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\begin{document}

\begin{titlepage}
    
	    \begin{center}
	        \huge
		\textbf{Riemann Surfaces and Modular Function Field Extensions}\\[0.75in]
		\normalsize
		Bryden R. Cais\\
		cais@fas.harvard.edu\\
		(617) 493--2628\\
	    \vspace*{0.3in}
	        Supervised by Noam D. Elkies\\
	    \vspace*{3.75in}
	        A thesis presented to the Department of Mathematics\\
		in partial fulfillment of the requirements\\
		for the degree of Bachelor of Arts with Honors\\
	    \vspace*{.25in}
	        Harvard University\\
		Cambridge, Massachusetts\\
		April 1, 2002
	    \end{center}

\end{titlepage}

\tableofcontents
\newpage

\section{Introduction}

	    Given any Riemann surface $X$, we can consider the field of meromorphic functions on $X$,
	    denoted $K(X)$.  This is always an extension field of $\C$ and is isomorphic to $\C(x,y)$ with
	    $F(x,y)=0$ for some rational function in $x$ and $y$.  Now if $p\,:\,X\longrightarrow Y$
	    is any nonconstant holomorphic mapping of Riemann surfaces, then we can view $K(Y)$
	    as a subfield of $K(X)$ by the injective map $f\mapsto f\circ p$, for any $f\,\in\,K(Y)$.
	    In fact, we have a correspondence between degree $n$ covering maps $p\,:\,X\longrightarrow Y$
	    and degree $n$ field extensions $K(X)/K(Y)$.  The goal of Chapter $1$ is to develop enough
	    of the theory of compact Riemann surfaces to prove part of this correspondence.

	    In Chapter $2$, we turn our attention to 
	    the group $\SL{\Z}$ of two by two integer matrices with determinant $1$, which has a natural action
	    on the complex vector space $\C^2$.  This action descends to an action on the Riemann sphere $\P$,
	    which in turn descends to an action on the complex upper half plane 
	    $\H$, given by
	    \begin{eqnarray*}
		\begin{pmatrix} a&b\\c&d\end{pmatrix}z\ :=\ \frac{az+b}{cz+d}.
	    \end{eqnarray*}
	    Clearly, any subgroup $\Gamma$ of $\SL{\Z}$ also acts on $\H$ by the same formula.  We may therefore
	    consider the quotient space $\H/\Gamma$.  By the appropriate construction, this space can be made into 
	    a compact Riemann surface $X$.  In particular, for each positive integer $N$, we study the Riemann surface 
	    $X(N)$ that arises as a
	    quotient of the upper half plane by the \textit{principal congruence subgroup of level $N$} of $\SL{\Z}$.
	    These are the normal subgroups $\G{N}$ of $\SL{\Z}$ given as the
	    kernel of the reduction modulo $N$ map $r_N\,:\,\SL{\Z}\longrightarrow \SL{\Z/N\Z}$.  
	    There is a natural map $\pi\,:\,X(N)\longrightarrow X(1)$ that realizes $X(N)$
	    as a covering space of $X(1)$.  We can then use the tools of the first chapter to
	    show that we have a field extension $K(X(N))/K(X(1))$.  
	    The fact that $\G{N}$ is normal
	    in $\G{1}$ for each $N$ will enable us to show that this is in fact a Galois field extension
	    with Galois group $\G{1}/\pm\G{N}\simeq \SL{\Z/N\Z}/\pm\{1\}$.
	    
	    In the last chapter, we study the Galois field extensions $K(X(N))/K(X(1))$.
	    For precisely $5$ values of $N$, namely $1\leq N\leq 5$,
	    the Riemann surface $X(N)$ is of genus $0$.  We will show that this implies that the field $K(X(N))$
	    is generated by a single element over $\C$.  We then turn to the theory of Elliptic functions and
	    construct such a generator for each $N$ of interest.  Using this generator, the group $\G{1}/\pm\G{N}$
	    can be made to act on the Riemann sphere.  We show how to interpret this action as the symmetries
	    of an inscribed solid.  Finally, we use the Galois correspondence to give a degree $N$ field
	    extension of $K(X(N))$ for each $2\leq N\leq 5$ corresponding to an index $N$ subgroup
	    of $\G{1}/\pm\G{N}$.  We know that such a field extension is given by adjoining a root
	    of some degree $N$ polynomial over $K(X(1))$.  Using the tools of the first two chapters, we 
	    construct such a polynomial.  As a consequence, when $N=4$ or $5$, we obtain a description
	    of the field extension $K(X(N))/K(X(1))$ as the spliting field of a degree $N$ polynomial.
	    
\chapter{Riemann Surfaces}

\section{Definitions and Notations}

	    A \textit{Riemann surface} $X$ is a one dimensional connected complex analytic
	    manifold.  That is, $X$ is a connected, Hausdorff topological space $S$ equipped 
	    with a \textit{complex structure}.  For every point $P\,\in\,S,$ there exists a neighborhood 
	    $U_P$ of $P$ and a \textit{complex chart} $\varphi_P$ from $U_P$ to the interior of the unit disc 
	    such that $\varphi_P$ is a homeomorphism, and $\varphi_P(P)=0.$  These complex charts are
	    required to be \textit{holomorphically compatible}; that is, for any charts
	    $\varphi_P,\varphi_Q$ the map     
	    \begin{eqnarray*}
	        \varphi_Q\varphi_P^{-1}:\varphi_P(U_P\cap U_Q)\longrightarrow 
		\varphi_Q(U_P\cap U_Q)
	    \end{eqnarray*}
	    is biholomorphic.  Notice that it is immaterial whether we require a complex chart
	    to map a neighborhood of a point homeomorphically to the interior of the unit disc
	    or simply onto an open subset of $\C$ as we can easily convert the latter situation
	    into the former by an appropriate Riemann mapping. 
	    
	    We remark that the usual topological terminology used in describing the 
	    space $S$ carries over to the Riemann surface $X$.  Thus, a point $P$ of
	    $X$ is just the point $P$ of $S$, a closed subset of $X$ is a closed subset of $S$,
	    and so on.  In particular, if the space $S$ is compact, we shall say that $X$ itself 
	    is compact.

\section{Maps Between Riemann Surfaces}

	    \begin{defn}
	    \label{holo}
	        Let $X$ and $X^{\prime}$ be two Riemann surfaces with
	        $f$ any map from $U\subset X$ into $X^{\prime}$. For a point $P$ of $X$, 
		set $Q=f(P)$ and let $\varphi_P,\varphi^{\prime}_Q$ be complex charts
		at $P,Q$ respectively.
	        Recall that $f$ is said to be \textit{analytic} at $P$ if the function 
	        $\varphi_Q^{\prime}f\varphi_P^{-1}$ which maps the interior of the unit disc
	        to itself is complex analytic at the origin, and that if $f$ is analytic at every point
	        of $U$ then it is an \textit{analytic map} from $U$ to $X^{\prime}.$  We also
		call $f$ a \textit{holomorphic mapping}.
	    \end{defn}

	    Let $X$ be any Riemann surface and $U\subset X$ a neighborhood of
	    a point $P_0\,\in\,X$.  Let $f:\,U\longrightarrow \C$ be a one to one analytic map such
	    that $f(P_0)=0$.  Then $t=f(P)$ is said to be
	    a \textit{locally uniformizing variable} at $P_0$.
	    Since $f$ is one to one, we will speak of a point $t$ of $X$, by which we
	    mean the point $P$ of $X$ such that $f(P)=t.$
	    Locally uniformizing variables generalize the idea of complex charts, since
	    for any point $P_0\,\in\,X,$ one may take $t=\varphi_{P_0}(P)$ as a locally
	    uniformizing variable, where $\varphi_{P_0}$ is a complex chart on $X$ at $P_0$.
	    This shows in particular that every point of $X$ has a locally uniformizing variable. 
	    Any two locally uniformizing variables are related to each other by a power series,
	    as the following theorem makes evident.
	    \begin{thm}
	    \label{uniform}
		Suppose that $t=f(P)$ and $s=g(P)$ are two locally uniformizing variables
		at some point $P_0$ of a Riemann surface $X$.  Then for all $P$ near $P_0$
		one has
		    \begin{eqnarray}
			s=a_1t+a_2t^2+\cdots,
		    \label{unif}
		    \end{eqnarray}
		where $a_1\not= 0.$  Conversely, if $t$ is any locally uniformizing variable
		at $P_0$, then any power series of the above form gives another locally 
		uniformizing variable at $P_0$.
	    \end{thm}
	    \begin{proof}
		Let $U,V$ be the domains where $f,g$ are defined.  Since both 
		$s,t$ are locally uniformizing variables, the functions $f,g$ are
		one to one and analytic.  Thus, the map $gf^{-1}:\,f(U\cap V)\longrightarrow g(U\cap V)$
		is one to one and analytic.  Since $gf^{-1}(0)=0$, we have
		$s=gf^{-1}(t)=a_1t+a_2t^2+\cdots$ for all $P$ near $P_0$.  Since this map
		is one to one, we must have $a_1\not=0$.  Conversely, any power series
		of the form \ref{unif} gives an analytic one to one function of $t$ and hence
		an analytic one to one function $g$ of $P$ near $P_0$ with $g(P_0)=0.$ 
	    \end{proof}
	    We now use Theorem \ref{uniform} to prove a key result about any analytic
	    map between two Riemann surfaces.
	    \begin{thm}
	    \label{ramif}
		Let $X,\ Y$ be any two Riemann surfaces and let $f:\,X\longrightarrow Y$ be analytic
		at $P_0$, with $f(P_0)=Q_0$ and $f$ not identically $0$.  Then there exist locally 
		uniformizing variables $t$ at $P_0$ and $s$ at $Q_0$ such that 
		  \begin{eqnarray}
		      s=t^n,\quad n\geq 1.
		  \label{branch}
		  \end{eqnarray}
	    \end{thm}
	    \begin{proof}
		Let $t=\varphi(P)$ and $t^{\prime}=\varphi^{\prime}(Q)$ for $P\,\in\,X$ near $P_0$ and 
		$Q\,\in\,Y$ near $Q_0$ be locally uniformizing variables.  The analyticity of $f$ at
		$P_0$ is equivalent to the condition that $t^{\prime}=\varphi^{\prime}f\varphi^{-1}(t)$
		be analytic at $t=0$.  This amounts to being able to write $t^{\prime}$ as a power
		series in $t$, viz.
		\begin{eqnarray*}
		    t^{\prime}=a_1t+a_2t^2+\cdots.
		\end{eqnarray*}
		Since $f$ is not identically zero, there exists some $n\geq 1$ such that $a_n\not=0$,
		so that 
		\begin{eqnarray*}
		    t^{\prime}=a_nt^n(1+\cdots).
		\end{eqnarray*}
		Since any power series with leading term $a_nt^n$ has an analytic $n^{\mathrm{th}}$
		root, there exists some power series
		\begin{eqnarray*}
		    t_1=b_1t+b_2t^2+\cdots,
		\end{eqnarray*}
		with $b_1\not=0$ and $t^{\prime}=t_1^n$.  The converse of Theorem \ref{uniform} tells
		us that $t_1$ is a locally uniformizing variable at $P_0$, and this completes the proof.
	    \end{proof}

	    The integer $n$ such that $s=t^n$ is in fact independent of the choice of uniformizing variable,
	    which follows from the prrof of Theorem \ref{ramif}.
	    \begin{thm}[Identity Theorem]
	    \label{identity}
		Let $X,\ Y$ be Riemann surfaces, and $f:\,X\longrightarrow Y$ a holomorphic mapping.  
		If $f=0$ on a set $A$ having a limit point $a\,\in\,X$, then $f\equiv0$
		on $X$.   
	    \end{thm}
	    \begin{proof}
		Let $U\subset X$ be the set of all points in $X$ having a neighborhood $W$ such that
		$f\equiv 0$ on $W$.  By definition, $U$ is open.  We show that it is closed.  Let
		$b$ be a limit point of $U$.  Since $f$ is continuous, $f(b)=0$.  Let 
		$\varphi_1\,:\,U_1\longrightarrow V_1$ and $\varphi_2\,:\,U_2\longrightarrow V_2$
		be locally uniformizing variables at $b$ and $f(b)$ with the property that $U_1$
		is connected.  The map
		\begin{eqnarray*}
		    \varphi_2\circ f\circ\varphi_1^{-1}\,:\,V_1\longrightarrow V_2\subset\C
		\end{eqnarray*}
		is holomorphic since $f$ is.  Moreover, $U\cap U_1\not=\emptyset$ since
		$b$ is a limit point of $U$ and $U_1$ is open.  By the identity theorem
		for holomorphic functions on domains in $\C$ \cite[pg. 228]{rem}, we see that $f$ is identically
		$0$ on $U_1$ and that $a\,\in\,U$.  Thus, $b\,\in\,U$ and $U$ is closed.  Since $X$ is connected,
		we must have $U=\emptyset$ or $U=X$.  Since $a\,\in\,U$, the first case is impossible and
		$f\equiv 0$ on $X$.
	    \end{proof}
	    
\section{Functions on a Riemann Surface}

	    Recall that a holomorphic function on an open subset $Y$ of a Riemann surface $X$ is 
	    just a holomorphic mapping $f:\,Y\longrightarrow \C.$
	    It is not difficult to see that the sum and product
	    of any two holomorphic functions on an open subset $Y$ of a  Riemann surface 
	    $X$ are again holomorphic.  Moreover, the constant functions are holomorphic. 
	    Therefore the set of all holomorphic functions on $Y$, denoted $\mathcal{O}(Y)$,
	    is endowed with the structure of a $\C$-algebra.  Every locally uniformizing
	    variable on $Y$ is clearly holomorphic.  Sometimes we will have a function
	    $f$ which is holomorphic on some deleted neighborhood of a point $a\,\in\,Y\subset X$
	    and we will want to extend $f$ to a holomorphic function on the entire (undeleted)
	    neighborhood.  The following well known theorem tells us when this is possible:
	    \begin{thm}[Riemann's Removable Singularities Theorem]
	    \label{RRS}
		Let $Y$ be an open subset of a Riemann surface $X$ and let $a\,\in\,Y.$  
		The function $f\,\in\,\mathcal{O}(Y\setminus \{a\})$ may be uniquely continued to
		a function $\tilde{f}\,\in\,\mathcal{O}(Y)$ precisely when $f$ is bounded 
		on some deleted neighborhood of $a$.
	    \end{thm}
	    We do not prove this theorem, but note that it follows easily from the analogous
	    principle for holomorphic functions on $\C$ \cite[pg. 5]{forster}.

	    Given a Riemann surface $X$, we would like for the $\C$-algebra $\mathcal{O}(X)$ 
	    to have the structure of a field, but this is not possible if we restrict ourselves 
	    to working only with holomorphic functions, as dividing by a function $f$ with a 
	    zero at some point $a$ introduces a pole at $a$.  Moreover, we will be working 
	    only with compact Riemann surfaces, and as the following lemma shows, holomorphic
	    functions on a compact Riemann surface are very uninteresting:
	    \begin{lem}
	    \label{boring}
		Every holomorphic function $f$ on a compact Riemann surface $X$ is constant.
	    \end{lem}
	    \begin{proof}
		By definition, $f$ is a holomorphic mapping $f:\,X\longrightarrow \C$.
		Suppose that $f$ is nonconstant.
		First, $f(X)$ is open.  This follows from Theorem \ref{ramif}, since
		given a point $a\,\in\,X$ with $f(a)=b,$ there exist locally uniformizing
		variables $t,\ s$ at $a,\ b$ with $s=t^k$, so that $f$ maps a neighborhood
		of $a$ to a neighborhood of $b$.  Second, $f(X)$ is compact since $X$ is.  
		Thus, $f(X)$ is a compact, open subset of $\C$, and therefore empty.  
		This is a contradiction.
	    \end{proof} 
	    The remedy to these problems is to allow functions to take the value $\infty$.
	    \begin{defn}
	    \label{mero}
		Let $Y$ be an open subset of a Riemann surface $X$.  A \textit{meromorphic function}
		$f$ on $Y$ is a holomorphic mapping(other than the constant mapping
		$f\equiv \infty$) to the Riemann sphere: $f:\,Y\longrightarrow \P$.
	    \end{defn}
	    Let $z=f(P)$ be a meromorphic function on $Y\subset X$ and suppose that $f$ is analytic
	    at the point $P_0\,\in\,Y$.  We then have two cases:
	    \begin{enumerate}
		\item $f(P_0)=z_0\not=\infty.$  Then $z-z_0$ is a locally uniformizing variable at
		$P_0$.  Now let $t$ be any locally uniformizing variably at $P_0$.  Then by Theorem 
		\ref{uniform} we may write $z=a_0+a_1t+a_2t^2+\cdots.$  The least integer $n$ such
		that $a_n\not=0$ is the \textit{order} of $f$ at $P_0$.  

