\section{Hyperelliptic Cryptosystems}
We will now work with a specific collection of examples of algebraic
schemes that can be used for cryptosystems---hyperelliptic curves. In
this case we can explicitly give a system of representatives for the
elements of the $K$-group, a mechanism for ``reducing'' any element to
one of these representing elements and also a bound for the size of
the group. We will see in the next section how such information about
a group can be used to make computations with the group efficient.
\subsection{Hyperelliptic curves}
Loosely speaking, hyperelliptic curves represent the solutions of the
equations of the form
\[ y^2 + a(x) y + b(x) = 0 \]
where $a$ and $b$ are polynomials in $x$. To put this in the language
of schemes developed earlier, we first restrict our attention to
schemes over $\Spec(\bbF)$ where $\bbF$ is a finite field called the {\em
ground field}. Next we consider the $d-tuple$ Veronese embedding of
$\bbP^1$ in $\bbP^d$; also known as the ``rational normal curve of
degree $d$''; this is given as the locus of $(1:x:x^2:\cdots:x^d)$ as
$(1:x)$ varies over $\bbP^1$. Alternatively, it is described by the
system of equations $X_pX_q=X_rX_s$ for all $p$, $q$, $r$, $s$ such
that $p+q=r+s$. Let us consider $\bbP^{d+1}$ with $(X_0:\cdots:X_d:Y)$
as its co-ordinates so that $\bbP^d$ is obtained by projecting from
the point (vertex) $v=(0:\cdots:0:1)$. Let $S_d$ denote the ``cone''
over the rational normal curve of degree $d$; it is the subvariety of
$\bbP^d$ defined by the same set of equations as above (in other words
the variable $Y$ is ``free''). Now suppose that $a(x)=\sum_i a_i x^i$
is a polynomial of degree at most $d$ and $b(x)=\sum_i b_i x^i$ is a
polynomial of degree at most $2d$. We consider the linear forms
\begin{eqnarray*}
A(X) & = & \sum_{i=0}^d a_i X_i \\
B(X) & = & \sum_{i=0}^d b_i X_i \\
C(X) & = & \sum_{i=1}^{d} b_{d+i} X_i
\end{eqnarray*}
and the quadratic equation $Y^2+A(X)Y+B(X)X_0+C(X)X_d=0$. The addition
of this equation to the equations for $S$ defines a subvariety $T$ of
$S$. It is clear that the vertex $v$ does not lie on $T$ so that
projection gives a morphism on $T$ which lands in the rational normal
curve of degree $d$. Thus, we have a morphism $T\to\bbP^1$. There is
an involution on $\bbP^{d+1}$ which fixes the $X$'s and sends $Y$ to
$A(X)-Y$. Clearly this involution $\iota$ sends $T$ to itself and
pairs of points that are involutes of each other are sent to the same
point in $\bbP^1$. The variety $T$ is called a hyperelliptic curve,
the involution is called the hyperelliptic involution and the morphism
$T\to\bbP^1$ is called the canonical morphism.
Now it is clear that a solution $(x,y)$ of the equation
$y^2+a(X)y+b(x)=0$ gives rise to the solution $(1:x:\cdots:x^d:y)$ of
the above system. Conversely, if we have a solution
$(X_0:X_1:\cdots:X_d:Y)$ of the system of equations with $X_0$ a unit,
then we can put $(x,y)=(X_1/X_0,Y/X_0)$ to obtain a solution of the
two variable equation. Similarly, if $(X_0:\cdots:X_d:Y)$ is a
solution of the system of equations and $X_d$ is a unit then consider
the pair $(u,v)=(X_{d-1}/X_d,Y/X_d)$; this pair satisfies a two
variable equation $v^2+a'(u)v+b'(u)=0$, where $a'(u)=u^da(1/u)$ and
$b'(u)=u^{2d}b(1/u)$. One sees from the above system that either $X_0$
or $X_d$ must be a unit so we have covered all cases. The Jacobian
criterion for regularity can be used to show that the curve defined by
$y^2+a(x)y+b(x)=0$ is regular when either,
\begin{enumerate}
\item the discriminant $a(x)^2-4b(x)$ has distinct roots, or
\item the field $\bbF$ has characteristic $2$, $a(x)$ has distinct roots
and for each point $(x_0,y_0)$ where $x_0$ is a root of $a(x)$, the
polynomial $b(x)-b(x_0)-y_0a(x)$ vanishes with multiplicity one at
$0$.
