a monic irreducible polynomial of even degree, or

f =p1p2,p1 andp2 being monic irreducible polynomials of odd degree.

– casep= 2: K=k(y)with

(2) y²+b(x)ⁿy+aN(x)b(x) = 0,

where aN ∈Fq[x]^∗,b ∈Fq[x]^∗ is a monic irreducible polynomial, gcd(N, b) = 1 and DegN <(2n−1) Degb.

Proof. — See [36] for p >2 and [31] for anyp. To prove this, one has to study the 2-rank of the ideal class group, which is related to the number of ambiguous ideals.

If one wants to study the ideal class number one problem in characteristic 2, then the solutions have a normal equation given by (2).

3. Real quadratic extensions in even characteristic

We will now focus on the characteristicp= 2 case. There are similar results in the odd characteristic case. In Section 3, we keep the following conventions or notations:

q= 2^e,k=Fq(x), andK/k is a real quadratic extension defined by a normal equation (3) F(x, y) =y²+B(x)y+C(x) = 0,

B=Qr

i=1 B_iⁿⁱ andC=aNQr

i=1 Bi, theBi’s are monic irreducible distinct polyno- mials, N ∈Fq[x]^∗ is monic, gcd(B, N) = 1, a ∈ F^∗_q and m= DegC < 2(g+ 1) = 2 DegB. Recall thatSx={∞1,∞2}is the set of (degree one) places ofKabove the infinite place∞ofk. We denote byν1(resp.ν2) the valuation at∞1(resp.∞2), and by val℘(u) (resp. ord℘(u) =|val℘(u)|) the valuation (resp. the order) of anyu∈K^∗ at a place℘ofK.

Remark 15. — In odd characteristic, an affine normal model of a real quadratic ex- tension has a unique point at infinity, which is singular. Further in that case, the pole divisor of y is div∞(y) = (g+ 1)(∞1+∞2). In characteristic 2, the situation is quite different. A normal affine modelChas one or two points at infinity (see [20]

for instance). If g+ 1 6 m < 2(g+ 1) (thus g > 2), C has one singular point at infinity (the pointP0 = (0 : 1 : 0) in homogeneous projective coordinates) and there are two places of degree one, ∞1 and∞2, aboveP0. If m < g+ 1,C has two points

at infinity: one is smooth, say∞1, andP0 is singular if g>2 and there is a unique place of degree one, say∞2, aboveP0.

3.1. Principal divisors ofK. — Ifα=u+vy∈K, we recall thatαe=u+(v+B)y and the norm ofuis

N(α) =ααe=u²+uvB+Cv².

Between the two places at infinity, we select a place, say∞2, such thatν2(y)6ν1(y).

In the next Lemma, we will see that this place is well defined. Since K/k is real, 1/xis a local parameter at∞i, i= 1,2. We consider (see [34, p. 143]) the∞2-adic completion of K, denoted byKb2, and the embedding of Kin Kb2

K−→Kb2

α7−→

Xn i=−∞

cixⁱ, ci∈Fq andcn6= 0.

Observe that n=−ν2(α) and, if α=P(x)∈Fq[x], n= DegP. We denote bybαc thepolynomial part of the∞2-adic power series expansion ofα,i.e.bαc=Pn

i=0 cixⁱ ifn>0, and = 0 otherwise.

Definition 16. — We say thatα∈Kisreduced with respect to∞2if∞2is a pole ofα and a zero ofα.e

Further we will say “reduced” instead of “reduced with respect to∞2”.

Lemma 17. — Let K/k be a real quadratic extension with normal equation given by (3).

(1) Let bi be the unique place ofK above the finite place (Bi)ofk. Then div(B(x)) = 2

Xr i=1

nibi−(g+ 1)(∞1+∞2).

(2) If DegN > 0 (otherwise l = 0), consider the factorization N(x) = Ql

j=1 Nj(x)^lj. Each finite place (Nj) of k splits in K and we denote by nj and e

nj the places ofK above (Nj). Then, div(y) =

Xr i=1

bi+ Xl j=1

ljnj+ (g+ 1−m)∞1−(g+ 1)∞2, and

div(ey) = Xr i=1

bi+ Xl j=1

ljenj+ (g+ 1−m)∞2−(g+ 1)∞1.

(3) The polynomial part of the ∞2-adic power series expansion ofy is monic and Degbyc=g+ 1 = DegB. If1 6m < g+ 1,byc=B and, if g+ 16m <2(g+ 1), byc 6=B but the coefficients ofx^g+1, . . . , x^m−g inbycandB are equal.

