text:
\section{Introduction}

Let $C$ be a smooth, projective, hyperelliptic curve of genus 3 defined over $\mathbb{Q}$ and let $J$ be its Jacobian variety.  Recall that $J$ is a 3-dimensional abelian variety whose points can be identified with elements of the zero Picard group of $C$, $\text{Pic}^{0}\left( C \right)$.  An affine model of such a curve is
\begin{center}
$C : y^{2} = f\left( x \right)$
\end{center}
where $f \left( x \right) \in \mathbb{Q} [x ] $  has degree 7 or 8, and no repeated roots.  The Mordell-Weil theorem states that $J\left( L \right)$ is a finitely generated group for any number field $L$; that is, $J\left( L \right) \cong J\left( L \right)_{\text{tors}} \oplus \mathbb{Z}^{r}$ where $J\left( L \right)_{\text{tors}}$ is the finite torsion subgroup and $r$ is the rank.
For a hyperelliptic curve we can compute a large part of the 2-torsion subgroup of $J$,  $J\left[ 2 \right] = \lbrace P \in J : 2P =0 \rbrace$. For any two roots of $f$, $x_{1}$ and $x_{2}$, the class of the divisor $\left( x_{1}, 0 \right) - \left( x_{2}, 0 \right) - \infty_{1} - \infty_{2}$ is a non-zero element of $J\left[2 \right]$, where $\infty_{1}, \infty_{2}$ are two marked points on the projective curve, and $\infty_{1} = \infty_{2}$ when $f$ has degree $7$. Moreover, all points of order 2 are of this from when $f$ has degree $7$, see \cite{HyperCurves} or \cite{HyperCurves2}.

The problem of finding a point of order 3 is not as straightforward.  In Section 2, we will show that all 3-torsion elements correspond to ways of expressing $f$, or a scalar multiple of $f$, as
\begin{center}
$f\left( x \right) \left( x + \alpha_{1} \right)^{2} + \alpha_{7} \left( x^{3} + \alpha_{8}x^{2} + \alpha_{9}x + \alpha_{10} \right)^{3} = \left( \alpha_{2}x^{4} + \alpha_{3}x^{3} + \alpha_{4}x^{2} + \alpha_{5}x + \alpha_{6} \right)^{2}$
\end{center}
when $f$ has degree $7$, and
\begin{align*}
&  \left( -x^{6} - \frac{a_{7}}{2}x^{5}  - \left( -\frac{a_{6}}{2} + \frac{a_{7}^{2}}{8} \right)x^{4} + \alpha_{1} \left( - x^{5} - \frac{a_{7}}{2}x^{4} \right) - \alpha_{2} x^{4} + \alpha_{3}x^{3} + \alpha_{4}x^{2} + \alpha_{5}x + \alpha_{6} \right)^{2}  \\ &  \ \ \ \ \ \ =  \alpha_{7} \left( x^{3} + \alpha_{8}x^{2} \alpha_{9}x + \alpha_{10} \right)^{3}  + \left( x^{2} + \alpha_{1}x + \alpha_{2}\right)^{2}  f\left( x \right)
\end{align*}
when $f$ has degree $8$, for some $\alpha_{1}, \ldots, \alpha_{10} \in \overline{\mathbb{Q}}$, where $a_{6}$ and $a_{7}$ are coefficients of $f$.  The above correspondence can be used to define schemes parametrising the 3-torsion points of $J$.  In Sections 3 we give a method of approximating the points of such schemes as complex numbers using homotopy continuation and the Newton-Raphson method.  These numerical analysis techniques are used to efficiently compute approximations with a large precision, around 5000 decimal places, and in Section 4 we explain how such approximations are used to find algebraic expressions for the 3-torsion points of $J$, using lattice reduction.  In Section 5,  we compute the 3-torsion subgroups of the modular Jacobians  $J_{0}\left( 30 \right)$ and $J_{0}\left( 40 \right)$.  A similar method of complex approximations and lattice reduction was used in \cite{2torsion} to compute the 2-torsion subgroup of some non-hyperelliptic modular Jacobians.

The second half of this paper will explain how the 3-torsion subgroup $J\left[ 3\right]$  can be used to determine the local conductor exponent of $C$ at 2.
Recall that the conductor of a curve $C/ \mathbb{Q}$ is a representation theoretic constant, defined as a product $N =  \prod_{p} p^{n_{p}}$ over the primes $p$ where $C$ has bad reduction. Thus the problem of computing the conductor of $C$ reduces to computing the local exponents $n_{p}$ for all primes of bad reduction $p$.  When $C$ is an elliptic curve,  the $n_{p}$ can be computed using Tate's algorithm (see \cite[Chapter 4]{Silverman2}). For hyperelliptic curves of arbitrary genus,  there are formulae for $n_{p}$ for all $p \ne 2$, see \cite{Conductorexp}.
For curves of genus $2$,  Dokchitser and Dorris \cite{DokDor} give an algorithm for $n_{2}$. In \cite{DokDor}, the authors take $C$ to be a non-singular projective curve of genus $2$, defined over a finite extension $K$ of $\mathbb{Q}_{2}$. Then,  $n_{2}$ is the sum of the tame and wild parts,
\begin{center}
$n_{2} = n_{\text{tame}} + n_{\text{wild}}$
\end{center}
where $n_{\text{tame}}$ can be deduced from a regular model of the curve and $n_{\text{wild}}$ is the Swan conductor of the 3-adic Tate module of the Jacobian of $C/K$, and it can be computed from the action of $\Gal \left( K \left( J[3] \right) / K \right)$ on $J[3]$.

In the final two sections,  we will assume $C$ to be a smooth, projective and hyperelliptic curve of genus 3, defined over $\mathbb{Q}_{2}$,  and following \cite{DokDor} we use the action of $\Gal \left( \overline{\mathbb{Q}} / \mathbb{Q} \right)$ on $J\left[ 3 \right]$ to compute $n_{wild}$ when $C$ is hyperelliptic of genus 3.  In Section 6, we give a brief theoretic overview of how the local conductor exponent at $2$ is calculated using a regular model of the curve and the 3-torsion subgroup of its Jacobian.  In Section 7,  we compute  the wild part of $n_{2}$ for the modular curves $X_{0}\left(30 \right)$ and $X_{0}\left (40 \right)$ using the 3-torsion subgroups computed in Section 5.

\begin{Ack}
I would like to thank my supervisors Samir Siksek and Damiano Testa for their continued support, the many helpful conversations and their invaluable suggestions throughout this project.  I would also like to thank Tim Dokchitser for the helpful conversation regarding the tame part of the conductor and for his computation of the regular model of $X_{0}\left( 30 \right)$.
\end{Ack}

\section{Scheme of 3-torsion points }

Let $C$ be a smooth, projective,  hyperelliptic curve of genus 3, defined over a number field $K$.  By possible passing to a quadratic extension of $K$, $C$ has an affine model of the form
\begin{center}
$ y^{2} = f\left( x \right) $
\end{center}
where $f\left( x \right) \in K\left[ x \right]$ is monic, has degree 7 or 8 and has no repeated roots.

The projective closure of $C$ in $\mathbb{P}^{2}$ is defined by
\begin{center}
$Y^{2}Z^{d -2 } = Z^{d}f\left( X/Z \right)$
\end{center}
where $d$ is the degree of $f$.
\begin{remark}
We refer to the points of $C$ no appearing on the affine model as the points at infinity.  These correspond to $Z=0$, and we observe that there is a single such point, namely $( 0 : 1 :0 ) $ when the degree of $f$ is 7; and $2$ points:  $( 1 : 1 :0)$ and $( 1 : -1:0)$ when the degree of $f$ is 8.
\end{remark}
Let $J$ be the Jacobian variety of $C$. Recall that $J$ is a 3 dimensional,  abelian variety over $K$, whose points can be identified with points of $\textbf{Pic}^{0}\left(C \right)$, the zero Picard group of $C$. From now on we simply regard points on $J$ as classes of divisors of degree $0$ on $C$. See \cite{HyperCurves} or \cite{HyperCurves2} for details on the arithmetic of hyperelliptic curves.

The 3-torsion subgroup of $J$ consists of all elements $\left[ D \right]\in \textbf{Pic}^{0}\left( C \right)$ such that $3D = \text{div}\left( h \right)$, where $h$ is a rational function on $C$.  To parametrise all such points, we treat the two degree cases separately. We begin with the following straightforward result, which is required throughout the remainder of the section.

\begin{lemma}
Let $C$ be a smooth, projective and hyperelliptic curve of genus $g$ over a number field $K$ and let $K\left( C \right)$ be its function field. Let $y^{2} = f\left( x \right)$ be an affine model of the curve with $f \in K \left[ x \right]$. Suppose $g\left( x \right)$ is any polynomial in $x$, which is also an element of $K\left(C\right)$ and its divisor of zeros is of the form $3D$, where $D$ is an effective divisor. Then $g\left( x \right)$ is a cube as an element of $\overline{K}\left[ x \right]$.
\begin{proof}
We can write $g$ as
\begin{center}
    $g\left( x \right) = \alpha \left( x - \beta_{1} \right)^{r_{1}} \ldots \left( x - \beta_{s} \right)^{r_{s}}\left( x - \gamma_{1} \right)^{t_{1}} \ldots \left( x - \gamma_{n} \right)^{t_{n}}    $
\end{center}
where $f\left( \beta_{i} \right) =0$ for all $i = 1 \ldots s$, $f \left( \gamma_{j} \right) \neq 0$ for all $j = 1 \ldots n$ and $\alpha \in K^{\times}$.The divisor of zero of $g$ is
\begin{center}
    $\displaystyle \sum_{i=1}^{s} 2r_{i} \left( \beta_{i}, 0 \right) + \displaystyle \sum_{j=1}^{n} t_{j} \left(\left( \gamma_{j}, \sqrt{f\left(\gamma_{j} \right)} \right) + \left( \gamma_{j},  - \sqrt{f\left(\gamma_{j} \right)} \right)   \right) $
\end{center}
By assumption, this must equal $3D$, and hence $3$ divides $2r_{i}$ and $t_{j}$ for all $i = 1 \ldots s$ and $j = 1 \ldots n$, and the result follows.
\end{proof}
\end{lemma}
\begin{proposition}
Let $C$ be an odd degree hyperelliptic curve of genus $3$, over a number field $K$, with an affine model
\begin{center}
$y^{2} = f\left( x \right) = x^{7} + a_{6}x^{6} + a_{5}x^{5} + a_{4}x^{4} + a_{3}x^{3} + a_{2}x^{2} + a_{1}x + a_{0} $
\end{center}
where $a_{i} \in K$ and $f$ has no repeated roots. Let $J$ be the Jacobian of $C$. Then any non-zero 3-torsion point of $J$ is the form $\left[ \frac{1}{3} \text{div} \left( h \right) \right]$ where
\begin{center}
$h = y \left( x + \alpha_{1} \right) + \alpha_{2}x^{4} + \alpha_{3}x^{3} + \alpha_{4}x^{2} + \alpha_{5}x + \alpha_{6}$
\end{center}
with $ \alpha_{1}, \ldots, \alpha_{6}  \in \overline{K}$ satisfying
\begin{center}
$f\left( x \right) \left( x + \alpha_{1} \right)^{2} + \alpha_{7} \left( x^{3} + \alpha_{8}x^{2} + \alpha_{9}x + \alpha_{10} \right)^{3} = \left( \alpha_{2}x^{4} + \alpha_{3}x^{3} + \alpha_{4}x^{2} + \alpha_{5}x + \alpha_{6} \right)^{2}$
\end{center}
for some $ \alpha_{7}, \alpha_{8}, \alpha_{9}, \alpha_{10} \in \bar{K}$. Furthermore this correspondence preserves the action of $G_{K} = \Gal  \left( \overline{K} / K \right)$.
\end{proposition}

\begin{proof}
Let  $\infty$ be the unique point at infinity on this model and  $\left[ D \right] \in J\left[ 3\right] \setminus 0$. By Riemann-Roch there exists a unique,  effective divisor $D_{0} = P_{1} + P_{2} + P_{3} $  such that
\begin{center}
$ D \sim D_{0} - 3 \infty$
\end{center}
As $3D$ is principal,  $3D_{0} -9 \infty= \text{ div} \left( h \right) $ , where $h$ is a rational function on $C$. Thus $h$ is in the Riemann-Roch space $L \left( 9 \infty \right)$  which has basis
\begin{center}
$ 1, x, x^{2}, x^{3}, x^{4}, xy, y$
\end{center}
Then, replacing  $h$ by a scalar multiple if necessary,  $h$ is either a polynomial in $x$ of degree at most 4,  or $h = y + k\left(  x \right)$ with $k\left( x \right) \in K \left[ x \right]$, $\text{deg} \left( x \right) \leq 4$, or $h = y \left( x + \alpha_{1} \right) + k\left( x \right)$ with $\alpha_{1} \in K$, $k\left( x \right) \in K \left[ x \right]$ and $\text{deg} \left( x \right) \leq 4$.