\end{enumerate}
To apply this to the equation $v^2+a'(u)v+b'(u)=0$, we note that
\[ a'(u)^2-4b'(u) = u^{2d} ( a(1/u)^2 - 4 b(1/u) ) \]
Thus, if $a(x)^2-4b(x)$ has distinct roots, then the only multiple
root of $a'(u)^2-4(b'(u)$ can be at $u=0$; moreover, this happens only
if $a(x)$ has degree less than $d-1$ and $b(x)$ has degree less than
$2d-1$. From now one we will assume the $T$ is regular or
non-singular; in fact we will assume that $b(x)$ has degree {\em
equal} to $2d-1$. The point $(0:\cdots:0:1:0)$ is a point on the
curve $T$ is called the ``point at infinity'' and denoted $\infty$.
The number $g=d-1$ is called the genus of the hyperelliptic curve. The
points on $T$ where $a(x)^2-4b(x)$ vanishes and the point at infinity
are called the {\em Weierstrass points} of the hyperelliptic curve;
these are precisely the fixed points of the Weierstrass involution.
\subsection{Closed points}
A proper closed reduced irreducible subscheme of $T$ (or $\bbP^1$) is
called a {\em closed point}. Let $P$ be a closed point of $T$ and $Q$
be its image in $\bbP^1$. If $U$, $V$ denote the coordinates on
$\bbP^1$ then $Q$ is defined as the vanishing locus of an irreducible
homogeneous polynomial $F(U,V)$. Thus either $F=V$ and $Q$ is the
point at infinity on $\bbP^1$ or $V$ does not divide $F$. In the
latter case $Q$ is contained in $\bbA^1$ which is the open subset of
$\bbP^1$ where $V$ is a unit (\ie the complement of the point at
infinity). The coordinate on $\bbA^1$ is given by $x=U/V$ and $Q$
defined by the irreducible polynomial $f(x)=F(U,V)/V^{\deg(F)}$. Now,
if $Q$ is the point at infinity then the description in the previous
paragraph shows that $P$ must be the point at infinity on $T$. In the
second case $P$ is an irreducible closed subscheme of the subscheme of
$\bbA^2$ defined by the equations
\begin{eqnarray*}
y^2 + a(x) y + b(x) &=& 0 \\
f(x) &=& 0
\end{eqnarray*}
In other words, let $E=\bbF[x]/(f(x))$ be the finite extension of the
ground field $\bbF$ and let $\alpha$ and $\beta$ be the images of $a(x)$
and $b(x)$ in $E$. The closed point $P$ is given by solving the
equation $y^2+\alpha y+\beta$ over $E$. Clearly, there are three cases
to consider. The case when this equation has multiple roots (when
$\alpha^2-4\beta=0$) is clear the case which corresponds to
Weierstrass points. The case when this equation is irreducible over
$E$ is the case case when $P$ is the full inverse image of $Q$ under
the morphism $T\to\bbP^1$. Finally, when the quadratic equation is
solvable in $E$, there is an element $\gamma$ in $E$ that corresponds
to the point $P$. Now $\gamma$ is the image in $E$ of a polynomial
$g(x)$ in $\bbF[x]$, we can further choose $g$ so that its degree is less
than the degree of $f$. To summarise, a closed point of $T$ takes one
of the following forms:
\begin{enumerate}
\item The point at infinity on $T$.