		\item $f(P_0)=\infty.$  Then $1/z$ is a locally uniformizing variable at $P_0$ and we have
		$1/z=a_1t+a_2t^2+\cdots.$  This enables us to write $z=b_{-n}t^{-n}+b_{-n+1}t^{-n+1}+\cdots$
		for some positive integer $n$.  The integer $-n$ is called the \textit{order} of 
		$f$ at $P_0$.
	    \end{enumerate}
	    Thus we see that a zero of $f$ at $P_0$ corresponds to $f$ having positive order at $P_0$
	    while a pole corresponds to $f$ having negative order.  A function that is identically $0$
	    is defined to have order $\infty$.  By the remark after the proof of Theorem \ref{ramif}, the order
	    of a meromorphic function $f$ at a point $P_0$ is well defined (that is, it does not depend 
	    on a choice of uniformizing variable).

	    Clearly, the sum and product of any two meromorphic functions on a Riemann surface $X$
	    is again meromorphic.  Moreover, by Theorem \ref{identity}, we see that any meromorphic 
	    function that is not identically $0$ has isolated zeroes so that its reciprocal is meromorphic.
	    Thus, the set of all meromorphic functions on $X$ is a field, denoted $K(X).$  Notice that
	    every constant function $f:\,X\longrightarrow \P$ excluding $f\equiv \infty$ is a meromorphic
	    function, so that $K(X)$ contains $\C$.  We record this important fact here:
	    \begin{thm}
	    \label{field}
		The set of all meromorphic functions on a Riemann surface $X$ forms a field, denoted
		$K(X)$, and is an extension field of $\C$.  
	    \end{thm}
	    The field $K(X)$ of meromorphic functions on a Riemann surface $X$ is an important 
	    invariant of $X$ (that is, under isomorphism) and, as we shall see, encodes many of the 
	    properties of $X$.

	    One might protest that we do not yet know that there are any nonconstant meromorphic
	    functions on a given Riemann surface.  In fact, given any $n$ distinct points 
	    $P_1,P_2,\ldots,P_n$ in a Riemann surface $Y$, there exists a function $f\,\in\,K(Y)$ such that
	    $f(P_j)\not= f(P_i)$ for $i\not=j$.  We will prove this assuming the following:
	    \begin{thm}\cite[pg. 122]{iwasawa}
	    \label{existence}
		Let $P_1,\ P_2$ be any two distinct points on a Riemann surface $X$.  Then there
		exists $f\,\in\,K(X)$ with a zero of order one at $P_1$ and a pole of order one 
		at $P_2$.
	    \end{thm}
	    We now show by induction on $n$ that there exists $f\,\in\,K(X)$ with $f(P_j)$ distinct
	    for $j=1,2,\ldots,n$.  By Theorem \ref{existence}, we already know the result for $j=2$, 
	    so assume that we have some
	    $f\,\in\,K(X)$ such that $f(P_1),\ldots ,f(P_k)$ are all distinct.  Either $f(P_{k+1})$
	    is distinct from these values or we may reindex the $P_j$ so that $f(P_k)=f(P_{k+1})$.  
	    Now we can find some fractional linear transformation
	    which moves the values $f(P_j)$ for $j=1,\ldots,k$ all away from $\infty$, with the
	    resulting composition still being meromorphic.  Thus, assume that $f(P_j)$ is finite for
	    each $j$.  By Theorem \ref{existence}, we have some $g\,\in\,K(X)$ with 
	    $g(P_k)\not=g(P_{k+1})$.  Now consider
	    \begin{eqnarray*}
		\varphi_c\ =\ f+cg
	    \end{eqnarray*}
	    with $c\,\in\,\C$.  Clearly, $\varphi_c(P_i)=\varphi_c(P_j)$ if and only if 
	    \begin{eqnarray*}
		c\ =\ \frac{f(P_i)-f(P_j)}{g(P_i)-g(P_j)}.
	    \end{eqnarray*}
	    Since there are a finite number of pairs $P_i,\ P_j$, we can find a $c\,\in\,\C$ such that
	    $\varphi_c(P_i)\not=\varphi_c(P_j)$ for $i\not=j$, as claimed.

	    As an example of working with function fields on a Riemann surface, we give an explicit description
	    of the field $K(\P)$.  	
	    \begin{thm}\cite[pg. 11]{forster}
	    \label{sphere}
		$K(\P)\simeq \C(x)$.
	    \end{thm}
	    \begin{proof}
		Any $f\,\in\,K(\P)$ is a holomorphic mapping $f:\,\P\longrightarrow \P.$  Since $\P$ is compact,
		we see that $f$ has at most a finite number of poles, for if not, then the set of all the poles
		of $f$ would have a limit point in $\P$ and by Theorem \ref{identity}, $f$ would be identically 
		$\infty,$ and hence not a meromorphic function.  Without loss of generality, suppose that $\infty$
		is not a pole of $f$ (otherwise replace $f$ by $1/f$).  Now let $p_1,\ldots,p_n$ denote the poles
		of $f$ and let
		\begin{eqnarray*}
		    R_j(z)\ =\ \sum_{i=-k_j}^{-1}a_{i,j}(z-p_j)^i
		\end{eqnarray*}
		be the principal part of $f$ at $p_j$.  The function $f-(R_1+R_2+\ldots+R_n)$ is then
		a pole-free meromorphic function; that is, a holomorphic function on $\P$.  From
		Lemma \ref{boring}, it must be a constant.  Since each $R_j$ is a rational function
		of $z$, we conclude that $f$ must be rational also.
	    \end{proof}

\section{Branched Covers of a Riemann Surface}

	    \begin{defn}
	    \label{cover}
		Let $X$, $Y$ be topological spaces.  A map $p\,:\,X\longrightarrow Y$ is a
		\textit{covering map} if for any $y\,\in\,Y$, there exists a neighborhood
		$U$ of $y$ such that
		\begin{eqnarray*}
		    p^{-1}(U)=\coprod_{i=1}^n V_i
		\end{eqnarray*}
		where each $V_i$ is mapped homeomorphically by $p$ to $U$ and the $V_i$
		are disjoint.  If $X,\ Y$ are Riemann surfaces, we require that the map
		$p$ be holomorphic.
	    \end{defn}

	    As it turns out, the requirement that \textit{every} point of $Y$ should have $n$
	    distinct inverses in $X$ is rather rigid, and excludes most of the holomorphic
	    mappings between Riemann surfaces that interest us.  For example, if $n>1$ then $x\mapsto x^n$ 
	    is a holomorphic self-mapping of $\P$, but it is not a covering map in the above sense
	    since $0,\ \infty$ have only one inverse image each.  Moreover, Theorem \ref{ramif} tells us that
	    any holomorphic mapping between Riemann surfaces looks locally like $x^n$ for some $n$.  
	    This leads naturally to the idea of \textit{branch points} and \textit{branched coverings}.
	    \begin{defn}
		Let $X,\ Y$ be Riemann surfaces, $p\,:\,X\longrightarrow Y$ a holomorphic mapping,
		and $P_0,Q_0$ points of $X,Y$ respectively with $Q_0=p(P_0)$.
		By Theorem \ref{ramif}, there exist locally uniformizing variables $t$ at $P_0$ and
		$s$ at $Q_0$ with $s=t^n$ for some $n\geq 1$.  If $n\not=1$, the point $P_0\,\in\,X$ is called
		a \textit{branch point} and $Q_0\,\in\,Y$ a \textit{branch value}.
		The integer $n-1$ is called the \textit{branch number} of $p$ at $P_0$ and will
		be denoted $b_p(P_0)$.
	    \end{defn}
	    We shall also refer to branch points as \textit{ramification points}, and will often
	    say that $P_0$ has \textit{order} $n$ over $Q_0$ or that $Q_0$ is \textit{ramified}
	    of order $n$.  We can now
	    modify our definition of a covering map of Riemann surfaces to include branch points:
	    \begin{defn}
		Let $X,\ Y$ be Riemann surfaces and $p\,:\,X\longrightarrow Y$ a holomorphic mapping.
		If there is a discrete set $A\subset Y$ such that
		\begin{eqnarray*}
		    p\,:\,p^{-1}(Y\setminus A)\longrightarrow Y\setminus A
		\end{eqnarray*}
		is a covering map, then the map $p\,:\,X\longrightarrow Y$ is a \textit{branched covering map}.
	    \end{defn}
	    With this definition, we readily have:
	    \begin{thm}
	    \label{branchcov}
		Let $X$, $Y$ be compact Riemann surfaces and $p\,:\,X\longrightarrow Y$ a non-constant 
		holomorphic mapping.  Then $p$ is a branched covering map.
	    \end{thm}
	    \begin{proof}
		Let $p$ be as above.
		We have shown in the proof of Lemma \ref{boring} that any holomorphic mapping
		between compact Riemann surfaces is surjective.  Moreover, since $p$ is non-constant, 
		we see by Theorem \ref{identity} that the critical points of $p$ form a discrete
		and therefore finite set $B\subset X$.  Thus, $p$ is a local homeomorphism
		on $X\setminus B$.  Moreover, the inverse image under $p$ of any point in $Y$ must
		be finite since it is discrete.  Let $y\,\in\,Y$ and put 
		$p^{-1}(y)=\{x_1,\ldots,x_n\}$.  Let $U_j$ be a neighborhood of $x_j$ and $V_j$ a 
		neighborhood of $y$ such that $p\,:\,U_j\longrightarrow V_j$ is biholomorphic and
		the $U_j$ are disjoint.  Now let $V\subset \bigcap_{j=1}^n V_j$ such that 
		$p^{-1}(V)\subset \bigcup_{j=1}^n U_j$ and put $W_j=U_j\cap P^{-1}(V)$.  Then
		the $W_j$ are disjoint and $p\,:\,W_j\longrightarrow V$ is biholomorphic for
		each $j$.
	    \end{proof}

	    \begin{thm}
		Let $X,\ Y$ be compact Riemann surfaces and $f\,:\,X\longrightarrow Y$
		a non-constant holomorphic mapping.  There exists a positive integer $m$
		such that every point of $Y$ is taken precisely $m$ times by $f$, counting
		multiplicities.  
	    \end{thm}
	    \begin{proof}
		For each integer $n\geq 1$, define
		\begin{eqnarray*}
		    B_n\ =\ \left\{Q\,\in\,Y\,:\,\sum_{P\in f^{-1}(Q)}(b_f(P)+1)\geq n\right\}.
		\end{eqnarray*}
		By Theorem \ref{ramif}, we see that $B_n$ is open.  We show
		that it is also closed.  Let $Q_k\,\in\,B_n$ and suppose that 
		$\lim_{k\rightarrow\infty} Q_k=Q$.  The proof of Theorem \ref{branchcov} shows that
		the set of branch points of $f$ is finite.
		Thus, without loss of generality we may assume that $b_f(Q_k)=0$ for all $k$, i.e.
		that $f^{-1}(Q_k)$ has cardinality $n$ or greater.  Let $P_{1k},P_{2k},\ldots,P_{nk}$
		be $n$ distinct points of $f^{-1}(Q_k)$.  Again, since $X$ is compact, for each 
		$j=1,\ldots,n$, there exists a subsequence of $\{P_{jk}\}$ that converges to a 
		limit $P_j$, and $f(P_j)=Q$.  Of course, the $P_j$ might not be distinct, but 
		since $f(P_{jk})=Q_k$ for all $k$, we must have
		\begin{eqnarray*}
		    \sum_{P\in f^{-1}(Q)}(b_f(P)+1)\geq n.
		\end{eqnarray*}
		Therefore, since
		$B_n$ is both open and closed for any $n\geq 1$, we see that $B_n$ is either
		empty of all of $X$ for each $n$.  Now pick some $Q_0\,\in\,Y$ and set
		$\sum_{P\in f^{-1}(Q_0)}(b_f(P)+1)=m$.  Then by compactness again, $m$ is finite
		and we have $B_m=X$ and $B_n$ empty for all $n>m$.
	    \end{proof}

	    \begin{defn}
		Let $X,\ Y,\ f,$ and $m$ be as above.
		Since $f$ is a branched covering map by Theorem \ref{branchcov},
		we call the integer $m$ the \textit{degree} of the cover $f$, or simply the
		\textit{degree} of $f$.  We also refer to $f$ as an \textit{m-sheeted} branched 
		cover of $Y$ by $X$.  
	    \end{defn}
	    Geometrically, we view $X$ as consisting of $m$ copies of $Y$, each copy mapping biholomorphically
	    (after the branch points have been excluded) to $Y$.  The branch points are intuitively where
	    the copies of $Y$ (``sheets'') are glued together to form $X$.

\section{Deck Transformations}

	    \begin{defn}
		Let $X,Y$ be Riemann surfaces and $p\,:\,X\longrightarrow Y$ a branched covering map.
		A \textit{deck transformation} is a fiber preserving biholomorphic map, that is, a map
		$f$ such that the diagram
		\begin{eqnarray*}
	        \begin{CD}
		  X   @>f>>  X\\
		@VpVV      @VVpV\\
		  Y   @=     Y
	        \end{CD}
	        \end{eqnarray*}
		commutes.
		It is not difficult to see that the set of all deck transformations of the cover
		$p$ forms a group under composition.  We denote this group $\deck(X/Y)$.
	    \end{defn}
	    \begin{defn}
		Let $X,Y$ be Riemann surfaces, $p\,:\,X\longrightarrow Y$ a branched covering map,
		and $A\subset Y$ the set of branch values of $p$.
		The covering is \textit{normal} if the Deck group $\deck(X/Y)$ acts transitively
		on the fiber $p^{-1}(Q)$ for all points $Q\,\in\,Y\setminus A$.
	    \end{defn}
	    The notation $\deck(X/Y)$ is suggestive.  Namely, the group of Deck transformations
	    of a cover is intuitively very much like the Galois group of a field extension since
	    mappings $f\,\in\,\deck(X/Y)$ are required to ``fix'' the base space, $Y$, while
	    permuting the points in any given fiber $p^{-1}(Q)$.  The analogy in fact turns out 
	    to be a correspondence.

