On L-Functions of Cyclotomic Function Fields
Bruno Angl\` es
Universit\'e de Caen
Laboratoire Nicolas Oresme
CNRS UMR 6139
Campus II
Boulevard Mar\'echal Juin
BP 5186
14032 Caen Cedex
France.
We study two criterions of cyclicity for divisor class groups of function fields, the first one involves Artin L-functions and the second one involves ''affine'' class groups. We show that, in general, these two criterions are not linked.
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\title{On L-Functions of Cyclotomic Function Fields}

\author{
Bruno Angl\`  es \\
Universit\'e de Caen, \\
Laboratoire Nicolas Oresme, CNRS UMR 6139, \\
Campus II, Boulevard Mar\'echal Juin, \\
BP 5186, 14032 Caen Cedex, France.\\
E-mail: angles@math.unicaen.fr}
\date{}


\maketitle
\begin{abstract} We study two criterions of cyclicity for divisor class groups of function fields, the first one involves Artin L-functions
and the second one involves ''affine'' class groups. We show that, in general, these two criterions are not
linked.
\end{abstract}\par
 Let $P$ be a prime of $\mathbb F_q[T]$ of degree $d$ and let $K_P$ be the $P$th cyclotomic function field. In this paper
we study the relation between the $p$-part of $Cl^0(K_P)$ and the zeta function of $K_P,$ where $p$ is the characteristic of $\mathbb
F_q.$\par Let $\chi$ be an even character of the Galois group of $K_P/\mathbb F_q(T),$ $\chi \not =1.$ Let $g(X,\overline{\chi })$ be the 
''congruent to one modulo $p$'' part of the L-function of $K_P/\mathbb F_q(T)$ associated to the character $\overline{\chi}.$ We have two
criterions of cyclicity (\cite{GOS}, chapter 8): if $\rm{deg}_Xg(X,\overline{\chi })\leq 1$ then $Cl^0(K_P)_p(\chi )$ is a cyclic $\mathbb
Z_{p}[\mu_{q^d-1}]$-module, and if $Cl(O_{K_P})_p(\chi)=\{ 0\}$ then  $Cl^0(K_P)_p(\chi )$ is a cyclic $\mathbb
Z_{p}[\mu_{q^d-1}]$-module. David Goss has obtained   that if $Cl(O_{K_P})_p(\chi)$ is trivial  
then 
$g(X,\overline{\chi })$ is of degree at most one  (\cite{GOS}, Theorem 8.21.2).  Unfortunately,
there is a gap in the proof of this result. In fact,  we show that in general  
$Cl(O_{K_P})_p(\chi)=\{ 0\}$ does not imply  $\rm{deg}_Xg(X,\overline{\chi })\leq 1$ (Proposition
\ref{Proposition3}). We also prove that if $i$ is a $q$-magic number and if $\omega_P$ is the
Teichm\"uller character at
$P,$ then
$g(X,\omega_P^i)$ has simple roots when $i\equiv 0\pmod{q-1}$ (Proposition
\ref{Proposition7}).\par Note that  Goss conjectures that if $i$ is a $q$-magic number then
$\rm{deg}_Xg(X,\omega_P^i)\leq 1.$ This problem is still open and can be viewed as an analogue of
Vandiver's Conjecture for function fields (see section 5).\par
${}$\par
The author thanks David Goss (the proof of Lemma \ref{Lemma7} was communicated to the author by David Goss) and Philippe Satg\'e for several
fruitfull discussions.\par

\section{Notations}\par
${}$\par
Let $\mathbb F_q$ be a finite field having $q$ elements, $q=p^s$ where
$p$ is the characteristic of $\mathbb F_q.$ Let $T$ be an indeterminate
over $\mathbb F_q$ and set $A=\mathbb F_q[T],$ $k=\mathbb F_q(T).$ We
denote the set of monic elements of $A$ by $A^+.$ A prime of $A$ is a
monic irreducible polynomial in $A.$ We fix $\overline{k}$ an algebraic
closure of $k.$ We denote the unique place of $k$ which is a pole of $T$
by $\infty .$\par
${}$\par
Let $L/k$ be a finite geometric  extension of $k,$ $L\subset \overline{k}.$ We
set:\par
\noindent - $O_L:$ the integral closure of $A$ in $L,$\par
\noindent - $O_L^*:$ the group of units of $O_L,$\par
\noindent - $S_{\infty}(L):$ the set of places of $L$ above $\infty ,$\par
\noindent - $Cl^0(L):$ the group of divisors of degree zero of $L$ modulo
the group of principal divisors,\par
\noindent - $Cl(O_L):$ the ideal class group of $O_L,$\par
\noindent - $R(L):$ the groupe of divisors of degree zero with supports in $S_{\infty}(L)$ modulo the group of principal divisors
with supports in $S_{\infty}(L).$ \par
If $d$ is the greatest common divisor of the degrees of the elements in $S_{\infty}(L),$ we have the following exact sequence:
$$0\rightarrow R(L)\rightarrow  Cl^0(L)\rightarrow Cl (O_L) \rightarrow \frac{\mathbb Z}{d\mathbb Z}\rightarrow 0\, .$$
\par

Let $P$ be a prime of $A$ of degree $d.$ We denote the $P$th cyclotomic
function field by $K_P$ (see \cite{GOS}, chapter 7, and \cite{HAY}).
