Sharedwww / Tables / artin.texOpen in CoCalc
% artin8.tex
% by Kevin Buzzard and William Stein

\documentclass{article}
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\title{A mod five approach to modularity of icosahedral 
       Galois representations}
\author{Kevin Buzzard and William A. Stein}


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\begin{document}
\maketitle
\begin{abstract}
We give eight new examples of icosahedral Galois representations that
satisfy Artin's conjecture on holomorphicity of their $L$-function.
We give in detail one example of an icosahedral representation of
conductor ${\bf 1376}=2^5\cdot 43$ that satisfies Artin's conjecture.
We also briefly explain the computations behind seven additional
examples of conductors ${\bf 2416}=2^4\cdot 151$, ${\bf 3184}=2^4\cdot
199$, ${\bf 3556}=2^2\cdot 7\cdot 127$, ${\bf 3756}=2^2\cdot 3\cdot
313$, ${\bf 4108}=2^2\cdot 13\cdot 79$, ${\bf 4288}=2^6\cdot 67$, and
${\bf 5373}=3^3\cdot 199$.
\end{abstract}

\section*{Introduction}
Consider a continuous irreducible Galois representation
$$\rho:\galq\ra\GL_n(\C)$$ 
with $n > 1$. 
Inspired by his reciprocity law,
Artin conjectured in~\cite{artin:conjecture} that 
$L(\rho,s)$ has an analytic continuation to the whole complex plane.
Many of the known cases of this conjecture were obtained by
proving the apparently stronger assertion that~$\rho$ is \defn{automorphic}, 
in the sense that the $L$-function of~$\rho$ is equal to the $L$-function
of a certain automorphic representation (whose $L$-function is known to have
analytic continuation). In the special case where $n=2$ and $\rho$ is in
addition assumed to be odd, the automorphic representation in question
should be the one associated to a classical weight~$1$
modular eigenform, and in fact there is conjectured to be a
bijection between such~$\rho$ and the set of all weight~$1$
cuspidal newforms, which should
preserve $L$-functions. It is this bijection
that we are concerned with in this paper, so assume for the rest
of the paper that $n=2$ and~$\rho$ is odd.

In this special case, the construction
of~\cite{deligne-serre} shows how to construct a continuous irreducible
odd 2-dimensional representation from a weight~$1$ newform, and the problem
is to go the other way. Say that a representation is \defn{modular}
if it arises in this way.

If the image of~$\rho$ is solvable, 
then~$\rho$  is known to be modular
\cite{langlands:basechange, tunnell:artin};
if the image is not solvable, then $\im(\rho)$ in $\PGL_2(\C)$
is isomorphic to the 
alternating group~$A_5$, and the modularity of~$\rho$ 
is, in general, unknown. We call such a 2-dimensional representation an
``icosahedral representation''.
The published literature contains only eight examples (up to twist) 
of odd icosahedral Galois representations that are known to satisfy Artin's 
conjecture: one of conductor $800=2^5\cdot 5^2$ 
(see \cite{buhler:thesis}), and seven of conductors:
$2083,\, 2^2\cdot 487,\, 2^2\cdot 751,\,
  2^2 \cdot 887,\, 2^2\cdot 919,\, 
  2^5\cdot 73,\,\text{ and } 2^5\cdot 193$
(see \cite{freyetal}).

After the first draft of this paper was written, the
preprint~\cite{bdsbt} appeared, which contains a general theorem that
yields infinitely many (up to twist) modular icosahedral representations.
However, we feel that our work, although much less powerful, is still
of some worth, because it gives an effective computational approach to
proving that certain mod~5 representations are modular, without
computing any spaces of weight~1 forms or using effective versions of
the Chebotar\"ev density theorem. We also note that the
main theorem of~\cite{bdsbt} does not apply to any of the examples
considered in the present paper.  Very recently,
the preprint~\cite{taylor:artin2}
appeared, which gives local conditions under which an icosahedral
representation is modular.  In particular, \cite{taylor:artin2} also 
proves that the first three 
examples in the present paper, of conductors 1376, 2416, 3184, 
are modular; these correspond to the first, third, and fourth 
equations at the end of~\cite{taylor:artin2}. 
However, \cite{taylor:artin2} does not apply to 
our remaining five examples.


In this paper we give eight new examples of modular icosahedral
representations that were computed 
by applying the main theorem  of~\cite{buzzard-taylor} to 
the mod~$5$ reduction of~$\rho$.
We verify modularity mod~$5$ on a case-by-case basis. Later we shall
explain our approach more carefully, but let us briefly summarise it here.
By~\cite{buzzard-taylor},
the problem is to show that the mod~5 reduction of~$\rho$ is modular.
We do this by finding a candidate mod~5 modular form at weight~5
and then, using the table of icosahedral extensions of $\Q$ in~\cite{freyetal}
and what we know about the 5-adic representation attached to our candidate
form, we deduce that the mod~5 representation attached to our candidate
form must be the reduction of~$\rho$. In particular, this paper gives
a computational methods for checking the modularity of certain mod~5
representations whose conductors are not too large. We now give
more details.

In each of our examples it is easy to compute a few Hecke operators 
and be morally convinced that a mod~$5$ representation should be modular; 
it is far more difficult to prove this.
Effective variants of the Chebotarev density theorem require
that we check vastly more traces of Frobenius than is practical.
Instead we use the Local Langlands theorem for $\GL_2$, the
theory of companion forms, and Table~2 of~\cite{freyetal},
to provide proofs of modularity
in certain cases.

More precisely, let~$K$ be an icosahedral extension of~$\Q$ that is not
totally real, and consider a minimal lift $\rho:\GQ\ra \GL_2(\C)$
of 
   $$\GQ\ra \Gal(K/\Q)\ncisom{}A_5\subset \PGL_2(\C);$$
the lift is minimal in the sense  that its conductor is minimal.
Assume that~$5$ does not ramify in~$K$, and that 
a Frobenius element at~$5$ in $\Gal(K/\Q)$ does not have order~$1$ or~$5$.
Inspired by the possibility that~$\rho$ is modular,
we search for a mod~$5$ modular form of weight~$5$ whose existence would 
be forced by modularity of~$\rho$.  Indeed, we find
a candidate mod~$5$ form~$f$, and then prove that the fixed field 
of the kernel of the projective mod~$5$ representation 
associated to a certain twist of~$f$  must be~$K$.  
This proves that the mod~$5$ reduction of a twist
of~$\rho$ is modular, and the main theorem 
of \cite{buzzard-taylor} then implies 
that~$\rho$ is modular. 
We carried out this program for icosahedral representations
of the following conductors: 
${\bf 1376} = 2^5\cdot 43$,
${\bf 2416}=2^4\cdot 151$, 
${\bf 3184}=2^4\cdot 199$, 
${\bf 3556}=2^2\cdot 7\cdot 127$, 
${\bf 3756}=2^2\cdot 3\cdot 313$,
${\bf 4108}=2^2\cdot 13\cdot 79$, 
${\bf 4288}=2^6\cdot 67$, and 
${\bf 5373}=3^3\cdot 199$.


We choose an icosahedral field~$K$ and representation~$\rho$, 
then proceed as follows:
\vspace{.5ex}
\begin{numlist}
\item Search for a form~$f \in S_5(N,\eps;\Fbar_5)$ whose
      associated mod~$5$ Galois representation looks like 
      it is the mod~$5$ reduction of~$\rho$.
\item Twist~$f$ to obtain an eigenform~$g$ with coefficients in~$\F_5$.
\item Prove that~$\rho_g$ is unramified at~$5$ by finding a companion form.
\item Prove that the image of $\proj\rho_g$ is~$A_5$ by ruling out all 
      other possibilities.
\item Prove that the fixed field~$L$ of $\proj\rho_g$ has 
      root field of discriminant at most $2083^2$,
      so~$L$ is in Table~2 of~\cite{freyetal}; deduce that~$L=K$.
\item Apply the main theorem of~\cite{buzzard-taylor}
      to a lift of $\rhobar=\rho_g$
      to conclude that~$\rho$ is modular.
\end{numlist}



\section{Modularity of an icosahedral representation of 
conductor~$1376=2^5\cdot 43$}\label{sec:1376}
In this section we prove the following theorem.
\begin{theorem}\label{thm:1376}
The icosahedral representations whose corresponding 
icosahedral extension
is the splitting field of $x^5 + 2x^4+6x^3+8x^2+10x+8$
are modular.
\end{theorem}

Let~$K$ be the splitting field of $h=x^5 + 2x^4+6x^3+8x^2+10x+8$.
The Galois group of~$K$ is~$A_5$, so we obtain a homomorphism
$G_\Q\ra{}A_5\subset \PGL_2(\C)$;
let $\rho:G_\Q\ra\GL_2(\C)$ be a minimal lift, minimal
in the sense that the Artin conductor of~$\rho$ is minimal. 
By Table~$A_5$ of~\cite{buhler:thesis}, the conductor of~$\rho$ 
is $N=1376=2^5\cdot 43$.  Since 
$h\con (x-1)(x^2-x+1)(x^2-x+2)\pmod{5}$, 
and ${\rm disc}(h)$ is coprime to~$5$, 
any Frobenius element at~$5$ in $\Gal(K/\Q)$ has order~$2$.

We use the notation of Tables 3.1 and 3.2 of~\cite[pg. 46]{buhler:thesis};
from Table 3.2 we see that the type of~$\rho$ at~$2$ 
is~$17$ and the type at~$43$ is~$2$.
The mod~$N$ Dirichlet character~$\eps=\det(\rho)$
factors as~$\eps=\eps_2\cdot \eps_{43}$ where~$\eps_2$ is 
a character mod~$2^5$ and~$\eps_{43}$ is a character mod~$43$.
Corresponding to each type in Buhler's table, there is a character,
and fortunately Buhler's level $800$ example also was of type~$17$ at~$2$
(see the first line of~\cite[Table~3.2]{buhler:thesis}).
By~\cite[pg.~80]{buhler:thesis} $\eps_2$ is the unique
character of conductor~$4$ and order~$2$.  
A local computation shows that the image 
of~$\eps_{43}$ has order~$3$. 