\section{The Main Correspondence}

	    Given any branched covering map $p\,:\,X\longrightarrow Y$, any function $f\,\in\,K(Y)$
	    gives rise to a function $g\,\in\,K(X)$ via the \textit{pullback}
	    $p^*(f)=fp=g$.  We in fact have an injection $p^*\,:\,K(Y)\longrightarrow K(X)$
	    so that we may view $K(Y)$ as a subfield of $K(X)$, and we will often exploit this fact.
	    \begin{thm}[Main Theorem]
	    \label{main}
		Let $X,Y$ be compact Riemann surfaces and $p\,:\,X\longrightarrow Y$ an $n$-sheeted
		branched covering map.  Then $K(X)/p^*K(Y)$ is a degree $n$ field extension. 
		Conversely, let $Y$ be a Riemann surface and $L/K(Y)$ a degree $n$ field extension.
		Then there exists a Riemann surface $X$,
		an $n$-sheeted branched covering map $p\,:\,X\longrightarrow Y$, and $f\,\in\,K(X)$
		such that $L\simeq K(X)=p^*K(Y)(f)$.
		In both cases, the Deck group $\deck(X/Y)$ is isomorphic to $\aut(K(X)/p^*K(Y))$. 
	    \end{thm}
	    We will only prove the first assertion.  For proofs of the rest, see \cite{forster}.
	    \begin{proof}
		Let $f\,\in\,K(X)$ and denote the set of branch values of $p$ by $A$.  We know
		that $A$ is finite and that the inverse image
		under $p$ of a point $Q\,\in\,Y\setminus A$ consists of $n$ distinct 
		points $P_1,\ldots P_n$.  Since $p$ is a branched covering map, for each such $Q$ we have
		a neighborhood $U_Q$ of $Q$ such that 
		\begin{eqnarray*}
		    p^{-1}(U_Q)\ =\ \coprod_{i=1}^n V_i
		\end{eqnarray*}
		with $V_i$ a neighborhood of $P_i$, the neighborhoods $V_i,V_j$ disjoint for $i\not=j$, and
		$p\,:\,V_i\longrightarrow U_Q$ biholomorphic.  Denote the restriction of $p$
		to $V_i$ by $p_i$.  Since $p_i^{-1}$ is well defined on $Y$,
		\begin{eqnarray*}
		    (p_i^{-1})^*f\ =\ fp_i^{-1}
		\end{eqnarray*}
		is a meromorphic function on $U_Q$ for each $i$.  Therefore, the elementary symmetric
		functions
		\begin{eqnarray*}
		    s_j\ =\ \sum_{1\leq i_1<\cdots<i_j\leq n}\ \prod_{k=1}^j (p_{i_k}^{-1})^*f
		\end{eqnarray*}
		for $j=1,2,\ldots,n$ are meromorphic functions on $U_Q$.  Repeathing this argument
		for each $Q\,\in\,Y\setminus A$, we see that these functions piece together
		to form meromorphic functions $s_j$ defined on all of $Y\setminus A$.  Call
		these functions the \textit{elementary symmetric functions} of $f$ with respect
		to the covering $p$.  We now show that the elementary symmetric functions of
		$f$ may be continued meromorphically to all of $Y$.  Let $a\,\in\,A$.  Since $X$ is
		compact, $p^{-1}(a)$ is finite, say $p^{-1}(a)=\{b_1,\ldots,b_k\}$.  Since $f$ is
		meromorphic, it has isolated poles, so that we may take a neighborhood $U$ of $a$
		so that the only possible poles of $f$ in $V=p^{-1}(U)$ occur at the $b_j$.  Let
		$t$ be a locally uniformizing variable on $U$ at $a$.  Then
		$t(a)=0$, so that the function $p^*t=t p\,\in\,\mathcal{O}(V)$
		vanishes at each $b_j$.  Since $f$ is meromorphic, it has \textit{finite} order
		at each of the $b_j$ so that we can find an integer $k$ such that 
		\begin{eqnarray*}
		    (p^*t)^k f
		\end{eqnarray*}
		is holomorphic, and hence bounded, on $V$.  Thus, if $r_j$ are the elementary
		symmetric functions of $(p^*t)^k f$, then the $r_j$ are bounded on $U\setminus\{a\}$.
		By Theorem \ref{RRS}, the $r_j$ can be holomorphically extended to all of $U$.  Since
		$p^*t$ is a meromorphic function on $V\subset X$, the elementary symmetric functions
		of $p^*t$ are just $\bigl(\begin{smallmatrix}n\\j\end{smallmatrix}\bigr) t^j$, so that
		we have
		\begin{eqnarray*}
		    r_j\ =\ t^{kj} s_j
		\end{eqnarray*}
		for each $j$.  Since $t$ is a meromorphic function on $U\subset Y$ and $r_j$ may
		be holomorphically continued to all of $U$, we see that the $s_j$ can be
		meromorphically continued to $U$ for all $j$.  This shows that the elementary
		symmetric functions of $f$ with respect to $p$ are in $K(X)$.  Therefore, for
		every point $P\,\in\,X$, we have
		\begin{eqnarray*}
		    f^n-p^*s_1f^{n-1}+p^*s_2f^{n-1}+\cdots+(-1)^n p^*s_n\ =\ 0.
		\end{eqnarray*}
		This shows that the minimal polynomial of $f$ over $K(Y)$ has degree at most $n$.
		
		We showed after Theorem \ref{existence} that given $n$ distinct points in $X$, 
		there exists some $f\,\in\,K(X)$ that separates them.  Thus, for a point
		$Q_0\,\in\,Y$ with $n$ distinct preimages $p^{-1}(Q_0)=\{P_1,\ldots,P_n\}$, we have
		a $f\,\in\,K(X)$ with $f(P_i)\not=f(P_j)$ for $i\not=j$.  Moreover, since
		$f$ is continuous, there exists a neighborhood $U$ of $Q_0$ such that
		$f$ takes on $n$ distinct values for every $P\,\in\,p^{-1}(U)$.  Let $m<n$ and
		suppose that the minimal polynomial of $f$ over $K(Y)$ is
		\begin{eqnarray*}
		    f^m+c_1f^{m-1}+\cdots+c_m
		\end{eqnarray*}
		where $c_j\,\in\,K(Y)$ for each $j$.  Then the polynomial
		$f^m+c_1(Q)f^{m-1}+\cdots+c_m(Q)\,\in\,\C[f]$ has $n$ distinct roots for \textit{every}
		$Q\,\in\,U$, which implies that the $c_j$ are identically zero on $U$, and by 
		Theorem \ref{identity}, identically zero on $K(Y)$.  This is a contradiction, and
		the minimal polynomial of $f$ has degree $n$.

		Since $K(Y)$ is of characteristic zero, the primitive element theorem applies and
		$K(X)$ is generated as a field extension of $K(Y)$ by a single element.  We claim
		that
		\begin{eqnarray}
		\label{algfunfield}
		    K(X)\ =\ K(Y)(f).
		\end{eqnarray}
		To see this, notice that since the minimal polynomial of $f$ over $K(Y)$ has
		degree $n$, we have $[K(Y)(f):K(Y)]=n$.  Now let $g\,\in\,K(X)$.  Again, by
		the primitive element theorem we have $K(Y)(f,g)=K(Y)(h)$ for some $h\,\in\,K(X).$
		But we have shown that the minimal polynomial of $h$ has degree at most $n$.
		We then have
		\begin{eqnarray*}
		    n=[K(Y)(f):K(Y)]\leq[K(Y)(f,g):K(Y)]\leq n,
		\end{eqnarray*}
		from which we conclude that $g\,\in\,K(Y)(f)$ already.  This gives our claim and 
		completes the proof.
	    \end{proof}
	    Theorem \ref{main} is a very deep result.  From it we obtain 
	    \begin{thm}
	    \label{unique}
		Let $X,\ Y$ be compact Riemann surfaces.  Then $X\simeq Y$ if and only if
		$K(X)\simeq K(Y)$.
	    \end{thm}
	    In the particular case of branched covering maps $p\,:\,X\longrightarrow \P$ from a compact 
	    Riemann surface to the Riemann Sphere, we see by \ref{algfunfield} and Theorem \ref{sphere} that 
	    \begin{align}
	    \label{twogens}
		K(X)&\ =\ K(\P)(f)\ \simeq\ \C(x,s)\\
	    \intertext{with} 
	    F(x,s)&\ =\ 0,
	    \end{align}
	    where $f\,\in\,K(X)$ and $F$ is a rational function of two variables and degree $[K(X):p^*K(Y)]$.

\chapter{The Group $\SL{\Z}$ and its Subgroups}

\section{Definitions and Properties}
	
	    Given a ring $R$, we let $\SL{R}$ denotes the group of two by two matrices with entries
	    in $R$ of determinant $1$ in $R$.
	    Now $\SL{\Z}$ has a natural action on $\C^2$ which descends to $(\C^2\setminus\{0\})/\C^*$.  
	    Since $(\C^2\setminus\{0\})/\C^*\simeq \P$, we see that $\SL{\Z}$ acts on $\P$, and the
	    resulting action is called \textit{fractional linear transformation}.  Explicitly,
	    write $z\,\in\,\P$ as $x/y$ for $(x,y)\,\in\,\C^2\setminus\{0\}$.  
	    For any $\alpha=\bigl(\begin{smallmatrix}a & b\\c & d\end{smallmatrix}\bigr)\,\in\,\SL{\Z}$
	    we have 
	    \begin{eqnarray*}
	        \alpha\begin{pmatrix}x\\y\end{pmatrix}\ =\ \begin{pmatrix}ax+by\\cx+dy\end{pmatrix}, 
	    \end{eqnarray*}
	    which gives 
	    \begin{eqnarray*}
	        \alpha z&=&\frac{ax+by}{cx+dy}\\
		&=&\frac{az+b}{cz+d}.
	    \end{eqnarray*}
	    This action in fact descends to $\H$, which is verified by the formula
	    \begin{eqnarray}
	    \label{impart}
		\Im(\alpha(z))\ =\ \frac{\Im(z)}{|cz+d|^2},
	    \end{eqnarray}
	    where $\alpha$ is as before.  In fact, since 
	    \begin{eqnarray*}
		\frac{-az-b}{-cz-d}\ =\ \frac{az+b}{cz+d},
	    \end{eqnarray*}
	    we see that $\SL{\Z}/\pm 1=\PSL{\Z}$ acts on $\H$, where $1$ denotes the identity matrix.
	    Given any subgroup $\Gamma\subset \SL{\Z}$, we will denote by $\bar{\Gamma}$ the image
	    of $\Gamma$ in $\SL{\Z}/\{\pm 1\}$.
	    We now define an important class of normal subgroups of $\SL{\Z}$.
	    \begin{defn}
		For any positive integer $N$ define $\G{N}$ to be the subgroup of $\SL{\Z}$ 
		consisting of those matrices that are congruent modulo $N$ to the identity.
		That is, 
		\begin{eqnarray*}
		    \G{N}\ =\ \left\{
		    \begin{pmatrix}
	            a & b\\
	            c & d
	            \end{pmatrix}\,\in\,\SL{\Z}\,:\,a\equiv d\equiv 1,\ 
		    b\equiv c\equiv 0\ \mod N\right\}.
		\end{eqnarray*}
	    \end{defn}
	    We call $\G{N}$  the \textit{principal congruence subgroup of level} $N$.
	    Notice that that $\G{N}$ acts on $\H$ since it is a subgroup of $\G{1}=\SL{\Z}$.

\section{The Structure of $\G{N}$}
	    
	    Let $r_N:\,\SL{\Z}\longrightarrow \SL{\Z/N\Z}$ denote the reduction modulo
	    $N$ map.  Clearly, $\G{N}$ is the kernel of $r_N$ and is hence a normal
	    subgroup of $\G{1}$.  More is true, however, as the following theorem makes evident:
	    \begin{thm}\cite[pg. 61]{lang}
	    \label{exact}
	    The sequence
	        \begin{eqnarray*}
		    \begin{CD}
		        1@>>>\G{N}@>>>\SL{\Z}@>r_N>>\SL{\Z/N\Z}@>>>1
		    \end{CD},
	        \end{eqnarray*}
	    where the first two maps are the obvious inclusions and the last map is
	    trivial, is exact.
	    \end{thm}
	    \begin{proof}
		The only thing that is not obvious is the surjectivity of $r_N$.  Let
		\begin{eqnarray*}
		    \alpha\ =\ 
		    \begin{pmatrix}
	            a & b\\
	            c & d
	            \end{pmatrix}
		\end{eqnarray*}
		be an element of $\GL{\Z}$ with $ad-bc\equiv 1\ \mod N$.  Recall \cite[pg. 61]{lang}
		that there exist $\gamma,\ \delta\,\in\,\PSL{\Z}$ such that
		$\gamma\alpha\delta$ is diagonal.  We may therefore assume that $\alpha$ is
		diagonal since $r_N$ is a homomorphism.  Put
		\begin{eqnarray*}
		    \alpha\ =\ 
		    \begin{pmatrix}
	            a & 0\\
	            0 & d\\
	            \end{pmatrix}
		\end{eqnarray*}
		with $ad\equiv 1\ \mod N$.  Since $d$ has an inverse modulo $N$, 
		we have $(d,N)=1$, and hence there exist integers $u,\ v$ such that
		\begin{eqnarray*}
		    ud+vN=1.
		\end{eqnarray*}
		Let $ad=1+rN$ for some integer $r$.  Then the matrix
		\begin{eqnarray*}
		    \beta\ =\ 
		    \begin{pmatrix}
	            a-ruN & rvN\\
	            N & d\\
	            \end{pmatrix}
		\end{eqnarray*}
		satisfies $r_N(\beta)=\alpha$ and has determinant $1$ since 
		$ad-ruNd-rvN^2=1+rN(1-ud-vN)=1$.  Thus $r_N$ is surjective.
	    \end{proof}
	    We can now determine the structure of $\G{1}/\G{N}$.
	    \begin{cor}
		For any positive integer $N$, we have $\G{1}/\G{N}\simeq \SL{\Z/N\Z}$.
	    \end{cor}
	    \begin{proof}
		This follows from an application of the first isomorphism theorem
		to the exact sequence given in Theorem \ref{exact}.
	    \end{proof}

\section{Fundamental Domains}

	    Since $\G{N}$ acts on $\H$, we may consider the quotient space $\H/\G{N}$
	    consisting of the $\G{N}$ orbits on $\H$.  
	    \begin{defn}
	    \label{fundamental}
		Given a subgroup $\Gamma$ of $\G{1}$, a \textit{fundamental domain} for
		$\Gamma$ is a connected open subset $F\subset \H$ such that any point
		$z$ of $\H$ is equivalent modulo $\Gamma$ to some point of the closure
		of $F$ and no two points of $F$ are equivalent under $\Gamma$.
	    \end{defn}
	    As an example, we determine a fundamental domain for $\G{1}$.
	    \begin{thm}
	    \label{fun}
		A fundamental domain $F$ for $\G{1}$ is the open set in $\H$ bounded
		by the lines $\Re(z)=-\frac{1}{2}$, $\Re(z)=\frac{1}{2}$
		and the unit circle $\{z\,:\,|z|=1\}$.  Moreover,
		\begin{eqnarray*}
		    T\ :=\ 
		    \begin{pmatrix}
	            1 & 1\\
	            0 & 1
	            \end{pmatrix}\quad \text{and} \quad
		    S\ :=\ 
		    \begin{pmatrix}
	            0 & -1\\
	            1 & 0
	            \end{pmatrix}
		\end{eqnarray*}
		generate $\G{1}$. 
	    \end{thm}
	    \begin{proof}
	        (Adapted from \cite[pg. 16]{shimura} and \cite[pg. 30]{lang})
		Let $\Gamma^{\prime}$ be the subgroup of $\G{1}$ generated by $S,\ T.$
		First we show that every $z\,\in\,\H$ is equivalent under $\Gamma^{\prime}$ 
		to some $z^{\prime}$ in the closure of $F$.  Let $z\,\in\,\H$ and put 
		$\sigma=\bigl(
		    \begin{smallmatrix}
	            a & b\\
	            c & d
	            \end{smallmatrix}\bigr)\,\in\,\Gamma^{\prime}$.  
		Since the set
		\begin{eqnarray*}
		    \left\{cz+d\,:\,c,\ d\,\in\,\Z\right\}
		\end{eqnarray*}
		is a lattice in $\C$, the quantity $|cz+d|$ for $(c,d)\,\in\,\Z^2\setminus \{0\}$ is 
		bounded below.  By \ref{impart}, the set $\left\{\Im(\sigma(z))\,:\,\sigma\,\in\,\Gamma^{\prime}\right\}$
		is bounded above, say by
		$\sigma_0(z)=w$.  Since $\sigma_0,\ S\,\in\,\Gamma^{\prime}$, we have
		\begin{eqnarray*}
		    \Im(S\sigma_0(z))\ =\ \Im\left(\frac{-1}{w}\right)\ =\ 
		    \frac{\Im(w)}{|w|^2}\ \leq\ \Im(w),
		\end{eqnarray*}
		so that $|w|\geq 1.$  Now there exists some $n\,\in\,\Z$ such that
		\begin{eqnarray*}
		    -\frac{1}{2}\,\leq\,\Re(T^nw)\,\leq\,\frac{1}{2}.
		\end{eqnarray*}
		and obviously $\Im(T^nw)=\Im(w)$, so that $z^{\prime}=T^n\sigma_0(z)$
		lies in the closure of $F$ and is $\Gamma^{\prime}$ equivalent to $z$.  Since
		$\Gamma^{\prime}\subset \G{1}$, we trivially have that every point of $\H$ is equivalent
		under $\G{1}$ to a point in $F$.