Recall that $K_P/k$ is the maximal abelian extension of $k$ contained in
$\overline{k}$ such that:\par
\noindent - $K_P/k$ is unramified outside of $P, \infty ,$\par
\noindent - $K_P/k$ is tamely ramified at $P,\infty ,$\par
\noindent - for every place $v$ of $K_P$ above $\infty ,$ the completion
of $K_P$ at $v$ is equal to $\mathbb
F_q((\frac{1}{T}))(^{q-1}\sqrt{-T}).$\par
We recall that $\rm{Gal}(K_P/k)\simeq (A/PA)^*,$ and that the
decomposition group of $\infty$ in $K_P/k$ is equal to its inertia group
and is isomorphic to $\mathbb F_q^*.$\par
${}$\par
Let $E/\mathbb F_q$ be a global function field and let $F/E$ be a finite
geometric abelian extension. Set $G=\rm{Gal}(F/E)$ and $\widehat
{G}=\rm{Hom}(G,\mathbb C^*).$\par
 Let $\chi \in \widehat {G},$ $\chi \not
=1,$ we set:
$$L(X,\chi )=\prod_{v\, \rm{place\, of}\, E}(1-\chi
(v)X^{\rm{deg}v})^{-1},$$
Where $\chi (v)=0$ if $v$ is ramified in $F^{\rm{Ker}(\chi)}/E,$ and if
$v$ is unramified in $F^{\rm{Ker}(\chi)}/E,$ $\chi(v)=\chi
((v,F^{\rm{Ker}(\chi)}/E)),$ where  $(.,F^{\rm{Ker}(\chi)}/E)$ is the
global reciprocity map. If $\chi =1,$ we set $L(X,\chi )=L_E(X)$ where
$L_E(X)$ is the numerator of the zeta function of $E.$\par
Therefore, if $L_F(X)$ is the numerator of the zeta function of $F,$ we
get:
$$L_F(X)=\prod_{\chi \in \widehat{G}}L(X,\chi )\, .$$\par
Let $\Delta $ be a finite abelian group and let $M$ be a $\Delta $-module. Let $\ell$ be a prim�e
number such that $\mid \Delta \mid \not \equiv 0\pmod{\ell }.$ We fix an embedding of
$\overline{\mathbb Q}$ in $\overline{\mathbb Q_{\ell}}.$  Let
$W=\mathbb Z_{\ell }[\mu_{\mid \Delta \mid}].$ For $\chi \in \widehat{\Delta},$ we set:
$$e_{\chi}=\frac{1}{\mid \Delta \mid}\sum_{\delta \in \Delta}\chi(\delta)\delta^{-1}\in
W[\Delta],$$
and:
$$M_{\ell}(\chi )=e_{\chi}(M\otimes_{\mathbb Z}W).$$
Thus, we have:
$$ M\otimes_{\mathbb Z}W =\bigoplus_{\chi \in \widehat{\Delta}}M_{\ell}(\chi )\, .$$

\section{Cyclotomic Function Fields and Artin-Schreier Extensions}
${}$\par
Let $Q$ be a prime of $A$ of degree $n,$ write $Q(T)=T^n+\alpha T^{n-1}+\cdots ,$ $\alpha \in \mathbb F_q.$ We
set: $i(Q)=Tr_{\mathbb F_q/\mathbb F_p}(\alpha ).$ Let $a\in A,$ $a\not =0,$ we set:
$$i(a)=\sum_{Q\, \rm{prime \, of}\, A}v_Q(a)i(Q)\in \mathbb F_p,$$
where $v_Q$ is the normalized $Q$-adic valuation on $k.$\par
${}$\par
Let $\theta \in\overline{k}$ such that $\theta^p-\theta =T.$ Set $\widetilde {A}=\mathbb F_q[\theta ],$
$\widetilde{k}=\mathbb F_q(\theta)$ and $G=\rm{Gal}(\widetilde {k}/k).$ Note that $\widetilde{k}/k$ is
unramified outside $\infty $ and totally ramified at $\infty .$ Let $\widetilde{\infty }$ be the unique place
of $\widetilde {k}$ above $\infty .$\par
\newtheorem{Lemma1}{Lemma}[section]
\begin{Lemma1} \label{Lemma1}
Let $(.,\widetilde {k}/k)$ be the usual Artin symbol. For $a\in A\setminus \{ 0 \} :$
$$(a,\widetilde{k}/k)(\theta )=\theta -i(a).$$
\end{Lemma1}
\noindent{\sl Proof} By the classical properties of the Artin symbol, it is enough to prove the Lemma when $a$
is a prime of $A.$ Thus, let $P$ be a prime of $A$ of degree $d.$ We have:
$$(P,\widetilde{k}/k)(\theta )\equiv \theta ^{q^d}\pmod{P}.$$
But, for $n\geq 0,$ we have:
$$\theta^{p^n}=\theta +T+T^p+\cdots +T^{p^{n-1}}.$$
Therefore:
$$\theta^{q^d}\equiv \theta -i(P)\pmod{P}.$$
The Lemma follows. $\diamondsuit$\par
\newtheorem{Lemma2}[Lemma1]{Lemma}
\begin{Lemma2} \label{Lemma2}
Let $P$ be a prime of $A$ of degree $d$ such that $i(P)\not =0.$ Then $P$ is a prime of $\widetilde {A}$ of
degree $pd.$ Let $\widetilde{K_P}$ be the $P$th cyclotomic function field for the ring $\widetilde{A},$ then
$K_P\subset \widetilde{K_P}.$
\end{Lemma2}
\noindent{\sl Proof}  We have $-T=-\theta^p(1-\theta^{1-p}).$ Note that:
$$1-\theta^{1-p}\in \mathbb (F_q((\frac{1}{\theta }))^*)^{q-1}.$$
Therefore:
$$^{q-1}\sqrt{-T}\in F_q((\frac{1}{\theta }))(^{q-1}\sqrt{-\theta }).$$
Thus:\par
\noindent - $\widetilde{k}K_P/\widetilde {k}$ is unramified outside $P, \widetilde{\infty },$\par
\noindent - $\widetilde{k}K_P/\widetilde {k}$ is tamely ramified at $P, \widetilde {\infty},$ \par
\noindent - for every place $w$ of $\widetilde{k}K_P$ above $\widetilde{\infty },$ the completion of 
$\widetilde{k}K_P$  at $w$ is contained in  $F_q((\frac{1}{\theta }))(^{q-1}\sqrt{-\theta }).$\par
The Lemma follows by class field theory. $\diamondsuit$\par
${}$\par
Let $P$ be a prime of $A,$ $\rm{deg}_TP(T)=d$ and $i(P)\not =0.$ Let $L=\widetilde{k}K_P\subset
\widetilde{K_P}.$ Let $\Delta=\rm{Gal}(K_P/k)\simeq \rm{Gal}(L/\widetilde{k}).$ We have an isomorphism
compatible to class field theory: $\widehat {\Delta}\rightarrow \widehat {\rm{Gal}(L/\widetilde {k})},$ $\chi
\mapsto \widetilde {\chi}=\chi \circ N_{\widetilde{k}/k}.$ We fix $\zeta_p\in \overline{\mathbb Q}$ a primitive
$p$th root of unity.