Proof. — (1) Recall that, for all 1 6i6 r, the finite place (Bi) of k is ramified in K. Letbi be the unique place ofK above (Bi). Its degree is degbi= DegBi and its conorm in the constant field extensionK/Fq ofK/Fq is

Conorm_K/K(bi) = X

B_i(a)=0

(a,0), thusbiis a zero fory. Since DegB=g+ 1, one has

div(B(x)) = 2 Xr i=1

nibi−(g+ 1)(∞1+∞2).

(2) Using yey =C(x), we see that the finite zeroes of y (resp.ey) are among the zeroes of C(x) and the only possible poles for y (resp.y) are the poles ofe x, i.e.∞1

and∞2.

(a) The placebi is a zero ofy of order one and also a zero of ey =y+B of order one, sinceBi is a simple factor ofC.

(b) IfN =Ql

j=1 N_j^lj, withl>1, the places (Nj) ofksplit inK. We denote bynj andenj the two places above (Nj). Their respective degree equals DegNj. Leta∈Fq be a zero ofNj, then

Conorm_K/K(nj) = X

N_j(a)=0

(a,0), and Conorm_K/K(enj) = X

N_j(a)=0

(a, B(a)).

Thus, nj is a zero of y and enj is a zero of y+B. Since y(y+B) = C and DegC=m, one obtains

(4) div(y) + div(y+B) = Xl j=1

lj(nj+enj) + 2 Xr i=1

bi−m(∞1+∞2).

Thus, nj is a zero of y of order lj and enj is a zero of ey of order lj. We have obtained all the finite zeroes ofy andyeand the degree of the finite zero divisor ofy (resp.y) ise

deg X^r

i=1

bi+ Xl j=1

ljnj

= deg X^r

i=1

bi+ Xl j=1

ljenj

=m.

(c) Observe that [K : Fq(y)] = max{m, g + 1} = deg(div∞(y)) = deg(div0(y)). We set v = y/B(x) and consider equation (1). Then νj(v)<0, forj= 1 or 2, is impossible, since

νj(v) +νj(v+ 1) =νj

N(x) Qr

i=1 Bi(x)²ⁿⁱ⁻¹

= 2(g+ 1)−m >0.

Ifνj(v) = 0, thenνj(v+ 1) = 2(g+ 1)−m. Ifνj(v)>0, thenνj(v+ 1) = 0, thus νj(v) = 2(g+1)−m. Sinceνj(y) =νj(v)+νj(B(x)) =νj(v)−(g+1), this implies νj(y) =−(g+ 1) orνj(y) = (g+ 1)−m >−(g+ 1). Ifg+ 16m <2(g+ 1), one obtains deg(div∞(y)) =m, soνj(y) =−(g+ 1) andνj⁰(y) = (g+ 1)−m60 for

j⁰6=j. If 16m < g+ 1, one obtains deg(div∞(y)) =g+ 1, soνj(y) =−(g+ 1) and νj⁰(y) = m−(g+ 1) > 0 for j⁰ 6=j. Thus νj(y) = −(g+ 1) for at least one j, say j = 2. It can be shown that this is coherent with the notations in Remark 15. To conclude, we have obtained that

div(y) = Xr i=1

bi+ Xl j=1

ljnj+ (g+ 1−m)∞1−(g+ 1)∞2, and, using (4),

div(ey) = Xr i=1

bi+ Xl j=1

ljen+ (g+ 1−m)∞2−(g+ 1)∞1.

(3) Setd(x) =byc ∈k[x]. Then, Degd=−ν2(y) =g+ 1 = DegB andν2(y+B) = (g+ 1)−m. If 16m < g+ 1, we obtain thatν2(y+B)>0, sobyc=d(x) =B(x) and, since B is monic, d is monic too. If g+ 1 6 m < 2(g+ 1), we obtain that ν2(y+B)60. Thusd+B6= 0 and 06Deg(d+B) =m−(g+ 1)< g+ 1 = DegB.

Sodis monic and the coefficients ofx^g+1, . . . , x^m−g indandB are equal.

3.2. Quadratic irrationals. — We consider some particular elements inK, which are related to the representation of primitive ideals (see Section 2.3.2).

Definition 18. — We say thatα∈Kis aquadratic irrational ofK, ifα= (y+P)/Q, with (Q, P)∈k[x]^∗×k[x] and QdividesN(y+P),

(5) N(y+P) = (y+P)(y+P+B) =P²+BP+C.

Ifα= (y+P)/Qis a quadratic irrational,αis reduced with respect to∞2, or for short,αis reduced if and only if

−ν2(y+P+B)<DegQ <−ν2(y+P).