		Now we show that no two elements of $F$ are equivalent under $\G{1}$ (and hence under 
		$\Gamma^{\prime}$).  Let $z,\ z^{\prime}$ be distinct points of $F$ and suppose that
		there exists $\sigma=\bigl(
		    \begin{smallmatrix}
	            a & b\\
	            c & d
	            \end{smallmatrix}\bigr)\,\in\,\G{1}$ with $z^{\prime}=\sigma(z)$.  Without loss of
		generality, assume that $\Im(z)\leq \Im(z^{\prime})=\Im(z)/|cz+d|^2$.
		We therefore have 
		\begin{eqnarray*}
		    |c|\,\Im(z)\ \leq\ |cz+d|\ \leq\ 1.	
		\end{eqnarray*}
		Notice that if $c=0$ then $d=\pm 1$ which forces $a=d$ and $z^{\prime}=z\pm b$
		which is impossible since $b$ is a nonzero integer.  Hence, $c\not=0$.  From the definition
		of $F$, we see that $\Im(z)\geq \sqrt{3}/2$.  Therefore,
		$|c|\,\sqrt{3}/2\leq 1$ so that $c=\pm 1$ and $|z+d|\leq 1$.  Again from the definition
		of $F$ we see that if $|d|\geq 1$ then $|z+d|>1$ for any $z\,\in\,F$, so that $d=0$
		and $|z|\leq 1$.  This contradicts $z\,\in\,F$.  In fact, we have shown that if 
		$z,\ z^{\prime}$ in the closure of $F$ are $\G{1}$ equivalent, then 
		$z^{\prime}=T^{\pm 1}z$ or $z^{\prime}=Sz$.  In both cases, $z,\ z^{\prime}$ lie
		on the boundary of $F$ and are equivalent under $\Gamma^{\prime}$.

		We have shown that $F$ is a fundamental domain for both $\Gamma^{\prime}$ and $\G{1}$.
		Together with the fact that $\Gamma^{\prime}\subset \G{1}$, this implies that 
		$\Gamma^{\prime}=\G{1}$.  Indeed, let $\alpha\,\in\,\G{1}$ and
		$z\,\in\, F$.  Since $F$ is a fundamental domain for $\Gamma^{\prime}$, there exists 
		$\beta\,\in\,\Gamma^{\prime}$ with $w=\beta\alpha(z)$ in the closure of $F$.  In fact,
		we have shown above that $w$ must be in $F$ and hence $w=z$ (since $F$ is a fundamental
		domain for $\G{1}$).  Therefore, $\beta\alpha=1$ so that $\alpha\,\in\,\Gamma^{\prime}$.
	    \end{proof}

	    By Theorem \ref{fun}, we see that 
	    \begin{eqnarray*}
		F\cup \{z\,\in\,\H\,:\,\Re(z)=-1/2,\ |z|\geq 1\}\cup
		\{z\,\in\,\H\,:\,-1/2\leq \Re(z)\leq 0,\ |z|=1\}
	    \end{eqnarray*}
	    is a set of representatives for $\H/\G{1}$.  Similarly, one can find an explicit set
	    of representatives for $\H/\G{N}$.  We would like to put a topology on the resulting set
	    so that the space is compact.  To do this, however, we need to add some points.
	    The situation is completely analogous to that of the one point compactification of $\C$ by 
	    adding the point at infinity.

\section{Cusps}
	
	    It is obvious that any subgroup $\Gamma$ of $\G{1}$ acts on $\Q\cup\{\infty\}=\P(\Q)$ by the 
	    same formula that gives its action on $\H$.
	    \begin{defn}
		The \textit{cusps} of a subgroup $\Gamma$ of $\G{1}$ are the $\Gamma$ orbits
		of $\P(\Q)$.  We will denote by $C_{\Gamma}$ any complete set of 
		representatives of cusps of $\Gamma$. By abuse of terminology, will often refer
		to a single point as a cusp.
	    \end{defn}
	    The following theorem describes the set $C_{\G{N}}$ for each positive integer $N$.
	    \begin{thm}\cite[pg. 23]{shimura}
	    \label{cuspclass}
		With the convention that $\infty=\pm 1/0,$ two points $a/b,\ c/d$ of $\Q$ with
		$(a,b)=(c,d)=1$ are equivalent under $\G{N}$ iff $\pm \V{a}{b}\equiv \V{c}{d}\mod N.$ 
	    \end{thm}
	    \begin{proof}
		In one direction, suppose that $\V{a}{b}\equiv \V{c}{d}\mod N.$
		Since $(c,d)=1,$ we have integers $r,\ s$ such that $rc-ds=1$.  Then 
		$\tau=\bigl[\begin{smallmatrix}c & s\\ d & r\end{smallmatrix}\bigr]\,\in\,\G{1}$ 
		satisfies $\tau\V{1}{0}=\V{c}{d}$, so that
		$\tau^{-1}\V{a}{b}\equiv\V{1}{0}\mod N$.  Thus, if we can find some 
		$\sigma\,\in\,\G{N}$ with $\sigma\V{1}{0}=\tau^{-1}\V{a}{b}$, then we have
		$\tau\sigma\tau^{-1}\V{c}{d}=\V{a}{b}$.  Since $\G{N}$ is normal in $\G{1}$, 
		this shows that $a/b$ and $c/d$ are equivalent under $\G{N}$.  
		It therefore remains to show the result when $\V{c}{d}=\V{1}{0}$. 
		Since $(a,b)=(a,N)=1,$ there exist integers $p,q$ so that
		$ap+bq=(1-a)/N$.  Now let 
		$\sigma=\bigl[\begin{smallmatrix}a & -Nq\\ b & 1+Np\end{smallmatrix}\bigr]$.
		Then clearly $\sigma\,\in\,\G{N}$ and $\sigma\V{c}{d}=\V{a}{b}$.  This shows
		in any case that $a/b$ and $c/d$ are equivalent under $\G{N}$.  Conversely, 
		suppose that there exists $\sigma=\bigl[\begin{smallmatrix}p & q\\ r & s\end
		{smallmatrix}\bigr]\,\in\,\G{N}$ with $a/b=\sigma(c/d)$.
		Then $a/b=(cp+dq)/(rp+sq)$, so that there exists some $\lambda=m/n\,\in\,\Q$ with $(m,n)=1$
		and $\lambda\V{a}{b}=\bigl[\begin{smallmatrix}p & q\\ r & s\end{smallmatrix}\bigr]\V{c}{d}$.
		Equivalently, we have $ma=n(pc+qd)$ and
		$mb=n(rc+sd)$, from which we conclude (since $(m,n)=1$) that $n|a$ and $n|b$.  But
		$(a,b)=1$ so that $n=\pm 1$.  Similarly, since $\sigma$ has determinant $1$ and
		$(c,d)=1$, we have $m=\pm 1.$  Therefore, since $\sigma\,\in\,\G{N}$ we have
		$\pm\V{a}{b}\equiv\V{c}{d}\mod N$.
	    \end{proof}
	    Using this theorem, we easily see that $C_{\G{1}}=\{\infty\}$.
	    \begin{defn}
		We shall let $\H^*$ denote the \textit{extended complex plane}, that is, the subset
		of $\P$ given by $\H^*=\H\cup\Q\cup\{\infty\}$.
	    \end{defn}
	    From our remarks above, $\G{N}$ acts on $\H^*$ for any $N$, so that the quotient
	    $\H^*/\G{N}$ makes sense.  Note that since $\G{N}$ preserves $\P(\Q)$, we alwsys have 
	    $\H^*/\G{N}=\H/\G{N}\cup C_{\G{N}}$.  We now specify a topology on $\H^*/\G{N}$ so
	    that the resulting space is compact.

\section{Topology}

	    First, we topologize $\H^*$ by specifying a fundamental system of open neighborhoods of
	    any point $z\,\in\,\H^*$.  If $z\,\in\,\H$, then a fundamental system of open neighborhoods
	    of $z$ is just the usual one under the standard topology on $\C$.  If
	    $z\not=\infty$ is in $\Q$, we take as a fundamental system of open neighborhoods
	    all sets of the form  $\{z\}\cup S_z$, where $S_z$ is a circle of radius $r>0$ centered at $z+ir$.  
	    Finally, as a fundamental system of open neighborhoods of $\infty$, we take all sets of the form
	    $\{\infty\}\cup\{z\,\in\,\H\,:\,\Im(z)>c\}$, where $c>0$ is constant.
	    Clearly, this specifies a topology on $\H^*$.  Moreover, in this topology,
	    every $\sigma\,\in\,\G{N}$ is a homeomorphism $\sigma\,:\,\H^*\longrightarrow \H^*$. 
	    
	    We now endow $\H^*/\G{N}$ with the quotient topology.  That is, if 
	    $\pi\,:\,H^*\longrightarrow H^*/\G{N}$ is the quotient map, then a set $X\subset \H^*/\G{N}$
	    is open precisly when $\pi^{-1}(X)$ is open in $\H^*$.
	    It can be shown \cite[pg. 12]{shimura} that $\H^*/\G{N}$ with the above topology is Hausdorff
	    and locally compact, though we will not prove these facts.
	    From our discussion of a fundamental domain for $\H^*/\G{1}$ above, it is not difficult
	    to prove that $\H^*/\G{1}$ is compact.  Indeed, $\H^*/\G{1}=(\H/\G{1})\cup\{\infty\}$
	    is the one-point compactification of $\H/\G{1}$.  Using this fact, we readily show
	    that $\H^*/\G{N}$ is compact for each $N$.
	    \begin{thm}
		For every $N$, the topological space $\H^*/\G{N}$ is compact.
	    \end{thm}
	    \begin{proof}
		Since $\G{N}$ is of finite index in $\G{1}$, we may write 
		\begin{eqnarray*}
		    \G{1}\ =\ \bigcup_{i=1}^m g_i\G{N},
		\end{eqnarray*}
		where $g_i\,\in\,\G{1}$ and $m=[\G{1}:\G{N}]$.  Therefore, we have
		\begin{eqnarray*}
		    \H^*/\G{N}\ =\ \bigcup_{i=1}^m g_i(\H^*/\G{1}),
		\end{eqnarray*}
		which is a finite union of compact spaces, and hence compact.
	    \end{proof}
	    
\section{Fixed Points}

	    Now that we have made $\H^*/\G{N}$ into a compact topological space, we would like to
	    make it into a compact Riemann surface.
	    Before we can endow $\H^*/\G{N}$ with a smooth structure, however, we need to understand the points
	    of $\H$ that have nontrivial $\G{N}$ stabiliziers.
	    \begin{defn}
		By the \textit{isotropy subgroup} of a group $\Gamma$ acting on a topological space $S$ at a point $z\,\in\,S$ 
		we mean the stabilizer of $z$ in $\Gamma$, that is, the group 
		$\Gamma_z=\{\sigma\,\in\,\Gamma\,:\,\sigma(z)=z\}$.
	    \end{defn}

	    \begin{thm}
	    \label{isotrop}
		Every point $z\,\in\,\H$ with nontrivial stabilizer $\Gb{1}_z\subset\Gb{1}$ is equivalent
		under $\G{1}$ to $i$ or $\omega=e^{2\pi i/3}$.  Every $s\,\in\Q$ is $\G{1}$ equivalent
		to $\infty$.  Moreover, the isotropy subgroups 
		corresponding to $i$ and $\omega$ are $\langle S\rangle,\ \langle ST\rangle$ respectively, and the isotropy
		subgroup of $\Gb{1}$ at $\infty$ is generated by $T$.
	    \end{thm}

	    \begin{proof}
		Suppose that $z\,\in\,\H$ is fixed by some nontrivial $\alpha\,\in\,\Gb{1}.$  By Theorem \ref{fun}
		there exists some $\sigma\,\in\,\G{1}$ so that $\sigma(z)\,\in\,\bar{F}$.  We then have
		$\sigma\alpha(z)=\sigma(z)$, so that $\sigma\alpha\sigma^{-1}$ fixes $\sigma(z)\,\in\,\bar{F}$.
		Thus, it is enough to consider the fixed points of $\bar{F}$.  Let 
		$\gamma=\bigl[\begin{smallmatrix}a & b\\ c & d\end{smallmatrix}\bigr]\,\in\,\G{1}$ 
		fix $z\,\in\,\bar{F}$.  Then from \ref{impart}, we see
		that $|c|\,|z|\leq |cz+d|=1$.
		If $c=0$ then $a=d=\pm 1$ and $z\,\in\,\H$ is not fixed by $\gamma$.  Hence $c\not=0$.
		Since $z\,\in\,\bar{F}$ we have $|z|\geq 1$ so that $|c|=|z|=1$.  If $d\geq 1$ then
		$|cz+d|>1$ unless $d=1$ and $z=\omega$ or $d=-1$ and $z=\omega+1$.  If $d=0$ then 
		$b=\mp c=\pm 1$.  Since in this case we have $z=\pm a-1/z$, we conclude $a=1$ or $0$
		from the fact that $|z|=1$.  Thus we see that $z=i,\omega$ or $1+\omega$ and that
		the isotropy subgroups are
		\begin{eqnarray*}
		    \Gb{1}_i&=&\left\langle
		\begin{pmatrix}
	        0 & -1\\
	        1 & \phantom{-}0\\
	        \end{pmatrix}\right\rangle\ =\ \langle S\rangle,\quad \text{and}\\
		    \Gb{1}_{\omega}&=&\left\langle
		\begin{pmatrix}
	        0 & -1\\
	        1 & \phantom{-}1\\
	        \end{pmatrix}\right\rangle\ =\ \langle ST\rangle.
		\end{eqnarray*}
		Now by Theorem \ref{cuspclass}, every $s\,\in\,\Q$ is $\G{1}$ equivalent to $\infty$.
		Moreover if $z\mapsto (az+b)(cz+d)$ stabilizes $\infty$, we must have
		$c=0$.  This forces $a=d=\pm 1$, so that every $\gamma\,\in\,\Gb{1}$ fixing $\infty$ has
		the form $\gamma=\bigl(\begin{smallmatrix}1&b \\0&1 \end{smallmatrix}\bigr)$ for
		some integer $b$.  Since $T^b=\gamma$ and $T$ also fixes $\infty$, we see that
		$T$ generates $\Gb{1}_{\infty}$, as claimed.
	    \end{proof}

\section{The Riemann Surface $\H^*/\G{N}$}

	    We can now specify a smooth structure on $\H^*/\G{N}$.  We will need the following
	    useful lemma, which we do not prove: 
	    \begin{lem}
	    \label{open}
		Let $z\,\in\,\H^*$, and as before let $\G{N}_z$ be the isotropy subgroup 
		of $\G{N}$ at $z$.  Then there exists an open neighborhood $U$ of $z$ with
		\begin{eqnarray*}
		    \G{N}_z\ =\ \{\gamma\,\in\,\G{N}\,:\,\gamma(U)\cap U\not=\emptyset\}.
		\end{eqnarray*}
	    \end{lem}
	    For a proof of this lemma, see \cite[pg. 17]{shimura}.
	    Now let $z,\G{N}_z,U$ be as in Lemma \ref{open}, and denote by $\pi$ the natural projection 
	    (i.e. the quotient map)
	    \begin{eqnarray*}
		\pi_N\,:\,\H^*\longrightarrow \H^*/\G{N}.
	    \end{eqnarray*}
	    We know by Theorem \ref{isotrop} that $\Gb{N}_z$ is a finite cyclic subgroup of $\G{N}$
	    if $z\,\in\,\H$, and that if $z$ is a cusp, then there exists $g\,\in\,\G{1}$
	    with $gz=\infty$ and hence $g\Gb{N}_z g^{-1}=\langle T^k\rangle$ for some $k\geq 1$. 
	    From the definition of $U$, it is easy to see that the natural map (given by inclusion)
	    \begin{eqnarray*}
		U/\G{N}_z\longrightarrow \H^*/\G{N}
	    \end{eqnarray*}
	    is injective.  Moreover, $U/\G{N}_z$ is an open neighborhood of $\pi_N(z)$ in $\H^*/\G{N}$.
	    \begin{enumerate}
		\item If $\Gb{N}_z$ is trivial, then $\pi_N\,:\,U\longrightarrow U/\G{N}_z$ is
		a homeomorphism.  We therefore use $U/\G{N}_z$ as an open neighborhood
		of $z$ and $\pi_N^{-1}$ as a locally uniformizing variable at $\pi_N(z)$.