\newtheorem{Lemma3}[Lemma1]{Lemma}
\begin{Lemma3}\label{Lemma3}
${}$ \par
\noindent (1) Let $\chi \in \widehat{\Delta},$ $\chi \not =1.$ Let $L(X,\widetilde{\chi})$ be the Artin L-function
relative to $L/\widetilde{k}$ and to the character $\widetilde{\chi }.$ We have:
$$L(X,\widetilde{\chi })=\prod_{\phi \in \widehat{G}}L(X,\phi \chi ),$$
where $L(X,\phi \chi)$ is the Artin L-function relative to $L/k$ and the character $\phi \chi .$\par
\noindent (2) Let $\chi \in \widehat{\Delta},$ $\chi \not =1,$ $\chi$ even (i.e. $\chi (\mathbb F_q^*)=\{ 1\}$). Then:
$$\frac{L(X,\widetilde{\chi})}{L(X,\chi)}\equiv (1-X)^{p-1}L(X,\chi)^{p-1}\pmod{(1-\zeta_p)}.$$
\end{Lemma3}
\noindent{\sl Proof} Te assertion (1) is a consequence of the usual properties of Artin L-functions. Now, let $\phi \in
\widehat{G},$ $\phi \not =1.$ Since $\phi \chi$ is ramified at $\infty,$ we get:
$$L(X,\phi \chi )=\sum_{n\geq 0}(\sum_{a\in A^+,\, \rm{deg}(a)=n}\phi (a)\chi(a))X^n.$$
Thus:
$$L(X,\phi \chi)\equiv \sum_{n\geq 0}(\sum_{a\in A^+,\, \rm{deg}(a)=n}\chi (a)))X^n\pmod{(1-\zeta_p)}.$$
But, since $\chi$ is even, we have $\chi (\infty )=1.$  Therefore:
$$L(X,\phi \chi)\equiv (1-X)L(X,\chi)\pmod{(1-\zeta_p)}.$$
The Lemma follows. $\diamondsuit$\par
let $i\in\mathbb F_p$ and let $\sigma_i\in G$ such that $\sigma_i(\theta )=\theta -i.$ Let $\psi \in \widehat {G}$ given
by $\psi(\sigma_i)=\zeta_p^i.$
\newtheorem{Lemma4}[Lemma1]{Lemma}
\begin{Lemma4}\label{Lemma4}
Let $\chi \in \widehat{\Delta},$ $\chi$ even and non-trivial.\par
\noindent (1) Let $\phi \in \widehat{G},$ $\phi \not =1.$ Let $\sigma \in \rm{Gal}(\mathbb Q(\zeta_p)/\mathbb
Q)$ such that $\phi =\psi^{\sigma}.$ Then:
$$L(X,\phi \chi)=L(X,\psi \chi )^{\sigma}.$$
Furthermore $\rm{deg}_XL(X,\phi \chi )=d.$\par
\noindent (2) We have:
$$L(1,\psi \chi)\equiv (\sum_{a\in A^+, \rm{deg}(a)\leq d}i(a)\chi (a))(\zeta_p-1)\pmod{(1-\zeta_p)^2}.$$
\end{Lemma4}
\noindent{\sl Proof} Let $\mathbb Q (\chi )$ be the abelian extension of $\mathbb Q$ obtained by adjoining to
$\mathbb Q$ the values of $\chi .$ Let $\mathbb Z[\chi ]$ be the ring of integers of $\mathbb Q( \chi ).$ Note 
that $p$ is unramified in $\mathbb Q(\chi )$ and:
$$\rm{Gal}(\mathbb Q (\chi )(\zeta_p)/\mathbb Q(\chi ))\simeq \rm{Gal}(\mathbb Q(\zeta_p)/\mathbb Q).$$
Since $L(X,\phi \chi)$ is a polynomial in $\mathbb Z[\chi ][\zeta_p][X],$ we have:
$$L(X,\phi \chi)=L(X,\psi \chi )^{\sigma}.$$
Since $\chi$ and $\widetilde \chi$ are non-trivial even characters, we have:
$$\rm{deg}_XL(X,\widetilde {\chi})=pd-2,$$
and:
$$\rm{deg}_XL(X,\chi)=d-2.$$
Therefore $\rm{deg}_XL(X,\phi \chi)=d.$\par
Now, we have:
$$L(X,\psi \chi)=\sum_{n=0}^d(\sum_{a\in A^+\, \rm{deg}(a)=n}\zeta_p^{i(a)}\chi (a) )X^n.$$
But recall that:
$$\zeta_p^{i(a)}\equiv 1+i(a)(\zeta_p-1)\pmod{(1-\zeta_p)^2}.$$
Thus, since $\chi$ is even and  non-trivial, we get:
$$L(X,\psi \chi) \equiv L(X,\chi )(1-X) +(\zeta_p -1)(\sum_{n=1}^d(\sum_{a\in A^+\, \rm{deg}(a)=n}i(a)\chi (a))
X^n) \pmod{(1-\zeta_p)^2}.$$
The Lemma follows. $\diamondsuit$\par
We are now ready to prove the main result of this section:
\newtheorem{Proposition1}[Lemma1]{Proposition}
\begin{Proposition1}  \label{Proposition1}
Let $\chi \in \widehat {\Delta},$ $\chi \not =1,$ $\chi$ even. Let $W=\mathbb Z_p[\mu_{q^d-1}].$  We have:
$$\rm{Long}_W(\frac{Cl(O_L)_p(\widetilde {\chi})}{Cl(O_{K_P})_p(\chi )})\geq 1 \Leftrightarrow \sum_{a\in A^+\, \rm{deg}(a)\leq d}
i(a)\overline{\chi }(a) \equiv 0\pmod{p}.$$
\end{Proposition1}
\noindent{\sl Proof} Fix $\tau$ a generator of $G\simeq \rm{Gal}(L/K_P).$ Let $\varepsilon \in O_L^*.$ Since $L/K_P$ is totally
ramified at any prime above $\infty,$ there exists $\zeta \in \mathbb F_q^*$ such that $\tau (\varepsilon) =\zeta \varepsilon .$
But $\tau^p(\varepsilon)=\zeta^p\varepsilon =\varepsilon.$ Since we are in characteristic $p,$ we deduce that $\varepsilon \in
O_{K_P}^*.$ Therefore:
$$O_L^*=O_{K_P}^*.$$
Let $I$ be an ideal of $O_{K_P}$ such that $IO_L=\alpha O_L$ for some $\alpha \in O_L.$ Then, there exists $\varepsilon \in O_L^*$
 such that $\tau (\alpha )= \varepsilon \alpha .$ Since $O_L^*=O_{K_P}^*$ and since $\tau$ is of order $p,$ we deduce that $\alpha
\in O_{K_P}.$ This implies that:
$$Cl(O_{K_P})\hookrightarrow Cl(O_L).$$
One can also show that:
$$Cl^0(K_P)\hookrightarrow Cl^0(L).$$
Set $\Delta ^+=\frac{\Delta }{\mathbb F_q^*}.$  Let $\cal I$ be the augmentation ideal of $\mathbb F_p[\Delta^+].$ One sees that 
we have the following isomorphism of $\Delta $-modules:
$$\frac{R(L)}{R(K_P)}\otimes_{\mathbb Z}\mathbb Z_p\simeq \cal I.$$