Remark 19. — Let us show that, ifα= (y+P)/Qis reduced, one has DegP < g+ 1 and DegQ < g+ 1 (compare with [37, p. 567]). Remember thatB andbycare monic of degree g+ 1 = −ν2(y) and thus B+byc= 0 or Deg(B+byc) < g+ 1. If α is reduced, one has DegQ <−ν2(y+P) thusbyc+P 6= 0. Then

(6) αis reduced ⇐⇒ −ν2(y+P+B)<DegQ <Deg(byc+P).

(1) If P 6= byc+B, α is reduced if and only if Deg(byc+P +B) < DegQ <

Deg(P+byc). One obtains DegP < g+ 1 and DegQ < g+ 1.

(2) IfP =byc+B, then−ν2(y+P+B)<0, DegP = Deg(byc+B)< g+ 1 and αis reduced if and only if DegQ < g+ 1.

Lemma 20. — y is a quadratic irrational which is reduced if and only if byc = B, i.e.16m < g+ 1. Moreover, ify is reduced, all quadratic irrationalsα= (y+P)/Q such that DegP <DegQ < g+ 1 are reduced.

Proof. — Forα=y, we haveP = 0,Q= 1 andQdividesN(y) =y(y+B) =C, thus y is a quadratic irrational. By Lemma 17,y is reduced if and only if 16m < g+ 1 and if and only ifbyc=B. Note that, ify is not reduced, then y+bycis reduced.

Using (6), we see that, ifyis reduced, all quadratic irrationalsα= (y+P)/Qsuch that DegP <DegQ < g+ 1 are reduced.

We prove the following result, which is a generalization of the caseα=y seen in Proposition 17. It is the equivalent result for characteristic 2 of results in [2] (see also [28, prop. 9] and [3]).

Proposition 21. — Let α= (y+P)/Qbe a quadratic irrational. Set a0=bαcand let Q⁰∈Fq[x]be such that

(7) QQ⁰ =N(y+P) = (y+P)(y+P+B) =P²+BP+C.

(1) Let I (resp.I⁰) be the set of i, 1 6i 6r, such that Bi divides Q (resp.Q⁰).

Then I∩I⁰=∅and the factorizations of QandQ⁰ are Q(x) =Y

j∈J

Uj(x)^njY

i∈I

Bi(x), Q⁰(x) = Y

j∈J⁰

Uj⁰(x)^n0j Y

i∈I⁰

Bi(x),

where all places (Uj)(resp.(U_j⁰)), if any, are split. We denote byuj andeuj (resp.u⁰_j andeu⁰_j) the two finite conjugated places above (Uj)(resp.(U_j⁰)).

(2) We consider the following quasi-reduced divisors DandD⁰ of respective degree DegQandDegQ⁰

D=X

i∈I

bi+X

i∈J

njuj, D⁰ =X

i∈I⁰

bi+X

i∈J

n⁰_ju⁰_j.

The finite pole (resp. zero) divisors ofαandαe are div∞,f(α) =De div0,f(α) =De⁰ div∞,f(α) =e Ddiv0,f(α) =e D⁰. (3) Assume moreover that αis reduced. Then

(a) Dega0=g+ 1−DegQ >0,DegQ⁰ < g+ 1.

(b) ν2(α) =ν1(α) = Dege Q−(g+ 1) = −Dega0 <0 andν1(α) =ν2(α) =e g+ 1−DegQ⁰>0.

(c) The principal divisors ofαandαe are

div(α) =De⁰+ (g+ 1−DegQ⁰)∞1−De−(Dega0)∞2, div(α) =e D⁰+ (g+ 1−DegQ⁰)∞2−D−(Dega0)∞1.

The pole divisor (resp. zero divisor) of α and of αe are non-special divisors of degreeg+ 1.

Proof. — (1) Assume Bi divides Q. Using (7) and the fact that Bi is a simple factor of C, we have that Bi is a simple factor of Q and it is not a factor of Q⁰. Let I ⊂ {1, . . . , r} (resp.I⁰) be the set of i such that Bi divides Q (resp.Q⁰). Of

course, the set I or I⁰ may be empty. We have I∩I⁰ = ∅. Notice that, if Bi

divides Q, Bi is also a factor of P and, since Bi is ramified, valb_i(P) > 2. Let a ∈ k be such that Q(a) = 0 and B(a) 6= 0. Then a is a zero of a Uj and there are exactly two places above the finite place (x−a) of k(x), which are (a, b) and (a, b) = (a, b] +B(a)), b = y(a) ∈ k being such that b²+bB(a) +C(a) = 0. The places (a, b) and(a, b) = (a, b] +B(a)) are zeroes of Q of order nj. Since Q divides (P+y)(P+B+y), (a, b) or(a, b) = (a, b] +B(a)) is a zero ofP+y,i.e.b=P(a) or b=P(a) +B(a). If (a, b) is a zero of P+y, then it is not a zero of P +y+B and (a, b) is a zero of] P+B+yand not a zero ofP+y. Thus there exists a finite placeuj

ofKsuch that Conorm_K/K(uj) =P

U_j(a)=0(a, P(a)) which is a zero ofP+yand not a zero ofP+y+B, euj is a zero ofP+y+B and not a zero ofP+y. This proves also that (Uj) is split. Finally

div(Q) =X

j∈J

nj(uj+euj) + 2X

i∈I

bi−(DegQ)(∞1+∞2).