\begin{case1}
Suppose $h \in K\left[ x \right]$ and $d = \text{deg}\left( h \right) \leq 4$. Let $\theta_{1}, \ldots , \theta_{d}$ be the roots of $h$.  The divisor of zeros of $h$ is $3D_{0} = 3P_{1} + 3P_{2} + 3P_{3}$ since $ \text{ div} \left( h \right) = 3D_{0} - 9 \infty$. We can also compute the divisor of zeros directly,  and find it to be
\begin{center}
$\displaystyle \sum_{i=1}^{d} \left( \left( \theta_{i} , \sqrt{f\left( \theta_{i} \right)} \right) +  \left( \theta_{i} ,  - \sqrt{f\left( \theta_{i} \right)} \right) \right) $
\end{center}
The above divisor has degree at most $8$,  whilst $\text{deg}\left( 3D_{0} \right) =9$, and hence they cannot be equal. Thus $h$ cannot be a polynomial in $x$ of degree at most $4$.
\end{case1}
\begin{case1}
Suppose $h = y + g\left( x \right)$ where $g \in K \left[ x \right]$ and $\text{deg} \left( g \right) \leq 4$, and let $\tilde{h} = -y + g\left( x \right)$. As before, the divisor of zeros of $h$ is $3D_{0}$, and the divisor of zeros of $\tilde{h}$ is
\begin{center}
$3 \iota \left( D_{0} \right) = 3\iota\left( P_{1} \right) + 3 \iota \left( P_{2} \right) + 3 \iota \left( P_{3} \right)$
\end{center}
where $\iota : C \longrightarrow C$ denotes the hyperelliptic involution on $C$. The divisor of zeros of $h\tilde{h}$ is $3D_{0} + 3\iota \left( D_{0} \right)$, and hence $h \tilde{h} = - f\left( x \right) + g \left( x \right)^{2}$ is necessarily a cube as an element of $\overline{K} \left[ x \right]$ by Lemma 1. However, this is a contradiction since $ - f\left( x \right) + g \left( x \right)^{2}$ has degree 7 or 8.
\end{case1}
\begin{case1} This is the only remaining case.
Suppose $h = y \left( x + \alpha_{1} \right) + g \left( x \right)$ where $\alpha_{1} \in K $, $g\left( x \right) \in K \left[ x \right] $ and $g$ has degree at most 4,  and let $\tilde{h} = -  y \left( x + \alpha_{1} \right) + g \left( x \right)$. Arguing as before,  the divisor of zeros of $h \tilde{h}$ is $3D_{0} + 3\iota \left( D_{0} \right) $ and hence by Lemma 1, $h \tilde{h} \in K\left[ x \right]$ is necessarily a cube.  Hence
\begin{align*}
 h \tilde{h} &= \left( y \left( x + \alpha_{1} \right) + g \left( x \right) \right) \left( -y \left( x + \alpha_{1} \right) + g \left( x \right) \right) \\
 &= -f\left( x \right) \left( x + \alpha_{1} \right)^{2} + g\left( x \right)^{2} \\
 &= \ \alpha_{7} \left( x^{3} + \alpha_{8} x^{2} + \alpha_{9} x + \alpha_{10} \right)^{3}
\end{align*}
for some $\alpha_{7}, \ldots , \alpha_{10} \in \overline{K}$, where $g \left( x \right) = \alpha_{2} x^{4} + \alpha_{3} x^{3} + \alpha_{4}x^{2} + \alpha_{5}x + \alpha_{6}$ for some $\alpha_{2}, \ldots, \alpha_{6} \in \overline{K}$.
\end{case1}
\end{proof}

Equating coefficients in this expression
\begin{center}
$f\left( x \right) \left( x + \alpha_{1} \right)^{2} + \alpha_{7} \left( x^{3} + \alpha_{8}x^{2} + \alpha_{9}x + \alpha_{10} \right)^{3} = \left( \alpha_{2}x^{4} + \alpha_{3}x^{3} + \alpha_{4}x^{2} + \alpha_{5}x + \alpha_{6} \right)^{2}$
\end{center}
gives 10 equations in $\alpha_{1}, \ldots, \alpha_{10}$, where $\left( \alpha_{1}, \ldots, \alpha_{6} \right)$ define a 3-torsion point.  We will refer to the scheme defined by these 10 equations as the scheme of 3-torsion points.

\begin{proposition}
Let $C$ be an even degree hyperelliptic curve of genus $3$, over a number field $K$, with an affine model
\begin{center}
$y^{2} = f\left( x \right) =x^{8} +  a_{7}x^{7} + a_{6}x^{6} + a_{5}x^{5} + a_{4}x^{4} + a_{3}x^{3} + a_{2}x^{2} + a_{1}x + a_{0} $
\end{center}
where $a_{i} \in K$ and $f$ has no repeated roots. Let $J$ be the Jacobian of $C$. Then any non-zero 3-torsion point of $J$ is the form $\left[ \frac{1}{3} \text{div} \left( h \right) \right]$ where
\begin{align*}
 h  & =  x^{2}y - x^{6} -\frac{a_{7}}{2} x^{5} + \left( -\frac{a_{6}}{2} + \frac{a_{7}^{2}}{8} \right)x^{4} + \alpha_{1} \left( xy - x^{5} - \frac{a_{7}}{2} x^{4} \right) +   \alpha_{2} \left( y - x^{4} \right)  \\ &  \ \ \ \ \ + \alpha_{3}x^{3} + \alpha_{4} x^{2} + \alpha_{5} x + \alpha_{6}
\end{align*}
for some $ \alpha_{1}, \ldots, \alpha_{6} \in \overline{K}$ satisfying
\begin{center}
$-f \left( x \right) l \left( x \right)^{2} + g \left( x \right) ^{2} = \alpha_{7} \left( x^{3} + \alpha_{8}x^{2} + \alpha_{9}x + \alpha_{10} \right)^{3} $
\end{center}
for some $\alpha_{7},\ldots, \alpha_{10} \in \overline{K} $ where
\begin{align*}
l\left( x \right) &= x^{2} + \alpha_{1} x + \alpha_{2} \\
g \left( x \right) &= - x^{6} + \left( -\frac{a_{7}}{2} - \alpha_{1} \right) x^{5} + \left( - \frac{a_{6}}{2} + \frac{a_{7}^{2}}{8} - \frac{\alpha_{1}a_{7}}{2} - \alpha_{2} \right) x^{4} + \alpha_{3}x^{3} + \alpha_{4}x^{2} + \alpha_{5}x + \alpha_{6}
\end{align*}
Furthermore this correspondence preserves the action of $G_{K} = \Gal  \left( \overline{K} / K \right)$.
\end{proposition}

\begin{proof}
Let  $\infty_{+}$ and $ \infty_{-}$ be the two points at infinity on this model and  $\left[ D \right] \in J\left[ 3\right] \setminus 0$. By Riemann-Roch there exists a unique,  effective divisor $D_{0} = P_{1} + P_{2} + P_{3} $  such that
\begin{center}
$ D \sim D_{0} - \infty_{+} - 2 \infty_{-}$
\end{center}
As $3D$ is principal,  $3D_{0} -3 \infty_{+} - 6 \infty_{-} = \text{ div} \left( h \right) $ , where $h$ is a rational function on $C$. Thus $h$ is in the Riemann-Roch space $L \left( 3 \infty_{+} +  6 \infty_{-}     \right)$  which has basis
\begin{center}
$ 1, x, x^{2}, x^{3},  \ y - x^{4}, \  xy - x^{5} - \frac{a_{7}}{2}x^{4},  \  x^{2}y - x^{6} - \frac{a_{7}}{2}x^{5} + \left( -\frac{a_{6}}{2} + \frac{a_{7}^{2}}{8} \right) x^{4}$
\end{center}
By possibly replacing $h$ by a scalar multiple,  $h$ will necessarily fall in one of the following four cases.
\begin{case2}
Suppose $h$ is a polynomial in $x$ of degree $d \leq 3$.  Let $\theta_{1}, \ldots, \theta_{d}$ be the roots of $h$.  The divisor of zeros of $h$ is $3D_{0}$ since  $ \text{ div} \left( h \right) = 3D_{0} - 9 \infty$. We can also compute the divisor of zeros directly,  and find it to be
\begin{center}
$\displaystyle \sum_{i=1}^{d} \left( \left( \theta_{i} , \sqrt{f\left( \theta_{i} \right)} \right) +  \left( \theta_{i} ,  - \sqrt{f\left( \theta_{i} \right)} \right) \right) $
\end{center}
The above divisor has degree at most $6$,  whilst $\text{deg}\left( 3D_{0} \right) =9$, and hence they cannot be equal. Thus $h$ cannot be a polynomial in $x$ of degree at most $3$.
\end{case2}

