Now the morphism
*T*^{1} is flat and so we get a group
homomorphism
*K*(^{1})*K*(*T*). In particular, in the various cases
enumerated above, for closed points *P* in *T* that lie over closed
points *Q* in
^{1} we have:

- The image of the element () under this homomorphism is 2().
- If
*Q*is the closed point corresponding to an irreducible factor of*a*(*x*)^{2}- 4*b*(*x*), then the image of (*Q*) is 2(*P*). - 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*). - 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'*).

Thus, elements of
Pic^{0}(*T*) can be written in the form
*n*_{i}[*P*_{i}] + [*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* )() 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 *not* unique. We can
further assume that any irreducible factor of *f* (*x*) that divides
*a*^{2}(*x*) - 4*b*(*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.

- Assume that
*f*_{1}(*x*) and*f*_{2}(*x*) are co-prime. We find*h*_{1}(*x*) and*h*_{2}(*x*) so that*h*_{1}*f*_{1}+*h*_{2}*f*_{2}= 1. Let*g*(*x*) be the reduction of*h*_{1}*f*_{1}*g*_{2}+*h*_{2}*f*_{2}*g*_{1}modulo*f*_{1}*f*_{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*)). - 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*) - 4*b*(*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. - Assume that the common factor
*h*(*x*) of*f*_{1}(*x*) and*f*_{2}(*x*) is co-prime to*a*^{2}(*x*) - 4*b*(*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. - Assume that the common factor
*h*(*x*) of*f*_{1}(*x*) and*f*_{2}(*x*) is co-prime to*a*^{2}(*x*) - 4*b*(*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*)^{k1}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*)^{k}*f'*_{1}(*x*)*f'*_{2}(*x*). Reducing this solution modulo*f*_{1}*f*_{2}=*n*_{1}*n*_{2}*f'*_{1}*f'*_{2}, we obtain the required pair (*f*(*x*),*g*(*x*)).

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
^{d + 1}. We have
noted earlier that if *L* is any
^{d} sitting linearly in
^{d + 1}, then we have an exact sequence

0(^{1}×*L*)_{d + 1}^{1}×^{d + 1}*H* 0

Now the restriction of
(
Now, any collection of *d* + 1 points in
^{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 2*d*. In particular, given any divisor *D* of degree
*d*, we can find an *L* that contains
*D* + (), so that *L*
intersects *T* in
*D* + () + *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} - 4*b*(*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), 2*d* - 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*) = *a*_{i}*x*^{i} have degree at
most *d* and put
*G*(*X*) = *a*_{i}*X*_{i}. Then
(*f* (*x*)*f'*(*x*), *g*(*x*))
represents the divisor *L* *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
^{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