This implie that we have the following exact sequence of $W$-modules:
$$0\rightarrow \frac{W}{pW}\rightarrow \frac{Cl^0(L)_p(\widetilde {\chi })}{Cl^0(K_P)_p(\chi )}\rightarrow 
\frac{Cl(O_L)_p(\widetilde{\chi })}{Cl(O_{K_P})_p(\chi )}\rightarrow 0\, .$$
Now, by the results of Goss and Sinnott (\cite{GOS&SIN}):
$$\rm{Long}_WCl^0(L)_p(\widetilde{\chi })= v_p(L(1,\overline{\widetilde{\chi }})),$$
and
$$\rm{Long}_WCl^0(K_P)_p(\chi )=v_p(L(1,\overline{\chi })).$$
Thus by Lemma \ref{Lemma3}:
$$\rm{Long}_W(\frac{Cl(O_L)_p(\widetilde {\chi})}{Cl(O_{K_P})_p(\chi )})=(p-1)v_p(L(1,\psi \overline{\chi }))-1.$$
It remains to apply Lemma \ref{Lemma4}. $\diamondsuit$\par
\section{Derivatives of L-functions}
${}$\par
Let $P$ be a prime of $A$ of degree $d.$ We fix an embedding of $\overline{\mathbb Q}$ in $\overline{\mathbb Q_p}.$ Set $\Delta
=\rm{Gal}(K_P/k)$ anf
$W=\mathbb Z_p[\mu_{q^d-1}].$ We fix an isomorphism
$\Phi_P:\, A/PA\rightarrow W/pW.$ Then $\Phi_P$ induces an isomorphism:
$$\omega_P:\, \Delta \rightarrow \mu_{q^d-1}\subset W^*.$$
The morphism $\omega_P$ is called  ''the'' Teichm\" uller character at $P.$
Note that $\widehat{\Delta }$ is a cyclic group and $\omega_P$ is a generator of this group.\par
Let $i\in \mathbb N,$ set:\par
\noindent - $\beta (0)=1,$\par
\noindent - $\beta (i)=\sum_{a\in A^+}a^i$ if $i\geq 1,$ $i\not \equiv 0\pmod{q-1},$\par
\noindent - $\beta (i)=-\sum_{a\in A^+}\rm{deg}(a)a^i$if $i\geq 1,$ $i\equiv 0\pmod{q-1}.$\par
\noindent One can prove that for all $i\in \mathbb N,$ $\beta (i) \in A.$ We also see that:
$$\forall i\in \mathbb N,\, 0\leq i\leq q^d-2,\, \Phi_P(\beta (i))\equiv L(1, \omega_P^i) \pmod{p}.$$
Therefore, if $1\leq i \leq q^d-2,$ by the results of Goss and Sinnott (\cite{GOS&SIN}), we have:
$$\rm{Long}_WCl^0(K_P)_p(\omega_P^{-i})\geq 1 \Leftrightarrow \beta(i) \equiv 0\pmod{P}.$$
The numbers $\beta(i)$ are called the Bernoulli-Goss polynomials.\par
 Recall that we have a surjective morphism of $\Delta$-modules:
$$W[\Delta^+]\rightarrow R(K_P)\otimes_{\mathbb Z}W,$$
where $\Delta^+=\Delta /\mathbb F_q^*.$ Thus for $\chi \in \widehat{\Delta},$ $\chi$ even, $R(K_P)_p(\chi)$ is a cyclic $W$-module.
But,  for such a character, we have the exact sequence of $W$-modules:
$$0\rightarrow R(K_P)_p(\chi )\rightarrow Cl^0(K_P)_p(\chi) \rightarrow Cl(O_{K_P})_p(\chi )\rightarrow 0\, .$$
This implies that, if $Cl(O_{K_P})_p(\chi )=\{ 0\},$ $Cl^0(K_P)_p(\chi )$ is a cyclic $W$-module.\par
David Goss has shown (\cite{GOS}, Corollary 8.16.2) that for  $\chi$ is even, $\chi \not =1,$  if $L'(1,\overline{\chi })\not
\equiv 0 \pmod{p}$ (here $L'(1,\overline{\chi})$ is the derivative of $L(X,\overline{\chi })$ taken at $X=1$), then
$Cl^0(K_P)_p(\chi )$ is a cyclic $W$-module.\par
Therefore a natural question arise. Let $\chi \in \widehat{\Delta},$ $\chi \not =1,$ $\chi$ even. Assume that $L(1,\overline{\chi
})\equiv 0 \pmod{p}.$ Do we have: 
$$Cl(O_{K_P})_p(\chi)=\{ 0\} \Rightarrow L'(1, \overline{\chi })\not \equiv 0 \pmod{p}?$$
Our aim in this section is to show that in general the answer is no.\par
${}$\par
Let $d$ be an integer, $d\geq 1.$ For $i\in \{ 1,\cdots, q^d-2\},$ we set:
$$\gamma (d,i)=\sum_{a\in A^+,\, \rm{deg}(a)\leq d}i(a)a^i.$$
\newtheorem{Lemma5}{Lemma}[section]
\begin{Lemma5} \label{Lemma5}
Let $\tau \in \rm{Gal}(\mathbb F_q(T)/\mathbb F_q(T^p-T))$ such that $\tau (T)=T+1.$ Let $i\in \{ 1,\cdots ,q^d-2\},$ $i\equiv 0\pmod{q-1}.$
Recall that $q=p^s.$ We have:
$$\tau(\gamma (d,i))=\gamma(d,i)+s\beta (i).$$
\end{Lemma5}
\noindent{\sl Proof} Let $Q$ be a prime of $A$ of degree $n.$ Write $Q=T^n+\alpha T^{n-1}+\cdots,$ where $\alpha \in \mathbb F_q.$ Then $\tau
(Q)=T^n+(\alpha +n)T^{n-1}+\cdots$ . Therefore $i(\tau (Q))=i(Q)+s\rm{deg}(Q).$ This implies that:
$$\forall a\in A\setminus \{ 0\},\, i(\tau (a))=i(a)+s\rm{deg}(a).$$
Now:
$$\tau (\gamma (d,i))=\sum_{a\in A^+,\, \rm{deg}(a)\leq d}i(a) \tau (a)^i.$$
Therefore:
$$\tau (\gamma (d,i))=\sum_{a\in A^+,\, \rm{deg}(a)\leq d}(i(\tau(a))-s\rm{deg}(a))\tau (a)^i.$$
Thus:
$$\tau(\gamma(d,i))=\sum_{a\in A^+,\, \rm{deg}(a)\leq d}i(\tau (a))\tau (a)^i -s\sum_{a\in A^+,\, \rm{deg}(a)\leq d}\rm{deg}(\tau (a)) \tau
(a)^i.$$
Observe that $\sum_{a\in A^+,\, \rm{deg}(a)\leq d}i(\tau (a))\tau (a)^i =\gamma (d,i)$ and $-\sum_{a\in A^+,\, \rm{deg}(a)\leq d}\rm{deg}(\tau
(a)) \tau (a)^i=\beta (i).$ $\diamondsuit$\par
\newtheorem{Proposition2}[Lemma5]{Proposition}
\begin{Proposition2} \label{Proposition2}
Let $P$ be a prime of $A$ of degree $d$ such that $i(P)\not =0.$ Set $Q(T)=P(T^p-T).$ Then $Q$ is a prime of $A$ of degree $pd.$ Let $i$ be