\begin{case2}
Suppose $h$ is of the form
\begin{align*}
h &  = y - x^{4} + \alpha_{1}x^{3} + \alpha_{2}x^{2} + \alpha_{3}x + \alpha_{4}  \\ & =  \ y + g\left( x \right)
\end{align*}
for some $\alpha_{1}, \ldots, \alpha_{4} \in \overline{K}$, where $g\left( x \right) = - x^{4} + \alpha_{1}x^{3} + \alpha_{2}x^{2} + \alpha_{3}x + \alpha_{4} $. Let $\tilde{h} = -y + g\left( x \right)$.  Arguing as in the proof of the previous proposition, the divisor of zeros of $h$ is $3D_{0}$; and the divisor of zeros of $\tilde{h}$ is $3 \iota \left( D_{0} \right)$. The divisor of zeros of $h\tilde{h} \in K \left[ x \right]$ is $3 D_{0} + 3 \iota \left( D_{0} \right)$, and thus by Lemma 1, $h\tilde{h} \in K \left[ x \right]$ is necessarily a cube.
We find that
\begin{align*}
h\tilde{h} &= \left( y + g\left( x \right) \right) \left( -y + g \left( x \right) \right) \\
               &= -f\left( x \right) + g\left( x \right)^{2}
\end{align*}
has degree at most 7, and hence it has degree 6 or 3 if it is indeed a cube.   Suppose $h\tilde{h}$ has degree 6, so $h\tilde{h} = q^{3}$ where $q \in K\left[ x \right]$ is a quadratic polynomial.  Let $\theta_{1}, \theta_{2}$ be the roots of $q$. Then the divisor of zeros of $h\tilde{h}$ is
\begin{center}
$\displaystyle 3  \sum_{i=1}^{2} \left( \left( \theta_{i}, \sqrt{f\left( \theta_{i} \right) } \right) + \left( \theta_{i} , - \sqrt{f \left( \theta_{i} \right) }\right) \right) $
\end{center}
and by considering the degree of this divisor,  it cannot equal $3D_{0}$.  A very similar argument shows that the $\text{deg} \left( h \tilde{h} \right) = 3$ also leads to a contradiction. Thus $h$ cannot be of the stated form.
\end{case2}
\begin{case2}
Suppose $h$ is of the form
\begin{align*}
h &= xy - x^{5} - \frac{a_{7}}{2}x^{4} + \alpha_{1} \left( y - x^{4} \right) + \alpha_{2}x^{3} + \alpha_{3}x^{2} + \alpha_{4}x + \alpha_{5} \\
&= l\left( x \right)y + g \left( x \right)
\end{align*}
for some $\alpha_{1}, \ldots, \alpha_{5} \in \overline{K}$, where $l\left( x \right) = x + \alpha_{1}$, $g\left( x \right) = -x^{5} -\frac{a_{7}}{2}x^{4} - \alpha_{1}x^{4} + \alpha_{2}x^{3} + \alpha_{3}x^{2} + \alpha_{4}x + \alpha_{5}$. Let $\tilde{h} = -  l\left( x \right) y + g \left( x \right)$.  Arguing as before, $h\tilde{h} \in K \left[ x \right]$ is a cube. We find that
\begin{align*}
h\tilde{h} &= \left( l\left( x \right)y + g\left( x \right) \right) \left( -l\left( x \right)y + g \left( x \right) \right) \\
               &= -l\left( x \right)^{2} f\left( x \right) + g\left( x \right)^{2}
\end{align*}
has degree at most $8$, and hence it must have degree $3$ or $6$ if it is a cube.  As in case 2, both possible degrees lead to a contradiction.  Hence $h$ cannot be of the stated form.
\end{case2}
\begin{case2}
Suppose $h$ is of the form
\begin{align*}
h &= x^{2}y - x^{6} - \frac{a_{7}}{2}x^{5} + \left( -\frac{a_{6}}{2} + \frac{a_{7}^{2}}{8} \right) x^{4} +   \alpha_{1}\left(  xy - x^{5} - \frac{a_{7}}{2}x^{4} \right)  + \alpha_{2} \left( y - x^{4} \right)  \\ & \ \ \ \ + \alpha_{3}x^{3} + \alpha_{4}x^{2} + \alpha_{5}x + \alpha_{6} \\
&= l\left( x \right)y + g \left( x \right)
\end{align*}
for some $\alpha_{1}, \ldots, \alpha_{6} \in \overline{K}$, where $l\left( x \right) = x^{2} + \alpha_{1}x + \alpha_{2}$, $g\left( x \right) = -x^{6} - \frac{a_{7}}{2}x^{5}  - \left( -\frac{a_{6}}{2} + \frac{a_{7}^{2}}{8} \right)x^{4} + \alpha_{1} \left( - x^{5} - \frac{a_{7}}{2}x^{4} \right) - \alpha_{2} x^{4} + \alpha_{3}x^{3} + \alpha_{4}x^{2} + \alpha_{5}x + \alpha_{6}$.
Following previous arguments, set $\tilde{h} = -l\left( x \right)y + g\left( x \right)$, then by considering the divisor of zeros of $h\tilde{h}$ we find that $h\tilde{h} \in K \left[ x \right]$ must be a cube.  In general,
\begin{align*}
h\tilde{h} &= \left( l\left( x \right)y + g\left( x \right) \right) \left( -l\left( x \right)y + g \left( x \right) \right) \\
               &= -l\left( x \right)^{2} f\left( x \right) + g\left( x \right)^{2}
\end{align*}
has degree 9, and so it must be the cube of a degree 3 polynomial; and so there exist $\alpha_{7}, \ldots, \alpha_{10} \in \overline{K}$ such that
\begin{center}
$ -l\left( x \right)^{2} f\left( x \right) + g\left( x \right)^{2} = \alpha_{7} \left( x^{3} + \alpha_{8}x^{2} \alpha_{9}x + \alpha_{10} \right)^{3}$
\end{center}
Thus such $h$ define 3-torsion points on $J$.
\end{case2}
\end{proof}

Equating coefficients in the expression
\begin{align*}
&  \left( -x^{6} - \frac{a_{7}}{2}x^{5}  - \left( -\frac{a_{6}}{2} + \frac{a_{7}^{2}}{8} \right)x^{4} + \alpha_{1} \left( - x^{5} - \frac{a_{7}}{2}x^{4} \right) - \alpha_{2} x^{4} + \alpha_{3}x^{3} + \alpha_{4}x^{2} + \alpha_{5}x + \alpha_{6} \right)^{2}  \\ &  \ \ \ \ \ \ =  \alpha_{7} \left( x^{3} + \alpha_{8}x^{2} + \alpha_{9}x + \alpha_{10} \right)^{3}  + \left( x^{2} + \alpha_{1}x + \alpha_{2}\right)^{2}  f\left( x \right)
\end{align*}
gives 10 equations in $\alpha_{1}, \ldots, \alpha_{10}$, where $\left( \alpha_{1}, \ldots, \alpha_{6} \right)$ define a 3-torsion point.  We will refer to the scheme defined by these 10 equations as the scheme of 3-torsion points.

\section{Complex Approximations and Homotopy Continuation}

Let $e_{1}, \ldots ,   e_{10}$ be ten equations in $\alpha_{1}, \ldots, \alpha_{10}$ defining a scheme of 3-torsion points as in the previous section.  The solutions of $e_{1}, \ldots, e_{10}$ can be approximated as complex points using the Newton-Raphson method.  We give a brief overview of this, a detailed explanation can be found in \cite[Page 298]{NA}

Let $E = \left( e_{1}, \ldots, e_{10} \right)$ and view it as a function $\mathbb{C}^{10} \longrightarrow \mathbb{C}^{10}$. Let $dE$ be the Jacobian matrix of $E$. Suppose $\mathbf{x}_{0}$ is an approximate solution to $E$ with $dE\left( \mathbf{x}_{0} \right)$ invertible.  For $k \geq 1$, define
\begin{center}
$\mathbf{x}_{k} = \mathbf{x}_{k-1} - dE\left( \mathbf{x}_{k-1} \right)^{-1}E\left( \mathbf{x}_{k-1} \right)$
\end{center}
Provided the initial approximation $\mathbf{x}_{0}$ is a good enough approximation,  the resulting sequence $\lbrace \mathbf{x}_{k} \rbrace_{k \geq 0 }$ converges to a root of $E$, with each iterate having increased precision.

This method requires initial complex approximations to the solutions of $E$. These can be obtained using homotopy continuation and its implementation in \texttt{Julia} (see \cite{julia}).

\subsection{Homotopy Continuation}
Homotopy continuation is a method for numerically approximating the solutions of a system of polynomial equations by deforming the solutions of a similar system whose solutions are known.  We give a brief sketch of the idea, but a more detailed explanation if this theory can be found in \cite{julia} or \cite{HC} .

The total degree of $E$ is $\text{deg} \left( E \right) =  \prod_{i} \text{deg} \left( e_{i} \right)$, where $\text{deg} \left( e_{i} \right)$ is the maximum of the total degrees of the monomials of $e_{i}$.

Let $F$ be a system of $10$ polynomials in $\alpha_{1}, \ldots, \alpha_{n}$,  which has exactly $\text{deg}\left( E \right)$ solutions and these solutions are known.  The system $F$ will be known as a start system.  The standard homotopy of $F$ and $E$ is a function
\begin{align*}
 & H :  \mathbb{C}^{10} \times \left[ 0, 1 \right] \longrightarrow \mathbb{C}^{10} \\
    & H \left( \mathbf{x}, t \right) \  = \  \left( 1 - t \right) F\left( \mathbf{x} \right) + t E \left( \mathbf{x} \right)
\end{align*}

Fix $N \in \mathbb{N}$, and for any $s \in \left[ 0, N \right] \cap \mathbb{N}$ define $H_{s}\left( \mathbf{x} \right) = H \left( \mathbf{x}, s/N  \right)$, this is a system of $10$ polynomials in $\alpha_{1}, \ldots, \alpha_{n}$.

For $N$ large enough, the solutions of $H_{s} \left( \mathbf{x} \right)$ are good approximations of the solutions of $H_{s+1}\left( \mathbf{x} \right)$, and using the Newton-Raphson method we can increase their precision.  The solutions of $H_{0} \left( \mathbf{x} \right) = F\left( \mathbf{x} \right)$ are known, and they can be used to define solution paths to approximate solutions of $H_{N}\left( \mathbf{x} \right) = E \left( \mathbf{x} \right)$.

There are two important things to highlight.
\begin{itemize}
\item[1.] Given any $E$,  a start system ( and its solutions) can always be computed.
\item[2.] A start system can be modified to ensure solutions paths are non-overlapping and converging to approximate solutions of $E$.
\end{itemize}

Homotopy Continuation  is implemented in the \texttt{Julia} package HomotopyContinuation.jl (see \cite{julia}).

\begin{remark}
The implementation of homotopy continuation in \texttt{Julia} gives approximates to solutions of $E$ which are accurate to 16 decimal places.  For our computations we used the approximate solutions and 1000 iterations of Newton-Raphson to obtain an accuracy of 5000 decimal places.
\end{remark}

\section{Algebraic Expressions}
Suppose $  \left( \alpha_{1}, \ldots , \alpha_{10} \right)$ is a point on a scheme of 3-torsion points defined by $E = \left( e_{1}, \ldots, e_{10} \right)$,  which has a complex approximation $\left( a_{1}, \ldots,  a_{10} \right)$,  accurate to $k$ decimal places.  We use the short vector algorithm to find the minimal polynomials of the $\alpha_{i}$ and define the corresponding 3-torsion point.

Throughout this section we'll assume $K = \mathbb{Q}$ to simplify notation.  At the end of the section,  we explain how to deal with a general number field $K$.
\subsection{Minimal Polynomials}
Fix $i$, $1 \leq i \leq 10$ and let $\alpha = a_{i}$, $\theta = \alpha_{i}$. As $\alpha$ is an algebraic number, there exists $d \in \mathbb{N}$ and $c_{0}, \ldots, c_{d} \in \mathbb{Z}$ such that
\begin{center}
$c_{d}\alpha^{d} + \ldots + c_{1}\alpha + c_{0} = 0$
\end{center}
Suppose $\theta \in \mathbb{R}$. Fix a constant $C \thicksim  10^{k}$ and let $\mathcal{L}_{k}$ be the lattice generated by the columns of the $\left( d+1 \right) \times \left( d + 1 \right)$ matrix
\begin{center}
$A_{k} = \begin{pmatrix}
1 & \ldots & 0 & 0  \\
0 & \ldots & 0 & 0 \\
\vdots & \ddots & \vdots & \vdots \\
0 & \ldots & 1& 0 \\
\left[ C \theta^{d} \right]   & \ldots & \left[ C \theta \right]  & \left[ C \right]

\end{pmatrix}
 = \left( v_{d} , \ldots, v_{1}, v_{0} \right)
$
\end{center}
where $\left[ a \right]$ denotes the integer part of $a$. As $c_{0}, \ldots, c_{d} \in \mathbb{Z}$
\begin{center}
$  \mathbf{c}_{k} = \begin{pmatrix}
c_{d} \\
\vdots \\
c_{1} \\
a
\end{pmatrix} = c_{d}v_{n} + \ldots + c_{0}v_{0}  \in \mathcal{L}_{k}
$
\end{center}
where $a = c_{d}\left[ C \theta^{d} \right]  + \ldots + c_{1} \left[ C \theta \right]  + c_{0}  \left[ C \right]  $. We can recover $\mathbf{c}_{\infty} = \left( c_{d} , \ldots , c_{0} \right)$ from $\mathbf{c}_{k}$ by setting
\begin{center}
$ c_{0} =  a - \left(  c_{d}\left[ C \theta^{d} \right]  + \ldots + c_{1} \left[ C \theta \right]  \right)  $
\end{center}

For any $k \geq 1 $
\begin{center}
\begin{align*}
\vert \vert \mathbf{c}_{k} \vert \vert &  = \sqrt{ c_{d}^{2} + \ldots + c_{1}^{2} + a^{2} } \\
 &  \leq \sqrt{ c_{d}^{2} + \ldots + c_{1}^{2} + \left( c_{d}^{2} + \ldots + c_{1}^{2} + c_{0}^{2} \right)^{2} } \\
 &  \ \sim \  \vert \vert \ \mathbf{c}_{\infty} \vert \vert
\end{align*}
\end{center}
so $\vert \vert \mathbf{c}_{k} \vert \vert \ \sim \  \vert \vert \ \mathbf{c}_{\infty} \vert \vert $  and $c_{k} \in \mathcal{L}_{k}$ for all $k \geq 1 $. Thus for $k$ large enough, we expect $\mathbf{c}_{k}$ to be the shortest vector in the lattice $\mathcal{L}_{k}$.