If~$\rho$ is modular,  then there is a weight~$1$ 
newform $f_?\in S_1(N,\eps;\Qbar)$ that gives rise to~$\rho$.
Suppose for the moment that~$\rho$ is modular, so that~$f_?$ exists.
Choose a prime of~$\overline{\Z}$ lying
over~$5$, and denote by~$\fbar_?$ the reduction 
of $f_?$ modulo this prime. The Eisenstein series 
$E_4\in M_4(1;\F_5)$  is congruent to~$1$ modulo~$5$, so
$E_4\cdot{}\fbar_?\in S_5(N,\eps;\Fbar_5)$ has the same $q$-expansion 
as $\fbar_?$.  Using a computer, we can search for a 
form $f\in S_5(N,\eps,\Fbar_5)$ that has the same 
$q$-expansion as the conjectural form $E_4\cdot{}\fbar_?$.

Instead of multiplying $\fbar_?$ by~$E_4$, we could have multiplied 
it by an  Eisenstein series of weight~$1$, level~$5$, and character $\eps'$.
We used $E_4$ because the dimension of $S_5(N,\eps;\Fbar_5)$ 
is~$696$ whereas the dimension of the relevant space 
$S_2(5\cdot 1376, \eps_{43})$ of weight~$2$ cusp forms is~$1040$.

\subsection{Searching for the newform~$f$}
Using modular symbols (see Section~\ref{sec:modsym}) we 
compute (at least up to semi-simplification) the space
$S_5(1376,\eps;\F_{25})$. Note that there is injective map
from the image of~$\eps$ into $\F_{25}^*$.  By computing
the kernels of various Hecke operators on this space, we find~$f$.
In the following computations, we represent nonzero elements of~$\F_{25}$ 
as powers of a generator~$\alp$ of~$\F_{25}^*$, which satisfies
$$\alp^2 + 4\alp + 2=0.$$
Our character $\eps_{43}$ was represented 
as the map sending $(1,3)\in(\Z/2^5\Z)^*\cross(\Z/43\Z)^*$ to
$2\alp+1$. Note that~3 is a primitive root mod~43, and that $2\alp+1$
has order~3.

If the least common multiple of the degrees of the factors of  
the polynomial~$h$ modulo an 
unramified prime~$p$ is~$2$, then $\Frob_p\in\Gal(K/\Q)$ 
has order~$2$.  The minimal polynomial of $\rho(\Frob_p)\in\GL_2(\C)$ 
is then $x^2-1$, so $\rho(\Frob_p)$ has trace~$0$.  
The first three primes $p \nmid 5\cdot 1376$ such 
that $\rho(\Frob_p)$ has  order~$2$ are $p= 19,31,97$. 
We computed the mod~$5$ reduction $\sS_5(1376,\eps;\F_{25})^{+}$
of the $\Z_5[\zeta_3]$-lattice of 
modular symbols of level~$1376$ and 
character~$\tilde{\eps}$ where 
complex conjugation acts as $+1$.
Here~$\tilde{\eps}$ denotes the Teichm\"uller lift of~$\eps$. 

Let~$V$ be the intersection of the kernels of $T_{19}$, $T_{31}$, and 
$T_{97}$ inside of the space $\sS_5(1376,\eps;\F_{25})^{+}$ of mod~5
modular symbols.
The space~$V$ is $8$-dimensional, 
and no doubt all the eigenforms in this space give rise to~$\rho$ or one
of its twists. One of the eigenvalues of~$T_3$ on this space
is~$\alp^{16}$, and the kernel $V_1$ of $T_3-\alp^{16}$ is $2$-dimensional
over $\F_{25}$. The Hecke operator~$T_5$ acted as a diagonalisable matrix on
$V_1$, with eigenvalues $\alp^{10}$ and $\alp^{22}$, so the corresponding
two systems of eigenvalues must correspond to mod~$5$ modular eigenforms,
and furthermore we must have found all mod~$5$ modular eigenforms
of this level, weight and character,
such that $a_{19}=a_{31}=0$ and $a_3=\alp^{16}$.

\begin{remark}
The careful reader might wonder how we know that the
systems of mod~$5$ eigenvalues really do correspond to mod~$5$ modular
forms, and not to perhaps some strange mod~$5$ torsion in the space of
modular symbols. However, we eliminated this possibility by
computing the dimension of the full space of mod~$5$ modular 
symbols where complex conjugation acts as~$+1$, and checking that it 
equals $696$, the dimension of $S_5(1376,\tilde{\eps},\C)$, which we 
computed using the formula in \cite{cohen-oesterle}.
\end{remark}

Let~$f$ be the eigenform in~$V_1$ that satisfies
$a_5=\alp^{22}$; the $q$-expansion of~$f$ begins
$$f=q + \alp^{16}q^3 + \alp^{22}q^5 + \alp^{14}q^7 
 + \alp^{14}q^9 + 4q^{11}+\cdots.$$
Further eigenvalues are given in Table~\ref{table:1376}.
The primes~$p$ in the table such that~$a_p=0$ are 
exactly those
predicted by considering the splitting behavior of~$h$. 
This is strong evidence that~$\rho$ is modular,
and also that our modular symbols algorithm have been correctly
implemented. 

\begin{table}
\caption{\label{table:1376}Eigenvalues of~$f$}
\begin{center}
$$\begin{array}{|rl|}\hline
2&0\\
3&\alpha^{16}\\
5&\alpha^{22}\\
7&\alpha^{14}\\
11&4\\
13&\alpha^{14}\\
17&\alpha^{14}\\
19&0\\
23&\alpha^{16}\\
29&\alpha^{8}\\
31&0\\
37&\alpha^{10}\\
41&1\\
43&\alpha^{10}\\
47&1\\
53&\alpha^{22}\\
\hline\end{array}
\begin{array}{|rl|}\hline
59&4\\
61&\alpha^{14}\\
67&\alpha^{4}\\
71&\alpha^{20}\\
73&\alpha^{2}\\
79&\alpha^{20}\\
83&\alpha^{4}\\
89&\alpha^{10}\\
97&0\\
101&\alpha^{8}\\
103&\alpha^{14}\\
107&0\\
109&\alpha^{10}\\
113&2\\
127&0\\
131&2\\
\hline\end{array}
\begin{array}{|rl|}\hline
137&0\\
139&\alpha^{22}\\
149&\alpha^{4}\\
151&1\\
157&\alpha^{14}\\
163&0\\
167&\alpha^{22}\\
173&4\\
179&\alpha^{2}\\
181&\alpha^{14}\\
191&\alpha^{10}\\
193&4\\
197&0\\
199&3\\
211&0\\
223&0\\
\hline\end{array}
\begin{array}{|rl|}\hline
227&\alpha^{10}\\
229&0\\
233&\alpha^{14}\\
239&0\\
241&\alpha^{2}\\
251&\alpha^{2}\\
257&3\\
263&\alpha^{16}\\
269&2\\
271&\alpha^{8}\\
277&0\\
281&\alpha^{16}\\
283&0\\
293&3\\
307&\alpha^{4}\\
311&\alpha^{22}\\
\hline\end{array}
\begin{array}{|rl|}\hline
313&0\\
317&0\\
331&\alpha^{14}\\
337&0\\
347&\alpha^{16}\\
349&\alpha^{4}\\
353&0\\
359&0\\
367&\alpha^{22}\\
373&0\\
379&3\\
383&3\\
389&1\\
397&\alpha^{16}\\
401&0\\
409&2\\
\hline\end{array}
\begin{array}{|rl|}\hline
419&3\\
421&\alpha^{20}\\
431&4\\
433&\alpha^{4}\\
439&\alpha^{20}\\
443&0\\
449&0\\
457&0\\
461&0\\
463&\alpha^{10}\\
467&0\\
479&0\\
487&\alpha^{8}\\
491&\alpha^{2}\\
499&\alpha^{20}\\
503&\alpha^{2}\\
\hline\end{array}
\begin{array}{|rl|}\hline
509&\alpha^{8}\\
521&\alpha^{10}\\
523&\alpha^{14}\\
541&\alpha^{20}\\
547&\alpha^{22}\\
557&3\\
563&1\\
569&\alpha^{16}\\
571&\alpha^{22}\\
577&\alpha^{14}\\
587&\alpha^{20}\\
593&0\\
599&\alpha^{22}\\
601&0\\
607&\alpha^{16}\\
613&2\\
\hline\end{array}
\comment{\begin{array}{|rl|}\hline
617&0\\
619&\alpha^{20}\\
631&\alpha^{20}\\
641&4\\
643&1\\
647&4\\
653&1\\
659&\alpha^{14}\\
661&2\\
673&\alpha^{8}\\
677&4\\
683&0\\
691&\alpha^{16}\\
701&\alpha^{14}\\
709&4\\
719&\alpha^{4}\\
\hline\end{array}
}
$$
\end{center}
\end{table}

\subsection{Twisting into $\GL(2,\F_5)$}
Although there is a representation
$\rho_f:\GQ\ra\GL(2,\F_{25})$ attached to $f$,
it is difficult to say anything about its image without further
work. We use a trick to show that the image of $\rho_f$ is small.
Firstly, for a character~$\chi:\GQ\to\Fbar_5$, let~$\tilde\chi$
denote its Teichm\"uller lift to~$\Qbar_5$. By a result of Carayol,
there is a characteristic 0 eigenform
$\tilde{f}\in S_5(N,\tilde{\eps};\Qbar_5)$ lifting $f$. 
The twist $\tilde{g}=\tilde{f} \tensor \tilde{\eps}_{43}$ is, by
\cite[Prop. 3.64]{shimura:intro}, an eigenform in
$S_5(43N, \tilde{\eps}_2; \Qbar_5)$, and its reduction is
a form $g\in S_5(43N,\eps_2,\F_{25})$.
The eigenvalues $a_p(g) = a_p(f) \eps_{43}(p)$, for the
first few
$p\nmid 5N$, are given in Table~\ref{table:1376twist}.

