e^{150}$ therefore has a class number $\geq 4$. On the other hand, there is no imaginary quadratic field of discriminant $d$ and class number 3 for $907<|d|<10^{2500}$ \cite {12}. Therefore (after an examination of a table of class numbers of the first quadratic fields): \begin{theorem} The imaginary quadratic fields of class number 3 are the 16 fields of discriminant: $-23,-31,-59.-83,-107,-139,-211,-283,-307,-331,-379,-499,-547,-643,-883,-907$. \end{theorem} \section{Application to a conjecture of Serre} Let $\r$ be a continuous representation of $Gal(\overline{\Q}/\Q)$ in $GL_2(V)$ where $V$ is a dimension 2 vector space over a finite field $\F_q$ of characteristic $p$. Assume this is an odd representation, i.e., that $\r(c)$ the image of the complex conjugation, seen as an element of $Gal(\overline{\Q}/\Q)$ has eigenvalues 1 and -1. In that case put $G=Im\r$. In \cite{17} Serre defines the level, the character and the weight of such a representation: \be \item The level. Let $l$ be a prime number different from $p$. Write $G_i$ ($i=0,\ldots$) the groups of ramifications of $\r$ at $l$. Let \[n(l)=\sum_{i=0}^\infty \frac{g_i}{g_0}\codim V^{G_i},\]where $g_i=|G_i|$. The conductor of the representation $\r$ is defined as \[N=\prod_{l\neq p}l^{n(l)}.\] \item The character. The determinant of $\r$ yields a character of $Gal(\overline{\Q}/\Q)$ in $\F_q^*$, for which the conductor divides $pN$. Therefore, one can write \[\det\r=\e\chi^{k-1},\]where $\chi$ is the cyclotomic character of conductor $p$ and $\e$ is the character $(\Z/N\Z)^*\to\F_q^*.$ The integer $k$ is defined mod $(p-1)$, and the fact that the representation is odd implies that $\e(-1)=(-1)^k$. By definition, $\e$ is the character of the representation $\r$. \item The weight. The integer $k$ above is defined mod $(p-1)$. Read Serre's article for the definition of the weight $k\in\Z$ of the representation $\r$. As the conductor $N$ depends only on the behavior of $\r$ ar places coprime with $p$, the definition of weight only uses the local properties at $p$ of the representation $\r$. \ee Then Serre's conjecture is: \begin{conjecture} Let $\r$ be a representation as above, of weight $k$, level $N$ and character $\e$. Assume this representation is irreducible. Then it comes from a cusp form $\m p$ of weight $k$, level $N$ and character $\e$. \end{conjecture} This conjectures, if true, has numerous consequences: it implies the Taniyama-Weil conjecture and Fermat's theorem. Many such representations $\r$ are modular, either by construction, or because they are part of classical conjectures (Langlands, Artin, $\ldots$) that carry on the conjecture (but sometimes in a weak form, i.e., with a weight or conductor bigger than those defined in \cite{17}.) In order to verify (or contradict) Serre's conjecture, we need to find the extensions $K/\Q$ of Galois group subgroup of $GL_2(\F_q)$ of odd determinant and $p\neq 2$. It is in general not difficult to calculate, for $l$ prime and not too large, the trace $a_l$ of $\Frob_l$ in $GL_2(\F_q)$: if $P(x)$ is a polynomial whose roots generate $K$ the decomposition of $P \m l$ usually will suffice. It is however, much harder to find modular forms $\m p$, if they exist, that correspond to the representation $\r$ given by the field $K$: the discriminant of $K$ is