an integer such that $1\leq i\leq q^d-2,$ $i\equiv 0 \pmod{q-1}$ and $Cl(O_{K_P})_p(\omega_P^{-i})=\{ 0\} .$ Then:
$$\rm{Long}_W Cl(O_{K_Q})_p(\omega_Q^{-i(q^{pd}-1)/(q^d -1)})\geq  1 \Leftrightarrow \gamma (d,i)\equiv 0 \pmod{P}.$$
\end{Proposition2}
\noindent{\sl Proof} We have:
$$\Phi_P(\gamma (d,i))\equiv \sum_{a\in A^+,\, \rm{deg}(a)\leq d}i(a) \omega_P^i(a)\pmod{p}.$$
It remains to apply Proposition \ref{Proposition1}. $\diamondsuit$ \par
\newtheorem{Lemma6}[Lemma5]{Lemma}
\begin{Lemma6} \label{Lemma6}
Assume $p\not = 2.$ Let $d\geq 1$ be an integer.  There exists a prime $P$ in $A,$ $\rm{deg}(P)=d,$  such that
$i(P(T))i(P(T+1))\not =0.$
\end{Lemma6}
\noindent{\sl Proof} Let $Q$ be a prime of $A$ of degree $d$ such that $i(Q)\not = 0.$ Such a prime exists by the normal basis Theorem. Fix
$\overline{\mathbb F_q}$ an algebraix closure of $\mathbb F_q.$ We assume that $i(Q(T+1))=0.$ Write $Q=T^d+\alpha T^{d-1}+\cdots$ . Then 
$Tr_{\mathbb F_q/\mathbb F_p}(\alpha)=-sd.$ Therefore $sd\not \equiv 0 \pmod{p}.$Let $\theta \in\overline{\mathbb F_q}$ such that $Q(\theta
)=0.$ We observe that:
$$\forall \zeta \in \mathbb F_p,\, Tr_{\mathbb F_{q^d}/\mathbb F_p}(\zeta \theta )=-\zeta sd.$$
Since $p\geq 3,$ we can find $\zeta \in \mathbb F_p^*$ such that $-\zeta sd \not = -sd.$ Set $P(T)=\rm{Irr}(\zeta \theta , \mathbb F_q ; T).$
Then $P$ is a prime of degree $d$ such that $i(P)i(\tau (P))\not = 0.$ $\diamondsuit$ \par
\newtheorem{Proposition3}[Lemma5]{Proposition}
\begin{Proposition3} \label{Proposition3}
Assume that $p\not =2$ and $s\not \equiv 0\pmod{p}.$ Let $d$ be an integer, $d\geq 2,$ and let $P$ be a prime of degree $d$ such that
$i(P(T))i(P(T+1))\not =0.$ Set $Q(T)=P(T^p-T).$ Then:\par
\noindent - $L(1,\omega_Q^{-(q-1)(q^{pd}-1)/(q^d-1)})\equiv 0\pmod{p},$\par
\noindent - $L'(1,\omega_Q^{-(q-1)(q^{pd}-1)/(q^d-1)})\equiv 0\pmod{p},$\par
\noindent - $Cl(O_{K_Q})_p(\omega_Q^{-(q-1)(q^{pd}-1)/(q^d-1)})=\{ 0\}.$
\end{Proposition3}
\noindent{\sl Proof} Set $R=P(T+1)$ and $Z=R(T^p-T).$ We observe that we have an isomorphism:
$$Cl(O_{K_Q})_p(\omega_Q^{-(q-1)(q^{pd}-1)/(q^d-1)})\simeq Cl(O_{K_Z})_p(\omega_Z^{-(q-1)(q^{pd}-1)/(q^d-1)}).$$
Not also that $\beta (q-1)=1.$ Thus:
$$Cl(O_{K_P})_p(\omega_P^{-(q-1)})=Cl(O_{K_R})_p(\omega_R^{-(q-1)})=\{ 0\}.$$
We have:
$$L(1,\omega_Q^{-(q-1)(q^{pd}-1)/(q^d-1)})\equiv L(1,\omega_Z^{-(q-1)(q^{pd}-1)/(q^d-1)})\equiv 0\pmod{p}.$$
And, by Lemma \ref{Lemma3}, since $p\geq 3$: 
$$L'(1,\omega_Q^{-(q-1)(q^{pd}-1)/(q^d-1)})\equiv L'(1,\omega_Z^{-(q-1)(q^{pd}-1)/(q^d-1)})\equiv 0\pmod{p}.$$
Suppose that we have $Cl(O_{K_Q})_p(\omega_Q^{-(q-1)(q^{pd}-1)/(q^d-1)})\not =\{ 0\}.$ Then by Proposition\ref{Proposition2}:
$$\gamma(d,q-1)\equiv 0\pmod{P},$$
 and also:
$$\gamma (d,q-1)\equiv 0\pmod{R}.$$
Thus:
$$\tau (\gamma (d,q-1))\equiv 0\pmod {\tau (P)}.$$
Now, by Lemma \ref{Lemma5}, and the fact that $\tau (P)=R,$  we get:
$$\gamma(d,q-1)+s\beta (q-1)\equiv 0\pmod{R}.$$
Therefore we get $s\equiv 0\pmod{p}$ which is a contradiction. The Proposition follows. $\diamondsuit$ \par
\section{Cyclicity of Class Groups and L-Functions}
${}$\par
Let $E/\mathbb F_q$ be a global function field and let $F/E$ be a finite geometric abelian extension. Set $\Delta =\rm{Gal}(F/E).$ Let $\ell$
be a prime number. Let's recall some well-known facts about $L$-functions.\par
${}$\par
Set $T_{\ell }=\rm{Hom}(\mathbb Q_{\ell }/\mathbb Z_{\ell } , J)$ where  $J$ is the inductive limit of the $ Cl^0(\mathbb F_{q^n}F),$ $n\geq
1.$ We fix an embedding of $\overline{\mathbb Q}$ in $\overline{\mathbb Q_{\ell}}.$ Let $\gamma$ be the Frobenius of $\mathbb F_q.$ Then
$\gamma$ and $\Delta$ act on $T_{\ell }.$\par
If $\ell \not =p,$ we have (see \cite{ROS},chapter 15):
$$\rm{Det}(1-\gamma X\mid_{T_{\ell}})=L_F (X),$$
 where $L_F (X)$ is the numerator of the zeta function of $F.$\par
If $\ell =p,$ write $L_F(X)=\prod_{i}(1-\alpha_i X)$ and set $L_F^{nr}(X)=\prod_{v_p(\alpha_i)=0}(1-\alpha_i X).$ Then (see \cite{CRE} and
also \cite{GOS&SIN}):
$$\rm{Det}(1-\gamma X\mid_{T_{p}})=L_F^{nr} (X).$$\par
Now assume that $\ell$ does not divide the cardinal of $\Delta,$ then the above results are also valid character by character. More
precisely, if $\ell \not = p,$ we have:
$$\forall \chi \in \widehat{\Delta},\, \rm{Det}(1-\gamma X\mid_{T_{\ell}(\chi)})=L (X, \overline{\chi }).$$
If $\ell =p,$ for $\chi \in \widehat {\Delta},$ write $L(X,\chi)=\prod_i(1-\alpha_i(\chi )X)$ and set
$L^{nr}(X,\chi)=\prod_{v_p(\alpha_i(\chi )=0}(1-\alpha_i (\chi )X).$ Then:
$$\forall \chi \in \widehat{\Delta},\, \rm{Det}(1-\gamma X\mid_{T_{p}(\chi)})=L^{nr} (X, \overline{\chi }).$$\par