Similarly, we can show that div(Q⁰) =X

j∈J⁰

n⁰_j(u⁰_j+eu⁰_j) + 2X

i∈I⁰

bi−(DegQ⁰)(∞1+∞2).

(2) The finite poles ofα= (y+P)/Q(resp.α) are among the zeroes ofe Q.

(a) Ifbi is a zero ofQ, we have seen that valb_i(Q) = 2, valb_i(P)>2 andbi

is a simple zero ofy (see Lemma 17). So valb_i(P+y) = 1 andbi is a pole forα of order 1.

(b) Ifuj is a zero ofQ, we have seen that it is a zero ofP+yand not a zero ofP+y+B. Moreover, 0<valu_j(Q)6valu_j(P+y) anduj is not a pole forα.

But theneuj (resp.uj) is a pole forα(resp.α) of ordere nj.

Finally, div∞,f(α) = De and div∞,f(α) =e D. Considering the quadratic irrational α⁰ = (y+P)/Q⁰, we have a similar result, div∞,f(α⁰) = De⁰ and div∞,f(αe⁰) = D⁰. But since by (7), α⁰ = (y+P)/Q⁰ = Q/(y+P+B) = 1/α, we have dive ∞,f(α⁰) = div0,f(α) and dive ∞,f(αe⁰) = div0,f(α), so the result follows.

(3) Ifαis reduced, DegP < g+ 1 = Degbycand DegQ < g+ 1 by Remark 19.

(a) Thus, since Degbyc = g + 1 > DegP, a0 = b(P+y)/Qc = bbyc/Qc, Dega0= (g+ 1)−DegQ >0 and, since DegC <2(g+ 1), DegQ⁰< g+ 1.

(b) Sinceααe=Q⁰/Q, for all place℘ofK,

(8) val℘(α) + val℘(α) = vale ℘(Q⁰)−val℘(Q).

Since DegP < DegB = g+ 1 = −ν2(y), one hasν2(y+P) =−(g+ 1) and

∞2 is a pole for αsuch that ν2(α) = DegQ−(g+ 1) = −Dega0. Similarly, ν1(y+P +B) = −(g + 1), thus ∞1 is a pole for αe and ν1(α) = Dege Q− (g+ 1) = −Dega0. Using (8), we obtain ν1(α) = g+ 1−DegQ⁰ > 0 and ν2(α) =e g+ 1−DegQ⁰>0.

(c) The pole (resp. zero) divisors ofαandαeare reduced of degreeg+ 1 and by Proposition 13 these divisors are non-special.

From the proof of the preceding Proposition, we can deduced the following trivial observation.

Lemma 22. — Let Q∈Fq[x] be monic and irreducible. Then the following assertions are equivalent:

(1) There exists P ∈Fq[x] such thatQ dividesN(y+P).

(2) The equationT²+BT +C= 0 modQhas at least one solution in Fq[x].

(3) eitherQ|B and then the place(Q)ofk ramifies inK (andP = 0)

or gcd(Q, B) = 1 and then the place (Q) of k splits in K (this is the case for instance if Q|C).

No finite place (Q) of k, which is inert in K, is such that Q|N(y +P) for some P ∈Fq[x].

3.3. Reduced integral ideals. — As said before, an integral ideal is of the fol- lowing form A = (S)[Q, y+P], where S, Q, P ∈ Fq[x] and Q divides N(y+P) = P²+BP+C. Without loss of generality, it can be assumed thatQ, S are monic and DegP <DegQand then the representation ofAis unique.

Lemma 23. — LetA= [Q, y+P]be a primitive ideal ofOx. ThenAis quasi-reduced.

Moreover,Ais reduced if and only if degQ6g.

Proof. — (see also [37, Th. 12]). First notice that an integral ideal can be quasi- reduced only if it is primitive. LetA=Q

i∈I℘^e_iⁱ be the factorization of a primitive ideal A = [Q, y+P]. Then, each ℘i is a common zero of Q and y+P. None of the ℘i’s are equal to ∞j, j = 1 or 2, and none of them are inert (cf.Lemma 22).