\begin{table}
\caption{\label{table:1376twist}Eigenvalues of~$g=f\tensor\eps_{43}$}
\begin{center}
$$
\begin{array}{|rl|}\hline
2&*\\%0\\
3&1\\
5&*\\%3\\
7&2\\
11&4\\
13&2\\
17&2\\
19&0\\
23&1\\
29&1\\
31&0\\
37&3\\
41&1\\
43&*\\%0\\
47&1\\
53&2\\
\hline\end{array}
\begin{array}{|rl|}\hline
59&4\\
61&2\\
67&4\\
71&4\\
73&3\\
79&4\\
83&4\\
89&3\\
97&0\\
101&1\\
103&2\\
107&0\\
109&3\\
113&2\\
127&0\\
131&2\\
\hline\end{array}
\begin{array}{|rl|}\hline
137&0\\
139&2\\
149&4\\
151&1\\
157&2\\
163&0\\
167&2\\
173&4\\
179&3\\
181&2\\
191&3\\
193&4\\
197&0\\
199&3\\
211&0\\
223&0\\
\hline\end{array}
\begin{array}{|rl|}\hline
227&3\\
229&0\\
233&2\\
239&0\\
241&3\\
251&3\\
257&3\\
263&1\\
269&2\\
271&1\\
277&0\\
281&1\\
283&0\\
293&3\\
307&4\\
311&2\\
\hline\end{array}
\begin{array}{|rl|}\hline
313&0\\
317&0\\
331&2\\
337&0\\
347&1\\
349&4\\
353&0\\
359&0\\
367&2\\
373&0\\
379&3\\
383&3\\
389&1\\
397&1\\
401&0\\
409&2\\
\hline\end{array}
\begin{array}{|rl|}\hline
419&3\\
421&4\\
431&4\\
433&4\\
439&4\\
443&0\\
449&0\\
457&0\\
461&0\\
463&3\\
467&0\\
479&0\\
487&1\\
491&3\\
499&4\\
503&3\\
\hline\end{array}
\begin{array}{|rl|}\hline
509&1\\
521&3\\
523&2\\
541&4\\
547&2\\
557&3\\
563&1\\
569&1\\
571&2\\
577&2\\
587&4\\
593&0\\
599&2\\
601&0\\
607&1\\
613&2\\
\hline\end{array}
\begin{array}{|rl|}\hline
617&0\\
619&4\\
631&4\\
641&4\\
643&1\\
647&4\\
653&1\\
659&2\\
661&2\\
673&1\\
677&4\\
683&0\\
691&1\\
701&2\\
709&4\\
719&4\\
\hline\end{array}
\comment{
\begin{array}{|rl|}\hline
727&4\\
733&0\\
739&2\\
743&2\\
751&4\\
757&3\\
761&3\\
769&0\\
773&0\\
787&4\\
797&1\\
809&3\\
811&1\\
821&2\\
823&3\\
827&3\\
\hline\end{array}
\begin{array}{|rl|}\hline
829&2\\
839&0\\
853&2\\
857&0\\
859&0\\
863&4\\
877&1\\
881&1\\
883&0\\
887&2\\
907&0\\
911&1\\
919&1\\
929&0\\
937&3\\
941&4\\
\hline\end{array}
\begin{array}{|rl|}\hline
947&2\\
953&1\\
967&4\\
971&3\\
977&0\\
983&0\\
991&3\\
997&3\\
&\\
&\\
&\\
&\\
&\\
&\\
&\\
&\\
\hline\end{array}}
$$
\end{center}
\end{table}

\begin{proposition}\label{prop:1376-g}
Let $g=f\tensor \eps_{43}$.  Then $a_p(g)\in \F_5$ 
for all~$p\nmid \ell N$.
\end{proposition}
\begin{proof}
Consider an eigenform $\tilde{f}\in S_5(N,\tilde{\eps};\Qbar_5)$
lifting~$f$ as above.
Associated to~$\tilde{f}$ there is an automorphic
representation~$\pi=\tensor_v'\pi_v$ of $\GL(2,\bA)$, where~$\bA$
is the ad\`{e}le ring of~$\Q$.
Because $43\mid\mid N$, and~$43$ divides the conductor
of $\eps$, we see that the local component $\pi_{43}$ of $\pi$ at
$43$ must be ramified principal series. By Carayol's theorem,
$\rho_{\tilde{f}}|_{D_{43}} \sim
          \abcd{\Psi_1}{0}{0}{\Psi_2}$
with, without loss of generality,~$\Psi_2$ unramified.  We have
$(\Psi_1\cdot \Psi_2)|_{I_{43}}=\tilde{\eps}|_{I_{43}}=\tilde{\eps}_{43}$,
therefore, $\rho_{\tilde{f}}|_{I_{43}} \sim
          \abcd{\tilde{\eps}_{43}}{0}{0}{1}$.

Now twist~$\tilde{f}$ by $\tilde{\eps}_{43}^{-1}$; we find that
$\rho_{\tilde{f}\tensor\tilde{\eps}_{43}^{-1}}|_{I_{43}} \sim
          \abcd{1}{0}{0}{\tilde{\eps}^{-1}_{43}}$.
In particular, there is an
eigenform~$\tilde{f}'\in S_5(N,\tilde{\eps}_2\tilde{\eps}^{-1}_{43},\Qbar_5)$
whose associated Galois representation is the twist by $\tilde{\eps}^{-1}_{43}$
of that of $\tilde{f}$ (recall that $N=1376$ and so~$43$ divides~$N$
exactly once). Let~$f'$ denote the mod~$5$ reduction of~$\tilde{f}'$. Then
one checks easily that $f'\in S_5(N,\eps_2\eps^{-1}_{43},\F_{25})=S_5(N,\eps^5,\F_{25})$.

For all primes $p\nmid5N$ we have $a_p(f')=\eps_{43}(p)^{-1}a_p(f)$.
In particular, we have $a_p(f')=0$ for
$p=19,31$.
Also, $\eps_{43}(3)=\alp^8$ and $\eps_{43}(5) =\alp^8$, so
$$a_3(f')=\alp^{16}/\alp^8 = \alp^8 = (\alp^{16})^5$$
$$a_5(f')=\alp^{22}/\alp^8 = \alp^{14} = (\alp^{22})^5.$$
Now if $\sigma$ is the non-trivial automorphism of $\F_{25}$,
then $\sigma(f')$ and $f$ both lie in 
$S_5(1376,\eps;\F_{25})$ and have same~$a_p$ for
$p=3,5,19,31$, so they are equal because we found~$f$
by computing the unique eigenform with given~$a_p$ for $p=3,5,19,31$.
So $g = f\tensor\eps_{43} = \sigma(f)\tensor\eps_{43}^2$.
Thus for all $p\nmid 5N$, we see that
$a_p(g) = a_p(f)^5 \eps_{43}^2$ has fifth power
$a_p(g)^5 = a_p(f)^{25} \eps_{43}^{10} 
          = a_p(f) \eps_{43} = a_p(g)$.
\end{proof}

\subsection{Proof that~$\rho_g$ is unramified at~$5$}

We begin with a generalisation of~\cite{sturm:cong}.
Let $M>4$ be an integer, and let $h=\sum_{n\geq1}c_nq^n$ be a 
normalised cuspidal eigenform
of some weight~$k\geq1$, level~$M$ and character~$\chi$, defined over some
field of characteristic not dividing~$M$. Even though the base field
might not have characteristic zero, we may still define the conductor
of $\chi$ to the the largest divisor $f$ of $M$ such that~$\chi$
factors through $(\Z/f\Z)^\times$.
Let~$I$ be a set of primes, with the property that for all~$p$
in~$I$, one of the following conditions hold:

(i) $p$ divides~$M$ but~$p$ does not divide~$M/\cond(\chi)$, or

(ii) $p$ divides~$M$ exactly once, and $h$ is $p$-new, in the sense
that there is no eigenform $h'$ of level $M/p$ such
that the $T_n$-eigenvalues of $h$ and $h'$ agree for all $n$
prime to $p$.

Let~$C$ denote the orbit of the cusp~$\infty$ in $X_1(M)$
under the action of the group generated by $w_p$ for $p\in I$, and
the Diamond operators $\langle d\rangle_M$. The orbit of~$\infty$
under the Diamond operators has size $\phi(M)/2$, and each
$w_p$ increases the size of the orbit by a factor of~2. In this
situation, we have

\begin{lemma} The first~$t$ terms of the $q$-expansion
of~$h$ at any cusp in~$C$ are determined by~$M$,~$k$, $\chi$, $c_p$
for~$p$ in~$I$, and $c_n$ for $1\leq n\leq t$.

\end{lemma}

\begin{remark} Our proof is just a translation of Corollary~4.6.18
of \cite{miyake} into the language of moduli problems (Miyake's argument
technically is only valid over the complex numbers).
\end{remark}
\begin{proof}
If $J\subseteq I$ is any subset, and $w_J$ denotes the product
of $w_p$ for $p\in J$, then $h|w_J$ is an eigenform for all the
Diamond operators, and this observation reduces the proof
of the lemma to showing that for $p\in I$, if $h|w_p=\sum_n d_nq^n$,
then $d_j$ for $1\leq j\leq n$ and $d_q$ for all $q\in I$
are determined by $M$, $k$, $\chi$, $p$, $c_j$ for $1\leq j\leq n$
and $c_q$ for all $q\in I$.

We first deal with primes $p$ of the form (i).
Say $M=p^mR$, where $R$ is prime to $p$.
Thinking of~$h$ as a rule for attaching
$k$-fold differentials to elliptic curves equipped with points
of order $p^m$ and $R$, we have by definition that
$$h(\G_m/q^\Z,\zeta,\zeta_R)=\bigg(\sum c_nq^n\bigg)(dt/t)^k,$$
where $\zeta=\zeta_{p^m}$ and $\zeta_R$ are fixed $p^m$th and $R$th roots
of unity in $\G_m$ which correspond to the cusp~$\infty$,
and $dt/t$ is the canonical differential on the Tate
curve $\G_m/q^\Z$. We normalise things such that
$$h(\G_m/q^{p^m\Z},q,\zeta_R)=\bigg(\sum d_nq^n\bigg)(dt/t)^k,$$
and remark that because $h$ is an action for the diamond operators,
we do not have to worry too much about whether this corresponds to
the standard normalisation of the $w_p$-operator.