We use Hermite's theorem to determine when the shortest vector in $\mathcal{L}_{k}$ is a suitable candidate for the vector of coefficients.
\begin{theorem}{(Hermite)}
Let $\mathcal{L}$ be an $n$ dimensional lattice and $M$ the length of the shortest non-zero vector in $\mathcal{L}$. There exist constant $\mu_{n} \in \mathbb{R}_{\ge 0}$ depending only on $n$ such that
\begin{center}
$M^{n} \leq \mu_{n} d \left( \mathcal{L} \right)^{2} $
\end{center}
where $d \left( \mathcal{L} \right)$ is the discriminant of $\mathcal{L}$.
\end{theorem}
There are bounds on these $\mu_{n}$ given in \cite[Page 66]{smart}.  For a general lattice of full rank, we expect this bound to be close to the actual size of the shortest non-zero vector in the lattice.
\begin{proof}
See \cite[ Page 66]{smart}
\end{proof}
\begin{heu}
For a full rank lattice $\mathcal{L} \subset \mathbb{R}^{n}$, the length of the shortest vector in $\mathcal{L}$ is approximately $ d \left( \mathcal{L} \right)^{1/n}$
\end{heu}

Hermite's theorem suggests that the length of the shortest vector in $\mathcal{L}_{k}$ is approximately  $d \left( \mathcal{L}_{k} \right)^{\frac{1}{d+1 }}$, so for $k$ large enough, $\vert \vert \mathbf{c}_{k} \vert \vert $ should be much smaller than this bound.  This helps in identifying $\mathbf{c}_{k}$ and the corresponding vector $\mathbf{c}_{\infty}$.  We use the short vector algorithm implemented in \texttt{Magma} (see \cite{shortvectors}) to find short vectors in $\mathcal{L}_{k}$.

\begin{remark}
When the imaginary part of $\theta$ is not 0,  the same method can be used but with $\mathcal{L}_{k}$ being generated by the columns of

\begin{center}
$A_{k} = \begin{pmatrix}
1 & \ldots & 0 & 0   \\
0 & \ldots & 0 & 0   \\
\vdots & \ddots & \vdots & \vdots\\
0 & \ldots & 1 & 0 \\
\left[ C \text{Re}\left(  \theta^{d} \right)  \right]   & \ldots  &   \left[ C  \text{Re} \left( \theta \right)  \right] &  \left[ C \right]  \\
\left[ C \text{Im}\left(  \theta^{d} \right)  \right]   & \ldots   &   \left[ C  \text{Im} \left( \theta \right)  \right] &  0
\end{pmatrix}
$
\end{center}

where $\text{Re} \left( \theta \right)$ and $\text{Im} \left( \theta \right)$ denote the real and imaginary parts of $\theta$.
\end{remark}

To summarise, the strategy for finding the coefficients of the minimal polynomial of $\alpha$ is as follows.

\begin{itemize}
\item[1.]Choose $d$.
\item[2.] Define the lattice  $\mathcal{L}_{k}$.
\item[3.] In  $\mathcal{L}_{k}$ look for vectors which are shorter than, say  $1/1000 d \left(  \mathcal{L}_{k} \right)^{\frac{1}{d+1}} $.  If such a vector doesn't exists,  either increase the precision k and start again, or choose a different degree and start again.
\item[4.] If such a vector exists, verify that $\theta$ is an approximate solution of the corresponding polynomial. If this is not the case,  choose a different degree and start again.
\end{itemize}

\subsection{Coefficient Relations}
Suppose $\left( \alpha_{1}, \ldots, \alpha_{6} \right)$ define a rational function $h$ on $C$,  which corresponds to a 3-torsion point as in Section 2. Let $f_{i}$ be the minimal polynomial of $\alpha_{i}$ and set $d_{i} = \text{deg} \left( f_{i} \right)$.  For a fixed root $\alpha = \alpha_{1} $ of $f_{1}$, we want to determine the roots of $f_{2}, \ldots, f_{6}$ defining $h$, and thus the corresponding 3-torsion point.

Firstly, we can try to express $\alpha_{2}, \ldots, \alpha_{6}$ in terms of powers of $\alpha$. Let $K_{1} = \mathbb{Q} \left( \alpha \right)$ be the number field defined by $\alpha$.  If $f_{2}$ has a root over $K_{1}$, we can write it as
\begin{center}
$b_{d_{1}}\alpha_{2} = b_{d_{1}-1} \alpha^{d_{1} -1} + \ldots + b_{1} \alpha + b_{0}$
\end{center}
for some $b_{0}, \ldots, b_{d_{1}} \in \mathbb{Z}$.  Let $a_{1}$, $a_{2}$ be complex approximations of $\alpha$, $\alpha_{2}$ correct to $k$ decimal places.  If $a_{1}, a_{2} \in \mathbb{R}$, we search for $b_{0}, \ldots, b_{d_{1}}$ by looking for short vectors in the lattice generated by the columns of

\begin{center}
$A_{k} = \begin{pmatrix}
1 & \ldots & 0  & 0   & 0 \\
0 & \ldots & 0 & 0 & 0  \\
\vdots & \ddots & \vdots & \vdots& \vdots  \\
0 & \ldots &  0 & 1 &  0 \\
\left[ C a_{1}^{d_{1}-1} \right]   & \ldots & \left[ C a_{1}  \right]   & \left[ C a_{2}  \right] & \left[ C \right]

\end{pmatrix}
$
\end{center}

where $C$ is a constant of order $10^{k}$. If $a_{1}, a_{2} \not \in \mathbb{R}$, we instead search for short vectors in the lattice generated by the columns of

\begin{center}
$A_{k} = \begin{pmatrix}
1 & \ldots & 0 & 0 & 0   \\
0 & \ldots & 0 & 0 & 0  \\
\vdots & \ddots & \vdots & \vdots & \vdots \\
0 & \ldots & 0 & 1 & 0 \\
\left[ C \text{Re}\left(  a_{1}^{d_{1} -1} \right)  \right]   & \ldots & \left[ C  \text{Re} \left( a_{1} \right)  \right]  &   \left[ C  \text{Re} \left( a_{2}  \right)  \right] &  \left[ C \right]  \\
\left[ C \text{Im}\left(  a_{1}^{d_{1}-1} \right)  \right]   & \ldots & \left[ C  \text{Im} \left( a_{1} \right)  \right]  &   \left[ C  \text{Im} \left( a_{2} \right)  \right] & 0
\end{pmatrix}
$
\end{center}

\begin{remark}
If no relations as above exist, we can use a similar lattice method to look for higher order relations, that could help to identify the corresponding the root of $f_{2}$.
\end{remark}

When the degrees $d_{i}$ are large,  coefficient relations can be difficult to find.  We can instead factorise the minimal polynomials.  Suppose $a_{i}$ is a complex approximation of $\alpha_{i}$, and $f_{i}$ is the minimal polynomial of $\alpha_{i}$. Over $\mathbb{C}$, $f_{i}$ can be factorised into linear factors.
\begin{center}
$f_{i} = s_{1} \ldots s_{d_{i}}$
\end{center}
For $k$ large enough,  there is an $n$ such that $s_{n}\left( a_{i} \right)$ is almost zero. Thus $s_{n}$ corresponds to the required root of $f_{i}$.

\subsection{Algebraic Expressions over Number Fields}

Let $K$ be a number field of degree $n$. Then, by the Primitive Element theorem,  $K = \mathbb{Q} \left( \theta \right)$ for some $\theta \in K $. Suppose we want to find the minimal polynomial of $\alpha \in \overline{K}$ given a complex approximation $a \in \mathbb{C}$ of $\alpha$, accurate to $k$ decimal places.  Suppose $a \in \mathbb{R} $ and $\theta \in \mathbb{R}$.

As $\alpha \in \overline{K}$, there exist $d \in \mathbb{N}$ and $c_{0}, \ldots, c_{d} \in K$ such that
\begin{center}
$c_{d}\alpha^{d} + \ldots + c_{1}\alpha + c_{0} = 0$
\end{center}
As $c_{i} \in K$, there exist $c_{i,j} \in \mathbb{Q}$ with
\begin{center}
$c_{i} = \displaystyle \sum_{j=0}^{n-1} c_{i,j} \theta^{j}$
\end{center}
We can further assume $c_{i,j} \in \mathbb{Z}$ by clearing denominators in the above expression.  Therefore, there exist $c_{i,j} \in \mathbb{Z}$ such that
\begin{center}
$\left( \sum_{j=0}^{n-1} c_{d,j} \theta^{j} \right) \alpha^{d} + \ldots + \left( \sum_{j=0}^{n-1} c_{1,j} \theta^{j} \right)  \alpha +   \sum_{j=0}^{n-1} c_{0,j} \theta^{j}  = 0$
\end{center}
As before, to find $c_{i,j} \in \mathbb{Z}$, we take $C$ to be a constant of order $10^{k}$ and look for short vectors in the lattice generated by the $n\left( d + 1 \right) \times n\left(d + 1 \right)$ matrix

\begin{center}
$ \begin{pmatrix}
1 & \ldots & 0 & 0 & \ldots & 0 & \ldots & 0 & 0  \\
0 & \ldots & 0 & 0 & \ldots & 0 & \ldots & 0 & 0 \\
\vdots & \ddots & \vdots & \vdots & \ddots & \vdots & \ddots & \vdots & \vdots \\
0 & \ldots & 0& 0 & \ldots & 0 & \ldots & 1 & 0   \\
\left[ C \theta^{n-1} a^{d} \right]   & \ldots & \left[ C \theta a^{d} \right]  & \left[ C a^{d} \right] &\ldots &  \left[ C \theta^{n-1} \right]   & \ldots & \left[ C \theta \right]  & \left[ C \right]

\end{pmatrix}$
\end{center}

where we view $\theta$ as a real number.  When the imaginary part of $a$ or $\theta$ (or both) is not $0$, we generalise the above as in the Section 4.1.  As before,  we look for relations amongst the roots of each minimal polynomial to determine the 3-torsion points.

\begin{remark}
This method is extremely inefficient for a number field of large degree.
\end{remark}

\section{Examples}
\allowdisplaybreaks
Using the method described in sections 2-4,  we computed the 3-torsion subgroup of the modular Jacobians $J_{0}\left( 30\right)$ and $J_{0} \left( 40 \right)$.

\subsection{$J_{0}\left(30 \right) \left[ 3 \right]$}

We work with the model of the modular curve $X_{0}\left( 30\right)$ given by \texttt{Magma}
\begin{center}
$ y^{2} + \left( -x^4 -x^3 -x^2 \right) y = 3x^7 + 19x^6 + 60x^5 + 110x^4 + 121x^3 + 79x^2 + 28x + 4$
\end{center}

Completing the square gives a model of the form required by section 2,
\begin{center}
$ y^{2} = x^8 + 14x^7 + 79x^6 + 242x^5 + 441x^4 + 484x^3 + 316x^2 + 112x  + 16$
\end{center}