		\item If $\Gb{N}_z$ is cyclic of order $n>1$, (where in the cases that we are dealing
		with, $n=2$ or $3$), then let $\lambda\,:\,\H\longrightarrow \Delta$ be a 
		biholomorphic mapping of the upper half plane to the unit disc with $\lambda(z)=0$.
		Notice that $\lambda\Gb{N}_z\lambda^{-1}$ is a cyclic group of automorphisms
		of the disc preserving $0$.  As such, $\lambda\Gb{N}_z\lambda^{-1}$ consists
		of the mappings $w\longrightarrow \z{n}^k w$, where $\z{n}=e^{2\pi i/n}$ is a
		primitive $n^{\text{th}}$ root of unity.  Therefore, the map 
		$p\,:\,U/\G{N}_z\longrightarrow\C$ given by
		\begin{eqnarray*}
		    p(\pi_N(z))\ =\ \lambda^n(z)
		\end{eqnarray*}
		is a homeomorphism of $U/\G{N}_z$ with an open subset of $\C$, so we have an open
		neighborhood $U/\G{N}_z$ and a locally uniformizing variable $p$ for the point
		$\pi_N(z)$.
		
		\item Finally, if $z$ is a cusp, then we have seen in Theorem \ref{isotrop} that
		$\Gb{N}_z$ is $\G{1}$ conjugate to $\Gb{N}_{\infty}$ and that $\Gb{N}_{\infty}$
		as a subgroup of $\Gb{N}$ is generated by some power $T^k$.  (It is, in fact, not
		difficult to see that $k=N$ since $N$ is the smallest power of $T$ such that
		$T^N\,\in\,\Gb{N}$).  We therefore reduce everything to the case $z=\infty$ by 
		letting $g\,\in\,\G{1}$ be some element taking $z$ to $\infty$.  Now the map
		$p\,:\,U/\G{N}_z\longrightarrow \C$ given by
		\begin{eqnarray}
		    p(\pi_N(z))\ =\ e^{2\pi i g(z)/k}
		\label{cuspuniform}
		\end{eqnarray}
		certainly maps $U/\G{N}_z$ onto an open neighborhood of $\C$.  That it is an injection
		follows from the fact that $e^{2\pi i g(z_1)/k}=e^{2\pi i g(z_2)/k}$ if and only if
		$(T^k)^m(g(z_1))=g(z_2)$ for some $m$, that is, if and only if $\pi_N(g(z_1))=\pi_N(g(z_2))$.
		We therefore have an open neighborhood and a uniformizing variable corresponding to 
		the cusp $z$.  
	    \end{enumerate}
	    It may readily be checked that the complex charts specified above are compatible, and
	    that we have thus defined a Riemann surface.
	    \begin{defn}
		We define the \textit{modular curve} $X(N)$ to be the Riemann surface $\H^*/\G{N}$.
	    \end{defn}

\section{The Natural Map $\H^*/\G{N}\longrightarrow \H^*/\G{1}$}
\label{natmap}

	    Consider the diagram
	    \begin{eqnarray*}
	    \begin{CD}
		    \H^*      @=      \H^*\\
		@V{\pi_N}VV      @VV{\pi_1}V\\
		    X(N)  @>f>>       X(1)
	    \end{CD}
	    \end{eqnarray*}
	    where $\pi_N,\ \pi_1$ are the natural quotient maps and $f$ is the natural
	    map which makes the diagram commute.  That is, for any point $\pi_N(z)\,\in\,X(N)$, the value
	    $f(\pi_N(z))$ is defined to be $\pi_1(z)$.  
	    Using the complex structures on $X(N)$ and $X(1)$, it can be shown that
	    $f$ is in fact a holomorphic mapping of Riemann surfaces and is the quotient mapping
	    \begin{eqnarray}
	    \label{modcover}
		f\,:\,X(N)\longrightarrow X(N)/G,
	    \end{eqnarray}
	    where $G=\Gb{1}/\Gb{N}$.  Theorem \ref{branchcov} tells
	    us that $f$ is a branched covering map.  Moreover, we see that
	    $f$ is of degree $[\Gb{1}:\Gb{N}]$ since every point in $X(1)$ has $[\Gb{1}:\Gb{N}]$
	    inverse images under $f$ (counting multiplicities).
	    We are now in precisely the situation analyzed in Chapter $1$, and we shall use the
	    techniques developed there to study the function fields $K(X(N))$ for certain
	    value of $N$.  First, however, we determine the genus of the surface $X(N)$.

\section{Genus}
\label{Genus}

	    Let $X,Y$ be compact Riemann surfaces of genus $g,g^{\prime}$ respectively,
	    and $p\,:\,X\longrightarrow Y$ a degree $n$ branched covering map.  Let $b_p(P)$ denote 
	    the branch number of $p$ at $P\,\in\,X$.  Then the \textit{Hurwitz formula} 
	    \cite[pg. 19]{shimura} tells us that
	    \begin{eqnarray}
	    \label{hurwitz}
		2g^{\prime}-2=n(2g-2)+\sum_{P\in X}b_p(P).
	    \end{eqnarray}

	    We have shown in the proof of Theorem \ref{isotrop} that the branch points of the covering 
	    $f\,:\,X(N)\longrightarrow X(1)$ are all equivalent under $\G{1}$ to one of $\omega=e^{2\pi i/3},\ i,\ \infty$.
	    Moreover, it is not difficult to determine the branch number for each point.  Explicitly,
	    since $S=\bigl(\begin{smallmatrix}\phantom{-}0&1\\-1&0\end{smallmatrix}\bigr)$ and 
	    $ST=\bigl(\begin{smallmatrix}0&-1\\1&\phantom{-}1\end{smallmatrix}\bigr)$, we see that $S,ST\,\not\in\,\G{N}$
	    for any $N>1$.  Therefore, $S,ST\,\in\,\Gb{1}/\Gb{N}=G$, so that by \ref{modcover} we have:
	    \begin{enumerate}
	        \item Every point $P\,\in\,X(N)$ that is equivalent under $\G{1}$ to $\omega$
		has $b_f(P)=2$.

		\item Every point $P\,\in\,X(N)$ that is equivalent under $\G{1}$ to $i$ 
		has $b_f(P)=1$.
	    \end{enumerate}
	    Now let $\mu_N=[\Gb{1}:\Gb{N}]$ be the degree of the branched covering map $f$.  Then we clearly have
	    \begin{enumerate}
		\item The number of \textit{distinct} points $P\,\in\,X(N)$ that are equivalent
		under $\G{1}$ to $\omega$ is $\mu_N/|\Gb{N}_{\omega}|=\mu_N/3$.

		\item The number of \textit{distinct} points $P\,\in\,X(N)$ that are equivalent
		under $\G{1}$ to $i$ is $\mu_N/|\Gb{N}_i|=\mu_N/2$.
	    \end{enumerate}
	    Finally, we must compute the branch number at infinity and the size of $C_{\G{N}}$.  Notice that
	    $T^N=\bigl(\begin{smallmatrix}1&N\\0&1\end{smallmatrix}\bigr)$ is the smallest power of $T$
	    contained in $\Gb{N}$.  Therefore, $G_{\infty}$ as a subgroup of $G$ is cyclic of order $N$ 
	    so that $X(N)$ has precisely $\mu_N/N$ inequivalent cusps, that is, $\left|C_{\G{N}}\right|=\mu_N/N$.
	    Finally, since each cusp has stabilizer conjugate to $G_{\infty}$, we see that $b_f(s)=N-1$
	    for any cusp $s$.  Let $g_N$ denote the genus of the Riemann surface $X(N)$.  Then
	    putting all of our information together and using \ref{hurwitz}, we find
	    \begin{align}
		2g_N-2&=\mu_N (2g_1-2)+2\mu_N/3+\mu_N/2+(N-1)\mu_N/N\notag\\
		&=\mu_N\left(2g_1+\frac{N-6}{6N}\right),\notag\\
	    \intertext{so that}
	        g_N&=1+\mu_N\left(g_1+\frac{N-6}{12N}\right).\label{genus}
	    \end{align}
	    In the next chapter, we will show that $g_1=0$.
	    
\chapter{Modular Functions}

\section{Definitions}
	
	    \begin{defn}
		By a \textit{modular function} of level $N$ we shall mean a meromorphic function on the modular curve
		$X(N)$.  
	    \end{defn}
	    Clearly, any modular function $f$ of level $N$ may be extended to a meromorphic function $\tilde{f}$ on $\H^*$
	    by the pullback $\tilde{f}=\pi_N^*f$.  We then see that $\tilde{f}$ is a meromorphic function on $\H^*$, invariant
	    under the action $g\tilde{f}(z)=\tilde{f}(g\tau)$ for all $g\,\in\,\G{N}$ and $\tau\,\in\,\H^*$.
	    Such functions may be viewed as meromorphic functions on $\H$ invariant under $\G{N}$ with the
	    following additional property:  For each $g\,\in\,\G{1}$, the function $g\tilde{f}(\tau)=\tilde{f}(g\tau)$ admits a 
	    laurent expansion in the variable $q^{1/N}:=e^{2\pi i\tau/N}$ with only finitely many negative powers of $q^{1/N}$. 
	    That $\tilde{f}$ admits such an expansion may be
	    seen as follows:
	    \begin{enumerate}
		\item $\tilde{f}$ is invariant under the action of $\G{N}$, and in particular the transformation $T^N$,
		so that by standard results from fourier analysis, $f(\tau)$ has such an expansion for all $\tau\,\in\H$.

		\item At the cusps, \ref{cuspuniform} tells us that such an expansion exists since $e^{2\pi i h(z)/N}$
		is a uniformizing variable at $z$ where $h$ takes $z$ to $\infty$.

		\item The expansion has only finitely many negative powers of $q^{1/N}$ because $\tilde{f}$ is meromorphic on
		$\H^*$.
	    \end{enumerate}

\section{The Field Extension $K(X(N))/K(X(1))$}
\label{galoisext}

	    We have seen in section \ref{natmap} that the natural map $f\,:\,X(N)\longrightarrow X(1)$ is
	    a branched covering map of degree $[\Gb{1}:\Gb{N}]$, and therefore that $K(X(N))/K(X(1))$ is 
	    degree $[\Gb{1}:\Gb{N}]$ field extension.  In fact, we have
	    \begin{thm}
	    \label{galoisextthm}
		The extension $K(X(N))/K(X(1))$ is Galois, with Galois group $\Gb{1}/\Gb{N}$.
	    \end{thm}
	    \begin{proof}
		Let $G=\Gb{1}/\Gb{N}$.  By \ref{modcover}, we have $X(1)=X(N)/G$.  
		We show that $G$ injects into $\aut(K(X(N))/K(X(1)))$.  
		Let $g,h\,\in\,G$ be distinct.  Then there exists some point $z\,\in\,X(N)$ with
		$gz\not=hz$.  By Theorem \ref{existence}, there exists $f\,\in\,K(X(N))$ with $f(gz)\not=f(hz)$,
		that is, $gf\not=hf$.  This completes the proof.
	    \end{proof}

\section{Elliptic Functions}
\label{elliptic}

	    Let $\omega_1,\omega_2\,\in\,\C$ be such that $\Im(\omega_1/\omega_2)>0$, fix $L\subset \C^2$
	    to be the lattice generated by $\omega_1,\omega_2$, and put $L^{\prime}=L\setminus\{0\}$  
	    Recall that the \textit{Weierstrass} function defined by
	    \begin{eqnarray}
		\wp(z,L)\ :=\ \frac{1}{z^2}+\sum_{\omega\in L^{\prime}}\left\{\frac{1}{(z-\omega)^2}-\frac{1}{\omega^2}\right\}
	    \label{wpdef}
	    \end{eqnarray}
	    is a meromorphic doubly periodic function of $z$ with periods $\omega_1$ and $\omega_2$.  The
	    function $\wp(z,L)$ admits the Laurent series expansion
	    \begin{eqnarray*}
		\wp(z,L)\ =\ \frac{1}{z^2}+\frac{1}{20}g_2(L)z^2+\frac{1}{28}g_3(L)z^4+\cdots,
	    \end{eqnarray*}
	    where
	    \begin{align*}
		g_2(L)&\ :=\ 60\sum_{\omega\in L^{\prime}}\frac{1}{\omega^4}\quad\text{and}\\
		g_3(L)&\ :=\ 140\sum_{\omega\in L^{\prime}}\frac{1}{\omega^6}
	    \end{align*}
	    are the \textit{Eisenstein series} of weights $4$ and $6$ \cite[pg. 10]{lang}.  Obviously,
	    $g_2(L)$ and $g_3(L)$ satisfy
	    \begin{eqnarray*}
		g_2(\lambda L)&=&\lambda^{-4}g_2(L)\\
		g_3(\lambda L)&=&\lambda^{-6}g_3(L)
	    \end{eqnarray*}
	    for any $\lambda\,\in\,\C^*$; that is, they are \textit{homogenous} of degrees $-4$ and $-6$, respectively.
	    The \textit{discriminant}
	    \begin{eqnarray*}
		\Delta(L)\ =\ g_2^3(L)-27g_3^2(L),
	    \end{eqnarray*}
	    so named because it \textit{is} the discriminant of the cubic polynomial $y^2=4x^3-g_2x-g_3$ satisfied
	    by $(x,y)=(\wp(z,L),\wp^{\prime}(z,L))$, is therefore homogenous of degree $-12$.  It is furthermore true
	    \cite[pg. 11]{lang} that $\Delta(L)\not=0$ for any lattice $L\subset \C^2$.  This fact will be crucial
	    when we define the modular function $J$.  More than this is true, however.  When $L$ is the lattice
	    generated by $1$ and $\tau\,\in\,\H$, we in fact have the product expansion \cite[pg. 249]{lang}
	    \begin{eqnarray}
	        \Delta(L)\ =\ (2\pi i)^{12}q\prod_{n=1}^{\infty}(1-q^n)^{24},
	    \label{Deltaprod}
	    \end{eqnarray}
	    where $q=e^{2\pi i\tau}$.
	    
	    Recall that the \textit{Weierstrass sigma} and \textit{zeta} functions are defined as \cite[pg. 239]{lang}
	    \begin{align}
		\sigma(z,L)&\ :=\ z\,\prod_{\omega\in L^{\prime}}\left(1-\frac{z}{\omega}\right)
		e^{z/\omega+\frac{1}{2}(z/\omega)^2}\label{sigpro}\\
	    \intertext{and}
		\zeta(z,L)&\ :=\ \frac{\sigma^{\prime}(z,L)}{\sigma(z,L)}.\label{zetadef}
	    \end{align}
	    Taking the logarithm of the product \ref{sigpro} and differentiating twice with respect to $z$,   
	    it may be seen using \ref{wpdef} and \ref{zetadef} that
	    \begin{eqnarray*}
		\zeta^{\prime}(z,L)\ =\ -\wp(z,L).
	    \end{eqnarray*}
	    It follows that
	    \begin{eqnarray*}
		\zeta(z+\omega,L)\ =\ \zeta(z,L)+\eta(\omega,L),
	    \end{eqnarray*}
	    for some constant $\eta(\omega,L)$ and any $\omega\,\in\,L$.  In fact, $\eta(\omega,L)$ extends
	    to a function $\eta(z,L)$ which is $\R$-linear in $z$ \cite[pg. 27]{K-L}.  Both $\zeta(z,L)$
	    and $\eta(z,L)$ are homogenous of degree $-1$, that is
	    \begin{align}
		\zeta(\lambda z,\lambda L)&\ =\ \lambda^{-1}\zeta(z,L),\notag\\
		\eta(\lambda z,\lambda L)&\ =\ \lambda^{-1}\eta(z,L),\label{etahomog}\\
	    \intertext{while $\sigma(z,L)$ is homogenous of degree $1$:}
	        \sigma(\lambda z,\lambda L)&\ =\ \lambda\sigma(z,L),\label{sigmahomog}
	    \end{align}
	    for any $\lambda\,\in\,\C^*$.  Furthermore, the sigma function satisfies \cite[pg. 28]{K-L}
	    \begin{eqnarray}
		\sigma(z+\omega,L)\ =\ \varepsilon(\omega)e^{\eta(\omega,L)(z+\omega/2)}\sigma(z,L),
	    \label{sigprop}
	    \end{eqnarray}
	    where $\omega\,\in\,L$ is arbitrary and $\varepsilon(\omega)$ is defined by
	    \begin{eqnarray*}
		\varepsilon(\omega)\ :=\ 
		\begin{cases}
		\phantom{-}1&\quad\text{if}\quad \omega\,\in\,2L\\
		-1&\quad\text{otherwise}
		\end{cases}.
	    \end{eqnarray*}

	    Since we will work exclusively with the lattice $L$ generated by $1,\tau$ with $\tau\,\in\,\H$,
	    set $W=\bigl(\begin{smallmatrix}\omega_1\\\omega_2\end{smallmatrix}\bigr)=
	    \bigl(\begin{smallmatrix}\tau\\1\end{smallmatrix}\bigr)$.  Further, let $a=(a_1,a_2)\,\in\,\Q^2$
	    and fix $z=a\cdot W=a_1\tau+a_2$.  We now define the \textit{Klein forms}
	    \begin{eqnarray}
		\k_a(W)\ =\ \k(z,L)\ :=\ e^{-\eta(z,L)z/2}\sigma(z,L).
	    \label{kleindef}
	    \end{eqnarray}
	    For fixed $a$, it is evident that $\k_a(W)$ is a function only of $\tau$, and we will often write
	    $\k_a(\tau)=\k_a(W)$.  
	    The Klein forms satisfy several properties:
	    \begin{enumerate}
	        \item They are homogenous of degree $1$:
	        \begin{eqnarray}
		    \k_a(\lambda W)\ =\ \lambda\k_a(W)
		\label{prop1}
	        \end{eqnarray}
		for any $\lambda\,\in\,\C^*$.  This follows from \ref{etahomog} and \ref{sigmahomog}.
		