We recall that the operator $pU_p$ in this setting can be thought
of as being defined by the rule:
$$(pU_ph)(E,P,Q)=\sum_C\pi^*h(E/C,\overline{P},\overline{Q}),$$
where $C$ runs through the subgroups of $E$ of order $p$ which have
trivial intersection with $\langle P\rangle$, and $\pi$ denotes the canonical
projection $E\to E/C$.
We see that
\begin{align*}
(pc_p)^m\big(\sum d_nq^n\big)(dt/t)^k&=(p^mU_{p^m}h)(\G_m/q^{p^m\Z},q,\zeta_R)\\
&=\sum_{c=0}^{p^m-1}\pi^c*h(\G_m/\langle q^{p^m},\zeta q^c\rangle,q,\zeta_R),
\end{align*}
where $\pi$ denotes the canonical projection from $\G_m/\langle q^{p^m}\rangle$
to the appropriate quotient. This last sum can be written as a
double sum
\begin{align*}
&\sum_{c\in(\Z/p^m\Z)^\times}\pi^*h(\G_m/\langle q^{p^m},\zeta q^c\rangle,q,\zeta_R)+\sum_{a=0}^{p^{m-1}-1}\pi^*h(\G_m/\langle q^{p^m},\zeta q^{pa}\rangle,q,\zeta_R)\\
=&\sum_{b\in(\Z/p^m\Z)^\times}\pi^*h(\G_m/\langle q^{p^m},\zeta^{-b}q\rangle,q,\zeta_R)+p^{m-1}\pi^*U_{p^{m-1}}h(\G_m/\langle q^{p^m},\zeta^{p^{m-1}}\rangle,q,\zeta_R)\\
=&\sum_{b\in(\Z/p^m\Z)^\times}\pi^*h(\G_m/\langle\zeta^{-b}q\rangle,\zeta^b,\zeta_R)+(pc_p)^{m-1}\pi^*h(\G_m/\langle q^{p^m},\zeta^{p^{m-1}}\rangle,q,\zeta_R)\\
=&\sum_{b}\chi_p(b)\sum_{n\geq1}c_n(\zeta^{-b}q)^n(dt/t)^k+p^k(pc_p)^{m-1}\pi^*h(\G_m/\langle q^{p^{m+1}}\rangle,q^p,\zeta_R^p),
\end{align*}
where we have written $\chi=\chi_R\chi_p,$ for $\chi_R$ a character of
level~$R$ and $\chi_p$ a character of level~$p^m$. We deduce that
\begin{align*}
&(pc_p)^m\big(\sum d_nq^n\big)(dt/t)^k-p^k(pc_p)^{m-1}\chi_R(p)\pi^*h(\G_m/\langle q^{p^{m+1}}\rangle,q^p,\zeta_R)\\
=&\bigg(\sum_n\big(\sum_b\chi_p(b)\zeta^{-bn}\big)c_nq^n\bigg)(dt/t)^k\\
=&W(\chi_p)\big(\sum_{p\nmid n}\chi_p(-n)^{-1}c_nq^n\big)(dt/t)^k
\end{align*}
where $W(\chi_p)=\sum_{b\in(\Z/p^m\Z)^\times}\chi_p(b)\zeta^b$ can be
checked to be nonzero because the conductor of $\chi_p$ is $p^m$.
Hence
$$(pc_p)^m\sum_n d_nq^n-p^k(pc_p)^{m-1}\chi_R(p)\sum_n d_nq^{np}=W(\chi_p)\chi_p(-1)\sum_{p\nmid n}\chi_p(n)^{-1}c_nq^n.$$
Equating coefficients of $q$ we deduce that $W(\chi_p)\chi_p(-1)=(pc_p)^md_1$,
and because $h|w_p$ is an eigenform for $T_n$ for all $n$ prime to $p$,
with eigenvalues determined by $\chi$ and $c_n$, we deduce that we can
determine $d_n$ for $n$ prime to $p$ from $c_n$. It remains to establish
what $d_p$ is, and equating coefficients of $q^p$ in the above equation
gives us that $(pc_p)^md_p=p^k(pc_p)^{m-1}\chi_R(p)d_1$ and hence
that $d_p$ is determined by $\chi$ and $c_p$.
Note that as a consequence we see that $d_p/d_1=p^{k-1}\chi_R(p)/c_p$,
a classical formula if the base field is the complexes.

Now we deal with primes of the form (ii) (note that we never use
this case in the rest of the paper). We think of $h$ as a rule associating
$k$-fold differentials to triples $(E,C,Q)$ where $C$ a cyclic subgroup of
order~$p$ and $Q$ a point of order~$R=M/p$. Because $h$ is $p$-new, the
trace of $h$ down to $X_1(M/p)$ must be zero, and hence we see
that for any elliptic curve $E$ equipped with a point $Q$ of order $R$,
$$\sum_C\pi^* h(E/C,E[p]/C,\overline{Q})=0.$$ As before, normalise things so
that
$$h(\G_m/q^\Z,\mu_p,\zeta_R)=\bigg(\sum_n c_nq^n\bigg)(dt/t)^k$$
and
$$h(\G_m/q^{p\Z},\langle q\rangle,\zeta_R)=\bigg(\sum_n d_nq^n\bigg)(dt/t)^k.$$
The fact that the trace of~$h$ is zero implies that
$$(pU_p)h(\G_m/q^{p\Z},\langle q\rangle,\zeta_R)+\pi^*h(\G_m/q^\Z,\mu_p,\zeta_R)=0,$$
and hence that
$$c_p\sum d_nq^n+p^{k-1}\sum c_nq^n=0$$
from which we deduce that the $d_n$ can be read off from $c_p$ and the $c_n$.
\end{proof}
\begin{remark} The size of~$C$ is
$\phi(M).2^{|I|-1}$, and the usefulness of this lemma is that
if $h_1$ and $h_2$ are two normalised eigenforms of the same level,
weight and character as above, both new at all primes in~$I$,
and the coefficients of $q^n$ in the
$q$-expansions of $h_1$ and $h_2$ agree for $n\in I$ and $n\leq t$,
then $h_1-h_2$ has a zero of order at least $t+1$ at all cusps in~$C$,
and in particular if
$\phi(M).2^{|I|-1}(t+1)>k/12[\SL_2(\Z):\Gamma_1(M)]=\deg(\omega^k)$ on
$X_1(M)$ then $h_1=h_2$. Using the fact that
$[\Gamma_0(M):\Gamma_1(M)]=\phi(M)/2$,
we deduce
\end{remark}
\begin{corollary}\label{cor:bound} 
Let $h_1$ and $h_2$ be two normalised eigenforms as above.
If the coefficients of $q^n$ in the $q$-expansions of $h_1$ and $h_2$ agree
for all primes in $I$ and for all
$n\leq\frac{k}{12}[\SL_2(\Z):\Gamma_0(M)]/2^{|I|}$ then $h_1=h_2$.
\end{corollary}
\begin{remark} One can certainly do better than this corollary
in many cases. For example, when $n>1$ and
$p^n$ exactly divides both the level
of an eigenform and the conductor of its character, then one can compute
the $q$-expansion of the eigenform at many ``middle cusps'' too,
and hence increase the size of $C$ in the result above.
\end{remark}

We now go back to the explicit situation we are concerned with.
Although~$g$ is an eigenform of level $59168=2^5\cdot 43^2$, 
we can still consider the corresponding representation
$\rho_g :\GQ\ra \GL(2,\F_5)$, and then directly analyze
its ramification.
\begin{proposition}
The representation~$\rho_g$ is unramified at~$5$.
\end{proposition}
\begin{proof}
Continuing the modular symbols computations as above, 
we find that~$V_1$ is spanned by the two eigenforms
\begin{align*}
f\,\,&=q + \alp^{16}q^3 + \alp^{22}q^5 + \alp^{14}q^7 
 + \alp^{14}q^9 + 4q^{11}+\cdots\\
f_1&=q + \alp^{16}q^3 + \alp^{10}q^5 + \alp^{14}q^7 
 + \alp^{14}q^9 + 4q^{11}+\cdots.
\end{align*}
For $p\neq 5$ and $p\leq 997$, we have $a_p(f_1)=a_p(f)$.
To check that $a_p(f) = a_p(f_1)$ for all $p\neq 5$,
it suffices to show that the difference~$f-f_1$ has
$q$-expansion involving only powers of~$q^5$; 
for this we use the $\theta$-operator
$q\frac{d}{dq}:S_5(1376,\eps,\F_{25})\ra S_{11}(1376,\eps;\F_{25})$.
Since~$\theta$ sends normalized eigenforms to normalized eigenforms,
it suffices to check that the subspace of 
$S_{11}(1376,\eps;\F_{25})$ generated by~$\theta(f)$ 
and~$\theta(f_1)$ has dimension~$1$.  
Corollary~\ref{cor:bound} implies that it suffices to verify that the 
coefficients $a_p(\theta(f))$ and $a_p(\theta(f_1))$ are equal for all 
$$p \leq \frac{11}{12}\cdot [\SL_2(\Z):\Gamma_0(1376)]\cdot \frac{1}{2} 
 = 968.$$ 
The eigenform~$f$ must be new because we computed it by finding
the intersections of the kernels of Hecke operators $T_p$ with 
$p\nmid 1376$; if~$f$ were an oldform then the intersection of the 
kernels of these Hecke operators
would necessarily have dimension greater than~$1$.
Because it takes less than a second
to compute each $a_p(\theta(f))$, we were easily able to verify that the
space generated by $\theta(f)$ and $\theta(f_1)$ has dimension~$1$.