The scheme of 3-torsion points is defined by 10 equations
\begin{align*}
  &  -\alpha_{2}^2 + \alpha_{6}^2 - \alpha_{7}\alpha_{10}^3 , \\
   &  -2\alpha_{1}\alpha_{2} - 14\alpha_{2}^2 + 2\alpha_{5}\alpha_{6} - 3\alpha_{7}\alpha_{9}\alpha_{10}^2 ,  \\
   &  -\alpha_{1}^2 - 28\alpha_{1}\alpha_{2} - 79\alpha_{2}^2 - 2\alpha_{2} + 2\alpha_{4}\alpha_{6} + \alpha_{5}^2 - 3\alpha_{7}\alpha_{8}\alpha_{10}^2 - 3\alpha_{7}\alpha_{9}^2 \alpha_{10} ,  \\
  &  -14\alpha_{1}^2 - 158\alpha_{1}\alpha_{2} - 2\alpha_{1} - 242\alpha_{2}^2 - 28\alpha_{2} + 2\alpha_{3}\alpha_{6} + 2\alpha_{4}\alpha_{5} - 6\alpha_{7} \alpha_{8} \alpha_{9} \alpha_{10} -
        \alpha_{7} \alpha_{9}^3 - 3\alpha_{7}\alpha_{10}^2 , \\
  &  -79\alpha_{1}^2 - 484\alpha_{1}\alpha_{2} - 112\alpha_{1}\alpha_{6} - 28\alpha_{1} - 441\alpha_{2}^2 - 2\alpha_{2}\alpha_{6} - 158\alpha_{2} + 2\alpha_{3}\alpha_{5} + \alpha_{4}^2 +
        2820\alpha_{6} - 3\alpha_{7}\alpha_{8}^2\alpha_{10}  \\
         & \ \ \ \ \ \ \ \ \ \ \ - 3\alpha_{7}\alpha_{8}\alpha_{9}^2 - 6\alpha_{7}\alpha_{9}\alpha_{10} - 1 ,  \\
   &  -242\alpha_{1}^2 - 882\alpha_{1}\alpha_{2} - 112\alpha_{1}\alpha_{5} - 2\alpha_{1}\alpha_{6} - 158\alpha_{1} - 484\alpha_{2}^2 - 2\alpha_{2}\alpha_{5} - 484\alpha_{2} +
        2\alpha_{3} \alpha_{4} + 2820\alpha_{5}  \\
        & \ \ \ \ \ \ \ \ \ \ \ \ \  - 112\alpha_{6} - 3\alpha_{7}\alpha_{8}^2 \alpha_{9} - 6\alpha_{7}\alpha_{8}\alpha_{10} - 3\alpha_{7}\alpha_{9}^2 - 14,  \\
   &  -441\alpha_{1}^2 - 968\alpha_{1}\alpha_{2} - 112\alpha_{1}\alpha_{4} - 2\alpha_{1}\alpha_{5} - 484\alpha_{1} - 316\alpha_{2}^2 - 2\alpha_{2} \alpha_{4} - 882\alpha_{2} + \alpha_{3}^2 +
        2820\alpha_{4} - 112\alpha_{5}  \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \- 2\alpha_{6} - \alpha_{7}\alpha_{8}^3 - 6\alpha_{7}\alpha_{8}\alpha_{9} - 3\alpha_{7}\alpha_{10} - 79, \\
 &    -484\alpha_{1}^2 - 632\alpha_{1}\alpha_{2} - 112\alpha_{1}\alpha_{3} - 2\alpha_{1}\alpha_{4} - 882\alpha_{1} - 112\alpha_{2}^2 - 2\alpha_{2}\alpha_{3} - 968\alpha_{2} +
        2820\alpha_{3} - 112\alpha_{4} - 2\alpha_{5}  \\ & \ \ \ \ \ \ \ \ \ \ \ - 3\alpha_{7}\alpha_{8}^2 - 3\alpha_{7}\alpha_{9} - 242, \\
   &  2820\alpha_{1}^2 - 112\alpha_{1}\alpha_{2} - 2\alpha_{1}\alpha_{3} - 158888\alpha_{1} - 3452\alpha_{2} - 112\alpha_{3} - 2\alpha_{4} - 3\alpha_{7}\alpha_{8} + 1987659, \\
  &  2820\alpha_{1} - 112\alpha_{2} - 2\alpha_{3} - \alpha_{7} - 158404,
\end{align*}
where the 3-torsion points are classes of divisors of the form $\frac{1}{3} \text{div} \left( h \right)$
\begin{align*}
h  & = x^2y - x^6 -7x^5 -15x^4 + \alpha_{1} \left( xy - x^5 -7x^4  \right) + \alpha_{2} \left( y - x^4 \right)  + \alpha_{3}x^3  \\ & \ \ \ \ \ \ \ \  + \alpha_{4}x^{2} + \alpha_{5}x + \alpha_{6}
\end{align*}

By approximating the solutions of the above system and then finding precise algebraic expressions for the 3-torsion points,  we find that $J_{0}\left(30 \right) \left[ 3 \right] \cong \left( \mathbb{Z} / 3 \mathbb{Z} \right)^{6}$ can be generated using 3 Galois orbits,  2 consisting of 8 points each, and 1 consisting of 6 points.

For each orbit, we give the minimal polynomial of $\alpha_{1}$ and expressions for $\alpha_{2}, \ldots, \alpha_{6}$ in terms of $\alpha_{1}$.

\begin{align*}
\hline
 & u^6 - 21u^5 + 184u^4 - 861u^3 + 2296u^2 - 3381u + 2439 \\
  \hline
 & \alpha_{1} = u \\
 & \alpha_{2} = u -2 \\
 & \alpha_{3} =  \left(1 /639 \right)  \left( 4u^5 - 70u^4 + 704u^3 - 3962u^2 - 3192u  - 10638 \right) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\
 & \alpha_{4}  = \left( 1 / 213 \right) \left( 4u^5 - 70u^4 + 704u^3 - 3962u^2 + 5541u  - 7230 \right) \\
 & \alpha_{5} = \left( 1 /213 \right) \left( 4u^5 - 70u^4 + 704u^3 - 3962u^2 + 8310u - 8934) \right) \\
 &\alpha_{6} = \left( 1 / 639 \right) \left( 4u^5 - 70u^4 + 704u^3 - 3962u^2 + 9588u - 10638 \right) \\
 \hline
\end{align*}

\begin{align*}
\hline
& u^8 - 28u^7 + 343u^6 - 2401u^5 + 10414u^4 - 28147u^3 + 45290u^2 -
39200u + 13925 \\
\hline
& \alpha_{1} = u \\
& \alpha_{2} = 2u -2 \\
& \alpha_{3} =  \left( 1 / 2169 \right) \left(16u^7 - 392u^6 + 4116u^5 - 24010u^4 + 83312u^3 - 168882u^2
+ 113309u - 54568 \right)  \\
& \alpha_{4} = \left( 1 /723 \right)\left(32u^7 - 784u^6 + 8232u^5 - 48020u^4 + 166624u^3 - 337764u^2
+ 326392u  - 119258 \right) \\
& \alpha_{5} = \left( 1 / 723 \right) \left(64u^7 - 1568u^6 + 16464u^5 - 96040u^4 + 333248u^3 -
675528u^2 + 699056u  - 279004 \right) \\
& \alpha_{6} =  \left(1/2169  \right)\left( 128u^7 - 3136u^6 + 32928u^5 - 192080u^4 + 666496u^3 -
1351056u^2 + 1427032u - 592712 \right) \\
\hline
\end{align*}

\begin{align*}
\hline
& u^8 - 86u^7 + 2449u^6 - 33383u^5 + 252436u^4 - 1109723u^3 + 2786294u^2 -
    3689116u + 2224811 \\
\hline
& \alpha_{1} = u \\
& \alpha_{2} =  \left( 1/1214905376480298255 \right) ( 29876018790328u^7 - 2417413903833052u^6 + 60684703080638118u^5  \\ &   - 674976608990629628u^4 + 3832952879194486442u^3 - 11087064205570838970u^2  \\ &
+ 16027124735004738752u  - 11008190935547438114)  \\
& \alpha_{3} = \left(-1/1214905376480298255 \right)(226884728945872u^7 - 18363364083540328u^6 + 460287793516793082u^5  \\ &   - 5069981080078429502u^4  + 28138121917331765018u^3 - 77651266046887373580u^2  \\ & + 119961145357139022083u  - 45446963859192685796 ) \\
&\alpha_{4} =  (-1/ 404968458826766085)(221239419854296u^7 - 17945665351388284u^6 + 451320054316335906u^5  \\ &  - 4984082896579474376u^4  + 27907810187789236094u^3 - 78911274653216131110u^2 \\ &   + 111314232875845983914u - 60056349012914418458 ) \\
& \alpha_{5} = (-1/1214905376480298255)(593735119981072u^7 - 48868278713945128u^6 + 1258524218508960012u^5  \\ &   -14170047695264405192u^4 + 81156089944126098548u^3 - 237313632893922545220u^2  \\ &   + 339196464518479391108u  - 198602211969557067116) \\
 & \alpha_{6} = (-1/242981075296059651)(35094171383296u^7 - 3004896812056480u^6 + 81787675076005272u^5 \\ &  - 944075033879118080u^4 + 5498949927080657672u^3 - 16418946803186159928u^2 \\  &   + 23935267946866848320u - 14729053581484018328 ) \\
\hline
\end{align*}
The field of definition of definition of all 3-torsion points defined by the above expressions is the degree 144 number field $L$ defined as follows. Let $K$ be the degree 48 number field defined by
\begin{center}
   $ x^{48} - 9x^{47} + 36x^{46} - 75x^{45} + 57x^{44} + 45x^{43} + 114x^{42} - 1134x^{41} + 2649x^{40} - 2694x^{39} - 9x^{38} + 3708x^{37} - 4208x^{36} - 549x^{35} - 477x^{34} + 24297x^{33} - 35388x^{32} - 15957x^{31} - 58908x^{30} + 587655x^{29} - 1095192x^{28} + 147498x^{27} + 2477835x^{26} - 4287114x^{25} + 2891076x^{24} + 570960x^{23} - 2932713x^{22} + 2692353x^{21} - 803187x^{20} - 889560x^{19} + 1287588x^{18} - 729954x^{17} + 58869x^{16} + 358671x^{15} - 388314x^{14} + 194094x^{13} - 21821x^{12} - 50094x^{11} + 63396x^{10} - 45024x^9 + 22035*x^8 - 8640x^7 + 2955x^6 - 684x^5 + 111x^4 - 24x^3 + 1$
\end{center}
then, $L$ is the degree 6 extension of $K$ defined by
\begin{center}
    $x^6 - 21x^5 + 184x^4 - 861x^3 + 2296x^2 - 3381x + 2439 $
\end{center}

\subsection{$J_{0}\left(40 \right) \left[ 3 \right]$}

We work with the model of the modular curve $X_{0} \left( 30 \right)$ given by \texttt{Magma}
\begin{center}
$ y^{2} + ( -x^4 - 1  ) y =  2x^6 - x^4 + 2x^2$
\end{center}

Completing the square gives a model of the form required by section 2,
\begin{center}
$ y^{2} = x^8 + 8x^6 -2x^4 + 8x^2 + 1 $
\end{center}

The scheme of 3-torsion points is defined by 10 equations
\begin{align*}
& \alpha_{2}^2 - \alpha_{6}^2 - \alpha_{9}^3\alpha_{10} , \\
  &  2\alpha_{1}\alpha_{2} - 2\alpha_{5}\alpha_{6} - 3\alpha_{8}\alpha_{9}^2\alpha_{10}, \\
  & \alpha_{1}^2 + 8\alpha_{2}^2 + 2\alpha_{2} - 2\alpha_{4}\alpha_{6} - \alpha_{5}^2 - 3\alpha_{7} \alpha_{9}^2\alpha_{10} -
        3\alpha_{8}^2\alpha_{9}\alpha_{10} , \\
        &  16\alpha_{1}\alpha_{2} + 2\alpha_{1} - 2\alpha_{3}\alpha_{6} - 2\alpha_{4}\alpha_{5} - 6\alpha_{7}\alpha_{8}\alpha_{9}\alpha_{10} -
       \alpha_{8}^3\alpha_{10} - 3\alpha_{9}^2\alpha_{10}, \\
 &   8\alpha_{1}^2 - 2\alpha_{2}^2 + 2\alpha_{2}\alpha_{6} + 16\alpha_{2} - 2\alpha_{3}\alpha_{5} - \alpha_{4}^2 + 8\alpha_{6}
        - 3\alpha_{7}^2\alpha_{9}\alpha_{10} - 3\alpha_{7}\alpha_{8}^2\alpha_{10} - 6\alpha_{8}\alpha_{9}\alpha_{10} + 1 , \\
  &  -4\alpha_{1}\alpha_{2} + 2\alpha_{1}\alpha_{6} + 16\alpha_{1} + 2\alpha_{2}\alpha_{5} - 2\alpha_{3}\alpha_{4} + 8\alpha_{5} -
        3\alpha_{7}^2\alpha_{8}\alpha_{10} - 6\alpha_{7}\alpha_{9}\alpha_{10} - 3\alpha_{8}^2\alpha_{10} , \\
   & -2\alpha_{1}^2 + 2\alpha_{1}\alpha_{5} + 8\alpha_{2}^2 + 2\alpha_{2}\alpha_{4} - 4\alpha_{2} - \alpha_{3}^2 + 8\alpha_{4}
        + 2\alpha_{6} - \alpha_{7}^3\alpha_{10} - 6\alpha_{7}\alpha_{8}\alpha_{10} - 3\alpha_{9}\alpha_{10} + 8 , \\
 &   16\alpha_{1}\alpha_{2} + 2\alpha_{1}\alpha_{4} - 4\alpha_{1} + 2\alpha_{2}\alpha_{3} + 8\alpha_{3} + 2\alpha_{5} -
        3\alpha_{7}^2\alpha_{10} - 3\alpha_{8}\alpha_{10} , \\
   & 8\alpha_{1}^2 + 2\alpha_{1}\alpha_{3} + 8\alpha_{2} + 2\alpha_{4} - 3\alpha_{7}\alpha_{10} - 18, \\
 &   8\alpha_{1} + 2\alpha_{3} - \alpha_{10}, \\
\end{align*}

where the 3-torsion points are classes of divisors of the form $\frac{1}{3} \text{div} \left( h \right)$,
\begin{center}
$ h   = x^2y - x^6 -4x^4  + \alpha_{1} \left(   xy - x^5  \right) + \alpha_{2} \left( y - x^4 \right)  + \alpha_{3}x^3  + \alpha_{4}x^{2} + \alpha_{5}x + \alpha_{6} $
\end{center}
By approximating the solutions of the above system and then finding precise algebraic expressions for the 3-torsion points,  we find that $J_{0}\left(40 \right) \left[ 3 \right] \cong \left( \mathbb{Z} / 3 \mathbb{Z} \right)^{6}$ can be generated using 3 Galois orbits,  2 consisting of 6 points each, and 1 consisting of 8 points.