		\item Using \ref{sigprop}, it can be shown \cite[pg. 28]{K-L} that 
	        \begin{eqnarray}
		    \k_{a+b}(W)\ =\ \epsilon(a,b)\k_a(W),
		\label{prop2}
	        \end{eqnarray}
	        where $b\,\in\,\Z^2$ is arbitrary and $\epsilon(a,b)$ is a $d^{\text{th}}$ root of unity with $d$
		the least common multiple of the denominators of the components of $a$.

		\item Finally, let $\alpha\,\in\,\SL{\Z}$.  Then
		\begin{eqnarray}
		    \k_a(\alpha W)\ =\ \k_{a\alpha}(W).
		\label{prop3}    
		\end{eqnarray}
		This last property follows from the Definition \ref{kleindef}.
	    \end{enumerate}

	    Our main use of Klein forms will be in the explicit construction of modular functions
	    of level $N$.  Let $a\,\in\,\frac{1}{N}\Z^2$ and suppose that $\alpha\,\in\,\G{N}$.  Then by 
	    \ref{prop2} and \ref{prop3}, we have
	    \begin{eqnarray*}
		\k_a(\alpha W)\ =\ \epsilon(\alpha)\k_a(W), 
	    \end{eqnarray*}
	    where $\epsilon(\alpha)$ is in fact a $(2N)^{\text{th}}$ root of unity.  Thus, the Klein forms may be used as
	    the ``building blocks'' of modular functions.  In section \ref{kleinproducts}, we will see
	    how this explicit construction is carried out.  We conclude this section by noting the $q$-product
	    expansion for the Klein forms and a useful corollary of this formula:
	    with $z=a\cdot W=a_1\tau+a_2$, put $q =e^{2\pi i\tau}$ as before and $q_z=e^{2\pi i z}$.  We then have
	    \begin{eqnarray}
		\k_a(\tau)\ =\ -\frac{q^{(1/2)(a_1^2-a_1)}}{2\pi i}e^{\pi i a_2(a_1-1)}(1-q_z)
		\prod_{n=1}^{\infty}\frac{(1-q^nq_z)(1-q^n/q_z)}{(1-q^n)^2}.
	    \label{qprod}
	    \end{eqnarray}
	    Notice that if we change $a$ to $a+b$ for some $b\,\in\,\Z^2$ then the $q$-series \ref{qprod}
	    changes by a root of unity.  Therefore, from now on we consider $a$ to be the representative of
	    its class modulo $\Z^2$ such that
	    \begin{eqnarray*}
		0\leq a_1<1\quad\text{and}\quad 0\leq a_2 <1.
	    \end{eqnarray*}
	    \begin{cor}
	    \label{kformorders}
		Let $a$ be as above.  Then the Klein form $k_a(\tau)$ is holomorphic on $\H$ and the order of $k_a(\tau)$ 
		at $\infty$ is 
		\begin{eqnarray}
		    \frac{1}{2}(a_1^2-a_1).
		\end{eqnarray}
	    \end{cor}

	    \begin{proof}
		This follows directly from the formula \ref{qprod}.
	    \end{proof}
	    We remark that we can use Corollary \ref{kformorders} to find the order of $\k_a(\tau)$ at any cusp $s$.
	    Explicitly, let $\alpha_s\,\in\,\G{1}$ take $s$ to $\infty$.
	    Then the order of $\k_a(\tau)$ at $s$ is simply the order of $\k_{a\alpha_s}(\tau)$ at $\infty$.

\section{The $J$ Function}
\label{J}

	    We now construct a modular function of level $1$.  As before, let $\tau\,\in\,\H$ and $L=\langle1,\tau\rangle$
	    be the lattice generated by $1$ and $\tau$.  Since $\Delta(L)$ and $g_2(L)^3$ are both homogenous of
	    degree $-12$, the function
	    \begin{eqnarray}
		J(\tau)&:=&\frac{g_2^3(L)}{\Delta}\\
		&=&\frac{g_2^3(L)}{g_2^3(L)-27g_3^2(L)}
	    \label{Jdef}
	    \end{eqnarray}
	    is homogenous of degree $0$.  Therefore, since the lattices $L=\langle1,\tau\rangle$ and 
	    $\mathcal{L}=\langle1,-1/\tau\rangle$ satisfy the obvious relation $L=\tau \mathcal{L}$,
	    we see that $J(\tau)$ is invariant under the transformation $\tau\rightarrow -1/\tau=S\tau$.
	    Moreover, since $\langle1,\tau+1\rangle=L$, we have $J(\tau+1)=J(\tau)$.  Finally, the nonvanishing
	    of $\Delta$ for any $\tau\,\in\,\H$ tells us that $J(\tau)$ is holomorphic on $\H$.  Expanding
	    \ref{Jdef} as  $q$-series, we see that
	    \begin{eqnarray}
		J(\tau)\ =\ \frac{1}{1728}\left(\frac{1}{q}+744+196884q+\cdots\right).
	    \label{Jser}
	    \end{eqnarray}
	    Therefore, $J(\tau)$ has a pole of order $1$ at $\infty$.  Since $J$ is invariant under
	    $S$ and $T$, by Theorem \ref{fun} we see that $J$ defines a meromorphic function on $X(1)$, 
	    that is, $J\,\in\,K(X(1))$.  We then have:
	    \begin{thm}
	    \label{Jgener}
	    \begin{eqnarray*}
		K(X(1))\ =\ \C(J).
	    \end{eqnarray*}
	    \end{thm}
	    \begin{proof}(See \cite[pg. 63]{lang})
		Let $f\,\in\,K(X(1))$.  If $f$ has a pole of order $r$ at $z_0\,\in\,\H$ then the function
		$f(J-J(z_0))^r$ is analytic at $z_0$.  Therefore, there exists some polynomial $Q\,\in\,\C(J)$
		such that $Qf$ is holomorphic on $\H$.  By Lemma \ref{boring}, if $Qf$ is not constant on 
		$X(1)$, it has a pole at infinity (the only cusp) of order $m$, say.  Since $J$ has a simple
		pole at infinity, there exists a constant $c_0\,\in\,\C$ so that $Qf-cJ^m$ has a pole
		of at most order $m-1$.  By descent, there exists some polynomial $P\,\in\,\C(J)$
		so that $Qf-P$ has no pole at infinity and no pole in $\H$.  But then $Qf-P$ is
		a holomorphic function on $X(1)$ and therefore constant by Lemma \ref{boring}.  We conclude
		that $f\,\in\,\C(J)$, as desired.  
	    \end{proof}
	    Notice that this proof only uses two facts:
	    \begin{enumerate}
		\item The surface $X(1)$ is compact.

		\item The function $J\,\in\,K(X(1))$ has only a single pole, of order one at infinity.
	    \end{enumerate}
	    Therefore, we see that the same proof will work for any compact Riemann surface $\mathcal{R}$ provided 
	    we can find a function $\mathcal{J}\,\in\,K(\mathcal{R})$ that has only a single pole of order one at
	    infinity (or equivalently at any cusp).

\section{Special Values of J}
	    
	    Since the only points in $\H^*$ with nontrivial stabilizer in $\Gb{1}$ are equivalent
	    under $\G{1}$ to $i,\omega,\infty,$ (Theorem \ref{isotrop}), these points are in some
	    sense ``special'' points.  We therefore compute the value of $J$ at each of them, since
	    we will use these values frequently.
	    \begin{prop}
    	    \label{Jvals}
	    We have
	    \begin{eqnarray*}
		J(\omega) &=& 0\\
		J(\infty) &=& \infty\\
		J(i) &=& 1.
	    \end{eqnarray*}
	    \end{prop}
	    \begin{proof}
		The value at $\infty$ follows from the $q$-series \ref{Jser}.  Let $L_{\tau}$ be the lattice
		generated by $1,\tau$.  Then
		\begin{align*}
		    g_3(L_i)&=\sum_{(m,n)\in\Z^2\setminus\{0\}} \frac{1}{\left(m+ni\right)^6}\\
		    &=i^6\sum_{(m,n)\in\Z^2\setminus\{0\}} \frac{1}{\left(mi-n\right)^6}\\
		    &=-g_3(L_i),\\
		\intertext{from which it follows that $g_3(L_i)=0$.  Similarly,}
		    g_2(L_{\omega})&=\sum_{(m,n)\in\Z^2\setminus\{0\}} \frac{1}{\left(m+n\omega\right)^4}\\
		    &={\omega}^4\sum_{(m,n)\in\Z^2\setminus\{0\}} \frac{1}{\left((m-n)\omega-n\right)^4}\\
		    &=\omega g_2(L_{\omega}),
		\end{align*}
		so that $g_2(L_{\omega})=0$.  The proposed values of $J$ now follow.
	    \end{proof}

\section{The Function Field $K(X(N))$}
\label{funfield}

	   Theorem \ref{Jgener} tells us that $K(\P)\simeq K(X(1))$.  By Theorem \ref{main}, we conclude
	   that $X(1)\simeq \P$, as was alluded to in section \ref{Genus}.  Then by \ref{genus}, we have that
	   the genus $g_N$ of $X(N)$ is
	   \begin{eqnarray}
	   \label{modgen}
		g_N\ =\ 1+\mu_N\frac{N-6}{12N},
	   \end{eqnarray}
	   where $\mu_N=[\Gb{1}:\Gb{N}]$ is the degree of the natural covering map $f\,:\,X(N)\longrightarrow X(1)$ of 
	   section \ref{natmap} and $N>1$.  
	   Therefore, for $1\leq N\leq 5$, $g_N=0$ so that the field $K(X(N))$ is \textit{rational}, that is,
	   generated by a single element over $\C$.  Moreover, it follows immediately from \ref{modgen} that
	   for $N>6$, we have $g_N >1$, while $g_6=1$.  Therefore, the five values of $N$ in the range
	   $1\leq N\leq 5$ are the \textit{only} values of $N$ for which $X(N)$ has genus $0$ and the corresponding
	   function field $K(X(N))$ is rational.
	   
	   We have already constructed a generator for $K(X(1))$ over $\C$; namely, the function $J$ of section \ref{J}.
	   The goal of the next two sections is to determine such functions for the other four values of $N$ cited above.


\section{Products of Klein Forms}
\label{kleinproducts}

	    As mentioned in section \ref{elliptic}, we will use the Klein forms to explicitly construct
	    modular functions of level $N$.  To do this, we need to know when a product of Klein forms
	    (taken to both positive and negative integer exponents) is a modular function.  The following
	    theorem tells us when this is so:

	    \begin{thm}
	    \label{kleinprod}
		Fix a positive integer $N\geq 1$ and let $A\subset \Z^2$ be a finite set consisting
		of pairs of integers not both divisible by $N$.  Put
		\begin{eqnarray*}
		    \mathcal{A}\ =\ \left\{\frac{1}{N} a\,:\,a\,\in\,A\right\}
		\end{eqnarray*}
		and to $\alpha=(a_1/N,a_2/N)\,\in\,\mathcal{A}$ associate the integer $m(\alpha)$.  Suppose that
		\begin{eqnarray*}
		    \sum_{\alpha\in\mathcal{A}} m(\alpha)\ =\ 0,
		\label{kleinhomog}
		\end{eqnarray*} and let
		\begin{eqnarray*}
		    f(\tau)\ =\ \prod_{\alpha\in\mathcal{A}}\k_{\alpha}^{m(\alpha)}(\tau).
		\end{eqnarray*}
		Then if $N$ is odd, $f$ is a modular function of level $N$ if and only if
		\begin{align*}
		    \sum_{\alpha\in\mathcal{A}} m(\alpha)a_1^2&\equiv\ 
		    \sum_{\alpha\in\mathcal{A}} m(\alpha)a_2^2\ \equiv\ 
		    \sum_{\alpha\in\mathcal{A}} m(\alpha)a_1a_2\ \equiv\ 0\mod N,\\
		\intertext{while if $N$ is even, $f$ is a modular function of level $N$ if and only if}
		    \sum_{\alpha\in\mathcal{A}} m(\alpha)a_1^2&\equiv\ 
		    \sum_{\alpha\in\mathcal{A}} m(\alpha)a_2^2\ \equiv\ 0\mod 2N\quad\text{and}\\ 
		    \sum_{\alpha\in\mathcal{A}} m(\alpha)a_1a_2&\equiv\ 0\mod N.	
		\end{align*}
	    \end{thm}

	    \begin{proof}
		See \cite[pg. 68]{K-L}.
	    \end{proof}

\section{Generators for $K(X(N)),\ N\leq 5$}

	    \begin{thm}
		Set $\zeta_k=e^{2\pi i/k}$ and let the functions $J_N$ for $2\leq N\leq 5$ be given by
		\begin{align*}
		    J_2 &= \frac{\k^4_{(0,\frac{1}{2})}}{\k^4_{(\frac{1}{2},0)}} &  
		    J_3 &= \frac{\k^3_{(\frac{1}{3},0)}}{\k^4_{(0,\frac{1}{3})}}\\
		    &=16q^{1/2}\left(\prod_{n=1}^{\infty}\frac{1-q^{2n}}{1-q^{n/2}}\right)^8 &  
		    &= \frac{1}{i\sqrt{27}}q^{-1/3}\left(\prod_{n=1}^{\infty}\frac{1-q^{n/3}}{1-q^{3n}}\right)^3\\
		    J_4 &= \frac{\k^3_{(0,\frac{1}{4})}\k_{(\frac{1}{2},\frac{1}{4})}}
		    {\k^3_{(\frac{1}{4},0)}\k_{(\frac{1}{4},\frac{1}{2})}} & 
		    J_5 &= \frac{\k_{(\frac{2}{5},0)}\k_{(\frac{2}{5},\frac{1}{5})}\k_{(\frac{2}{5},\frac{2}{5})}
		    \k_{(\frac{2}{5},\frac{3}{5})}\k_{(\frac{2}{5},\frac{4}{5})}}
		    {\k_{(\frac{1}{5},0)}\k_{(\frac{1}{5},\frac{1}{5})}\k_{(\frac{1}{5},\frac{2}{5})}
		    \k_{(\frac{1}{5},\frac{3}{5})}\k_{(\frac{1}{5},\frac{4}{5})}}\\
		    &= \zeta_8^3\sqrt{8}q^{1/4}\prod_{n=1}^{\infty}
		    \frac{(1-q^{4n})^2(1-q^{n/2})}{(1-q^{n/4})^2(1-q^{2n})} & 
		    &= \zeta_5q^{-1/5}\prod_{n=1}^{\infty}\frac{(1-q^{5n-2})(1-q^{5n-3})}{(1-q^{5n-4})(1-q^{5n-1})}.
		\end{align*}
		Then for $2\leq N\leq 5$, $K(X(N))\ =\ \C(J_N)$. 
	    \end{thm}

	    \begin{proof}
		In section \ref{J}, we showed that any function on $X(N)$ with a single simple pole generates
		$K(X(N))$.  Thus, it is enough to show that $J_N\,\in\,K(X(N))$ and that $J_N$ has only a single
		pole of order one, for each $N$ with $2\leq N\leq 5$.  That each $J_N$ is a modular function of
		level $N$ follows after a short calculation from Theorem \ref{kleinprod}.  We then use Theorem
		\ref{cuspclass} to compute the set of cusps $C_{\G{N}}$ for the above values of $N$.  We find:
		\begin{eqnarray*}
		    C_{\G{2}}&=&\left\{0,1,\infty\right\},\\
		    C_{\G{3}}&=&\left\{0,1/2,1,\infty\right\},\\
		    C_{\G{4}}&=&\left\{0,1/3,1/2,2/3,1,\infty\right\},\\
		    C_{\G{5}}&=&\left\{0,2/9,1/4,2/7,1/3,2/5,1/2,5/8,2/3,3/4,1,\infty\right\}.
		\end{eqnarray*}
		Finally, we use Corollary \ref{kformorders} to compute the order of $J_N$ at each
		cusp and find that $J_N$ has only a single simple pole for each $N$ with
		$2\leq N\leq 5$.  This completes the proof.
	    \end{proof}

	    As a consequence of this theorem, we are now able to give an explicit isomorphism $X(N)\longrightarrow \P$.
	    In particular, since the parameter $x$ of $\C(x)$ induces the trivial isomorphism 
	    $\id\,:\, \P\longrightarrow \P$, the map
	    \begin{eqnarray}
	    \label{Jisom}
		\tau\longrightarrow J_N(\tau)
	    \end{eqnarray}
	    is an isomorphism of $X(N)$ with $\P$.  We shall use this in the next section to show how
	    $\Gb{1}/\Gb{N}$ acts on $\P$ for $2\leq N\leq 5$.