Moreover, ifQ(x) =Q

j∈J Uj(x)^njQ

i∈IBi(x), then (using the notations of the proof of Proposition 21)

A=Y

j∈J

uⁿ_j^jY

i∈I

bi,

andAis quasi-reduced. The polynomial norm ofAis NA=Qand degA= DegQ.

ThusAis reduced if and only if DegQ6g.

Without loss of generality, it can be assumed that any reduced ideal is such that A = [Q, y+P], with Q|P²+BP +C, DegP < DegQ < DegB and Q is monic.

Notice that, if α= (P+y)/Q is a reduced quadratic irrational, thenA= [Q, y+P] is reduced. But, conversely, ifA= [Q, y+P] is a reduced ideal, withQ|P²+BP+C, DegP <DegQ <DegB andQis monic, thenα= (P+y)/Q is not always reduced.

But, according to Remark 19, ify is reduced, thenα= (P+y)/Qis reduced.

3.4. Fundamental unit ε and regulator rx. — The unit group of Ox is O_x^∗ = F^∗_q× hεi, whereεis afundamental unit. Then the regulator ofK/k isrx=|ν1(ε)|=

|ν2(ε)|and

div(ε) =rx(∞1− ∞2).

Our purpose is now to compute the ideal class number hx of Ox. For that, we can apply Schmidt’s formula h=hxrx, compute the divisor class number using for instance the zeta function (h=L(1), whereL(t) is the numerator polynomial of the zeta function) and compute the regulator. This last task can be achieved using the continued fraction expansion ofy.

3.5. Continued fraction expansion (CFE) in characteristic 2. — For anyp, the results concerning the continued fraction expansion algorithm are very similar to the number field case. There are plenty of references for the odd characteristic case.

For the case p = 2, we refer to [37] (see also [24]). In this section, we recall basic results.

3.5.1. Definitions

Definition 24. — Letα0 = (y+P0)/Q0 ∈K be a quadratic irrational and set a0 = bα0c. Fori>1, define the i-th iterate αi recursively by

ai−1=bαi−1c, αi= 1 αi−1+ai−1.

The CFE ofα0 is the sequence [a0;a1, a2, . . .]. Fori>1, we consider the functions θei defined by

θe1= 1, and fori>1, θei+1 = Yi j=1

1 αej

.

If α = (y+P)/Q is a quadratic irrational, we will say “the CFE of α” or “the CFE ofA= [Q, y+P]”. The CFE is finite if and only ifα0= (y+P0)/Q0∈Fq[x].

If α0 ∈ Kr Fq[x], the CFE is quasi–periodic and periodic, the period τ (resp. the quasi-period ρ) is the least integer nsuch that αn =αn0 (resp.αn =cαn0, c ∈F^∗_q), with 06n0< n. The CFE ofα0 is obtained as follows. For alli>0,

αi=Pi+y Qi ,

where thePi’s and Qi’s are defined recursively as follows:

Pi+1=aiQi+Pi+B, Qi+1Qi=P_i+1² +BPi+1+C.

Another way to compute thePi’s andQi’s is the following. SetQ−1= (P²+BP+C)/Q andd=byc, then, for alli>0, compute recursivelyai, Pi+1, Qi+1using the following

formulae

ai =

Pi+d Qi

ri =Pi+d (modQi) Pi+1 =d+ri+B

Qi+1=Qi−1+ai(ri+ri−1).

Notice that

(9) Qi+1Qi=P_i+1² +BPi+1+C=N(Pi+1+y) = (Pi+1+y)(Pi+1+B+y).

Remark 25. — If αi is reduced, then αj is reduced for all j > i. Moreover, there exists i>0 such that αi is reduced (see [37, Th. 1]). If α0 =y is not reduced, then αi is reduced for all i >1 and then all αi’s, for i> 1, are reduced. In fact, ify is not reduced, then byc+B 6= 0. The first data of the CFE of y are : P0 = 0 and Q0 = 1, r0 = 0, Q−1 = C, a0 = d = byc, P1 = d+B, Q1 = d(d+B) +C and α1= (y+P1)/Q1. SinceQ1=P₁²+P1B+C= (y+d)(y+d+B) andν2(y+d)>0, we have

ν2(Q1) =−Deg(Q1) =ν2(y+d) +ν2(y+d+B)>−Deg(B) =g+ 1 thus Deg(Q1)< g+ 1. The result follows from Remark 19.

Ifαi is reduced, then DegPi and DegQi are< g+ 1, thus (10) ai=

Pi+d Qi

= d

Qi

,16Degai=g+ 1−DegQi6g+ 1.