Now, let $\chi \in \widehat {\Delta},$ write:
$$L(X,\chi )=\prod_i(1-\alpha_i(\chi )X),$$
and set:
$$g(X,\chi)=\prod_{v_{\ell}(\alpha_i(\chi) -1)>0}(1-\alpha_i(\chi )X).$$
Set:
$$g(X)=\prod_{\chi \in \widehat{\Delta }}g(X, \chi ).$$
We also set:
$$\forall \chi \in \widehat {\Delta},\, H(X, \chi) =(1+X)^{\rm{deg}_X\, g(X,\chi )}g((1+X)^{-1},\chi ),$$
and:
$$H(X)=\prod_{\chi \in \widehat{\Delta }}H(X, \chi).$$\par
For $n\geq 0,$ set $F_n=\mathbb F_{q^{\ell ^n}}F,$ and let $A_n$ be the $\ell$-Sylow subgroup of $Cl^0(F_n).$ Let $F_{\infty}=\cup_{n\geq
0}F_n$ and let $A_{\infty}$ be the inductive limit of the $A_n,$ $n\geq 0.$ We set:
$$Y=\rm{Hom}(\mathbb Q_{\ell}/\mathbb Z_{\ell} , A_{\infty}).$$
Set $\Gamma=\rm{Gal}(F_{\infty}/F),$then $\gamma $ is a topological generator of $\Gamma \simeq \mathbb Z_{\ell}.$\par
\newtheorem{Lemma7}{Lemma}[section]
\begin{Lemma7} \label{Lemma7}
${}$\par
\noindent (1) For all $n\geq 0,$ we have an isomorphism of $\Delta$-modules:
$$\frac{Y}{(\gamma^{\ell^n}-1)Y}\simeq A_n.$$
(2) Assume $\mid \Delta \mid \not \equiv 0\pmod{\ell }.$Then, $\forall \chi \in \widehat{\Delta},$ $\forall n\geq 0,$ we have:
$$\frac{Y(\chi)}{(\gamma^{\ell^n}-1)Y(\chi)}\simeq A_n(\chi ).$$
\end{Lemma7}
\noindent{\sl Proof} We prove assertion (1), and note that (2) is a consequence of (1). Recall that $A_{\infty}$ is a divisible group (see
\cite{ROS}, Proposition 11.16). We start with the following exact sequence:
$$0\rightarrow A_n\rightarrow A_{\infty}\rightarrow A_{\infty}\rightarrow 0,$$
where the middle map is  the multiplication by $\gamma^{\ell^n}-1.$ We apply $\rm{Hom}(\mathbb Q_{\ell}/\mathbb Z_{\ell}, .)$ to this
sequence, we get:
$$0\rightarrow Y\rightarrow Y\rightarrow \rm{Ext}^1(\mathbb Q_{\ell}/\mathbb Z_{\ell}, A_n)\rightarrow 0.$$we also have the following exact
sequence:
$$0\rightarrow \mathbb Z_{\ell }\rightarrow \mathbb Q_{\ell}\rightarrow \frac{\mathbb Q_{\ell}}{\mathbb Z_{\ell}}\rightarrow 0.$$
We apply $\rm{Hom}(., A_n)$ to this last sequence, using the fact that:
$$\rm{Ext}^1(\mathbb Q_{\ell}, A_n)=\{ 0 \},$$
we get:
$$\rm{Hom}(\mathbb Z_{\ell },A_n)\simeq \rm{Ext}^1(\mathbb Q_{\ell}/\mathbb Z_{\ell} , A_n).$$
The Lemma follows. $\diamondsuit$\par
\newtheorem{Proposition4}[Lemma7]{Proposition}
\begin{Proposition4} \label{Proposition4}
${}$\par
 \noindent (1) Let $\Lambda =\mathbb Z_{\ell}[[X]]$ be the Iwasawa algebra of $\Gamma$ over $\mathbb Z_{\ell}$ where $X$ acts like $\gamma
-1.$ 
 Then $Y$ is a finitely generatyed $\Lambda$-module and a torsion $\Lambda$-module. The characteristic polynomial of the $\Lambda$-module $Y$
is equal to $H(X).$\par
\noindent (2) Assume that $\ell$ does not divide the cardinal of $\Delta.$ Let $\Lambda=W[[X]]$ be the Iwasawa algebra of $\Gamma$ over
$W=\mathbb Z_{\ell}[\mu_{\mid \Delta \mid}]$ where $X$ acts like $\gamma -1.$ Then, for $\chi \in \widehat{\Delta},$ $Y(\chi )$ is a finitely
generated $\Lambda$-module and a torsion $\Lambda$-module. The characteristic polynomial of tha $\Lambda$-module $Y$ is equal to
$H(X,\overline{\chi }).$
\end{Proposition4}
\noindent{\sl Proof} We prove (1), the proof of (2) is essentially similar. For all $n\geq 0,$ we set $\omega_n(X)=(1+X)^{\ell^n}-1.$ By
Lemma \ref{Lemma7}, we have:
$$\forall n\geq 0,\, \frac{Y}{\omega_n Y}\simeq A_n.$$
Therefore $Y$ is a finitely generated $\Lambda$-module and a torsion $\Lambda$-module. Let $r\in \mathbb N$ such that we have an isomorphism
of groups:
$$Y\simeq \mathbb Z_{\ell}^r.$$
Then, there exists a constant $\nu \in \mathbb Z,$ such that, for all $n$ sufficiently large:
$$\mid \frac{Y}{\omega_n Y} \mid =\ell^{rn+\nu}.$$
But, for all $n\geq 0,$ we have:
$$\mid A_n\mid =\ell^{v_{\ell}(L_{F_n}(1))}.$$
Therefore, there exists a constant $\nu'\in \mathbb Z$ such that, for all $n$ sufficiently large:
$$\mid A_n\mid = \ell ^{\rm{deg}_X H(X) n+\nu'}.$$
Thus: $r=\rm{deg}_X H(X).$ But let $V(X)$ be the characteristic polynomial of the $\Lambda$-module $Y.$ We know that $r=\rm{deg}_X V(X),$ and
we also know that $V(X)$ divides $(1+X)^{\rm{deg} L_F(X)}L_F((1+X)^{-1}).$ But $V(X)$ is a distinguished polynomial, thus $V(X)$ divides
$H(X).$ The Proposition follows. $\diamondsuit$\par
\newtheorem{Proposition5}[Lemma7]{Proposition}
\begin{Proposition5} \label{Proposition5}
${}$\par
\noindent (1) If $A_0$ is a cyclic $\mathbb Z_{\ell}$-module then $g(X)$ has simple roots.\par
\noindent (2) Assume that $\mid \Delta\mid \not \equiv 0\pmod{\ell }.$ Let $\chi \in \widehat{\Delta}.$ If $A_0(\chi )$ is a cyclic
$W$-module then $g(X,\overline{\chi })$ has simple roots.