\section{The Action of $\G{1}$ on $J_N$}
\label{action}

	   We showed in section \ref{galoisext} that the field extension $K(X(N))/K(X(1))$ is
	   Galois with Galois group $\Gb{1}/\Gb{N}$.  Since by Theorem \ref{fun} the transformations 
	   $S=\bigl(\begin{smallmatrix}0&1\\-1&0\end{smallmatrix}\bigr)$ and
	   $T=\bigl(\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\bigr)$ generate $\G{1}$, we can
	   determine the action of the Galois group $\Gb{1}/\Gb{N}$ on $J_N$ by determining the
	   action of $S$ and $T$ on $J_N$.  
	   
	   Now if $f\,\in\,K(X(N))$
	   satisfies $K(X(N))=\C(f)$, then we certainly have $gf=R(f)$ for any $g\,\in\,\G{1}$, where 
	   $R\,\in\,\C(x)$.  However, it is not difficult to see that $f$ has only a single simple pole 
	   if and only if $gf$ has only a single simple pole.  Therefore, $\C(gf)=\C(f)$, so that we
	   may write $f=Q(gf)$ where $Q\,\in\,\C(x)$.  Then we have $f=Q\circ R(f)$ so that
	   $Q\circ R=1$ which implies that $Q,R$ are degree one rational maps.  Put more simply,
	   the automorphism group of $\P$ is just $\PSL{\C}$ \cite[pg. 12]{M-M}, and since $f\mapsto gf$
	   induces an automorphism of $K(\P)$, it follows that $f=\phi_g(gf)$ for some fractional linear transformation
	   $\phi_g\,\in\,\PSL{\C}$.  Moreover, it is not difficult to explicitly determine $\phi_g$.
	   Without loss of generality, we may suppose that $f$ has a simple pole at infinity (if not, replace
	   $f$ by $1/(f-c_1)$ where $c_1$ is the constant term in the $q$-expansion of $f$).  If $gf$ has a
	   simple pole at infinity, then there exists $r\,\in\,\C$ such that $f-r(gf)$ has no poles in 
	   $X(N)$ and is therefore constant.  Otherwise, let $c_g$ be the constant term in the $q$-expansion of $gf$.  
	   Then $1/(gf-c_g)$ has a simple pole at infinity and we are reduced to the previous case.  
	   We now use this process to determine the action of $S$ and $T$ on $J_N$ for $2\leq N\leq 5$.

	   With $\zeta_k=e^{2\pi i/k}$ as before, we have
	   \begin{align}
		T\circ J_2 &= \frac{-J_2}{1+J_2} & S\circ J_2 &= \frac{1}{J_2}\label{t2}\\
		T\circ J_3 &= \z{3}+\z{3}^2J_3 & S\circ J_3 &= \frac{-1}{J_3}\label{t3}\\
		T\circ J_4 &= \frac{\z{4} J_4}{1-J_4} & S\circ J_4 &= \frac{1}{\z{4} J_4}\label{t4}\\
		T\circ J_5 &= \z{5}^{-1}J_5 & S\circ J_5 &= \frac{\z{5}^2+(1+\z{5}+\z{5}^2)J_5}{J_5-(1+\z{5}+\z{5}^2)}\label{t5}
	   \end{align}
	   Now by \ref{Jisom}, the above action of $\Gb{1}/\Gb{N}$ on $J_N$ for $2\leq N\leq 5$
	   induces an action on $\P$.  That is for each $N$, for any point $z\,\in\,\P$, and any $g\,\in\,\Gb{1}/\Gb{N}$,
	   we define 
	   \begin{eqnarray*}
		gz\ :=\ J_N(gJ_N^{-1}(z)).
	   \end{eqnarray*}

	   We can give a simpler description of this action.  We begin with $N=2$, though this is in some
	   senses the least intuitive of the $4$ cases.  The $3$ cusps may be viewed as the $3$ vertices
	   of the equatorial triangle of a \textbf{double triangular pyramid} inscribed in the unit sphere.  Moreover,
	   the two preimages of $0$, i.e. the $\Gb{1}/\Gb{2}$ orbit of $\omega$, correspond to the two
	   polar tips of this double pyramid.  Projecting this solid to a triangulation of $\P$ by $6$
	   triangles, it is evident that $\Gb{1}/\Gb{2}\simeq \mathbf{S}_3$ acts on $\P$ via symmetries
	   of the double pyramid.  The group is generated by rotations of $2\pi/3$ fixing polar points
	   (corresponding to the order $3$ stabilizer of $\omega$ generated by $ST$) and by the order
	   $2$ symmetry that interchanges the two tips (which corresponds to the transformation $T$).
	   
	   Viewing
	   the $4$ cusps of $X(3)$ as the four vertices of a regular \textbf{tetrahedron} inscribed in $\P$, which has been
	   projected to a triangulation of $\P$ by $4$ triangles, we see that
	   $\Gb{1}/\Gb{3}$ acts on $\P$ by symmetries of the tetrahedron: the group
	   is generated by rotations of $2\pi/3$ about a vertex (corresponding to the transformation
	   $T$, which stabilizes $\infty$) and by rotations of $\pi$ about the midpoints of the edges
	   (corresponding to $S$).  We therefore see that the points in the orbit of $i$ (i.e. the
	   preimage of $1$ under $J$) correspond to the midpoints of the vertices of this tetrahedron.

	   Similarly, we view the $6$ cusps of $X(4)$ as the vertices of a regular \textbf{octrahedron} (projected
	   to a triangulation of $\P$).  
	   Then $\Gb{1}/\Gb{4}$ acts on $\P$ by symmetries of the octahedron.  As before, the 
	   group is generated by $T$ (rotations by $\pi/2$ about a vertex) and $S$ (rotations by $\pi$ about
	   the midpoint of an edge).  The $12$ midpoints of the vertices correspond to the points in $\P$ in 
	   the orbit of $i$.

	   Finally, the $12$ cusps of $X(5)$ give us the $12$ vertices of a regular \textbf{icosahedron}, and
	   $\Gb{1}/\Gb{5}$ acts on $\P$ via the $\mathbf{A}_5$ action on the icosahedron.  The
	   group is generated by rotation by $2\pi/5$ about a vertex (again corresponding to $T$---a fact that
	   is made clear by \ref{t5}) and rotation through $\pi$ about the midpoint of any edge.
	   The $20$ edge midpoints correspond, as before, to the points in the orbit of $i$.

\section{Index $N$ Subgroups of $\Gb{1}/\Gb{N}$ and the Associated Curves}

	    By Theorem \ref{exact}, we have the isomorphism
	    \begin{eqnarray*}
		\Gb{1}/\Gb{N}\ \simeq\ \SL{\Z/N\Z}/\{\pm 1\}.
	    \end{eqnarray*}
	    For small values of $N$, we can easily determine the size and structure of $\SL{\Z/N\Z}/\{\pm 1\}$.
	    Indeed, the description given in the previous section of the action of $\Gb{1}/\Gb{N}$ on $\P$
	    realizes the group $\Gb{1}/\Gb{N}$ as a subgroup of a permutation group.
	    Moreover, the four groups $\Gb{1}/\Gb{N}$ for $2\leq N\leq 5$ enjoy a special 
	    property: namely, for each of the above $N$, the group $\Gb{1}/\Gb{N}$ contains an index $N$ 
	    subgroup.  In summary, we have
	    \begin{enumerate}
	        \item $\mathbf{C}_3\subset \mathbf{S}_3\simeq \Gb{1}/\Gb{2}$.

		\item $\mathbf{V}_4\subset \mathbf{A}_4\simeq \Gb{1}/\Gb{3}$.

		\item $\mathbf{S}_3\subset \mathbf{S}_4\simeq \Gb{1}/\Gb{4}$.

		\item $\mathbf{A}_4\subset \mathbf{A}_5\simeq \Gb{1}/\Gb{5}$.
	    \end{enumerate}
	    Let $G_N$ denote an index $N$ subgroup of $\Gb{1}/\Gb{N}$.  
	    By Theorem \ref{galoisextthm} and the Galois correspondence, for each $G_N$
	    we obtain a degree $N$ extension of the field $\C(J)$ of rational functions in $J$.
	    One way to describe these extensions is to use Theorem \ref{main}, which tells us that
	    there is some Riemann surface $Y(N)$ and some $f\,\in\,K(Y(N))$ such that our degree $N$ extension
	    is just $K(Y(N))=\C(J)(f)$ with $R_N(J,f)=0$ for some degree $N$ rational map $R_N$ over $\C$.
	    In order to give as simple and complete a description of these extensions as possible,
	    we would like to explicitly find the maps $R_N$.  That is the goal of the next section.

\section{The Polynomials}
	   
	    As above, let $Y(N)$ be the Riemann surface corresponding to the index $N$ subgroup
	    $G_N$ of $\Gb{1}/\Gb{N}$.  Since $X(N)$ is rational over $\C$, so is $Y(N)$---that is, we have an 
	    isomorphism $f_N\,:\,Y(N)\longrightarrow \P$ with $K(Y(N))=\C(f_N)$.  Since
	    $Y(N)$ is a degree $N$ cover of $\P$, the map
	    \begin{eqnarray}
		J\,:\,Y(N)\longrightarrow \P
	    \label{Ncover}
	    \end{eqnarray}
	    is of degree $N$.  Viewing $Y(N)$ as $\P$ by the isomorphism $\tau\longrightarrow f_N(\tau)$,
	    we see that \ref{Ncover} is a degree $N$ map of $\P$ to itself.  By Theorem \ref{sphere},
	    there exists $R_N(x)\,\in\,\C(x)$ of degree $N$ with $R_N(f_N)=J$.  All this may be 
	    viewed in the following commutative diagram:
	    \begin{eqnarray*}
	    \begin{CD}
		  Y(N)  @>\sim>f_N>  \P\\
		@V{J}VV      @VV{R_N}V\\
		  \P   @=     \P
	    \end{CD}.
	    \end{eqnarray*}
	    Using that the automorphism group of $\P$ is $\PSL{\C}$, it is not difficult to see that the map $R_N$ is
	    unique up to fractional linear transformation.
	    Since $f_N\,:\,Y(N)\longrightarrow \P$ is an isomorphism, the map $f_N$ is unbranched.  
	    However, by Theorem \ref{isotrop} and Proposition \ref{Jvals}, we see that $J$ as a map from $Y(N)$ to $\P$ 
	    \textit{is} branched above $0,1,\infty$ (and only these points).  By composing $f_N$ with a 
	    fractional linear transformation, we can ensure that $f_N$ has a simple pole at $\infty$.  
	    Theorem \ref{isotrop} tells us that $\Gb{1}_{\infty}=\langle T\rangle$.  Moreover, it is
	    not difficult to see that $T$ has order $N$ in $\Gb{1}/\Gb{N}$.  Since $N$ divides the order of
	    $\Gb{1}/\Gb{N}$ only \textit{once}, for each $2\leq N\leq 5$, the group $G_N$
	    cannot contain an element of order $N$, the upshot being that $G_N$ 
	    has trivial intersection with $(\Gb{1}/\Gb{N})_{\infty}$ when $N$ is prime.  That this is so
	    for $N=4$ follows from the fact that the square of any order $4$ element in $\mathbf{S}_4$
	    is a double transposition and hence is not contained in $\mathbf{S}_3$.
	    
	    Since $X(N)$ is a degree $[\Gb{1}:\Gb{N}]$ cover of $X(1)$, the preimage of $\infty$ under the map
	    $J\,:\,X(N)\longrightarrow \P$ consists of $[\Gb{1}:\Gb{N}]/N$ points of order $N$.  Since
	    $G_N\cap (\Gb{1}/\Gb{N})_{\infty}=\{1\}$, the group $G_N$ acts transitively on these points.  Therefore,
	    the preimage of $\infty$ under the map $J\,:\,Y(N)\longrightarrow \P$ consists of a single point
	    of order $N$.  Since $f_N$ has a simple pole at $\infty$, we see that the map $R_N$ has a pole
	    of order $N$ at $\infty$.  Since $R_N$ is a degree $N$ rational map, we have shown that in fact
	    $R_N$ is a \textit{polynomial} of degree $N$. 

	    In the following analysis, especially for $N=4,5$, we shall need the following proposition:
	    \begin{prop}
	    \label{conj}
		Let $S$ be a topological space and let $G$ be a group acting on $S$.  If
		$z_1,z_2\,\in\,S$ are in the same $G$ orbit then their stabilizers in $G$
		are conjugate.
	    \end{prop}
	    The proof follows directly from the definition of the stabilizer of a point, so we omit it.
    	    We now proceed to determine the degree $N$ polynomial $R_N$ for $2\leq N\leq 5$.  We shall implicitly
	    use the fact from section \ref{genus} that $\Gb{1}_i\cap\G{N}=\Gb{1}_{\omega}\cap\G{N}=\{1\}$ for all $N\geq 2$

	\subsection{$N=2$}
	    
	    Here, $\Gb{1}/\Gb{2}\simeq\mathbf{S}_3$ and the (unique, normal) index $2$ subgroup of interest
	    is $G_2=\mathbf{C}_3$.  We first compute the preimages of the points $0,1$ of 
	    the map $J\,:\,X(2)\longrightarrow \P$.  
	    \begin{enumerate}
	        \item Since the stabilizer $\Gb{1}_{\omega}$ is generated by the 
		order $3$ element $ST$, we see that the preimage of $0$ consists of $2$ points of order $3$.

		\item Similarly, we have $\Gb{1}_i=\langle S\rangle$, which is of order $2$.  Therefore, the preimage
		of $1$ consists of $3$ points of order $2$.
	    \end{enumerate}
	    With this information, we can then determine the preimages of each of the points $0,1$ in $Y(2)$.
	    In particular,
	    \begin{enumerate}
		\item The group $\Gb{1}_{\omega}$ is isomorphic to $\mathbf{C}_3$.  Therefore, the preimages
		of $0$ in $X(2)$ are branch points of order $3$ above the preimages of $0$ in $Y(2)$.
		Thus, the preimage of $0$ in $Y(2)$ consists of two single points.  
		