usually large, thus so is the conductor of $\r$, which is related to it, so it is not easy to make the computations. \subsection{The case $SL_2(\mathbb{F}_4)$} A troubling case is that of $p=2$, because, since $-1\equiv 1 (\m 2)$ all representations are odd. The representations of $Gal(\overline{\Q}/\Q)$ in $GL_2(\F_2)=S_3$ (although altogether real, cf. \cite{17}) come from weight 1 modular forms; the group $S_3$ can be realized as a subgroup of $GL_2(\CC)$. One can hope that by multiplication with convenient Eisenstein series, one can obtain a modular form of weight and level predicted by the Serre conjecture (cf. \cite{17} for examples.) In order to obtain the most interesting case for characteristic 2, one considers the representations with values in $GL_2(\F_4)$. The isomorphism $A_5\simeq SL_2(\F_4)$ allows us to obtain several examples. Let $K$ be an extension of $\Q$ of Galois group $A_5$. Since $A_5$ ``immerses ?'' into $PGL_2(\CC)$, if the field is not completely real, the associated representation $\r$ comes from a weight 1 modular form (module Artin's conjecture, cf. \cite{2}). Suppose now that $K$ is real. None of the classical conjectures allow us to suspect that $\r$ comes from a modular form, even if of higher weight or level. It is this case that we will study in what follows. The method of graphs here is indispensable, the modular forms that we look at having a conductor too large to be studied with the Eichler-Selberg trace formula. Let $P(x)=x^5+a_1x^4+a_2x^3+a_3x^2+a_4x+a_5$ be a rational polynomial of discriminant $D$. In order that the field of roots of $P$ be $A_5$ it is sufficient and necessary that $P$ be irreducible, that $D$ be square-free, and that there exist a prime number $l$ not dividing $D$ so that $P \m l$ having exactly two roots in $\F_l$ (this last condition assuring that the group is all of $A_5$). It is clear that $\e=1$. If $p\mid{}D$, $p$ coprime with 30, $n(p)=1$ if it ``seulment si l'inertie en p ==?'' is of order 2, and thus the polynomial $P$ has at most double roots mod $p$. As far as the weight $k$ is concerned, it is either 2 or 4 according to the ramification of $K$ at 2. To simplify the computation, we have limited to searching examples among the representations of prime level and weight 2. On the other hand, since it is about representations in $SL_2(\F_4)$m the coefficient $a_2$ of the sought modular form, if it exists, cannot be in $\F_4$, but in $\F_{16}$. This comes from the fact that the coefficient $a_l$ of a modular form $\m l$ is equal to an eigenvalue of $\Frob_l$, and not to its trace. Now, if a matrix in $SL_2(\F_4)$ is of order 5, its eigenvalues are in $\F_{16}$ not in $\F_4$. The examples treated above were obtained by making a systematic search on a computer of convenient polynomials (totally real, of type $A_5$, for which the conductor of the associated representation is a prime $N$, and for which the weight is 2). Thereafter, for each such polynomial $P$, one computes the corresponding eigenvalue $a_2$ (in $\F_{16}$), and one tries to find whether there exists a modular form mod 2 of level $N$ and weight 2 so that $T_2$ has $a_2$ as an eigenvalue. In all the cases considered, we have thereafter found an eigenspace of dimension 1 or 2. Using the operators $T_3,T_5$, one calculates the coefficients $a_3,a_5$, and verifies that they correspond to the values predicted by the decomposition of $P$ in 3 and 5. Clearly, this doesn't really prove that the representation $\r$ associated to $P$ is modular: we have only exhibited a modular form mod 2 of proper level and weight for which the terms $a_2,a_3,a_5$ are convenient. But there is a good indication of the truthfulness of the conjecture of Serre in the considered cases: an exhaustive search over numerous primes $N$ of the coefficients $a_2$ of modular forms of weight 2 and level $N$ proves that it is rare that there are fields of small degree. (Actually, is seems that 2, and in general the small primes, are the most ``inert'' possible in the fields that appear in the Hecke algebra of modular forms, fields which themselves in general appear to have the largest degree possible, taking into account constraints such as the Atkin-Lehner involutions, primes of Eisenstein, etc. One gets that one has small factors, -- corresponding for example to elliptic curves with prime conductor -- but this is apparently rare.) \subsection{A few examples} \be \item $P(x)=x^5-10x^3+2x^2+19x-6$. The discriminant is $(2^3887)^2$. This polynomial is irreducible mod 5, thus irreducible over $\Q$. Its roots are all real (apply Sturm's algorithm). One has that \[P(x)\equiv x(x-1)(x^3+x^2-1) \m 3,\] which gives a cycle of order 3; the Galois group of $K$, the field of roots of $P$, is thus $A_5$. %% %% %% THESE COMPUTATIONS ARE NOT CORRECT. %% I AM REDOING THEM %% %% %ORIGINAL: From $P(x)\equiv (x-446)(x-126)^2(x-538)^2 \m 887$ one gets that From $P(x)\equiv (x-462)(x-755)^2(x-788)^2 \m 887$ one gets that the conductor $N$ of the associated representation is $N=887$. One can also prove that 2 is ``little ramified'' in the sense of \cite{17}, thus $\r$ has weight 2. Examining the reduction mod 2 of $P$ proves that the coefficients $a_2,a_3,a_5$ of the modular form mod 2 of level 887 (which must correspond to $\r$ via the Serre conjecture) are 1, 1, j (where $j\in\F_4$ has the property that $j^2+j+1=0$). One therefore applies the method of graphs: the space of modular forms mod 2 of weight 2 and level 887 has dimension 73, and computation shows that the eigenspace $G_1$ of $T_2$ corresponding to the eigenvalue 1 has dimension 2; $T_3$ acts as the identity on $G_1$, and $j,j^2$ are the eigenvalues of $T_5$ acting on $G_1$, from where get a basis of $G_1$ formed by $f_1=q+q^2+q^3+q^4+jq^5+\cdots$ and $f_2=q+q^2+q^3+q^4+j^2q^5+\cdots$, eigenvectors of Hecke operators. These corroborate the conjecture. \item $P(x)=x^5-23x^3+55x^2-33x-1$. Then $D=13613^2,P(x)\equiv (x-6308)(x-2211)^2(x-8248)^2 \m 13613$, $N=13613$; $P$ being irreducible mod 2, $\Frob_2$ is a cycle of order 5, and $a_2=\zeta_5$ is a fifth root of unity, viewed as an element of $\F_{16}$. Computation also shows that in the space of modular forms mod 2 of level 13613 and weight 2, which has dimension 1134, $\zeta_5$ is a simple eigenvalue of $T_2$. The coefficients $a_3,a_5$ are respectively equal to $1+\zeta_5^2+\zeta_5^3=j$ and $\zeta_5^2+\zeta_5^3=j^2$, which are the traces of $\Frob_3,\Frob_5$ in $SL_2(\F_4)$. \item We write the other found polynomials; in each case there exists a modular form of weight 2 and appropriate level, for which the first terms $a_n$ correspond to