\end{Proposition5}
\noindent{\sl Proof} We prove (1). By Nakayama's Lemma, $Y$ is pseudo-isomorphic to $\Lambda/H(X)\Lambda .$ But, by a result of Tate
(\cite{TAT}), we know that the action of $\gamma$ on $Y$ is semi-simple. This implies that $H(X)$ has simple roots. $\diamondsuit$\par
${}$\par
Let's give an application of this last Proposition.
\newtheorem{Proposition6}[Lemma7]{Proposition}
\begin{Proposition6} \label{Proposition6}
We assume that $q\geq 5.$ Let $E/\mathbb F_q(T)$ be a real quadratic field, i.e. $[E:\mathbb F_q(T)]=2$ and $\infty$ splits completely in
$E.$  If $O_E$ is a principal ideal domain then $L_E(X)$ has simple roots.
\end{Proposition6}
\noindent{\sl Proof} Let $g$ be the genus of $E$ and write:
$$L_E(X)=\prod_{i=1}^{2g}(1-\alpha_iX).$$
Let $K=\mathbb Q(\alpha_1 ,\cdots ,\alpha_{2g}),$ then $K$ is a CM-field. Let $\alpha \in \{
\alpha_1 ,\cdots \alpha_{2g} \}.$ Then:
$$(1-\alpha )(1-\overline{\alpha })\geq q+1-2\sqrt{q}>1.$$
Therefore:
$$N_{K/\mathbb Q}(1-\alpha )>1.$$
Thus $1-\alpha$ is not a unit of $K.$ Let $\infty_1$ and $\infty_2$ be tha places of $E$ above $\infty.$ Then $R(E)$ is a quotient of
$\mathbb Z (\infty_1-\infty_2)$ and we have an exact sequence:
$$0\rightarrow R(E) \rightarrow Cl^0(E)\rightarrow Cl(O_E)\rightarrow 0\, .$$
Therefore, if $O_E$ is a principal ideal domain then $Cl^0(E)$ is a cyclic group. It remains to apply Proposition \ref{Proposition5}.
$\diamondsuit$ \par
It is conjectured that there exists infinitely many real quadratic function fields $E/\mathbb F_q(T)$ such that $O_E$ is a principal ideal
domain. In view of this conjecture, it will be interesting to prove that there exists infinitely many real quadratic function fields
$E/\mathbb F_q(T)$ such that $L_E(X)$ has simple roots.\par
\section{A Conjecture of Goss}
${}$\par
Set $D_0=1$ and for $i\geq 1,$ $D_i=(T^{q^i}-T)D_{i-1}^q.$ The Carlitz exponential is defined by:
$$Exp(X)=\sum_{i\geq 0}\frac{X^{q^i}}{D_i}\in k[[X]].$$
Let $n\in\mathbb N,$ write $n=a_0+a_1q+\cdots +a_r q^r,$ where $a_0,\cdots , a_r\in \{ 0,\cdots ,q-1 \}.$  We set:
$$\Gamma_n=\prod_{i=0}^r D_i ^{a_i}.$$
The $i$th Bernoulli-Carlitz number, $B(i)\in k,$ is defined by:
$$\frac{X}{Exp(X)}=\sum_{i\geq 0}\frac{B(i)}{\Gamma_i}X^i.$$
Let $P$ be a prime of $A$ of degree $d$ and let $i\in\{ 1, \cdots , q^d-2\} ,$ $i\equiv 0\pmod{q-1}.$ We have the following result
(\cite{OKA}):
$$Cl(O_{K_P})_p(\omega_P^i)\not = \{ 0\} \Rightarrow B(i)\equiv 0\pmod{P}.$$
We fix an embedding of $\overline{\mathbb Q}$ in $\overline{\mathbb Q_p}.$ Let $i\in \{ 1,\cdots , q^d-2\}.$ Write:
$$L(X,\omega_P^i)=\prod_j(1-\alpha_j(i)X),$$
and set:
$$g(X,\omega_P^i)=\prod_{v_p(\alpha_j(i)-1)>0}(1-\alpha_j(i)X).$$\par
Let $i\in \mathbb N.$ We say that $i$ is a $q$-magic number if there exist  $c\in \{ 0, \cdots , q-2\} $ and an integer $n\in \mathbb N$
such that $i=cq^n+q^n-1.$\par
\newtheorem{Proposition7}{Proposition}[section]
\begin{Proposition7} \label{Proposition7}
Let $P$ be a prime of $A$ of degree $d.$ Let $i$ be a $q$-magic number, $1\leq i\leq q^d-2,$ $i\equiv 0\pmod{q-1}.$ Then $g(X,\omega_P^i)$
has simple roots.