For each orbit, we give the minimal polynomial of $\alpha_{1}$ and expressions for $\alpha_{2}, \ldots, \alpha_{6}$ in terms of $\alpha_{1}$.

\begin{align*}
\hline
& u^6 + 4u^4 - 8u^2 + 12 \\
\hline
& \alpha_{1} = u \\
& \alpha_{2} = u + 1 \\
& \alpha_{3} = \left( -1/9 \right)(u^5 + u^3 + 16u  + 18)  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \\
& \alpha_{4} = \left(-1/3 \right)(u^5 + u^3 + 4u  + 3) \\
& \alpha_{5} = \left(-1/3 \right)(u^5 + u^3 + u  - 6) \\
& \alpha_{6} =  \left(-1/9 \right)(u^5 + u^3 + 7u  + 9) \\
\hline
\end{align*}

\begin{align*}
\hline
&  u^6 - 6u^5 + 4u^4 + 24u^3 + 256u^2 - 576u + 324 \\
\hline
& \alpha_{1} = u \\
& \alpha_{2} = \left(  1/198 \right) (-u^4 + 4u^3 + 58u^2 - 124u + 126 ) \ \ \ \ \ \ \ \ \ \ \ \ \ \  \\
& \alpha_{3} =  \left(-1/99 \right)(u^4 - 4u^3 - 58u^2 + 322u  + 468) \\
& \alpha_{4} = \left( -1/99 \right)(u^4 - 4u^3 - 58*u^2 - 74u + 765) \\
& \alpha_{5}  =  \left(-1/99 \right)(u^4 - 4u^3 - 58u^2 - 173u  + 468) \\
& \alpha_{6} = \left(1/198 \right)(u^4 - 4u^3 - 58u^2 + 520u  - 522 ) \\
\hline
\end{align*}

\begin{align*}
\hline
&  u^8 - 126u^4 - 648u^2 - 1323 \\
\hline
& \alpha_{1} = u \\
& \alpha_{2} = -1 \\
& \alpha_{3} =  \left( 1/189 \right)(u^7 - 63u^3 - 648u )  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \  \ \ \ \ \ \ \ \\
& \alpha_{4} =  3 \\
& \alpha_{5}  = -u  \\
& \alpha_{6} = 1 \\
\hline
\end{align*}

The field of definition of the 3-torsion subgroup is the degree 48 number field defined by
\begin{center}
$ x^{48} - 22x^{47} + 220x^{46} - 1298x^{45} + 4840x^{44} - 10758x^{43} + 7848x^{42} + 30564x^{41} - 90644x^{40} - 54378x^{39} + 983934x^{38} - 3228430x^{37} + 6037118x^{36} - 6706868x^{35} + 3859158x^{34} - 6290682x^{33} + 41469355x^{32} - 151827480x^{31} + 375328308x^{30} - 727099012x^{29} + 1204881284x^{28} - 1812362612x^{27} + 2558319144x^{26} - 3402905364x^{25} + 4192192588x^{24} - 4669768140x^{23} + 4602283152x^{22} - 3939374364x^{21} + 2873125672x^{20} - 1738390504x^{19} + 830314684x^{18} - 275496188x^{17} + 30094447x^{16} + 31178478x^{15} - 22364652x^{14} + 5362086x^{13} + 2307708x^{12} - 2995626x^{11} + 1676724x^{10} - 615660x^9 + 121728x^8 + 25686x^7 - 31194x^6 + 9162x^5 + 1458x^4 - 2088x^3 + 738x^2 - 126x + 9$
\end{center}

\section{Local Conductor Exponent at 2}
Throughout this section,  $C/K$  will denote a smooth, projective,  hyperelliptic curve defined over $K$, a finite extension of $\mathbb{Q}_{2}$.
Let $J$ be the Jacobian variety associated to $C$, $T = T_{l}J$ the $l$-adic Tate module and $V = V_{l}J = T \otimes_{\mathbb{Z}_{l}} \mathbb{Q}_{l}$ the associated $l$-adic representation, where $l$ is any prime different from $2$.  The conductor exponent of such a representations,  as defined in \cite{DokDor} and \cite{ulmer} , is
\begin{center}
$n = \displaystyle \int_{-1}^{\infty} \text{codim}  V^{G_{K}^{u}} \ du$
\end{center}
where $G_{K} = \Gal \left( \overline{K} / K \right)$ is the absolute Galois group of $K$ and $ \lbrace G_{K}^{u} \rbrace_{u \ge -1}$ denote the ramification groups of $G_{K}$ in upper numbering.  The tame and wild parts are defined as
\begin{align*}
n_{\text{tame}} & =  \displaystyle \int_{-1}^{0} \text{codim}  V^{G_{K}^{u}} \ du \\
n_{\text{wild}} & =  \displaystyle \int_{0}^{\infty} \text{codim}  V^{G_{K}^{u}} \ du
\end{align*}

\begin{remark}
The definition is independent of the choice of prime $l$, see \cite{ulmer}.
\end{remark}
Our approach is to take $l =3$ and use the 3-torsion subgroup, computed as in Section 2 to 4.
\subsection{Tame Conductor}
From the above, the tame part of the conductor can be computed as
\begin{center}
$n_{\text{tame}} = 6 - \text{dim}V_{3}J^{I}$
\end{center}
where $I \leqslant G_{K}$ is the inertia subgroup.

Alternatively, we can also also deduce the tame part of the conductor from the regular model of $C$. From a regular model of $C$ over $\mathbb{Z}_{2}$ we can calculate
\begin{itemize}
\item the abelian part $a$,  equal to the sum of the genera of all components of the model
\item the toric part $t$,  equal to the number of loops in the dual graph of $C$
\end{itemize}

Then, the tame part of the exponent is equal to $6 -2a -t$, see \cite[Chapter 9]{NeronModels} for details. Regular models can often be computed using the method described in \cite{ModelsoverDVRs}, however, this is often a challenging problem.
\subsection{Wild Conductor}
Recall that we wish to compute
\begin{center}
$ n_{\text{wild}} =  \displaystyle \int_{0}^{\infty} \text{codim}  V^{G_{K}^{u}} \ du$
\end{center}
For $u \ge 0$, $G_{K}^{u}$ is pro-p and $\text{codim} V^{G_{K}^{u}} = \text{codim} \bar{V}^{G_{K}^{u}} = \text{codim} J\left[ 3 \right]^{G_{K}^{u}}$, see \cite{ulmer}. We may replace $G_{K}$ by $G = \Gal \left( K\left( J[3] \right) / K \right)$, and thus
\begin{center}
$ n_{\text{wild}} =  \displaystyle \int_{0}^{\infty} \text{codim}  J[3]^{G^{u}} \ du$
\end{center}
Alternatively,  using the definition of $G^{u}$ and $G_{u}$,  the ramification groups in upper numbering and lower numbering respectively, we find
\begin{center}
$ n_{\text{wild}} =  \displaystyle \int_{0}^{\infty} \frac{\text{codim}  J[3]^{G_{u}}}{[ G_{0} : G_{u} ]} \ du = \displaystyle \sum_{k=0}^{\infty} \frac{\text{codim}  J[3]^{G_{k}}}{[ G_{0} : G_{k} ]}  $
\end{center}
\begin{remark}
Using our presentation of $J\left[ 3 \right]$, the ramification groups $G_{u}$ and their action on $J\left[ 3 \right]$ are completely explicit; and as a result $n_{\text{wild}}$ is a straightforward computation.
\end{remark}

\section{Examples}
\subsection{$J_{0}\left( 30 \right)$}

A regular model of $X_{0}\left( 30 \right)$ at $2$ is
\begin{center}
$y^{2} + \left( -x^{4} - x^{3} - x^2 +2x + 4\right) y = 3x^7 + 19x^6 + 61x^5 + 113x^4 + 124x^3 + 80x^2 + 24$
\end{center}
and from the dual graph, we deduce that $a =2$ and $t=1$ in this case, thus $n_{\text{tame}} = 1$.
Recall that the 3-torsion subgroup $J_{0}\left( 30 \right) [3]$ is defined over the degree $144$ number field defined by the polynomial $f$, stated in Section 5.1. Over $\mathbb{Q}_{2}$, $f$ splits as the product of $6$ degree $24$ factors, and the splitting field of $f$, denoted by $L$,  is a degree $24$ Galois extension of $\mathbb{Q}_{2}$.
Let $G = \Gal \left( L / \mathbb{Q}_{2} \right)$. We find that $G_{0} $ is cyclic of order $3$ and $G_{n} = 1$ for all $n \geq 1$. Using the explicit generators given in section 5.1, we find
\begin{center}
$J_{0}\left( 30 \right)[3]^{G_{0}} \simeq \left( \mathbb{Z} / 3 \mathbb{Z} \right)^{6}$
\end{center}
and so $n_{\text{wild}} = 0/1 = 0 $. Thus $n_{2} = 1 +  0  = 1 $.

As $X_{0}\left( 30 \right)$ is a modular curves, we may compute its conductor from the isogenous decomposition of its Jacobian into a product of abelian varieties of smaller dimension. The modular Jacobian $X_{0}\left( N \right)$ is isogenous to $\directsum_{f} A_{f}$ where $A_{f}$ is the abelian variety associated to a newform $f \in S_{2}\left( M_{f} \right)$ of some level $M_{f}$, and the direct sum is over equivalence classes of newforms in $ S_{2}\left( N \right) $. These can be computed using Stein's modular symbols algorithms and the implementations in \texttt{Magma}, see \cite{Stein}. The conductor of $J_{0}\left(N \right)$ is equal to product, over equivalence classes of newforms in $ S_{2}\left( N \right) $, of the conductors of $A_{f}$. This is clear from the definition of the conductor, as given in Section 6,  if we take $l \neq p$ and not dividing the degree of the isogeny.

In this example, we find
\begin{center}
    $J_{0}\left( 30 \right) \simeq  E_{1} \directsum E_{2} \directsum E_{3}$
\end{center}
where $E_{i}$ all have dimension $1$, and conductors $30$,$15$ and $15$, respectively. Thus $n_{2} =1$, verifying our computation.