Using (9), one obtainsN(αei) = αiαei =Qi−1/Qi and 1/αei = (Pi+y)/Qi−1. Then N(eθi+1) =Qi/Q0and

(11) θei+1= 1

αei

θei= Pi+y Qi−1

eθi, for alli>1.

Notice that, fori>2,θei+1=ai−1θei+θei−1 and theθei can be computed recursively.

Lemma 26. — Consider the CFE of a primitive ideal A1= [Q0, y+P0]and set Ai= [Qi−1, y+Pi−1], for alli>1.

All theAi are equivalent toA1. Conversely, ifAandBare equivalent reduced ideals, then in the CFE of A, there exists n such thatB =An. In particular, in the CFE of y we have for alli>1,

Ai= (eθi)Ox

and we obtain all the principal reduced ideals of Ox.

Proof. — see [37, th. 13 and 17]. Notice that is easy to show, using (9), that for all i > 1, (Qi)Ai = (Pi +y+B)Ai+1. Thus all the Ai’s are equivalent to A1 and (Q0θi)Ai = (Qi)A1. ForA1=Ox= [1, y],Ai is reduced for alli>2 and, using (11), one obtainsAi= (θei)Ox.

3.5.2. The CFE ofy. — The CFE ofyhas a lot of nice properties (cf.[37]). We have seen that it produces all the principal reduced ideals of Ox. Now we want to show that the regulator of the extension can be computed from this CFE. First, we have the following result, which is an analogous result in characteristic 2 of [3, Lemma 3]

(cf.also [28]). The proof is very similar to the odd characteristic case.

Proposition 27. — We consider the CFE of a reduced quadratic irrational α0 = (P0+y)/Q0. Fori>1, the divisors of theαi’s are

div(α1) =De0+ (Dega0)∞1−De1−(Dega1)∞2

div(α2) =De1+ (Dega1)∞1−De2−(Dega2)∞2

...

div(αi) =Dei−1+ (Degai−1)∞1−Dei−(Degai)∞2

div(αi+1) =Dei+ (Degai)∞1−Dei+1−(Degai+1)∞2, ...

where, for all i>0, the divisorsDei are reduced and such thatdegDei= DegQi. Proof. — For alli>0, one has

αi =ai+ 1 αi+1.

Then the finite zeroes of αi+1 are the finite poles of αi. The result follows from Proposition 21 and (10). Sinceαi+1αei+1=Qi/Qi+1, we deduce that, fori>0,

(12) div

1 e αi+1

=Di+1+ (Degai+1)∞1−Di−(Degai)∞2. Proposition 21 gives the values of theDi’s.

Corollary 28. — Consider the CFE of a reduced quadratic irrationalα0= (P0+y)/Q0. Set

R0= 0, Ri= Xi j=1

Deg(aj), for alli>1.

Then div(eθ1) = 0 and, for alli>1,

(13) div(θei+1) =Di+Ri∞1−D0−(Ri−1+ Dega0)∞2,

whereDi is the finite zero divisor ofαei. In particular,Diis reduced of degreeg+ 1− Degai. Considering the CFE ofα0=y, one has, for all i>1,

(14) div(θei+1) =Di+Ri∞1−(Ri−1+g+ 1)∞2.

Proof. — To prove (13), use (12), the definition ofθei+1and Proposition 27. Ifα0=y, we have proved that αi is reduced at least fori >1. Thus we can apply (13) with D0 = 0 (since Q0 = 1). Notice that, if y is reduced, then d = a0 = B, θe2 = y, R1 =g+ 1−m >0, and div(eθ2) = div(y) =D1+R1∞1−(g+ 1)∞2. The value of

We obtain an analogous result in characteristic 2 of [3, Th. 2].

Proposition 29. — We assume thaty is reduced and consider the CFE of y.

(1) For all i,`((Ri+g+ 1)∞2)−`((Ri−1+g+ 1)∞2) = Degai.

(2) LetRbe an integer. Then`(R∞2) = 1if06R < g+1and`(R∞2) =R−g+1 ifR>g+ 1. AssumeR>g+ 1and letj>2be such thatRj−26R−(g+ 1)< Rj−1. Ak-basis ofL(R∞2)is composed of the following functions

– eθi, for16i6j,

– for 26i6j and if Degai−1>1,x^γθei, with 16γ6Degai−1−1, – ifRj−2< R−(g+ 1),x^γθej, with16γ6R−(g+ 1)−Rj−2. Proof

(1) According to Proposition 13, all divisorsN∞2are non-special ifN >g. Thus, for alli>1,

`((Ri+g+ 1)∞2)−`((Ri−1+g+ 1)∞2) =Ri−Ri−1= Degai. (2) By (14), we have thateθi∈ L((Ri−2+g+ 1)∞2) for alli>2.