\begin{remark}
It is possible to avoid appealing to Corollary~\ref{cor:bound} by using
one of the following two alternative methods:
\begin{enumerate}
\item Define~$\theta$ directly on modular symbols and compute it.
\item Compute the intersection
  $$\bigcap_{p\geq 2} \ker(T_p - pa_p(f))
      \subset S_{11}(1376,\eps;\F_{25}).$$
  Since~$\theta(f)$ and~$\theta(f_1)$ both lie
  in the intersection, the moment the dimension 
  of a partial intersection is~$1$, it follows 
  that $\theta(f-f_1)=0$.
\end{enumerate}
We successfully carried out both alternatives.
For the first, we showed that~$\theta$ on modular symbols is
induced by multiplication by
$X^5Y - Y^5X$.
For the second, we find that after intersecting 
kernels for $p\leq 11$, the dimension is already~$1$. 
The first of these two methods took much less 
time than the second. 
\end{remark}

Next we use that $\theta(f-f_1)=0$ to show that $\rho_g$ is unramified,
thus finishing the proof of the proposition.
Since~$f$ is ordinary, Deligne's theorem (see~\cite[\S12]{gross:tameness}) 
implies that
$$\rho_f|_{D_5}\sim 
   \mtwo{\alp}{*}{0}{\beta}\qquad\text{over $\Fbar_5$}$$
with both~$\alp, \beta$ unramified,
$\alp(\Frob_5)=\eps(5)/a_5=\alp^8/\alp^{22}=\alp^{10}$, and
$\beta(\Frob_5)=\alp^{22}$.
Since $a_p(f_1)=a_p(f)$, for $p\neq 5$, we have
$${\rho_f}|_{D_5}\sim {\rho_{f_1}}|_{D_5} \sim \mtwo{\alp'}{*}{0}{\beta'}$$
with 
$\alp'(\Frob_5)=\alp^8/\alp^{10}=\alp^{22}$ and
$\beta'(\Frob_5)=\alp^{10}$;
in particular, $\alp'=\beta$.
Thus $\rho_f|_{D_5}$ contains $\alp\oplus \beta$, so 
$\rho_f|_{D_5}\sim\alp\oplus\beta$ and hence there is a choice
of basis so that $*=0$.

\end{proof}



\subsection{The image of $\proj \rho_g$}
\begin{proposition}\label{prop:image_is_A5}
The image of $\proj \rho_g$ is $A_5$.
\end{proposition}
\begin{proof}

The image~$H$ of $\proj \rho_g$ in $\PGL_2(\F_5)$ is easily checked to
lie in $\PSL_2(\F_5)\cong A_5$ because of what we know about the
determinant of $\rho_g$. Hence $H$ is a subgroup of $A_5$ that
contains an element of order~$2$ (complex conjugation) and an element
of order~$3$ (for example, $\rho_g(\Frob_7)$ has characteristic
polynomial $x^2-2x-1$).  This proves that~$H$ is isomorphic to
either~$S_3$,~$A_4$, or~$A_5$.  Let $L$ be the number field cut out
by~$H$.  If~$L$ were an $S_3$-extension, then there would be a
quadratic extension contained in it which is unramified outside
$2\cdot 5\cdot 43$; it is furthermore unramified at~$5$ by the
previous section and unramified at $43$ because $I_{43}$ has
order~$3$.  Thus it is one of the three quadratic fields unramified
outside~$2$.
In particular, the trace of $\Frob_p$
would be zero for all primes in a certain congruence class
modulo~8.
However, there are primes~$p$ congruent to $3$, $5$, and $7$ 
mod $8$ such that $a_p(g)\neq 0$, e.g., $3$, $7$, and $13$. 


If $H$ were isomorphic to $A_4$, then let~$M$ denote the cyclic
extension of degree~3 over~$\Q$ contained in~$L$. Now~$M$ is unramified
at~2 and~5, and hence is the subfield of $\Q(\zeta_{43})$ of degree~3.
Choose $p\nmid 1376\cdot 5$ that is inert in~$M$, i.e., so that
$p$ is not a cube mod $43$.  The order of
$\rho_g(\Frob_p)$ in $\GL_2(\F_5)$ must be divisible by~$3$.  However,
a quick check using Table~\ref{table:1376twist} shows that this is
usually not the case, even for $p=3$.
\end{proof}


\subsection{Bounding the ramification at~$2$ and~$43$}
Let~$L$ be the fixed field of $\ker(\proj(\rho_g))$. We have just
shown that $\Gal(L/\Q)$ is isomorphic to $A_5$. 
By a root field for~$L$, we mean
a non-Galois extension of $\Q$ of degree~5 whose Galois closure is~$L$.
\begin{proposition}
The discriminant of a root 
field for~$L$ divides $(43\cdot 8)^2=344^2$, and
in particular,~$L$ must be mentioned in Table~1 
of \cite[pg 122]{freyetal}.\end{proposition}
\begin{proof}
The analysis of the local behavior of~$\rho_f$ at~$43$ given in
Proposition~\ref{prop:1376-g}
shows that the inertia group at~$43$ in $\Gal(L/\Q)$ has order~$3$. Using
Table~3.1 of~\cite{buhler:thesis}, we see that if 
$\Gal(L/\Q)\isom A_5$
then it must be ``type~$2$'' at 43, and hence the discriminant of a root
field of~$L$, that is, of a non-Galois extension of~$\Q$ of degree~$5$ 
whose Galois closure is~$L$, must be $43^2$ at~$43$.

At~$2$ the behavior of~$\rho$ is more subtle and we shall not analyze
it fully. But we can say that, because~$\rho$ has arisen from
a form of level $1376=2^5.43$, we must be either of type~$5$
or one of types~$14$--$17$. In particular, the discriminant at~$2$ of a root
field for~$L$ will be at most~$2^6$.

Finally,~$L$ is unramified at all other primes, because~$\rho$ is.
Hence the discriminant of a root field for~$L$, assuming that
$\Gal(L/\Q)\cong A_5$, divides $(43.8)^2=344^2$. 
\end{proof}

We know that~$L$ is an icosahedral extension of~$\Q$ with 
discriminant dividing $43^2\cdot 2^6$.  Table~1 of \cite[pg 122]{freyetal}
contains all icosahedral extensions, such that the discriminant
of a root field is bounded by $2083^2$.  The table
must contain~$L$; there is only one icosahedral extension with 
discriminant dividing $43^2\cdot 2^6$, so $L=K$. 

\subsection{Obtaining a classical weight one form}
We have shown that a twist of the icosahedral
representation $\rho:\GQ\ra\GL(2,\C)$,
nobtained by lifting $\GQ\ra \Gal(K/\Q)\ncisom A_5$,
has a mod~$5$ reduction $\rho_g:\GQ\ra \GL_2(\F_5)$ that
is modular.  Since~$\rho$ ramifies at only finitely many primes,
and~$\rho$ is unramified at~$5$ with distinct eigenvalues, 
\cite{buzzard-taylor} implies that~$\rho$ arises from
a classical weight~$1$ newform.



\section{More examples}
The data necessary to deduce modularity of each of our eight 
icosahedral examples is summarized in 
Tables~\ref{table:more1}--\ref{table:more4}. 

The notation in Table~\ref{table:more1} is as follows.
The first column contains the conductor.  
The second column contains a $5$-tuple $[a_4,a_3,a_2,a_1,a_0]$ such
that the $A_5$-extension is the splitting field of the polynomial
$h=x^5+a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x + a_0$. 
The column labeled $\ord(\Frob_5)$ contains the order of the image
of $\Frob_5$ in $A_5$.  The next column, which is labeled ``$p$ with
$a_p=0$'', contains the first few~$p$ such that $a_p$ is easily seen
to equal~$0$ by considering the splitting of~$h$ mod~$p$. 
The $\eps$ column contains the character of the representation, where the
notation is as follows. Write $(\Z/N\Z)^*$ as a product of cyclic groups
corresponding to the prime divisors of~$N$ in ascending order, and then 
the tuples give the orders of the images of these cyclic factors; when
$8\mid N$, there are two cyclic factors corresponding to the prime~$2$.
Finally, the last column records the dimension of $S_5(\Gamma_1(N),\eps)$.

The notation in Table~\ref{table:more2} is as follows.  The first column
contains the conductor.  The second column contains an eigenform that
was found by first intersecting the kernels of the Hecke operators 
$T_p$ with~$p$ as in Table~\ref{table:more1}, and then locating an
eigenform.
In each case, a companion form was found, by computing $a_p(f)$ for
$p\leq$ bound, where bound is the bound from Corollary~\ref{cor:bound}.

Table~\ref{table:more3} shows that the fixed field
of the image of each $\proj(\rho_g)$ is icosahedral.  
The first column contains the 
conductor~$N$.  The second column contains a twist~$g$ of~$f$ such that 
$a_p(g)\in\F_5$ for all $p\nmid 5N$.  The third column contains 
a $\Frob_p$ such that $\proj(\rho_g(\Frob_p))$ has order~$3$,
along with the characteristic polynomial of $\rho_g(\Frob_p)$.
As in the proof of Proposition~\ref{prop:image_is_A5},
the other two boxes give data that allows us to deduce 
that the fixed field of the image of $\proj(\rho_g)$ is icosahedral.
The case $5373$ must be treated separately, because there are
three possibilities $M_1$, $M_2$, and $M_3$ 
for the cubic field~$M$ of the analogue of 
Proposition~\ref{prop:image_is_A5}.
For $M_1$ we find a prime~$p$ such that
$$(p^2\!\!\!\mod 9, \,\,\,p^{66}\!\!\!\mod 199)\not\in\{(1,1),(4,1),(7,1)\}$$
with $\rho_g(\Frob_p)$ of order not divisible by~$3$; 
for this, $p=2$ suffices, since the characteristic polynomial
of $\rho_g(\Frob_2)$ is $(x+2)^2$
and $(p^2\!\!\!\mod 9, \,\,\,p^{66}\!\!\!\mod 199) = (4,106)$. 
For $M_2$ we find a prime~$p$ such that
$$(p^2\!\!\!\mod 9, \,\,\,p^{66}\!\!\!\mod 199)\not\in\{(1,1),(4,92),(7,106)\}$$
with $\rho_g(\Frob_p)$ of order not divisible by~$3$; 
again, $p=2$ suffices.
For $M_3$ we find a prime~$p$ such that
$$(p^2\!\!\!\mod 9, \,\,\,p^{66}\!\!\!\mod 199)\not\in\{(1,1),(4,106),(7,92)\}$$
with $\rho_g(\Frob_p)$ of order not divisible by~$3$; 
here, $p=13$ suffices, as the characteristic polynomial
of $\rho_g(\Frob_p)$ is $(x+4)^2$ and 
$(p^2\!\!\!\mod 9, \,\,\,p^{66}\!\!\!\mod 199) = (7,106)$.