those values predicted by the Serre conjecture. \[P(x)=x^5+x^4-16x^3-7x^2+57x-35,N=8311,\sqrt{D}=N\] \[P(x)=x^5+2x^4-43x^3+29x^2+2x-3,N=8447,\sqrt{D}=2^2N\] \[P(x)=x^5+x^4-13x^3-14x^2+18x+14,N=15233,\sqrt{D}=2N\] \[P(x)=x^5+x^4-37x^3+67x^2+21x+1,N=24077,\sqrt{D}=2^2N\] \ee \section{Appendix: The curves $X_0(p)$ of genus 0} In \cite{5}, it is proven that if $p$ is a prime number then the curve $X_0(p)$ over $\Z_p$ is formally isomorphic to the curve of equation $xy=p^k$, in the neighborhood of each point reducing mod $p$ to a supersingular point $S$, $k$ being one half the number of automorphisms of $S$. If $X_0(p)$ has genus 0 (i.e., $p=2,3,5,7,13$) one has such a model over $\Z$, given by the function \bean x=\left(\frac{\eta(z)}{\eta(pz)}\right)^\frac{24}{p-1},\eean where $\eta(z)=q^{1/24}\prod_{i=1}^\infty (1-q^n)$ and $q=e^{2\pi iz}$. This results from Fricke \cite{7}, who gives for each of the above $p$'s an expression of the ``oubli ?'' homomorphism $j:X_0(p)\to X_0(1)$, which associates to each point $(E,C)$ of $X_0(p)$ the point $(E)$ of $X_0(1)$, parametrized by the modular invariant $j$. In the following we recall these equations and give the expressions of the correspondences $T_2,T_3$ over these curves. The variable $x$ is the one given by equation (2), the involution $W_p$ switches $x$ and $y$ and the divisor of $x$ is $(0)-(\infty)$, where $0$ and $\infty$ are two points of $X_0(p)$. \be \item $p=2$ The equations given by Fricke (modified to give the model of $X_0(2)$ over $\Z$) are: \[xy=2^{12}\] \[j=\frac{(x+16)^3}{x}\] $T_2$ is given by \[y^2-y(x^2+2^43x)-2^{12}x=0\] (to each point $x$ is associated by $T_2$ the formal sum of points of coordinate $y$ that are roots of this polynomial.) $T_3$ is given by \[x^4+y^4-x^3y^3-2^33^2x^2y^2(x+y)-2^23^25^2xy(x^2+y^2)+2\cdot 3^21579x^2y^2-2^{15}3^2xy(x+y)-2^{24}xy=0\] \item $p=3$. \[xy=3^6\] \[j=\frac{(x+27)(x+3)^3}{x}\] \[T_2:x^3+y^3-2^33xy(x+y)-x^2y^2-3^6xy=0\] \[T_3:y^3-y^2(x^3+2^23^2x^2+2\cdot 3^25y)-3^6yx (x+2^23^2)-3^{12}x=0\] \item $p=5$. \[xy=5^3\] \[j=\frac{(x^2+10x+5)^3}{x}\] \[T_2:x^3+y^3-x^2y^2-2^3xy(x+y)-7^2xy=0\] \[T_3:x^4+y^4-x^3y^3-2\cdot 3^2x^2y^2(x+y)-3^4xy(x^2+y^2)-2\cdot 3^223x^2y^2-2250xy(x+y)-5^6xy=0\] \item $p=7$. \[xy=7^2\] \[j=\frac{(x^2+13x+49)(x^2+5x+1)^3}{x}\] \[T_2:x^3+y^3-x^2y^2-2^3xy(x+y)-7^2xy=0\] \[T_3:x^4+y^4-x^3y^3-2^23x^2y^2(x+y)-2\cdot 3\cdot 7xy(x^2+y^2)-3\cdot 53x^2y^2-2^23\cdot 7^2xy(x+y)-7^4xy=0\] \item $p=13$. \[xy=13\] \[j=\frac{(x^2+5x+13)(x^4+7x^3+20x^2+19x+1)^3}{x}\] \[T_2:x^3+y^3-x^2y^2-2^2xy(x+y)-13xy=0\] \[T_3:x^4+y^4-x^3y^3-2\cdot 3x^2y^2(x+y)- 3\cdot 5xy(x^2+y^2)-3\cdot 11x^2y^2-2\cdot 3\cdot 13xy(x+y)-13^2xy=0\] \ee The polynomials above that give $T_2,T_3$ are of simpler form than the classical modular equations $\Phi_2(j,j')$ and $\Phi_3(j,j')$ (that correspond to the action of $T_2$ and $T_3$ on $X_0(1)$). For comparison, we recall their expressions: \bea \Phi_2(j,j') &=& j^3+j'^3-j^2j'^2+2^43\cdot 31jj'(j+j')-2^43^45^3(j^2+j'^2)\\ && + 3^45^34027jj'+2^83^75^6(j+j')-2^{12}3^95^9 \eea \bea \Phi_3(j,j') &=& j^4+j'^4-j^3j'^3-2^23^39907jj'(j^2+j'^2)+2^33^231j^2j'^2(j+j')\\ &&-2^{16}5^33^517\cdot 263jj'(j+j')+2^{15}3^25^3(j^3+j'^3)+2\cdot 3^413\cdot 193\cdot 6367j^2j'^2\\ && - 2^{31}5^622973jj'+2^{30}3^35^6(j^2+j'^2)+2^{45}3^35^9(j+j') \eea \end{document}