\item An irreducible factor $f(x)$ of the discriminant $a(x)^-4b(x)$
is given. In this case there is a unique polynomial $g(x)$ of degree
less than $\deg(f)$ so that $y=g(x)$ represents the (unique)
solution of the equation $y^2+a(x)y+b(x)$ in the field
$E=\bbF[x]/(f(x))$.
\item We have an irreducible polynomial $f$ that is co-prime to the
discriminant and the quadratic equation $y^2+a(x)y+b(x)$ is
irreducible modulo $f(x)$.
\item We have an irreducible polynomial $f(x)$ that is co-prime to the
discriminant. Moreover, we are given a polynomial $g(x)$ of degree
less than $\deg(f)$ so that $y=g(x)$ represents one of the two
solutions of the equation $y^2+a(x)y+b(x)$ in the field
$E=\bbF[x]/(f(x))$.
\end{enumerate}
We note that the first two cases above correspond to Weierstrass
points on $T$.
One should not be misled by the term ``closed point''---when
considering solutions over general finite rings (in our case rings
that are finite dimensional vector spaces over $\bbF$ suffice), we can
find that each closed point has many ``elements''. In fact, let $\bbF(P)$
denote the field $E=\bbF[x]/(f(x))$ in cases (2) and (4). In case (3) let
$\bbF(P)$ be the quadratic extension of $E$ where the irreducible
quadratic polynomial $y^2+\alpha y+\beta$ has its roots. We note that
$\bbF(P)$ is a finite extension of the finite field $\bbF$ and hence is a
Galois extension; thus it contains {\em all} the roots of any
polynomial which has {\em one} of root in it. From this one sees that
$P(\bbF(P))$ is a finite set of cardinality equal to the degree
$[\bbF(P):\bbF]$; note that this is $\deg(f)$ in cases (2) and (4) and is
$2\deg(f)$ in case (3). This number $\bbF(P):P]$ is called the {\em
degree} of the closed point $P$ and denoted $\deg(P)$.
\subsection{Divisors}
Let $Z$ be a proper closed subscheme of $T$ and $W$ be its image in
$\bbP^1$. As before $W$ is defined as the vanishing locus of a
homogeneous polynomial $F(U,V)$. Let $F=V^kF_1^{k_1}\cdots F_r^{n_r}$
be a factorisation of $F$ with $F_i$ irreducible and distinct from
each other and $V$. Clearly $W$ is the disjoint union of closed
subschemes $W_i$ each defined by the vanishing of $F_i(U,V)^{k_i}$ and
the scheme $W_0$ defined by $V^k=0$. As before, we write
$f_i(x)=F_i(U,V)/V^{\deg(F_i)}$, where $x=U/V$; let $Q_i$ denote the
closed point in $\bbA^1$ defined by $f_i$ and $Q_0$ be the point in
$\bbP^1$ defined by $V=0$. We can decompose $Z$ into the components
$Z_i$ that lie over the component $W_i$ of $W$. We can then classify
$Z_i$ according to the classification of the polynomials $f_i$ as
above. In cases (1), (2) and (3) above there is exactly one closed
point that lies over $Q_i$, thus the schemes $Z_i$ are ``thickenings''
of the corresponding closed points $P_i$. In case (4) there are two
closed points corresponding to the distinct roots; we denote these by
$P_{i,1}$ and $P_{i,2}$. Let $P_{i,1}$ correspond to the solution
$y=g(x)$ or $y^2+a(x)y+b(x)=0$ in $\bbF[x]/(f_i(x))$. By Hensel's lemma
we can find $g_{k_i}(x)$ in $\bbF[x]/(f_i(x)^{k_i})$ which is a ``lift''
of the solution $g(x)$. Thus we have the closed subscheme $Z_{i,1}$ of
$Z_i$ defined by the solution $y=g_{k_i}(x)$. Similarly, we have
$Z_{i,2}$ and it is clear that $Z_i$ is the union of these two
schemes. Thus each proper closed subscheme of $T$ is the disjoint
union of ``thickened'' closed points.