Table~\ref{table:more4} gives upper bounds on the ramification of the
fixed field of the image of $\proj(\rho_g)$.  These bounds 
were deduced using Table~3.1
of~\cite{buhler:thesis} by restricting the possible ``types'' using
information about the character $\eps$.  Note that though
the bounds are not sharp, e.g., the discriminant of 
the representation of conductor $2416$ is $2^4\cdot 151^2$, they
are all less than $2083^2$, so the corresponding
field must appear in Table~2 of~\cite{freyetal}.

\begin{table}
\caption{\label{table:more1}Data on icosahedral representations mod~$5$}
\begin{center}
\begin{tabular}{|clclll|}\hline
$N$&\hspace{2em}$h$&$\hspace{-1.5em}\ord(\Frob_5)$&$p$ with $a_p=0$&
   \hspace{1em}$\eps$ &$\hspace{-1.5em}\dim S_5(N,\eps)$\\\hline
{\bf 1376}&$[2,6,8,10,8]$ & \hspace{-1.5em}$2$&$19,31,97$&$[2,1,3]$&$696$\\
{\bf 2416}&$[0,-2,2,5,6]$ & \hspace{-1.5em}$2$&$53,97,127$&$[2,1,3]$&$1210$\\
{\bf 3184}&$[5,8,-20,-21,-5]$& \hspace{-1.5em}$2$&$31,89,97$&$[2,1,3]$&$1594$\\
{\bf 3556}&$[3,9,-6,-4,-40]$&\hspace{-1.5em}$3$&$19,29,89$&$[1,2,3]$&$2042$\\
{\bf 3756}&$[0,-3,10,30,-18]$&\hspace{-1.5em}$3$&$17,61,67$&$[1,2,3]$&$2506$\\
{\bf 4108}&$[4,3,9,4,5]$& \hspace{-1.5em}$3$&$17,23,31,89$&$[1,3,2]$&$2234$\\
{\bf 4288}&$[4,5,8,3,2]$& \hspace{-1.5em}$3$&$19,23,47$&$[1,2,3]$&$2164$\\
{\bf 5373}&$[2,1,7,23,-11]$& \hspace{-1.5em}$2$&$7,23,37,79,89$&$[2,3]$&$2394$\\
\hline\end{tabular}
\end{center}
\end{table}

\begin{table}
\caption{\label{table:more2}The newform $f$ and the companion form bound}
\begin{center}
\begin{tabular}{|cll|}\hline
$N$&\hspace{7em}$f$ &\hspace{-.3em}bound \\\hline
{\bf 1376}&
    $q + \alp^{16}q^3 + \alp^{22}q^5 + \alp^{14}q^7 
      + \alp^{14}q^9 + 4q^{11}+ \alp^{14}q^{13} + \cdots $ 
    & $968$\\
{\bf 2416}& 
    $q +3q^3 + \alp^{22}q^5 + \alp^{16}q^7 + \alp^{4} q^{11} 
       + \alp^2 q^{13} +  \alp^{16}q^{15} + \cdots$ 
    & $1672$ \\
{\bf 3184}& 
    $q + \alp^{16}q^3 + 3q^5 + \alp^{22}q^7 + \alp^{14}q^9 + 3q^{11} 
       + \alp^{22}q^{13} + \cdots$ 
    & $2200$\\
{\bf 3556}& 
    $q + \alp^{16}q^{3} + \alp^{14}q^5 + \alp^{10}q^7 + \alp^{14}q^9 
       + \alp^{2}q^{11} + \alp^{22}q^{13} + \cdots$
    & $1408$ \\
{\bf 3756}& 
      $ q + \alp^{14}q^3 + \alp^{14}q^5 + 3q^7 + \alp^4q^9 + \alp^{16}q^{11} + \alp^{10}q^{13} + \cdots$
    & $1727$ \\
{\bf 4108}&
    $ q + \alp^{16}q^{3} + \alp^{11}q^{5} + \alp^{20}q^{7} + \alp^{14}q^{9} + \alp^{10}q^{11} + 4q^{13} + \cdots $
    & $1540$\\

{\bf 4288}& 
     $q + 3q^3 + \alp^{14}q^{5} + \alp^{20}q^{7} + 3q^9 + \alp^{20}q^{11} + \alp^{16}q^{13} + \cdots $
     & $2992$ \\


{\bf 5373}& 
    $q + \alp^{16}q^{2} + \alp^{14}q^{4} + 4q^5 + 3q^8 + \alp^{4}q^{10} + 2q^{11}+\cdots $
     & $3300$ \\
\hline\end{tabular}
\end{center}
\end{table}


\begin{table}
\caption{\label{table:more3}Verification that the image of $\proj(\rho_g)$ is $A_5$}
\begin{center}
 Find a Frobenius element with projective order $3$.\vspace{1ex}\\
\begin{tabular}{|c|l|ll|}\hline
$N$ & \hspace{1em} $g$ & proj. order $3$&\hspace{.7em}charpoly \\\hline
{\bf 1376} & $f\tensor \eps_{43}$ 
    &  $\quad\Frob_7$ & $x^2-2x-1$
 \\
{\bf 2416}& $f\tensor \eps_{151}$ 
    &  $\quad\Frob_{19}$ & $x^2+2x-1$
 \\
{\bf 3184}& $f\tensor \eps_{199}$ 
    & $\quad\Frob_7$ & $x^2+3x+4$
 \\
{\bf 3556}& $f\tensor \eps_{127}$
    &  $\quad\Frob_{13}$ & $x^2+3x+4$ 
  \\
{\bf 3756}&$f\tensor\eps_{313}$
    & $\quad\Frob_{23}$ & $x^2 + 2x + 4$
  \\
{\bf 4108} & $f\tensor\eps_{13}$
    & $\quad\Frob_{29}$ & $x^2+3x+4$
 \\
{\bf 4288}& $f\tensor\eps_{67}$
    & $\quad\Frob_{11}$ & $x^2+x+1$
\\

{\bf 5373}& $f\tensor\eps_{199}$
    & $\quad\Frob_{11}$ & $x^2+3x+4$
\\
\hline\end{tabular}\vspace{3ex}

Not $S_3$: For all $t\in T$, find unramified $p$ s.t.\ $t\not\equiv \Box \mod p$ and $a_p(g)\neq 0$. \vspace{1ex}\\
\begin{tabular}{|c|l|l|}\hline
$N$ & $\qquad\quad{}T$ & $\qquad{}p$ \\\hline
{\bf 1376}
    &  $\{-1,-2\}$ & $3$, $7$\\
{\bf 2416}
    &  $  \{-1, -2\}$ & $3$, $7$ \\
{\bf 3184}
    & $  \{-1, -2\}$ &  $3$, $7$ \\
{\bf 3556}
    &  $ \{-1, -2, -7, -14\}$ & $3$, $13$, $3$, $11$\\
{\bf 3756}
    & $ \{-1,-2,-3,-6\}$ &  $7$, $7$, $11$, $13$\\
{\bf 4108}
    & $  \{-1, -2, -79, -158\} $ &$3$, $7$, $3$, $7$ \\
{\bf 4288}
    & $ \{-1,-2\}$ &  $3$, $7$ \\
{\bf 5373}
    & $  \{-3\}$ & $11$\\
\hline\end{tabular}\vspace{3ex}\\
\comment{
function FindS3(t, aplist, N) 
   P:=[p : p in [2..97] |IsPrime(p)];
   for i in [1..#aplist] do 
      p := P[i];
      if (5*N mod p ne 0) and not IsSquare(GF(p)!t) and aplist[i] ne 0 then
         return p;
      end if; 
   end for;
end function;
}
    

Not $A_4$: Unramified~$p$, not cube mod $\ell$, 
order of $\rho_g(\Frob_p)$ not divisible by $3$.
\vspace{1ex}\\
\begin{tabular}{|c|c|cl|}\hline
$N$ & $\ell$ & $p$&\hspace{.7em}charpoly$(\rho_g(\Frob_p))$ \\\hline
{\bf 1376}
    & $43$  & $3$ & $\qquad(x+2)^2$ \\
{\bf 2416}
    &  $151$ & $7$ & $\qquad(x+2)^2$ \\
{\bf 3184}
    & $199$ &  $3$ & $\qquad(x+2)^2$ \\
{\bf 3556}
    &  $127$ & $3$ & $\qquad(x+2)^2$\\
{\bf 3756}
    & $313$ &  $11$ & $\qquad(x+2)^2$\\
{\bf 4108}
    & $13$ & $3$ & $\qquad(x+2)^2$\\
{\bf 4288}
    & $67$ & $7$ & $\qquad(x+3)^2$\\
{\bf 5373}
    & --- & &(see text)\\
\hline\end{tabular}\vspace{3ex}\\
\comment{
function IsCubeMod(p, ell) // is p a cube in F_ell
   R<x>:=PolynomialRing(GF(ell));
   return not IsIrreducible(x^3-p);
end function;

procedure FindA4(ell, aplist, N, e1, e2) 
   P:=[p : p in [2..97] |IsPrime(p)];
   for i in [1..#aplist] do 
      p := P[i];
      if (5*N mod p ne 0) then
//      if (5*N mod p ne 0) and not IsCubeMod(p,ell) then
         t := GF(5)!(Evaluate(e2,p)*aplist[i]);
         d := GF(5)!Evaluate(e1,p); 
         R<x> := PolynomialRing(GF(5));
         f := x^2 - t*x + d;
"p =",p;
"f =",f;
"factor(f) =",Factorization(f);
      end if; 
   end for;
end procedure;
}