\subsection{$J_{0}\left( 40 \right)$}
Recall that the 3-torsion subgroup $J_{0}\left( 40 \right)[3]$ is defined over a degree $48$ number field defined by the polynomial $f$, stated in Section 5.2. This polynomial remains irreducible over $\mathbb{Q}_{2}$, and defines a degree $48$ Galois extension of $\mathbb{Q}_{2}$, which we denote by $L$. Let $G = \Gal \left( L/ \mathbb{Q}_{2} \right)$. We find that $G$ can be generated by $\tau_{1}$,$\tau_{2}$,$\beta$, $\sigma_{1}$, $\sigma_{2}$, where $\tau_{i}$ have order $2$, $\beta$ has order $3$ and $\sigma_{j}$ have order 4. Then,
\begin{align*}
& G_{0} = \langle \tau_{1}, \beta, \sigma_{1}, \sigma_{2} \rangle \ \text{and} \ \vert G_{0} \vert = 24 \\
& G_{1} =  \langle \tau_{1}, \sigma_{1}, \sigma_{2} \rangle \ \text{and} \ \vert G_{1} \vert = 8 \\
& G_{2} = G_{3}  =  \langle \tau_{1} \rangle \ \text{and} \ \vert G_{2} \vert = \vert G_{3} \vert = 2 \\
& G_{n} = 1 \ \text{for all } \ n \ge 4 \\
\end{align*}

Using the explicit generators stated in section 5.2, we can compute the Galois invariants
\begin{align*}
& J_{0}\left( 40 \right)^{G_{0}} \cong \left( \mathbb{Z} / 3\mathbb{Z} \right)^{2} \\
& J_{0}\left( 40 \right)^{G_{1}} \cong \left( \mathbb{Z} / 3\mathbb{Z} \right)^{4} \\
& J_{0}\left( 40 \right)^{G_{2}} \cong \left( \mathbb{Z} / 3\mathbb{Z} \right)^{4} \\
\end{align*}
and thus
\begin{center}
$n_{\text{wild}} = 4/1 + 2/3 + 2/12 + 2/12 = 5$
\end{center}
As in the previous example, we can use Stein's algorithm to compute $n_{2}$,
\begin{center}
    $ J_{0}\left( 40 \right) \simeq E_{1} \directsum E_{2} \directsum E_{3}$
\end{center}
where $E_{i}$ are abelian varieties of dimension 1, and conductors $2^3\cdot5$, $2^2\cdot5$ and $2^2\cdot5$, respectively. This suggests that $n_{2} = 7$.
\begin{remark}
The above suggest that $n_{\text{tame}} =2$, but the author is yet to verify this directly.
\end{remark}

redpajama_set_name: RedPajamaArXiv


text:
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redpajama_set_name: RedPajamaCommonCrawl


Montana's health department was dealt a major blow in late 2017, when the agency was forced to cut almost $50 million to help balance the state budget.
The ongoing partial government shutdown has scuttled an upcoming meeting between Missoula County and the Environmental Protection Agency.
That meeting was meant to discuss ongoing concerns about the closed and contaminated Smurfit-Stone Container pulp and paper mill site in Frenchtown.
Missoula County is updating its future growth map.
Why? Maybe think about it like this: you go to a restaurant, open the menu and have to choose from 64 different entrées. Not only would it be a chore to simply decide what to order, your selection probably wouldn't even taste very good.


Why is this night different from all other nights – they will be in our new building!
Please join Rabbi Shlomo and Rebbetzen Lynndy and Rabbi Eli and Rebbetzen Lauren and Family for this special evening.
Alternatively please join Rabbi Shlomo & Rebbetzin Bonnie and family for the Young Professionals and Young Couples Seder.
You will need to book for catering and security purposes and we would welcome any donations to cover the costs of this evening which are £42 per adult and £24 per child.
Please put reference Seder/your Family Name. Please call the office to let them know if you do this as we will need to know your numbers and any dietary requirements.


Your religion is the set of beliefs passed down to you from the church. Your spirituality, your faith, is something you create based on your commitments and ultimate concerns. You have to know what you are committed to, in order to have your faith mean anything.
There were 30 postulants in the room, all under 20, and I suspect none of us knew what he was talking about. We had memorized the Baltimore Catechism and had our rote answers for everything, but this notion of spirituality being separate from religion was unnerving. We had been taught what to think, not how to think. How did he expect us to know what our ultimate concerns were?
It took Father Grabys the whole semester to teach us how to think for ourselves, mine our own depths, and identify what we were willing to live for, as much as to die for. It was grueling work. I wanted religion to be religion, like I'd always known it. I didn't want to create something on my own. I didn't like any of his ideas. But week after week, we entered the mines, scoured our histories, surfaced our hopes, and finally, by the end of the class, each of us proclaimed our ultimate commitments.
We were forced to look beyond the constructs of religion to find our own beliefs.
Fifty years later, I look back and see that resolve has not wavered. Everything I do springs from that well. Whatever I create is connected to who I am spiritually. And who I am spiritually is the one who acts for justice and peace. This is my living faith. It has little to do with religion and does not require a church. It is a spiritual force that is active wherever I am.
If someone asked me how to create a living faith, I'd suggest two steps. First, let go of beliefs you inherited that no longer suit you. See what comes up when you think about the Divine, or Infinity, or the Cosmos. Write out your own Apostle's Creed. See what it feels like to be the author of your own spiritual beliefs. There is no need to be right. There is nothing to defend. You are unique and your spirituality originates in that very uniqueness.
Your faith is your own personal creation. It is your private covenant with the Source of Life, the One Who Is Breathing You. Based on everything you know to be true from your lived experience, you decide what you are absolutely committed to. It's like a marriage vow you make to the Great Beloved, Spirit, God, or whoever and whatever you choose to follow. No matter what your career is, what religion you grew up in, where you live, what terrible things have happened to you, underneath everything is the faith that holds you up, based on the pillars of your ultimate concerns.
You don't have to know anything more, read any new books, talk to any priest, rabbi, or minister; you only need to proclaim to yourself what you are committed to. If you live from that, dedicate yourself to that, then that very thing will be your living faith. When the Dalai Lama says, "My religion is human kindness," we learn something about his ultimate commitment. And we see from his actions that he practices his faith daily.
You may create a living faith without a practice, but you cannot sustain it. Just as your body needs food, your spirit needs silence and solitude. It needs communion time with Mind-at-Large. Your practice is where you breathe life into your soul. It needs your time and loving attention to thrive.
I started my practice in 1990 at 20 minutes a day. I had a candle lit and I sat there staring at it until the time went by. No phones, papers, no computers. Just me and the candle. When my thoughts drifted, I came back to my breathing and paid attention to that. I had no goal but to start a practice, because my life was unbalanced and a spiritual teacher said I'd never find balance without a practice.
Just as your body needs food, your spirit needs silence and solitude.
Since then, the most incredible thoughts, creations, and experiences have occurred during my meditation time that I can't seem to get enough of it. I now dedicate an hour every day to complete silence and solitude. That may seem like a lot, but it's nothing compared to the Dalai Lama's six hours daily. We all get to decide what works best for us, but first comes the establishment of a new habit. And that's why I say give it five minutes, for a start. You're the one in charge of your time. If you can't dedicate five minutes to your own well-being, you really have to wonder: how committed are you to being well?
So that's my blueprint for creating a living faith. Nothing to seek. Nowhere to go. It's alive inside you now, waiting to be claimed, acknowledged, and practiced. You are a sacred vessel carrying a bright and holy light. It's time now to let it shine.
Originally posted on Spirituality and Health, Covering a broad range of topics under the umbrella of health and spirituality.


Community organizers and fellow leftists protested today at Walmart in Pico Rivera, California. Some "striking" Walmart employees and fellow leftists boarded a bus to start a 1,500-mile bus journey titled 'Ride for Respect'. The bus protest will end at the company's annual meeting at its headquarters in Bentonville, Arkansas.


I work a lot with the guys at Criminal Damage and they have just released the Mens Spring / Summer Range for 2011. It's about 95 pages long so thought I would just post the first three. It's always good working with them.


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Research / World Public Sector Reports
Sustainable Development Goal 16: Focus on public institutions
The 2030 Agenda and the Sustainable Development Goals (SDGs) prominently feature institutions, both as a cross-cutting issue in many of the goals and as a standalone goal (SDG 16). The World Public Sector Report 2019 looks at national-level developments in relation to several concepts highlighted in the targets of Goal 16, which are viewed as institutional principles: access to information, transparency, accountability, anti-corruption, inclusiveness of decision-making processes, and non-discrimination. The report surveys global trends in these areas, documenting both the availability of information on those trends and the status of knowledge about the effectiveness of related policies and institutional arrangements in different national contexts. It also demonstrates how the institutional principles of SDG 16 have been informing the development of institutions in various areas, including gender equality and women's empowerment (SDG 5). The report further examines two critical instruments that can support effective public institutions and public administration for the SDGs, namely national budget processes and risk management. The World Public Sector Report 2019 aims to inform the first review of SDG 16 at the United Nations high-level political forum on sustainable development in July 2019, and to contribute to future efforts to monitor progress on SDG 16. By reviewing key challenges and opportunities for public institutions in the context of the implementation of the 2030 Agenda at the national level, the report also aims to inform efforts by all countries to create effective institutions to deliver the Sustainable Development Goals.
Launch of the World Public Sector Report 2019 at the 2019 United Nations Public Service Forum (UNPSF) in the Republic of Azerbaijan on 24 June 2019
World Public Sector 2019
Individual chapters - Advance unedited version
Introduction: Institutions and the Sustainable Development Goals
Chapter 1: Progress on institutional dimensions of SDG 16
Chapter 2: Corruption and the Sustainable Development Goals
Chapter 3: Budget processes in support of SDG implementation
Chapter 4: Risk management in the context of the Sustainable Development Goals
Chapter 5: Institutions for gender equality
Working Together: Integration, Institutions and the Sustainable Development Goals
The World Public Sector Report 2018 (WPSR 2018) examines how governments, public institutions and public administration can foster integrated approaches to the implementation of the 2030 Agenda and the Sustainable Development Goals. The report examines key challenges and opportunities for integrated approaches from the perspective of public administration, highlighting experiences from past decades both at the sectoral and cross-sectoral levels. It also examines how governments across the world have chosen to address existing interlinkages among the SDGs, and the implications of this for public administration and public institutions. The report thus aims to produce a comprehensive empirical analysis of policy integration for the SDGs at the national level, with a view to drawing lessons on how emerging initiatives aiming to policy and institutional integration might lead to long-term success in achieving the SDGs, in different developmental and governance contexts. Arguments made in the report are illustrated by concrete examples in relation to SDG goals, targets or clusters thereof. The report is built around two structuring dimensions: first, the 2030 Agenda and the Sustainable Development Goals as an integrated and indivisible set of goals and targets; and second, the role of the government and public service, including the institutional aspect, in fostering sustainable development.
Public administration experts comment on the World Public Sector Report 2018
WPSR 2018
English | Chinese
WPSR 2018 Chapters
Chapter 1: "The Case for Working Together"
Chapter 2: "Horizontal integration in the context of SDG implementation"
Chapter 3: "Vertical integration for the implementation of the SDGs"
Chapter 4: "Engagement with non-State actors for integrated implementation of the SDGs"
Chapter 5: "Integrated approaches to international migration: policy and institutional perspectives"
Chapter 6: "Integrated approaches to health and well-being"
Chapter 7: "Realizing the SDGs in post-conflict situations: Challenges for the State"
WPSR 2018 Chapter Briefs
Chapter 1: "Integrated approaches for implementing the SDGs: the role of public institutions and public administration"
[English] [French] [Spanish]
Click here for WPSR 2018 Presentations and Background Materials
Responsive and Accountable Public Governance
The 2015 World Public Sector Report, titled Responsive and Accountable Public Governance, presents the need for public governance to become more responsive and accountable in order for the State to lead the implementation of a collective vision of 2030 sustainable development. Social and technical innovations are providing an opportunity for the social contract between the State and the citizenry to shift towards more collaborative governance, supported by effective, efficient, transparent, accountable, inclusive, equitable and responsive public institutions.
Reconstructing Public Administration after Conflict
The 2010 World Public Sector Report brings to the fore a very critical issue - how to reconstruct public administration in post-conflict situations so as to enable it to promote peace and development in countries that have been affected by civil war and destruction. It is a question that has remained unresolved for decades and has brought poverty, despair, and death to people in many corners of the world. The Report shows that no progress can be made in promoting peace, development and protection of human rights unless appropriate governance and public administration institutions are established, leadership and human resources capacities are re-built, citizens are engaged in the process of reconstruction through decentralized participatory mechanisms and basic public services are delivered. In fact, unless newly established governments are able to provide essential public services to the population, including safety, security, health, education, shelter, access to water and sanitation and job opportunities, there will be no durable peace.The report also emphasizes that because post-conflict situations are heterogeneous, there are no "one size fits all" solutions to governance challenges. In each country, public administration reforms should be tailored to local needs. Finally, the report highlights that contrary to commonly held belief, post-conflict situations not only present challenges, but also offer numerous opportunities to leapfrog stages of development by adopting innovative practices in public administration, particularly through the application of ICTs in government and service delivery.
People Matter: Civic Engagement in Public Governance
The 2008 World Public Sector Report, People Matter: Civic Engagement in Public Governance, highlights the importance of civic engagement in public governance and by profiling several case studies, demonstrates how such practices gain the capacity to strengthen governance, make it more transparent and accountable and most importantly, contribute to developmental outcomes that are more sustainable, equitable and just. The Report also highlights several challenges and cautions that adequate attention must be given to the issues of power relations, institutional capacities of the government as well as the civil society organizations and adaptation of methodologies and strategies that suit the local conditions and other factors crucial to the introduction of successful civic engagement practices in public governance.
Unlocking the Human Potential for Public Sector Performance
As recommended by the UN Committee of Experts on Public Administration (UNCEPA), the third World Public Sector Report will be published in 2005, with a particular thematic focus on human resources management (HRM). More specifically, the report will explore how the human potential can be unlocked to enhance public sector performance. UNCEPA, at its Second Meeting in April 2003, stressed that human resources capacity was critical to the quality of public administration. The increasing complexity of both policy-making and administrative processes, as well as the erosion of human resources capacity to carry out those functions, are making it difficult for many Member States to implement national goals and strategies to reduce poverty and to promote sustainable human development, as emphasized in the Millennium Development Goals (MDGs).
E-Government at the Crossroads
The World Public Sector Report 2003 presents a view of e-government as a tool for creating public value. It puts e-government development in the context of the United Nations Millennium Declaration, the Report claims that meaningful e-government applications are those that support the environment conducive to human development and suggests that such an environment can be created by a conscious effort "world making". It discusses the special cases of e-participation and privacy, all as part of the main message the ICT by itself will not result in a different, better government, or higher quality of life, but that thoughtful reform and change must precede or go hand-in-hand with in the introduction of ICT to public administrative operations. The Report identifies development of a networked government, management of information and creation of knowledge as the most important e-government application of the future.
Globalization and the State 2001
Globalization, although not a new phenomenon, is unquestionably of paramount significance for all countries, developed or developing, rich or poor, large or small. What is globalization? How is globalization affecting the role and functions of the nation- State? Is globalization "good" or "bad"? Is there a universal understanding of its potential or its costs? Can all societies benefit from globalization? Are all States adequately prepared to enable their people to seize the opportunities of globalization while minimizing its negative effects? How should public administration systems be redesigned in view of the changes occurring at the global level? This Report, written in two parts, attempts to answer these and other essential questions in an objective and clear fashion, based on observed experience and the views of prominent experts on the matter. Part One deals with globalization and the State, and comprises five chapters. Part Two focuses on defining and measuring the size of the State.