For any such closed subscheme $Z$ of $T$ we have a vector space scheme
given by $(\bbG_a)_Z$ extended by zero on the rest of $T$. We denote
this vector space scheme by $(P)$ when $Z$ is the subscheme associated
to the a closed point $P$. The vector space scheme associated with the
``thickened'' closed points is equivalent, in the $K$-group, to $n(P)$
for some integer $n$. This can be shown by a ``composition series
argument''. A similar Jordan--H\"older composition series can be used
to show that the $K$-group of $T$ is generated by $(\bbG_a)_T$ and the
elements $(P)$. Moreover, if we consider an element $D$ of the form
$\sum_i n_i(P_i)$ of the $K$-group then the number $\deg(D)=\sum
n_i\deg(P_i)$ can be shown to be well-defined (independent of the
representation of $D$). Thus the important group becomes the group of
``divisors of degree 0'' of the subgroup of the $K$-group consisting
of elements of the form $\sum_i n_i(P_i)$ where $\sum_i
n_i\deg(P_i)=0$. This group is denoted $\Pic^0(T)$. An important
theorem of Weil states that there is a group scheme $J$ (called the
Jacobian variety of $T$) such that $\Pic^0(T)$ can be naturally
identified with $J(\bbF)$. There is also a natural analogy of this with
the divisor class group for quadratic number fields that we will
consider in the next subsection.
To compute the group $\Pic^0(T)$ of divisors of degree 0, it enough to
work modulo $(\infty)$ which is of degree 1, since any divisor can be
converted to one of degree 0 by subtracting a suitable multiple of
$(\infty)$. Thus, we see that this group is generated by the elements
$[P]=(P)-\deg(P)(\infty)$. For a divisor $D$ of degree $d$ we
introduce the notation $[D]=D-d(\infty)$ to denote the corresponding
element in $\Pic^0(T)$.
\subsection{Computing with the divisor class group}
Let $Q$ be any closed point in $\bbP^1$ that is different from the
point $\infty$ at infinity. As we saw above $Q$ is given as a closed
subscheme of $\bbA^1=\bbP^1-\infty$ as the vanishing locus of an
irreducible polynomial $f(x)$. If $k=\deg(f)$ then consider the
$f$-fold Veronese embedding of $\bbP^1$ in $\bbP^k$. We see that $Q$ is
precisely the intersection of the image of $\bbP^1$ with the
hyperplane $V(a_0X_0+\cdots+a_kX_k)$ (if $f(x)=a_0+\cdots+a_kX^k$).
Moreover, $V(X_0)$ intersects the image of $\bbP^1$ in a $k$-tuple
thickening of $\infty$. From earlier remarks on the $K$-group we see
that $(Q)=k(\infty)$ in $K(\bbP^1)$.
Now the morphism $T\to\bbP^1$ is flat and so we get a group
homomorphism $K(\bbP^1)\to K(T)$. In particular, in the various cases
enumerated above, for closed points $P$ in $T$ that lie over closed
points $Q$ in $\bbP^1$ we have:
\begin{enumerate}
\item The image of the element $(\infty)$ under this homomorphism is
$2(\infty)$.
\item If $Q$ is the closed point corresponding to an irreducible
factor of $a(x)^2-4b(x)$, then the image of $(Q)$ is $2(P)$.
\item If $f(x)$ is an irreducible polynomial so that $y^2+a(x)y+b(x)$
is irreducible modulo $f(x)$, then the image of $(Q)$ is $(P)$.
\item If $f(x)$ is an irreducible polynomial so that $y^2+a(x)y+b(x)$
has distinct roots $g(x)$ and $h(x)$ modulo $f(x)$, then there are
two closed points $P$ and $P'$ that lie over $Q$ and the image of
$(Q)$ is $(P)+(P')$.
