9.4 Computing with the divisor class group

Let Q be any closed point in $\mathbb {P}$¹ that is different from the point $\infty$ at infinity. As we saw above Q is given as a closed subscheme of $\mathbb {A}$¹ = $\mathbb {P}$¹ - $\infty$ as the vanishing locus of an irreducible polynomial f (x). If k = deg(f ) then consider the f-fold Veronese embedding of $\mathbb {P}$¹ in $\mathbb {P}$^k. We see that Q is precisely the intersection of the image of $\mathbb {P}$¹ with the hyperplane V(a₀X₀ + ^... + a_kX_k) (if f (x) = a₀ + ^... + a_kX^k). Moreover, V(X₀) intersects the image of $\mathbb {P}$¹ in a k-tuple thickening of $\infty$. From earlier remarks on the K-group we see that (Q) = k($\infty$) in K($\mathbb {P}$¹).

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

1. The image of the element ($\infty$) under this homomorphism is 2($\infty$).
2. 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).
3. If f (x) is an irreducible polynomial so that y² + a(x)y + b(x) is irreducible modulo f (x), then the image of (Q) is (P).
4. If f (x) is an irreducible polynomial so that y² + 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').

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 Pic⁰(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 y² + 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 y² + 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 Pic⁰(T).

To summarise, each divisor class in Pic⁰(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 a²(x) - 4b(x) divides f (x) at most once. Moreover, it is clear that the inverse of this class in Pic⁰(T) is represented by (f (x), g₁(x)), where g₁(x) is the reduction modulo f (x) of a(x) - g(x).

If (f₁(x), g₁(x)) and (f₂(x), g₂(x)) are two such pairs, then we can form their sum in Pic⁰(T) as follows.

1. Assume that f₁(x) and f₂(x) are co-prime. We find h₁(x) and h₂(x) so that h₁f₁ + h₂f₂ = 1. Let g(x) be the reduction of h₁f₁g₂ + h₂f₂g₁ modulo f₁f₂. We see that g(x) reduces to g₁(x) modulo f₁ and to g₂(x) modulo f₂. Hence, by the Chinese Remainder Theorem it follows that g(x)² + a(x)g(x) + b(x) is divisible by f (x) = f₁(x)f₂(x). Thus the sum is (f (x), g(x)).
2. Now suppose that h(x) is a common factor of f₁(x) and f₂(x). We further write h(x) = h₁(x)h₂(x) where h₁(x) is the common factor of h(x) with a²(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 Pic⁰(T). In other words, let f'₁(x) and f'₂(x) be the quotients of f₁(x) and f₂(x) by h₁(x) respectively, and let g'₁(x) and g'₂(x) be the reductions of g₁(x) by f₁(x) and g₂(x) by f₂(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.
3. Assume that the common factor h(x) of f₁(x) and f₂(x) is co-prime to a²(x) - 4b(x). Let h₁ be the highest common factor of h with g₁ + g₂ - a and let h₁ = h/h₂. Now, both g₁ and g₂ a solutions of y² + a(x)y + b(x) = 0 modulo h₁(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 Pic⁰(T). As in the previous case, we can replace the pairs (f₁, g₁) and (f₂, g₂) by another pair with the same sum, with the property that the f₁, f₂ and g₁ + g₂ - a have no common factor.
4. Assume that the common factor h(x) of f₁(x) and f₂(x) is co-prime to a²(x) - 4b(x) and to g₁(x) + g₂(x) - a(x). Now, both g₁ and g₂ are solutions of y² + 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 g₁(x) and g₂(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 f₁(x) = n₁(x)f'₁(x) where f'₁(x) has no factor in common with h(x), moreover n₁(x) is the greatest common factor of f₁(x) with h(x)^k₁ for a suitable power k₁; similarly f₂(x) = n₂(x)f'₂(x). Let k be such that h(x)^k is divisible by n₁(x)n₂(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, g₁(x) modulo f'₁(x) and g₂(x) modulo f'₂(x) to a solution modulo h(x)^kf'₁(x)f'₂(x). Reducing this solution modulo f₁f₂ = n₁n₂f'₁f'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 $\mathbb {P}$^{d + 1}. We have noted earlier that if L is any $\mathbb {P}$^d sitting linearly in $\mathbb {P}$^{d + 1}, then we have an exact sequence

0$$\to$$($$\mathbb {V}$$¹×L)_{$\mathbb {P}$^{d + 1}}$$\to$$$$\mathbb {V}$$¹×$$\mathbb {P}$$^{d + 1}$$\to$$H$$\to$$ 0

Now the restriction of ($\mathbb {V}$¹×L)_{$\mathbb {P}$^{d + 1}} to T is ($\mathbb {V}$¹×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 K₀(T) of (L $\cap$ T) is independent of L. One such L is V(X₀) 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 Pic⁰(T).

Now, any collection of d + 1 points in $\mathbb {P}$^{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 Pic⁰(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 Pic⁰(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 Pic⁰(T). Now, let g(x) = $\sum_{i}^{}$a_ixⁱ have degree at most d and put G(X) = $\sum_{i}^{}$a_iX_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 Pic⁰(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 $\mathbb {P}$^{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 Pic⁰(T) that is represented by the pair (f (x), g(x)). One method is to let g₁(x) be the reduction modulo f (x) of a(x) - g(x) and take the pair (f (x), g₁(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 f₂(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 g₂(x) be the reduction modulo f₂(x) of a(x) - b(x). We see that the pair (f (x), g(x)) and the pair (f₂(X), g₂(x)) represent the same element in Pic⁰(T). Moreover, if f (x) has degree d + k for some k $\geq$ 0, then f₂(x) has strictly smaller degree. Thus we have a method to reduce all pairs representing elements of Pic⁰(T) to pairs (f (x), g(x)) where f (x) has degree at most d - 1.