(a) Since y is reduced,θe2 =y and y ∈ L((g+ 1)∞2). Moreover (g+ 1)∞2

is non-special, so`((g+ 1)∞2) = 2 and (1, y) is ak-basis forL((g+ 1)∞2). Of course 1 is a basis ofL(R∞2) for allR, 06R < R1=g+ 1.

(b) If R > g+ 1, then `(R∞2) = R−g+ 1. Let j be such that Rj−2 6 R−(g+ 1)< Rj−1. This condition means that

L((Rj−2+g+ 1)∞2)⊂ L(R∞2) L((Rj−1+g+ 1)∞2).

For 16i6j, one hasθei ∈ L(R∞2). Then, we proceed recursively, using that for anyi>2 andγ >0

div(x^γθei) =γdiv0(x) + Di−1+ (Ri−1−γ)∞1−(Ri−2+ (g+ 1) +γ)∞2. – Set

n=`(R∞2)−`((Rj−2+g+ 1)∞2) =R−(g+ 1)−Rj−2.

If n = 0, a basis of L(R∞2) is a basis of L((Rj−2+g + 1)∞2) and if n > 0, we obtain a basis of L(R∞2) adding to a basis of L((Rj−2+g+ 1)∞2) the functionsx^γθej for 16γ6n.

– Assume j >3. Then, for all i such that 2 6i 6 (j−1) and Deg(ai−1)>1, all the functionsx^γθei, for 16γ6Degai−1−1, are inL((Rj−2+g+ 1)∞2) thus inL(R∞2).

We have foundN functions inL(R∞2), N =j+

Xj−1 i=2

(Degai−1−1) + (R−(g+ 1)−Rj−2) =R−g+ 1 =`(R∞2), having pairwise distinct valuations at∞2, so the result follows. It is easy to see that the preceding functions have also pairwise distinct valuations at∞1.

Proposition 30. — In the CFE ofy, let ρbe the least integer such that Qi ∈F^∗_q, then ρis the quasi-period anddiv(θeρ+1) =Rρ(∞1− ∞2). Moreover θeρ+1 is a fundamental unit andRρ is the regulator.

Proof. — Equation (14) can be written

div(θei+1) =Di−(DegDi)∞2+Ri(∞1− ∞2).

Recall thatDiis a reduced divisor of degreeg+1−Degaiand thatRi=Ri−1+Degai. Thus the principal divisor ofθei+1shows the equivalence between the two zero-degree divisorsDi−(DegDi)∞2 andRi(∞2− ∞1) and more generally,

Di−(DegDi)∞2+n(∞1− ∞2)∼(Ri−n)(∞2− ∞1),

for all 06n6g−DegDi= Degai+ 1, soRi−1+ 16Ri−n6Ri. It is easy to show that, ifρis the least integer such that Qi ∈ F^∗_q, then for 1 6i 6ρ, all Di’s are pairwise distinct. Using Corollary 11, we obtain the result.

4. Ideal class number one problem and examples

Now we want to study the ideal class number problem in characteristic 2 for real quadratic extensions and genusg>1. According to Proposition 14, a normal equation ofK/k must be given by:

(15) y²+bⁿy+aN b= 0,

withb∈F2^e[x] monic irreducible of degreeβ, gcd(N, b) = 1, DegN <(2n−1)β.

Lemma 31. — Let K/k be defined by (15)with n>2. Consider the CFE ofy.

(1) There exists k>1 such that Ak+1= [b, y].

(2) Set Rk =ν1(eθk+1). The regulator isrx= 2(Rk+ Degb).

Proof. — The conditionn>2 tells us that β < g+ 1 =nβ. The finite place (b) is the only ramified place and we denote by b the place above it. Since b divides the norm ofy, B = [b, y] is a primitive ideal. It is reduced because DegB=β 6g and ambiguous. Moreover,B² is a principal ideal:

[b, y]²= [b², by, y²] = (b)[b, y, bⁿ⁻¹y+aN] = (b)[1, y]

(use gcd(b, N) = 1) and thus B is principal, since the ideal class number is odd.

Applying Lemma 26, there existsk6ρ−1 such that Ak+1 = [b, y] = (eθk+1)Ox and by (14)

div(θek+1) =Dk+Rk∞1−(Rk−1+g+ 1)∞2=Dk+Rk∞1−(Rk−Degak+g+ 1)∞2,

where degDk =g+ 1−Degak =β, sinceak =bd/bc=bⁿ⁻¹. In factDk =b(see Corollary 28). We obtain that

div eθ²_k+1 b

= 2(Rk+β)(∞1− ∞2),

thusrx|2(Rk+β). Using the formulae giving the dataPi andQi of a CFE, one sees that Pk+1 =Pk−1. Then it can be shown that the quasi-periodρ is even and that ε=θeρ+1=θe²_k+1/b, thusrx= 2(Rk+β) (see [37, Th. 7, 8]). We could say more about the regulator but we do not want to pursue here.