\end{enumerate}
From the relation $(Q)=\deg(Q)(\infty)$ we obtain relations in each
case as follows. In case (2) we see that $\deg(P)=\deg(Q)$ so that
$(Q)-\deg(Q)(\infty)$ has the image $2[P]=2(P)-2\deg(P)(\infty)$; thus
$[P]$ is a two torsion point in this case.
In case (3), we have $\deg(P)=2\deg(Q)$ and so that
$(Q)-\deg(Q)(\infty)$ has image $[P]=(P)-\deg(P)(\infty)$; thus $[P]$
is 0 in this case. In case (4) $\deg(P)=\deg(P')=\deg(Q)$ and the
image of $(Q)-\deg(Q)(\infty)$ is $[P]+[P']$ which gives us the
identity $[P]+[P']=0$.
Thus, elements of $\Pic^0(T)$ can be written in the form $\sum_i
n_i[P_i] + \sum_j [P_j]$ where the former $[P_i]$ are all of type (4)
and the latter $[P_j]$ are of type (2). As we saw above, Hensel's
lemma allows us to lift the solution $y=g(x)$ of the equation
$y^2+a(x)y+b(x)$ modulo $f(x)$ in case (4) to a solution $y=g_k(x)$
modulo $f(x)^k$ for any $k$. Combining this with the Chinese remainder
theorem, we see that divisors are characterised as solutions $y=g(x)$
of $y^2+a(x)y+b(x)$ modulo $f(x)$, where $f(x)$ is not necessarily
irreducible. Conversely, given such a solution, let $Z=V(y-g(x),f(x))$
and we have the divisor $(Z)-\deg(f)(\infty)$ in $\Pic^0(T)$.
To summarise, each divisor class in $\Pic^0(T)$ is represented by a
pair of polynomials $(f(x),g(x))$, where $g(x)$ has degree less than
that of $f(x)$ and $g(x)^2+a(x)g(x)+b(x)$ is divisible by $f(x)$; as
we shall see below this representation is {\em not} unique. We can
further assume that any irreducible factor of $f(x)$ that divides
$a^2(x)-4b(x)$ divides $f(x)$ at most once. Moreover, it is clear that
the inverse of this class in $\Pic^0(T)$ is represented by
$(f(x),g_1(x))$, where $g_1(x)$ is the reduction modulo $f(x)$ of
$a(x)-g(x)$.
If $(f_1(x),g_1(x))$ and $(f_2(x),g_2(x))$ are two such pairs, then we
can form their sum in $\Pic^0(T)$ as follows.
\begin{enumerate}
\item Assume that $f_1(x)$ and $f_2(x)$ are co-prime.
We find $h_1(x)$ and $h_2(x)$ so that $h_1f_1+h_2f_2=1$. Let $g(x)$
be the reduction of $h_1f_1g_2+h_2f_2g_1$ modulo $f_1f_2$. We see
that $g(x)$ reduces to $g_1(x)$ modulo $f_1$ and to $g_2(x)$ modulo
$f_2$. Hence, by the Chinese Remainder Theorem it follows that
$g(x)^2+a(x)g(x)+b(x)$ is divisible by $f(x)=f_1(x)f_2(x)$. Thus the
sum is $(f(x),g(x))$.
\item Now suppose that $h(x)$ is a common factor of $f_1(x)$ and
$f_2(x)$. We further write $h(x)=h_1(x)h_2(x)$ where $h_1(x)$ is the
common factor of $h(x)$ with $a^2(x)-4b(x)$. Since the corresponding
elements $[P]$ (in case (2) as above) are of order 2 it follows that
this factor disappears when the sum is taken in $\Pic^0(T)$. In
other words, let $f'_1(x)$ and $f'_2(x)$ be the quotients of
$f_1(x)$ and $f_2(x)$ by $h_1(x)$ respectively, and let $g'_1(x)$
and $g'_2(x)$ be the reductions of $g_1(x)$ by $f_1(x)$ and $g_2(x)$
by $f_2(x)$ respectively. The sum of the pairs $(f'_1(x),g'_1(x))$
and $(f'_2(x),g'_2(x))$ is the same as the sum we want to compute.