\end{center}
\end{table}


\begin{table}
\caption{\label{table:more4}Bounding the discrimant of the fixed field 
of $\proj(\rho_g)$}
\begin{center}
\begin{tabular}{|cl|}\hline
$N$ & Bound on discriminant\\
{\bf 1376}& $\qquad2^6\cdot 43^2$\\
{\bf 2416}& $\qquad 2^6\cdot 151^2$\\
{\bf 3184}& $\qquad 2^6\cdot 199^2$\\
{\bf 3556}& $\qquad 2^2\cdot 7^2 \cdot 127^2$\\
{\bf 3756}& $\qquad 2^2\cdot 3^2 \cdot 313^2$\\
{\bf 4108}& $\qquad 2^2\cdot 13^2 \cdot 79^2$\\
{\bf 4288}& $\qquad 2^6\cdot 67^2$\\
{\bf 5373}& $\qquad 3^4\cdot 199^2$\\
\hline\end{tabular}
\end{center}
\end{table}

\section{Computing mod~$p$ modular forms}
\subsection{Higher weight modular symbols}
\label{sec:modsym}
The second author developed software that computes the space of
weight~$k$ modular symbols $\sS_k(N,\eps)$, for $k\geq 2$ and 
arbitrary~$\eps$.  
See~\cite{merel:1585} for the standard facts about higher weight 
modular symbols, and~\cite{stein:phd} for a description of
how to compute with them.

Let $K=\Q(\eps)$ be the field generated by the values of~$\eps$. 
 The cuspidal modular symbols $\sS_k(N,\eps)$ are a
finite dimensional vector space over~$K$, which is generated by all
linear combinations of  higher weight modular symbols 
    $$X^i Y^{k-2-i}\{\alp,\beta\}$$ 
that lie in the kernel of an appropriate boundary map.  There is an
involution~$*$ that acts on $\sS_k(N,\eps)$, and
$\sS_k(N,\eps)^+\tensor_K\C$ is isomorphic, as a module over the Hecke
algebra, to the space $S_k(N,\eps;\C)$ of cusp forms.

Fix $k=5$.  In each case considered in this paper,
there is a prime ideal~$\lambda$ 
of the ring of integers $\mathcal{O}$ of~$K$ 
such that $\mathcal{O}/\lambda\isom \F_{25}$.
Let~$\cL$ be the $\mathcal{O}$-module generated by all modular
symbols of the form $X^iY^{3-i}\{\alp,\beta\}$,
and let 
 $$\sS_5(N,\eps;\F_{25})=(\cL\tensor_{\mathcal{O}}\F_{25})\cap \sS_5(N,\eps).$$
This is the space that we computed.
The Hecke algebra acts on $\sS_5(N,\eps;\F_{25})$, so when
we find an eigenform we find a maximal ideal of the Hecke algebra.

As an extra check on our computation of 
$\sS_5(N,\eps;\F_{25})$, we computed the dimension
of $S_5(N,\eps;\C)$ using both the formula of~\cite{cohen-oesterle}
and the Hijikata trace formula (see~\cite{hijikata:trace}) 
applied to the identity Hecke operator.


\comment{%it's all in my thesis and it's not that relevant.
The Manin symbols are
$[i, (c,d)]$ where $0\leq i\leq k-2=3$ and
$(c,d)$ vary over points in the projective plane.
The Manin symbol $[i,(c',d')]$ corresponds to the
modular symbol $(g.X^iY^{3-i})\{g(0),g(\infty)\}$
where $g=\abcd{a}{b}{c}{d}\in\SL_2(\Z)$ is a matrix whose lower
two entries are congruent to $(c',d')$ modulo $N$,
and $g.X^iY^{3-i} := (dX-bY)^i(-cX+aY)^{3-i}$. 
Let $\sigma=\abcd{0}{-1}{1}{0}$, $\tau={0}{-1}{1}{-1}$
and for $\gamma\in\SL_2(\Z)$, let
$[i,(c,d)]\gamma = [\gamma.X^iY^{3-i}, (c,d)\gamma]$.
Since there are only finitely many 
Manin symbols, we  can 
compute $\sS_5(N,\eps)$ as the quotient of the $\F$-vector
space generated by Manin symbols modulo 
the following relations:
\begin{align*}
    {[i,(c,d)] + [i,(c,d)]\sigma} &= 0\\
    {[i,(c,d)] + [i,(c,d)]\tau + [i,(c,d)]\tau^2} &= 0\\
    {[i,(n c,n d)]}&=\eps(n)[i,(c,d)]
       \qquad n\in (\Z/N\Z)^*
\end{align*}
The quotient was computed by using a fast ``hashing'' function
to quotient out by the $2$-term relations.  The quotient
by the $3$-term relations was then computed using sparse
Gauss elimination.  One important subtlety is that, e.g., $\sigma$
and~$\tau$ do not commute so, after modding out by
the~$\sigma$ relations, it is important to mod out by~$3$ 
term relations coming both from~$\tau$ and~$\sigma\tau$. 

The main result of~\cite{merel:1585} gives
a way to compute the action of $T_p$ directly
on the Manin symbols.
Suppose $f\in\sS_5(N,\eps;\F_{25})$ is an eigenvector; to 
naively compute the action of~$T_p$ on~$f$ requires computing
the action of~$T_p$ on each Manin symbol involved in~$f$, 
and then summing the result. This requires roughly
$\dim\sS$ times as long as computing~$T_p$ on a single
Manin symbol. 
In order to quickly compute a large number of
Hecke eigenvalues we use the following projection trick.
Let $\vphi\in\Hom(\sS_5(N,\eps;\F_{25}),\F_{25})$ be a (left) eigenvector for all
Hecke operators~$T_p$ having the same eigenvalues as~$f$.
Choose a Manin symbol $x=[i,(c,d)]$ such 
that $\vphi(x)\neq 0$.  Since~$x$ is of a very simple form, 
it is easy to compute~$T_p(x)$ quickly.  We have
 $\vphi(T_p(x)) = (T_p(\vphi))(x) = a_p \vphi(x)$,
so since $\vphi(x)\neq 0$ we divide and find 
$a_p = \vphi(T_p(x))/\vphi(x)$. 
In fact, we use a generalization of this trick to 
quickly compute the action of~$T_p$ on any Hecke stable subspace 
$V\subset \Hom(\sS(N,\eps;\F_{25}),\F_{25})$.
}


\subsection{Complexity}
We implemented the modular symbols algorithms mentioned above
in \magma{} (see \cite{magma}) because of its robust support
for linear algebra over small finite fields.

The following table gives a flavor of the complexity of the
machine computations appearing in this paper.
The table indicates how much 
CPU time on a Sun Ultra E450 was required to compute all data 
for the given level,
including the matrices $T_p$ on the $2$-dimensional spaces,
for $p<2000$.   For example, the total time for level $N=1376$
was~$6$ minutes and~$58$ seconds.
\begin{center}
\begin{tabular}{|cr|}\hline
\vspace{-2ex}&\\
 N & time (minutes)\\
\vspace{-2ex}&\\
 1376&  6:58\hspace{2.5em}\mbox{}\\
 2416&  10:42\hspace{2.5em}\mbox{}\\
 3184&  14:16\hspace{2.5em}\mbox{}\\
 3556&  19:55\hspace{2.5em}\mbox{}\\
 3756&  27:47\hspace{2.5em}\mbox{}\\
 4108&  23:11\hspace{2.5em}\mbox{}\\
 4288&  15:18\hspace{2.5em}\mbox{}\\
 5376&  24:49\hspace{2.5em}\mbox{}\\\hline
\end{tabular}
\end{center}

\subsection{Acknowledgment} 
Some of the computing equipment was purchased
by the second author using a UC Berkeley Vice Chancellor Research Grant.
Additional computer runs were 
made on the Sun Ultra E450 of the Computational Algebra Group at 
the University of Sydney.   Allan Steel was very helpful in optimizing our 
code.

\comment{\bibliographystyle{amsplain}
\bibliography{biblio}}

\providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace}
\begin{thebibliography}{10}

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  Univ. Hamburg \textbf{3} (1923/1924), no. 1, 89--108.

\bibitem{buhler:thesis}
J.\thinspace{}P. Buhler, \emph{Icosahedral \protect{G}alois representations},
  Springer-Verlag, Berlin, 1978, Lecture Notes in Mathematics, Vol. 654.

\bibitem{bdsbt}
K.~Buzzard, M.~Dickinson, N.~Shepherd-Barron, and R.~Taylor, \emph{On
  icosahedral {A}rtin representations}, in preparation.


\bibitem{buzzard-taylor}
K.~Buzzard and R.~Taylor, \emph{Companion forms and weight one forms}, Ann. of
  Math. (2) \textbf{149} (1999), no.~3, 905--919.

\bibitem{cohen-oesterle}
H.~Cohen and J.~Oesterl{\'e}, \emph{Dimensions des espaces de formes
  modulaires},  (1977), 69--78. Lecture Notes in Math., Vol. 627.

\bibitem{magma}
W.~Bosma, J.~Cannon, and C.~Playoust, \emph{The {M}agma algebra system {I}:
  {T}he user language}, J. Symb. Comp. \textbf{24} (1997), no.~3-4, 235--265,
  \\\protect{\sf http://www.maths.usyd.edu.au:8000/u/magma/}.

\bibitem{deligne-serre}
P.~Deligne and J-P. Serre, \emph{Formes modulaires de poids $1$}, Ann. Sci.
  \'Ecole Norm. Sup. (4) \textbf{7} (1974), 507--530 (1975).