		\item Since $\Gb{1}_{i}\simeq \mathbf{C}_2$ has trivial intersection with $\mathbf{C}_3$, 
		the latter acts transitively on the preimages of $1$ under $J$ in $X(2)$.  Therefore,
		the preimage of $1$ in $Y(2)$ is a single double point.    
	    \end{enumerate}
	    Item $1$ above enables us to write $R_2(x)=c(x-a)(x-b)$ for some $a,b,c\,\in\,\C$, while
	    item $2$ tells us that $R_2(x)-1$ has a double root.  Therefore, $R_2(x)-1$ has a root
	    in common with $R_2^{\prime}(x)$.  We have
	    \begin{eqnarray*}
		R_2^{\prime}(x)&=&c(x-a)(x-b)\left(\frac{1}{x-a}+\frac{1}{x-b}\right)\\
		&=&c(2x-a-b),
	    \end{eqnarray*}
	    so that
	    \begin{eqnarray*}
		-c\left(\frac{a-b}{2}\right)^2=1.
	    \end{eqnarray*}
	    Since a fractional linear transformation exists taking any three points to any other three points,
	    we may suppose that $b=0$ and $a=2$ (since we have already specified that the preimage of $\infty$
	    under $f_2$ should be $\infty$).  This gives $c=-1$ and we find that $f_N$ satisfies the polynomial
	    \begin{eqnarray*}
		-x(x-2)=J,
	    \end{eqnarray*}
	    which may be rewritten as 
	    \begin{eqnarray*}
		-(x-1)^2=J-1,
	    \end{eqnarray*}
	    that is, the degree two extension $K(Y(2))/K(X(1))$ is generated by a root of the polynomial
	    \begin{eqnarray}
		Z^2-(J-1).
	    \end{eqnarray}
	    This is, of course, expected: since $\C$ contains all roots of unity and $K(Y(2))/K(X(1))$ is
	    a Galois extension with cyclic Galois group (since $\mathbf{C_3}$ is normal in $\mathbf{S}_3$ 
	    and the quotient is $\mathbf{C}_2$), we know that $K(Y(2))=K(X(1))(\sqrt{g})$, for some 
	    $g\,\in\,K(X(1))$ which is not a square in $K(X(1))$.  In fact, using \ref{Jdef} and \ref{Deltaprod}, we 
	    see that
	    \begin{eqnarray*}
		J-1&=&\frac{g_2^3}{g_2^3-27g_3^2}\\
		&=&\frac{27g_3^2}{g_2^3-27g_3^2}\\
		&=&\frac{27g_3^2}{(2\pi i)^{12}q\prod_{n=1}^{\infty}(1-q^n)^{24}},
	    \end{eqnarray*}
	    where $q=e^{2\pi i\tau}$ as usual.  We now see explicitly that we can exrtract a square root of $J-1$
	    to obtain the function
	    \begin{eqnarray*}
		\sqrt{J-1}\ =\ \frac{3\sqrt{3}}{(2\pi i)^6}\frac{g_3}{q^{1/2}\prod_{n=1}^{\infty}(1-q^n)^{12}},
	    \end{eqnarray*}
	    which generates $K(Y(2))$.

	\subsection{$N=3$}
	    
	    Since $\Gb{1}/\Gb{3}\simeq \mathbf{A}_4$, we have the unique, normal index
	    $3$ subgroup $G_3=\mathbf{V}_4$.  As for $N=2$, we first determine the preimages in $X(3)$ of
	    $0$ for the map $J\,:\,X(3)\longrightarrow \P$ and then use this information to find the
	    preimage of $0$ in $Y(3)$.
	    \begin{enumerate}
	        \item The stabilizer $\Gb{1}_{\omega}$ is generated by an element of order $3$ corresponding to
		a $3$-cycle in $\mathbf{A}_4$.  Therefore, the preimage of $0$ in $X(3)$ consists of $4$ points of order $3$.
	    \end{enumerate}
	    Since every nontrivial element of $\mathbf{V}_4$ has order $2$, we see that $\Gb{1}_{\omega}\simeq \mathbf{C}_3$
	    has trivial intersection with $G_3$.
	    We have:
	    \begin{enumerate}
	        \item The $4$ order $3$ points in $X(3)$ above $0$ form a single orbit under $G_3$.  Therefore,
	        the preimage of $0$ in $Y(3)$ is a single point of order $3$.
	    \end{enumerate}
	    Thus, we can write $R_3(x)=c(x-a)^3$.  Therefore, since $\C$ is algebraically closed (and hence
	    the value of $c$ is immaterial), we find that the degree $3$ field extension $K(Y(3))/K(X(1))$ 
	    is the splitting field (since it is a Galois extension) of
	    \begin{eqnarray}
		Z^3-J.
	    \end{eqnarray}
	    As for $N=2$, this could have been predicted.  The Galois group of $K(Y(3))/K(X(1))$ is
	    $\mathbf{C}_3$ and therefore cyclic.  Since $\C$ contains all roots of unity, the extension
	    is obtained by extracting a cube root, in this case, of $J$.  As for $N=2$, we find
	    \begin{eqnarray*}
		\sqrt[3]{J}\ =\ \frac{1}{(2\pi i)^4}\frac{g_2}{q^{1/3}\prod_{n=1}^{\infty}(1-q^n)^8}.
	    \end{eqnarray*}

	\subsection{$N=4$}
	
	    The situation for $N=4$ and $5$ is somewhat different.  Most notably, in these cases the group
	    $G_N$ is not normal in $\Gb{1}/\Gb{N}$ so that the field extension $K(Y(N))/K(X(1))$ is not
	    Galois.  We proceed as above.
	    \begin{enumerate}
		\item We have $\Gb{1}/\Gb{4}\simeq\mathbf{S_4}$, so that the preimage of $0$ in $X(4)$ 
		consists of $8$ points of order $3$.

		\item Similarly, over $1$ we have $12$ points of order $2$.
	    \end{enumerate}
	    By considering the irreducible two dimensional representation of $\mathbf{S}_4$, it can
	    be seen that the stabilizer $(\Gb{1}/\Gb{4})_i$ corresponds to a $2$-cycle in $\mathbf{S}_4$ and not a double transposition.
	    Moreover, from Proposition \ref{conj}, it follows that each $2$-cycle in $\mathbf{S}_4$ (there
	    are six in total) generates the stabilizer of two of the $12$ points above $1$.  Fixing a copy 
	    of $\mathbf{S}_3=G_3$ in $\mathbf{S}_4$ shows that for $6$ of these $12$ points, the stabilizer
	    is contained in $G_3$, while $G_3$ acts transitively on the other $6$.  Since in any case the
	    stabilizer has order $2$, the six points whose stabilizer is contained in $G_3$ break up into
	    two orbits under $G_3$.

	    The situation for $0$ is similar.  The stabilizer $(\Gb{1}/\Gb{4})_{\omega}$ viewed as a subgroup
	    of $\mathbf{S}_4$ is generated by a $3$-cycle.
	    It follows from Proposition \ref{conj} that the $8$ points of order $3$ above $0$ have isotropy
	    subgroups generated by the $8$ elements of order $3$ in $\mathbf{S}_4$.  Since our particular
	    copy $G_3$ of $\mathbf{S}_3$ contains precisely $2$ of the $3$-cycles in $\mathbf{S}_4$, we see
	    that $2$ of the $8$ points in the preimage of $0$ have stabilizer contained in $G_3$, while
	    $G_3$ acts transitively on the other $6$ points.  We have therefore shown that
	    \begin{enumerate}
		\item The preimage of $0$ in $Y(2)$ consists of a single point and a triple point.

		\item The preimage of $1$ consists of a double point and two single points. 
	    \end{enumerate}
	    Therefore, the polynomial $R_4(x)$ is
	    \begin{eqnarray*}
		c(x-a)(x-b)^3,
	    \end{eqnarray*}
	    for some $a,b,c\,\in\,\C$.  Now since the preimage of $1$ contains a double point,
	    the polynomials 
	    \begin{eqnarray*}
		R_3(x)-1&=&c(x-a)(x-b)^3-1\quad\text{and}\\
		R_3^{\prime}(x)&=&R_3(x)\left(\frac{1}{x-a}+\frac{3}{x-b}\right)\\
		&=&c(x-b)^2(4x-b-3a)
	    \end{eqnarray*}
	    have a common root.  Since $b$ is a root of $R_3(x)$, it cannot be a root of $R_3(x)-1$, and
	    therefore we see that $(b+3a)/4$ must be a root of $R_3(x)-1$.  We are then led to the equation
	    \begin{eqnarray*}
		-\frac{27c}{2^8}(b-a)^4=1.
	    \end{eqnarray*}
	    As before, we can make an affine change of variable so that $a,b$ are any values we like, as long
	    as they are not equal.  So let $a=3,b=-1$.  Then we must have $c=-1/27$, so that the
	    field extension $K(Y(4))/K(X(1))$ is generated by a root of 
	    \begin{eqnarray}
		(Z-3)(Z+1)^3+27J.
	    \label{deg4poly}
	    \end{eqnarray}

	    The fact that the smallest normal subgroup of $\mathbf{S}_4$ containing (not necessarilly properly)
	    $\mathbf{S}_3$ is $\mathbf{S}_4$ itself tells us that the normal closure of the extension 
	    $K(Y(4))/K(X(1))$ is the field $K(X(4))$ and consequently that $K(X(4))/K(X(1))$ may be described
	    as the splitting field of the degree $4$ polynomial \ref{deg4poly}. 

	\subsection{$N=5$}

	    In this case we have $\Gb{1}/\Gb{5}\simeq \mathbf{A}_5$ and $G_5=\mathbf{A}_4$.  We at once see that
	    \begin{enumerate}
		\item The preimage of $0$ in $X(5)$ consists of $20$ points of order $3$.  Moreover, by Proposition \ref{conj},
		the $20$ isotropy subgroups at these points are generated by the $20$ $3$-cycles in $\mathbf{A}_5$.

		\item The preimage of $1$ consists of $30$ points of order $2$.  The stabilizer of any point
		is generated by a double transposition (since $\mathbf{A}_5$ contains no other elements of
		order $2$).  Since there are $15$ double transpositions in $\mathbf{A}_5$, each double 
		transposition generates the isotropy subgroup at precisely two points in the preimage of $1$.
	    \end{enumerate}
	    Now any copy of $\mathbf{A}_4$ inside $\mathbf{A}_5$ contains exactly $8$ $3$-cycles and
	    $3$ double transpositions.  Therefore, the $20$ points above $0$ break up into
	    $3$ orbits under $G_5$: one orbit consists of $12$ points (unramified above
	    the preimages of $0$ in $Y(5)$) and the other two orbits consist of $4$ points of order
	    $3$ above the preimages of $0$ in $Y(5)$.  Similarly, the $30$ points in $X(5)$ above
	    $1$ form $3$ orbits under $G_5$.  Six of the $30$ points have their stabilizers contained
	    in our copy of $\mathbf{A}_4$, so that they form a single orbit of points of order $2$ over
	    $Y(5)$.  The action of $G_5$ on the remaining $24$ points is therefore transitive, so that
	    we obtain $2$ orbits of points of order $1$ over $Y(5)$.  Thus, we see that
	    \begin{enumerate}
		\item The preimage of $0$ in $Y(5)$ consists of $2$ single points and one triple point.

		\item The preimage of $1$ in $Y(5)$ consists of $2$ double points and one single point.
	    \end{enumerate}
	    Therefore, the desired polynomial has the form
	    \begin{eqnarray*}
		d(x-a)(x-b)(x-c)^3
	    \end{eqnarray*}
	    for some $a,b,c,d\,\in\,\C$.  As before, we can make an affine change of variable to ensure
	    that $a=1,b=-1$, say.  The condition at $1$ above then tells us that the polynomial
	    \begin{eqnarray*}
		d(x^2-1)(x-c)^3-1
	    \end{eqnarray*}
	    has two double roots, that is, that is shares two roots with its derivative
	    \begin{eqnarray*}
		d(x-c)^2(5x^2-2cx-3).
	    \end{eqnarray*}
	    We therefore see that 
	    \begin{eqnarray*}
		x_1\ =\ \frac{c+\sqrt{c^2+15}}{5}\quad\text{and}\quad x_2\ =\ \frac{c-\sqrt{c^2+15}}{5}
	    \end{eqnarray*}
	    must both be roots of 
	    \begin{eqnarray*}
		d(x^2-1)(x-c)^3-1.
	    \end{eqnarray*}
	    This gives two equations in two unknowns, which we readily solve to find
	    \begin{eqnarray*}
		c &=& -\frac{1}{3^2}i\sqrt{15}\\
		d &=& \frac{3^4\cdot 5^2}{2^{11}}i\sqrt{15}.
	    \end{eqnarray*}
	    Therefore, the field extension $K(Y(5))/K(X(1))$ is generated by a root of
	    \begin{eqnarray*}
		\frac{3^4\cdot 5^2}{2^{11}}i\sqrt{15} \left(x^2-1\right)\left(x+\frac{1}{3^2}i\sqrt{15}\right)^3-J.
	    \end{eqnarray*}
	    We can simplify this equation by absorbing a factor of $i\sqrt{15}$ into $x$ to obtain the polynomial
	    \begin{eqnarray}
		(15Z^2+1)(9Z-1)^3+\left(\frac{2^{11}}{5^4}\right) J.
	    \label{deg5}
	    \end{eqnarray}
	    As for $N=4$, since the smallest normal subgroup of $\mathbf{A}_5$ is $\mathbf{A}_5$ itself, 
	    we see that the Galois closure of
	    $K(Y(5))/K(X(1))$ is $K(X(5))$ and therefore that $K(X(5))$ is the splitting field
	    of the degree $5$ polynomial \ref{deg5}.
	
\begin{thebibliography}{15}
	
	\bibitem{MSRI} Elkies, N.D.  \textit{The Klein Quartic in Number Theory.}
	``The Eightfold Way.'' Levy, S., Ed. MSRI vol. 35, 1998.

	\bibitem{F-K} Farkas, H., Kra, I.  \textit{Riemann Surfaces.}  2nd ed.
	Springer-Verlag, New York 1992.

	\bibitem{F-Ktheta} Farkas, H., Kra, I.  \textit{Theta Constants, Riemann 
	Surfaces, and the Modular Group.}  GSM vol. 37.
	American Mathematical Society, Providence RI. 2001.

	\bibitem{forster} Forster, O.  \textit{Lectures on Riemann Surfaces.}  
	Springer-Verlag, New York 1981.

	\bibitem{iwasawa} Iwasawa, K.  \textit{Algebraic Functions.}  Translations 
	of Mathematical Monographs vol. 118. 
	American Mathematical Society, Providence RI. 1991.

	\bibitem{knapp} Knapp, A.W.  \textit{Elliptic Curves.}  Mathematical Notes 40.
	Princeton University Press, Princeton NJ. 1992.

	\bibitem{K-L} Kubert, D.S., Lang, S.  \textit{Modular Units.}  
	Springer-Verlag, New York 1981.

	\bibitem{lang} Lang, S.  \textit{Elliptic Functions.}
	Addison-Wesley, Reading MA. 1973. 

	\bibitem{M-M} McKean, H., Moll, V.  \textit{Elliptic Curves.}
	Cambridge University Press, Cambridge 1997.

	\bibitem{rem} Remmert, R.  \textit{Theory of Complex Functions.}
	Springer-Verlag, New York 1991.

	\bibitem{serre} Serre, J.P.  \textit{A Course in Arithmetic.}
	Springer-Verlag, New York 1973.

	\bibitem{shimura} Shimura, G.  \textit{Introduction to the Arithmetic Theory of
	Automorphic Functions.}  Princeton University Press, Princeton NJ. 1994.

	\bibitem{W-W} Whittaker, E.T., Watson, G.N.  \textit{A Course of Modern Analysis.}
	Cambridge University Press, Cambridge 1996.

	\bibitem{kirsten} Wickelgren, K.  \textit{Galois Theory of Riemann Surfaces.}
	Harvard University, 2002. (Unpublished)

\end{thebibliography}

\end{document}