\item Assume that the common factor $h(x)$ of $f_1(x)$ and $f_2(x)$
is co-prime to $a^2(x)-4b(x)$. Let $h_1$ be the highest common
factor of $h$ with $g_1+g_2-a$ and let $h_1=h/h_2$. Now, both $g_1$
and $g_2$ a solutions of $y^2+a(x)y+b(x)=0$ modulo $h_1(x)$ and their
sum is $a(x)$. It follows that these are complementary solutions as
in case (4) above. Thus these cancel out when the sum is taken in
$\Pic^0(T)$. As in the previous case, we can replace the pairs
$(f_1,g_1)$ and $(f_2,g_2)$ by another pair with the same sum, with
the property that the $f_1$, $f_2$ and $g_1+g_2-a$ have no common
factor.
\item Assume that the common factor $h(x)$ of $f_1(x)$ and $f_2(x)$ is
co-prime to $a^2(x)-4b(x)$ and to $g_1(x)+g_2(x)-a(x)$. Now, both
$g_1$ and $g_2$ are solutions of $y^2+a(x)y+b(x)=0$ modulo $h(x)$
and they are not complementary modulo any factor of $h(x)$. By the
uniqueness part of Hensel's lemma it follows that $g_1(x)$ and
$g_2(x)$ is have the same reduction $m(x)$ modulo $h(x)$. Another
application of Hensel's lemma allows us to lift $m(x)$ to a solution
$m_k(x)$ of the above equation modulo $h(x)^k$, for every power $k$.
Now, we can write $f_1(x)=n_1(x)f'_1(x)$ where $f'_1(x)$ has no
factor in common with $h(x)$, moreover $n_1(x)$ is the greatest
common factor of $f_1(x)$ with $h(x)^{k_1}$ for a suitable power
$k_1$; similarly $f_2(x)=n_2(x)f'_2(x)$. Let $k$ be such that
$h(x)^k$ is divisible by $n_1(x)n_2(x)$. By using the Chinese
Remainder theorem as before, we can find $g'(x)$ which lifts the
solutions $m_k(x)$ modulo $h(x)^k$, $g_1(x)$ modulo $f'_1(x)$ and
$g_2(x)$ modulo $f'_2(x)$ to a solution modulo
$h(x)^kf'_1(x)f'_2(x)$. Reducing this solution modulo
$f_1f_2=n_1n_2f'_1f'_2$, we obtain the required pair $(f(x),g(x))$.
\end{enumerate}
Finally we need to ``reduce'' divisors to a bounded collection. For
this we use our original description of the hyperelliptic curve $T$ as
a closed subscheme of the cone $S_d$ in $\bbP^{d+1}$. We have
noted earlier that if $L$ is any $\bbP^{d}$ sitting linearly in
$\bbP^{d+1}$, then we have an exact sequence
\[
0 \to (\bbV^1\times{L})_{\bbP^{d+1}} \to
\bbV^1\times\bbP^{d+1} \to H \to 0
\]
Now the restriction of $(\bbV^1\times{L})_{\bbP^{d+1}}$ to $T$ is
$(\bbV^1\times{D})_T$ where $D$ is the divisor on $T$ given by the
intersection of $L$ and $T$. As remarked earlier, this shows that the
class in $K_0(T)$ of $(L\cap T)$ is independent of $L$. One such $L$
is $V(X_0)$ which intersects $T$ in $2d(\infty)$. Thus, we note
that if $(L\cap T)=\sum_i n_i(P_i)$ then $\sum_i n_i[P_i]=0$ in
$\Pic^0(T)$.