\bibitem{freyetal}
G.~Frey (ed.), \emph{On {A}rtin's conjecture for odd \protect{$2$}-dimensional
  representations}, Springer-Verlag, Berlin, 1994.

\bibitem{gross:tameness}
B.\thinspace{}H. Gross, \emph{A tameness criterion for \protect{G}alois
  representations associated to modular forms (mod \protect{$p$})}, Duke Math.
  J. \textbf{61} (1990), no.~2, 445--517.

\bibitem{hijikata:trace}
H.~Hijikata, \emph{Explicit formula of the traces of \protect{H}ecke operators
  for \protect{$\Gamma_0(N)$}}, J. Math. Soc. Japan \textbf{26} (1974), no.~1,
  56--82.

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R.\thinspace{}P. Langlands, \emph{Base change for \protect{${\rm {G}{L}}(2)$}},
  Princeton University Press, Princeton, N.J., 1980.

\bibitem{merel:1585}
L.~Merel, \emph{Universal \protect{F}ourier expansions of modular forms}, On
  {A}rtin's conjecture for odd \protect{$2$}-dimensional representations
  (Berlin), Springer, 1994, pp.~59--94. Lecture Notes in Math., Vol. 1585.

\bibitem{miyake}
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\end{thebibliography}


\end{document}






***

[8458981, 509]


// Cohen-Oesterle Dimension computations in MAGMA:

> G<a2,b2,c> := DirichletGroup(1376,CyclotomicField(EulerPhi(1376)));
> eps:=a2*(c^(Order(c) div 3));
> Order(eps);
6
> DimensionCuspForms(eps,5);



> G<a2,b2,c> := DirichletGroup(2416,CyclotomicField(EulerPhi(2416)));
> eps:=a2*(c^(Order(c) div 3));
> Order(eps);
6
> DimensionCuspForms(eps,5);
1210

> G<a2,b2,c> := DirichletGroup(3184,CyclotomicField(EulerPhi(3184)));
> eps:=a2*(c^(Order(c) div 3));
> Order(eps);
6
> DimensionCuspForms(eps,5);
1594


> G<a,b,c> := DirichletGroup(3556,CyclotomicField(EulerPhi(3556)));
> eps:=(b^(Order(b) div 2))*(c^(Order(c) div 3));
> Order(eps);
6
> DimensionCuspForms(eps,5);
2042;


> G<a,b,c> := DirichletGroup(3756,CyclotomicField(EulerPhi(3756)));
> eps:=(b^(Order(b) div 2))*(c^(Order(c) div 3));
> Order(eps);
6
> DimensionCuspForms(eps,5);


> G<a,b,c> := DirichletGroup(4108,CyclotomicField(EulerPhi(4108)));
> eps:=(b^(Order(b) div 3))*(c^(Order(c) div 2));
> DimensionCuspForms(eps,5);

> G<a,b,c> := DirichletGroup(4288,CyclotomicField(EulerPhi(4288)));
> eps:=(b^(Order(b) div 2))*(c^(Order(c) div 2));
> DimensionCuspForms(eps,5);


> G<a,b> := DirichletGroup(5373,CyclotomicField(EulerPhi(5373)));
> eps:=(a^(Order(a) div 2))*(b^(Order(b) div 3));
> DimensionCuspForms(eps,5);


///////////////////////////////////////////

procedure powfrob(p, e1, e2, aplist) 
   Primes := [p : p in [2..97] |IsPrime(p)];
   n := Index(Primes,p);
   a := GF(5)!(Evaluate(e2,p)*aplist[n]);
   b := GF(5)!Evaluate(e1,p); 
   R<x>:=PolynomialRing(GF(5));
   Q<y> := quo<R|x^2-a*x+b>;
   x^2 - a*x + b;
   for i in [1..24] do 
      f := MinimalPolynomial(y^i);
      if Degree(f) le 1 then 
         printf "rho_g(Frob_%o)^%o satisfies %o.\n", p, i, f;
      end if; 
   end for; 
end procedure;



> // 1376
> h := x^5+2*x^4+6*x^3+8*x^2+10*x+8;
> N := 2^5*43;
> F<alp> := GF(25);
> G<a2,b2,c>:=DirichletGroup(N, F);
> eps := a2 * (c^(Order(c) div 3));
> aplist := [0,alp^16,alp^22,alp^14,4,alp^14,alp^14,0, alp^16,alp^8,0,alp^10,1,alp^10,1,alp^22,4,alp^14,alp^4,alp^20,alp^2,alp^20,alp^4,alp^10];
> qEigenform(aplist,eps,5);
//q + alp^16*q^3 + alp^22*q^5 + alp^14*q^7 + alp^14*q^9 + 4*q^11 + alp^14*q^13 + alp^14*q^15 + alp^14*q^17 + O(q^20)



> // 2416
> h := x^5-2*x^3+2*x^2+5*x+6;
> N := 2^4*151;
> k:=5;
> F<alp> := GF(25);
> G<a2,b2,c>:=DirichletGroup(N, F);
> eps := a2 * (c^(Order(c) div 3));
> aplist:=[0,3,alp^22,alp^16, alp^4, alp^2, alp^22,3,alp^22,3,alp^16,alp^22,2,alp^8,alp^8,0,1,alp^8,3,alp^8,2,2,2,alp^20,0,2,alp^16,1];
> qEigenform(aplist,eps,k);
// q + 3*q^3 + alp^22*q^5 + alp^16*q^7 + alp^4*q^11 + alp^2*q^13 + alp^16*q^15 + alp^22*q^17 + 3*q^19 + O(q^20)

> // 3184
> h := x^5+5*x^4+8*x^3-20*x^2-21*x-5;
> N:=2^4*199;
> F<alp> := GF(25);
> G<a2,b2,c>:=DirichletGroup(N, F);
> eps := a2 * (c^(Order(c) div 3));
> aplist:=[0,alp^16,3,alp^22,3,alp^22,3,alp^16];
> qEigenform(aplist,eps,5);
// q + alp^16*q^3 + 3*q^5 + alp^22*q^7 + alp^14*q^9 + 3*q^11 + alp^22*q^13 + alp^10*q^15 + 3*q^17 + alp^16*q^19 + O(q^20)

> // 3556 
> h := x^5+3*x^4+9*x^3-6*x^2-4*x-40;
> N := 2^2*7*127;
> F<alp> := GF(25);
> G<a,b,c>:=DirichletGroup(N, F);
> eps := b^(Order(b) div 2) * (c^(Order(c) div 3));
> aplist := [0,alp^16,alp^14,alp^10,alp^2,alp^22,alp^14,0, alp^10,0,alp^16,alp^20];


> // 3756
> h := x^5-3*x^3+10*x^2+30*x-18;
> N := 2^2*3*313;
> F<alp> := GF(25);
> G<a,b,c>:=DirichletGroup(N, F);
> eps := b^(Order(b) div 2) * (c^(Order(c) div 3));
> aplist:=[0,alp^14,alp^14,3,alp^16,alp^10,0,3,3,alp^2,alp^22,alp^22,alp^20,alp^16,alp^4,4,alp^8,0];
> qEigenform(aplist,eps,5);
// q + alp^14*q^3 + alp^14*q^5 + 3*q^7 + alp^4*q^9 + alp^16*q^11 + alp^10*q^13 + alp^4*q^15 + 3*q^19 + alp^8*q^21 + 3*q^23 + alp^4*q^25 + 3*q^27 + alp^2*q^29 + alp^22*q^31 + 2*q^33 + alp^8*q^35 + alp^22*q^37 + q^39 + alp^20*q^41 + alp^16*q^43 + 3*q^45 + alp^4*q^47 + 3*q^49 + 4*q^53 + 2*q^55 + alp^8*q^57 + alp^8*q^59 + O(q^62)


// 4108
> h := x^5+4*x^4+3*x^3+9*x^2+4*x+5;
> N := 2^2*13*79;
> F<alp> := GF(25);
> G<a,b,c>:=DirichletGroup(N, F);
> eps := b^(Order(b) div 3) * (c^(Order(c) div 2));
> aplist := [0,alp^16,alp^11, alp^20,alp^10,4,0,alp^14,0, alp^22, 0, alp^22,alp^10,alp^2,3,4,alp^14,alp^2,alp^10];
> qEigenform(aplist,eps,5);
//q + alp^16*q^3 + alp^11*q^5  + alp^20*q^7 + alp^14*q^9 + alp^10*q^11 
//       + 4*q^13 + alp^3*q^15 + alp^14*q^19     + 4*q^21 + O(q^24)


// 4288
> h := x^5+4*x^4+5*x^3+8*x^2+3*x+2;
> N := 2^6*67;
> F<alp> := GF(25);
> G<a2,b2,c>:=DirichletGroup(N, F);
> eps := a2*b2^(Order(b2) div 2)*(c^(Order(c) div 3));
> aplist := [0,3,alp^14, alp^20, alp^20, alp^16, alp^16,0,0, alp^14,alp^20,alp^4,alp^8,2,0,4,1,alp^16,alp^4,alp^20,0,alp^20,0,2];
> qEigenform(aplist,eps,5);
//q + 3*q^3 + alp^14*q^5 + alp^20*q^7 + 3*q^9 + alp^20*q^11 + alp^16*q^13 + alp^8*q^15 + alp^16*q^17 + alp^14*q^21 + O(q^24)


// 5373
> h := x^5+2*x^4+x^3+7*x^2+23*x-11;
> N := 3^3*199;
> F<alp> := GF(25);
> G<a,b>:=DirichletGroup(N, F);
> eps := a^(Order(a) div 2) * (b^(Order(b) div 3));
> aplist := [alp^16, 0, 4, 0, 2, alp^22, 1, alp^16, 0];
> qEigenform(aplist,eps,5);
//q + alp^16*q^2 + alp^14*q^4 + 4*q^5 + 3*q^8 + alp^4*q^10 + 2*q^11 + alp^22*q^13 + q^17 + alp^16*q^19 + alp^2*q^20 + alp^22*q^22 + O(q^24)