Slow Rhythm, Waltz, Box Rumba and Cha Cha. Do you watch with envy as other couples hit the dance floor and wish it was you. Then this is the class for you. Learn the basics in these dances and you will soon be moving around the dance floor. Ideal starting point for those getting married or anyone that is just wanting to have fun.


I will be trading as scrapsy@oxford tattoo and handing out flyers promoting the studio, as well as selling my own hand made tattoo inspired greetings cards. i'll be making cards at the show and taking special orders, so if you are after something specific, be sure to come by.
i'll also be taking orders for sculptures and will have some on display for inspiration. every one will be unique and personally designed, if you have an idea in mind, give me the basics and let my imagination take over, i guarantee you won't be disappointed!!


Below is the dustcover and subtitle from my defunct and error-laced first book on Robert E. Howard, which I am correcting and rewriting a piece at a time for inclusion in Volume Two of A Well of Heroes.
The writer of today faces the same prejudices as the writer of Howard's day, with mainstream readers recoiling from any depiction of a nonwhite or female villain, just as reactionary whites and masculinity advocates cannot abide the depiction of a diverse racial cast or of striving female characters. For the reader interested in realism in fiction, or in the question of the monolithic political correctness currently smothering any attempts at realism in fiction, whether in print or on film, Dark Art of an Aryan Mystic will illuminate the white, black and red corners of our collective imagination with that most hated and rejected shade of illumination, the gray of truth, which is sought by few and accepted by none.
The selections for this book had been intended to illuminate Howard's treatment of race in storytelling, and will therefore slant this volume in that direction, with the Race theme dominant. As a reviewer I was tempted to spread the re-worked content or Dark Art of an Aryan Mystic out. But on second thought, an effort to enforce an unnatural balance would be antithetical to the subject of this quest to connect on a textual and subtextual level with my dominant literary influence.


text:
The 1986 United Nations Security Council election was held on 16 October 1986 during the Forty-first session of the United Nations General Assembly, held at United Nations Headquarters in New York City. The General Assembly elected Argentina, Italy, Japan, West Germany, and Zambia, as the five new non-permanent members of the UN Security Council for two-year mandates commencing on 1 January 1987.

Rules

The Security Council has 15 seats, filled by five permanent members and ten non-permanent members. Each year, half of the non-permanent members are elected for two-year terms. A sitting member may not immediately run for re-election.

In accordance with the rules whereby the ten non-permanent UNSC seats rotate among the various regional blocs into which UN member states traditionally divide themselves for voting and representation purposes, the five available seats are allocated as follows:

One for African countries (held by Madagascar)
One for countries from the Asian Group (now called the Asia-Pacific Group) (held by Japan)
One for Latin America and the Caribbean (held by Trinidad and Tobago)
Two for the Western European and Others Group (held by Australia and Denmark)

To be elected, a candidate must receive a two-thirds majority of those present and voting. If the vote is inconclusive after the first round, three rounds of restricted voting shall take place, followed by three rounds of unrestricted voting, and so on, until a result has been obtained. In restricted voting, only official candidates may be voted on, while in unrestricted voting, any member of the given regional group, with the exception of current Council members, may be voted on.

Endorsed candidates
Prior to the election, the Mexican delegate on behalf of the Latin American and Caribbean Group endorsed Argentina's candidacy, the West German delegate reiterated the Western European and Other Group's endorsement for Italy's candidacy as well as his country's, and the delegate of Senegal, speaking only for his own government, endorsed Zambia and clarified that his country's candidacy was for the following session.

Result
Voting was conducted on a single ballot. Ballots containing more states from a certain region than seats allocated to that region were invalidated.

See also
List of members of the United Nations Security Council
Germany and the United Nations
India and the United Nations
Japan and the United Nations

References

External links
UN Document A/59/881 Note Verbale from the Permanent Mission of Costa Rica containing a record of Security Council elections up to 2004

1986 elections
1986
Non-partisan elections
1986 in international relations

redpajama_set_name: RedPajamaWikipedia


text:
är en stad i Landkreis Landshut i Regierungsbezirk Niederbayern i förbundslandet Bayern i Tyskland. Staden ligger nära floden Große Laber 21 km nordväst om Landshut och har cirka  invånare.

Källor

Kommuner, köpingar och städer i Landkreis Landshut


That looks amazing! Those photos are delicious haha, by the way if you get the chance could you check my blog out? New blog-maker and posting things about my cake making business; everything I do/make and I keep track of all my projects and every endeavor on there too. Thank you 🙂 ps – nice blog!


text:
Subject: Compliance and Regulations
Subject: Conformidade Institucional e Regulamentos
Health Sciences(3)
Nursing(3)
Compliance and Regulations(3)
Fundamentals of Nursing(3)
Bill Rogers, fl. 2001(3)
Jon Frank, fl. 2014-2016(3)
Medcom(3)
Cumplimiento y Normas(3)
Documentation-Legal(3)
Frank, Jon, fl. 2014-2016(3)
Rogers, Bill, fl. 2001(3)
HIPAA for Nurses, An Overview
produced by Jon Frank, fl. 2014-2016, Medcom, in HIPAA for Nurses (Cypress, CA: Medcom, 2020), 17 mins
The Health Insurance Portability and Accountability Act (HIPAA) of 1996 was developed to improve the efficiency and effectiveness of the healthcare system, and many changes have been implemented since then within the field. This program provides a detailed overview of why HIPAA was implemented, the various roles o...
The Health Insurance Portability and Accountability Act (HIPAA) of 1996 was developed to improve the efficiency and effectiveness of the healthcare system, and many changes have been implemented since then within the field. This program provides a detailed overview of why HIPAA was implemented, the various roles of workers throughout an organization, and how its implementation affects you and your organization. Penalties for violating HIPAA and i... The Health Insurance Portability and Accountability Act (HIPAA) of 1996 was developed to improve the efficiency and effectiveness of the healthcare system, and many changes have been implemented since then within the field. This program provides a detailed overview of why HIPAA was implemented, the various roles of workers throughout an organization, and how its implementation affects you and your organization. Penalties for violating HIPAA and inclusion of the alphanumeric Healthcare Common Procedure Coding System (HCPCS) have been included in this updated program.•	Describe the overall purpose and goals of the Health Insurance Portability and Accountability Act.•	Explain the role of the HIPAA Advisory Team, as well as the role of the individual healthcare worker in ensuring compliance with HIPAA reforms.•	Delineate the penalties for violating HIPAA.•	Describe healthcare access, portability and reliability.•	Describe the five rules of administrative simplification. Show more Show less
Jon Frank, fl. 2014-2016, Medcom, Bill Rogers, fl. 2001
HIPAA for Nurses
Speaker / Narrator
Bill Rogers, fl. 2001
Legal Aspects of Nursing, Documentation-Legal
Copyright © 2020 Medcom
HIPAA for Nurses, The Privacy Rule
The Health Insurance Portability and Accountability Act (HIPAA) of 1996 was developed to improve the efficiency and effectiveness of the healthcare system, and many changes have been implemented since then within the field. The primary focus of this program is the Privacy Rule, from the covered entities who must c...
The Health Insurance Portability and Accountability Act (HIPAA) of 1996 was developed to improve the efficiency and effectiveness of the healthcare system, and many changes have been implemented since then within the field. The primary focus of this program is the Privacy Rule, from the covered entities who must comply to the rights that patients have under the Privacy Rule. This program also covers regulations that have resulted from the enactme... The Health Insurance Portability and Accountability Act (HIPAA) of 1996 was developed to improve the efficiency and effectiveness of the healthcare system, and many changes have been implemented since then within the field. The primary focus of this program is the Privacy Rule, from the covered entities who must comply to the rights that patients have under the Privacy Rule. This program also covers regulations that have resulted from the enactment of the Privacy Rule.After viewing this program, the learner should be able to:•	Describe the Privacy Rule that exists within the Health Insurance Portability and Accountability Act.•	List covered entities who must comply with the Privacy Rule.•	Describe the rule that requires that patients be give written notice of an institution's privacy rules and practices.•	Explain patient's rights regarding the privacy of their medical and personal information•	Describe of key exceptions to the Privacy Rule.•	Describe the most common violations of the Privacy Rule and the consequences for healthcare workers and business associates. Show more Show less
Privacy rights, Health insurance industry, Documentation-Legal, Practice Compliance, Legal Aspects of Nursing
HIPAA for Nurses, The Security Rule
The need for attention to data security came about largely because of past abuses that occurred in many places in healthcare. While regulations associated with the Security Rule have been part of the Health Insurance Portability and Accountability Act (HIPAA) since it was enacted in 1996, they were modified and en...
The need for attention to data security came about largely because of past abuses that occurred in many places in healthcare. While regulations associated with the Security Rule have been part of the Health Insurance Portability and Accountability Act (HIPAA) since it was enacted in 1996, they were modified and enhanced with enactment of the Health Information Technology for Economic and Clinical Health Act (HITECH Act) of 2009. Additional change... The need for attention to data security came about largely because of past abuses that occurred in many places in healthcare. While regulations associated with the Security Rule have been part of the Health Insurance Portability and Accountability Act (HIPAA) since it was enacted in 1996, they were modified and enhanced with enactment of the Health Information Technology for Economic and Clinical Health Act (HITECH Act) of 2009. Additional changes to HIPAA, known as the Final Rule, were announced in January 2013 and signed into law in February 2013. This program will address what nurses and other healthcare workers need to know about the security of electronic data under HIPAA.After viewing this program, the learner should be able to:•	Describe the overall purpose and goals of the Health Insurance Portability and Accountability Act.•	Explain the difference between administrative, physical and technical safeguards.•	Explain the role of the privacy officer in a healthcare institution.•	Describe the various actions and precautions to take to ensure security of patient informationin an electronic format. Show more Show less
Privacy rights, Practice Compliance, Legal Aspects of Nursing, Documentation-Legal, Confidentiality