Now, any collection of $d+1$ points in $\bbP^{d+1}$ lie on an $L$
which contains them. More generally, on can show the same for a
divisor of degree $d+1$ on $T$. Now any $L$ intersects $T$ in a
divisor of degree $2d$. In particular, given any divisor $D$ of degree
$d$, we can find an $L$ that contains $D+(\infty)$, so that $L$
intersects $T$ in $D+(\infty)+E$ where $E$ has degree $d-1$. Thus,
we see that $[D]+[E]=0$ in $\Pic^0(T)$. The inverse of $[D]$ for a
divisor $D$ of degree $d$ is thus represented by $[E]$ where $E$ has
degree $d-1$. This is the basic geometric idea behind the reduction of
divisors. The algebraic steps for this reduction are described below.
As we saw above, elements of $\Pic^0(T)$ are represented by pairs
$(f(x),g(x))$, where $g(x)$ has degree less than the degree of $f(x)$
and $h(x)=g(x)^2+a(x)g(x)+b(x)$ is divisible by $f(x)$. Moreover, we can
also assume that $f(x)$ is divisible at most once by any irreducible
factor that it has in common with $a(x)^2-4b(x)$. Now if $f(x)$ has
degree $d+k$, then $h(x)$ has degree at most the
maximum of $\{2(d+k-1),(d-1)+(d+k-1),2d-1\}$. Thus writing
$h(x)=f(x)f'(x)$ we see that $f'(x)$ has degree at most the maximum of
$\{d+k-2,d-1\}$. Moreover, if $g'(x)$ is the reduction of $g(x)$
modulo $f'(x)$, then $(f'(x),g'(x))$ is another pair representing an
element of $\Pic^0(T)$. Now, let $g(x)=\sum_i a_i x^i$ have degree at
most $d$ and put $G(X)=\sum_i a_i X_i$. Then $(f(x)f'(x),g(x))$
represents the divisor $L\cap T$ where $L=V(Y-G(X))$, thus we see that
$(f'(x),g'(x))$ represents the inverse of the element of $\Pic^0(T)$
that is represented by $(f(x),g(x))$ in this case. This argument can
be generalised to the case $g$ has degree more than $d$ as well (by
using the $k$-tuple Veronese embedding of $\bbP^{d+1}$ and using
linear subspaces from there) to show the same result.
To summarise, we have two ways of representing the inverse of an
element of $\Pic^0(T)$ that is represented by the pair
$(f(x),g(x))$. One method is to let $g_1(x)$ be the reduction modulo
$f(x)$ of $a(x)-g(x)$ and take the pair $(f(x),g_1(x))$. The other
method is to take $f'(x)$ to be the quotient of $g(x)^2+a(x)g(x)+b(x)$
by $f(x)$ and $g'(x)$ to be the reduction of $g(x)$ modulo
$f(x)$. Combining these let $f_2(x)$ be the quotient by $f(x)$ of
\[
(a(x)-g(x))^2+a(x)(a(x)-g(x))+b(x) = g(x)^2-a(x)g(x)+b(x)
\]
and $g_2(x)$ be the reduction modulo $f_2(x)$ of $a(x)-b(x)$. We see
that the pair $(f(x),g(x))$ and the pair $(f_2(X),g_2(x))$ represent
the same element in $\Pic^0(T)$. Moreover, if $f(x)$ has degree $d+k$
for some $k\geq 0$, then $f_2(x)$ has strictly smaller degree. Thus
we have a method to reduce all pairs representing elements of
$\Pic^0(T)$ to pairs $(f(x),g(x))$ where $f(x)$ has degree at most
$d-1$.
\subsection{Frobenius Endomorphism}
Since all our varieties are defined over a finite field $\bbF$, there
is a special endomorphism to consider. Let $q$ be the number of
elements of the field, then for any element of the field $a=a^q$.