\end{Proposition7}
\noindent{\sl Proof} We have $i=q^n -1$ for some integer $n,$ $1\leq n \leq d-1.$ By a result of Carlitz (\cite{GOS}, Lemma 8.22.4):
$$B(q^d-1-i)=\frac{(-1)^{d-n}}{L_{d-n}^{q^n}},$$
where $L_0=1$ and for $j\geq 1,$ $L_j=(T^{q^j}-T)L_{j-1}.$ Therefore:
$$Cl(O_{K_P})_p(\omega^{-i})=\{ 0\} .$$
It remains to aplly Proposition \ref{Proposition5}. $\diamondsuit$\par
${}$\par
In \cite{GOS}, David Goss makes the following conjecture:\par
\noindent let $P$ be a prime of degree $d$ and let $i$ be a $q$-magic number, $1\leq i\leq q^d-2.$ Then $\rm{deg}_Xg(X,\omega_P^i)\leq 1.$
\par
It is natural to ask if there exist primes $P$ and $q$-magic numbers $i$, $ 1\leq i\leq q^{\rm{deg}P}-2,$ such that
$\rm{deg}_Xg(X,\omega_P^i)\geq 1.$ This is the case.\par
\newtheorem{Proposition8}[Proposition7]{Proposition}
\begin{Proposition8} \label{Proposition8}
Let $c\in \{ 0,\cdots ,q-2\}.$ There exist infinitely many primes $P$ such that:
$$\prod_{n=1}^{\rm{deg}P-1}\beta(cq^n+q^n-1)\equiv 0\pmod{P}.$$
\end{Proposition8}
\noindent{\sl Proof} We prove this Proposition for $c\not =0.$ The proof for $c=0$ is very similar. If we apply the results in \cite{SHE}, we
get:
$$\forall n\geq 0,\, \rm{deg}_T\, \beta(cq^n+q^n-1)=n(c+1)q^n-\frac{q^{n+1}-q}{q-1}.$$
Let $S$ be the set of primes $P$ in $A$ such that:
$$\prod_{i=1}^{\rm{deg}P-1}\beta(cq^n+q^n-1)\equiv 0\pmod{P}.$$
Let's assume that $S$ is a finite set. We set:
$$D=\prod_{P\in S}\rm{deg}P,$$
and $D=1$ if $S=\emptyset.$ Note that:
$$\forall P\in S,\, q^D\equiv 1\pmod{q^{\rm{deg}P}-1}.$$
Therefore, since $\beta(c)=1,$ we have:
$$\forall P\in S,\, \beta(cq^D+q^D-1)\equiv 1\pmod{P}.$$
But $\rm{deg}_T\beta(cq^D+q^D-1)\geq 1,$ thus we can select a prime $Q$ of $A$ such that $\beta (cq^D+q^D-1)\equiv 0\pmod{Q}.$ Note that
$Q\not \in S.$ Set $d=\rm{deg}Q.$ Since $d$ does note divide $D,$ there exists  an integer $r,$ $1\leq r\leq d-1,$ such that $D\equiv
r\pmod{d}.$ Therefore:
$$\beta(cq^D+q^D-1)\equiv \beta (cq^r+q^r-1)\equiv 0\pmod{Q}.$$
But this implies that $Q\in S,$ which is a contradiction. $\diamondsuit$\par
${}$\par
Let $P$ be a prime of $A$ of degree $d.$ Let $J$ be the jacobian of $K_P,$ i.e. $J$ is the inductive limit of the $Cl^0(\mathbb F_{q^n}K_P),$
$n\geq 1.$ Set $\mathbb F_{q^{p^{\infty}}}=\cup_{n\geq 0}\mathbb F_{q^{p^n}}\subset \overline{\mathbb F_q},$ where $\overline{\mathbb F_q}$
is the algebraic closure of $\mathbb F_q$ in $\overline{k}.$ We consider the $\Delta=\rm{Gal}(K_P/k)$-module:
$${\cal{A}}_P=\frac{J[p]^{\rm{Gal}(\overline{\mathbb F_q}/\mathbb F_{q^{p^{\infty}}})}}{Cl^0(K_P)[p]}.$$
As a consequence of the results in section 4, we get:
\newtheorem{Proposition9}[Proposition7]{Proposition}
\begin{Proposition9} \label{Proposition9}
Let $W=\mathbb Z_{p}[\mu_{q^d-1}]$ and let $\chi \in \widehat{\Delta}.$ We have:
$$\rm{dim}_{\frac{W}{pW}}{\cal A}_P(\chi )=\rm{deg}_X g(X,\overline{\chi })-\rm{dim}_{\frac{W}{pW}}Cl^0(K_P)_p(\chi ).$$
\end{Proposition9}\par
Note that in general, by Proposition \ref{Proposition3}, we do not have ${\cal A}_P=\{ 0\}.$ But  Goss conjecture implies the
following:\par
\noindent let $P$ be a prime of $A$ of degree $d$ and let $i$ be a $q$-magic number, $1\leq i\leq q^d-2,$ then ${\cal A}_P(\omega_P^{-i})=\{
0\}.$\par
It would be interesting to prove (or find a counter-example) to this weak form of Goss
conjecture.\par









 











































 




\begin{thebibliography}{8}
\bibitem{CRE} R. Crew, Geometric Iwasawa Theory and a Conjecture of Katz,
in Conf. Proc. Canad. math. Soc. Vol. 7, eds: H. Kisilevsky and J.
Labute, Amer. Math. Soc. (1987), 37-53.
\bibitem{GOS} D. Goss, Basic Structures of Function Fields Arithmetic,
Springer-Verlag (1996).
\bibitem{GOS&SIN} D. Goss and W. Sinnott, Class Groups of Function
Fields, Duke Math. J. 52 (1985), 507-516.
\bibitem{HAY} D. Hayes, Explicit Class Field Theory for Rational Function
Fields, Trans. Amer. Math. Soc. 189 (1974), 71-91.
\bibitem{OKA} S. Okada, Kummer's Theory for Function Fields, J. Number
Theory 38 (1991), 212-215.
\bibitem{ROS} M. Rosen, Number Theory in Function Fields, Springer-Verlag
(2002).
\bibitem{SHE} J. Sheats, The Riemann Hypothesis for the Goss Zeta
Function for $\mathbb F_q [T],$ J. Number Theory 71 (1998), 121-157.
\bibitem{TAT} J. Tate, Endomorphisms of Abelian Varieties over Finite
Fields, Invent. Math. 2 (1966), 134-144.


\end{thebibliography}

\end{document}
math0502130.extracted_bib
\begin{thebibliography}{8}
\bibitem{CRE} R. Crew, Geometric Iwasawa Theory and a Conjecture of Katz,
in Conf. Proc. Canad. math. Soc. Vol. 7, eds: H. Kisilevsky and J.
Labute, Amer. Math. Soc. (1987), 37-53.
\bibitem{GOS} D. Goss, Basic Structures of Function Fields Arithmetic,
Springer-Verlag (1996).
\bibitem{GOS&SIN} D. Goss and W. Sinnott, Class Groups of Function
Fields, Duke Math. J. 52 (1985), 507-516.
\bibitem{HAY} D. Hayes, Explicit Class Field Theory for Rational Function
Fields, Trans. Amer. Math. Soc. 189 (1974), 71-91.
\bibitem{OKA} S. Okada, Kummer's Theory for Function Fields, J. Number
Theory 38 (1991), 212-215.
\bibitem{ROS} M. Rosen, Number Theory in Function Fields, Springer-Verlag
(2002).
\bibitem{SHE} J. Sheats, The Riemann Hypothesis for the Goss Zeta
Function for $\mathbb F_q [T],$ J. Number Theory 71 (1998), 121-157.
\bibitem{TAT} J. Tate, Endomorphisms of Abelian Varieties over Finite
Fields, Invent. Math. 2 (1966), 134-144.


\end{thebibliography}