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9.2 Closed points

A proper closed reduced irreducible subscheme of T (or 1) is called a closed point. Let P be a closed point of T and Q be its image in 1. If U, V denote the coordinates on 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 1 or V does not divide F. In the latter case Q is contained in 1 which is the open subset of 1 where V is a unit (i. e. the complement of the point at infinity). The coordinate on 1 is given by x = U/V and Q defined by the irreducible polynomial f (x) = F(U, V)/Vdeg(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 2 defined by the equations
 y2 + a(x)y + b(x) = 0 f (x) = 0

In other words, let E = [x]/(f (x)) be the finite extension of the ground field and let and be the images of a(x) and b(x) in E. The closed point P is given by solving the equation y2 + y + over E. Clearly, there are three cases to consider. The case when this equation has multiple roots (when - 4 = 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 T1. Finally, when the quadratic equation is solvable in E, there is an element in E that corresponds to the point P. Now is the image in E of a polynomial g(x) in [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:
1. The point at infinity on T.
2. 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 y2 + a(x)y + b(x) in the field E = [x]/(f (x)).
3. We have an irreducible polynomial f that is co-prime to the discriminant and the quadratic equation y2 + a(x)y + b(x) is irreducible modulo f (x).
4. 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 y2 + a(x)y + b(x) in the field E = [x]/(f (x)).
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 suffice), we can find that each closed point has many elements''. In fact, let (P) denote the field E = [x]/(f (x)) in cases (2) and (4). In case (3) let (P) be the quadratic extension of E where the irreducible quadratic polynomial y2 + y + has its roots. We note that (P) is a finite extension of the finite field and hence is a Galois extension; thus it contains all the roots of any polynomial which has one of root in it. From this one sees that P((P)) is a finite set of cardinality equal to the degree [(P) : ]; note that this is deg(f ) in cases (2) and (4) and is 2 deg(f ) in case (3). This number (P) : P] is called the degree of the closed point P and denoted deg(P).

Next: 9.3 Divisors Up: 9 Hyperelliptic Cryptosystems Previous: 9.1 Hyperelliptic curves
Kapil Hari Paranjape 2002-10-20