4.1. CaseN = 1. — First, we compute the regulator of a real quadratic extension K=F2^e(x, y)/F2^e(x) having a normal equation (3) withN = 1.

Proposition 32. — Letq= 2^eand letK/kbe the real quadratic extension with normal equation (3), with

B(x) = Yr i=1

Bi(x)ⁿⁱ, C(x) =a Yr i=1

Bi(x), a∈F^∗_q. Then the regulator is rx=Pr

i=1(2ni−1) DegBi. Proof. — Notice thatm=Pr

i=1DegBi<2(g+ 1) =Pr

i=12niDegBi.

– Ifm>g+1, obviouslym=g+1,ni= 1 for all 16i6randaB(x) =C(x). yis not reduced and its polynomial part is (see Lemma 17)d=byc=Pg+1

j=0cjx^j=B+a.

Then

Q−1=C=aB

A1= [1, y] Q0= 1 P0= 0 a0=B+a r0= 0 A2= [a², y+a] Q1=a² P1=a a1= _a¹2B

The regulator isrx= dega1=g+ 1 and the fundamental unit isε=θe2=y+a.

– If 16m < g+ 1,yis reduced and its polynomial part isd=B. The CFE ofyis Q−1=C

A1= [1, y] Q0= 1 P0= 0 a0=B r0= 0 A2= [C, y] Q1=C P1= 0 a1= ¹_aQt

r=1Br(x)^nr−1 r1= 0 A3= [1, y] Q2= 1 P2= 0 a2=B

The period and quasi-period equal 2 and the regulator is rx = dega1+ dega2 = 2(g+ 1)−m. Moreover, the fundamental unit isε=θe3=_a¹y+ 1.

In the case whereN = 1, we see that the regulator grows slower with the genus than the divisor class number. Thus, there are only a finite number of real extensions such thathx= 1. To be more precise, we will use the following Lemma.

Lemma 33. — Let K/Fq be a function field of genusg>2. Denote byAi the number of effective divisors of degreei and byπi the reciprocal roots ofL(t), whereL(t)is the numerator polynomial of the zeta function of K/Fq. Then

(16)

g−2

X

i=0

Ai+

g−1

X

i=0

q^g−1−iAi=h Xg i=1

1

|1−πi|² 6h(g+ 1)(q+ 1)−A1

(q−1)² . In particular, we have the following lower bounds for the divisor class number h

(17) h> q^g−1(q−1)²

(g+ 1)(q+ 1)−A1

(18) h> (q−1)²(q^g−1+ 1) +A1(q−1)(q^g−1−1) (g+ 1)(q+ 1)−A1

.

Proof (cf.[18, Th. 1 and Th. 2]). — Notice thatA0= 1,A1 is the number of degree one places and Ai > A1 for all i > 0. Moreover (g+ 1)(q+ 1)−A1 > 0 by the Hasse-Weil bound. Set ΣK =Pg−2

i=0 Ai+Pg−1

i=0 q^g−1−iAi. The first lower bound is obtained from

ΣK >q^g−1.

In caseg= 1, one hash=A1, but the bound (17) is also true.

The second lower bound follows from ΣK>1 +q^g−1+A1

g−1

X

i=1

q^g−1−i= 1 +q^g−1+A1

q^g−1−1 q−1 .

Lemma 34. — LetK=F2^e(x, y), withy²+bⁿy+ab= 0, a∈(F2^e)^∗,b∈F2^e[x]monic irreducible of degree β andg+ 1 =nβ>2. The ideal class number is equal to one if and only if

– n= 1 and all finite places ofk of degree6g are inert inK,

– n >1 and all finite places ofkof degree6g are inert inK, except (b)which is ramified.

Proof. — Lemma 26 tells us that all reduced principal ideals are obtain in the CFE of y. In the proof of Proposition 32, we have seen that no non-zero reduced ideals are principal ifn= 1 and ifn >1 the only one is [b, y]. Then using Corollary 11 we obtain the result.

Theorem 35. — Letq= 2^e,K=Fq(x, y), withy²+bⁿy+ab= 0, a∈(Fq)^∗,b∈Fq[x]

monic irreducible of degree β and g+ 1 = nβ >2. The regulator is equal to rx = (2n−1)β. The only real quadratic extensions K/Fq(x) such that hx = 1 are (up to