Thus for any polynomial $f(t_1,\dots,t_r)$ with coefficients in $\bbF$
we have
\[ f(t_1,\dots,t_r) ^q = f(t_1^q,\dots,t_r^q) \]
Now consider the endomorphism of $\bbP^k$ given by
$(X_0:\cdots:X_k)\mapsto(X_0^q:\cdots:X_k^q)$. If
$X=V(F_1,\dots,F_p;G_1,\dots,G_q)$ is a subscheme of $\bbP^k$ and the
polynomials $F_i$ and $G_j$ have coefficients in $\bbF$, then this
endomorphisms sends $X$ to itself. This endomorphism of $X$ is called
the Frobenius Endomorphism $F:X\to X$.
If $A$ is any finite dimensional $\bbF$-algebra, then $a\mapsto a^q$
gives a ring homomorphism from $A$ to itself. Moreover if $A$ is
local, then the only elements of $A$ that are fixed under this
homomorphism are elements of $\bbF$. From this one can show that the
intersection of the diagonal $\Delta_X$ with the graph $\Gamma_{F}$ of
the Frobenius in $X\times X$ is precisely $X(\bbF)$; the points of $X$
over $\bbF$.
Now for any regular scheme $X$ over $\bbF$, the Frobenius $F$ is a
flat morphism and thus gives an endomorphism of $K_0(X)$. The latter
group thus acquires some ``structure'' in addition to being an abelian
group. In the case when $X$ is a curve (or more specifically a
hyperelliptic curve) this has additional consequences. As we remarked
above $K_0(X)$ is decomposed as the free group on $\bbG_a$ plus the
free group on $[\infty]$ (which can be any $\bbF$ point of $X$) and
the group $\Pic^0(X)$. Moreover, there is a group scheme $J$ so that
$\Pic^0(X)=J(\bbF)$. Thus, one way to determine the order of the group
$\Pic^0(X)$ is to determine the fixed points for the action of
Frobenius on this group scheme. Now, let $\ell$ be a prime that is
invertible the field $\bbF$. On can show that the points of order
$\ell$ in $J(E)$ for a large enough field extension $E$ of $\bbF$ form
a vector space of rank $2g$ over $\bbZ/\ell\bbZ$ (here $g$ is the
genus of the curve $X$). Moreover, there is a polynomial $P(T)$ of
degree $2g$ with {\em integer coefficients} that is satisfied by the
automorphism of this vector space that is given by the Frobenius
endomorphism; the important point is that this polynomial is {\em
independent} of $\ell$. Another important fact is that this
polynomial has roots that are complex numbers of absolute value
$q^{1/2}$. Finally, given $P(T)$ one can determine the number of
elements in $J(E)$ for any finite extension of $\bbF$. These results
were proved by Weil and were generalised to other varieties in the
form of the ``Weil conjectures'' which were proved by Grothendieck,
Deligne and others.
This approach was used by Schoof to calculate the order of $\Pic^0(T)$
in the case $T$ is an elliptic curve (or a hyperelliptic curve of
genus 1). In this case $P(T)$ is a quadratic polynomial of the form
$T^2+aT+q$; moreover, $J=T$ in this case. One can write polynomials
$f_{\ell}(x)$ that are satisfied by the $x$ co-ordinates of points of
order $l$. Thus we can use the action of the Frobenius on this
polynomial to determine $a$ modulo $\ell$ for a number of primes
$\ell$. The additional inequality $|a|\leq q^{1/2}$, can then we used
to determine $a$. One could attempt to generalise this to other
hyperelliptic curves. One must write down the equations that define
the $\ell$-torsion in the Jacobian $J$. From the action of the
Frobenius on this we can write down the coefficients of $P(T)$ modulo
$\ell$. The inequalities resulting from the knowledge of the absolute
value of the complex roots can then be used to bound the number of
$\ell$ for which this needs to be done in order to determine the
coefficients uniquely.
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