9.2 Closed points

A proper closed reduced irreducible subscheme of T (or $\mathbb {P}$¹) is called a closed point. Let P be a closed point of T and Q be its image in $\mathbb {P}$¹. If U, V denote the coordinates on $\mathbb {P}$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 $\mathbb {P}$¹ or V does not divide F. In the latter case Q is contained in $\mathbb {A}$¹ which is the open subset of $\mathbb {P}$¹ where V is a unit (i. e. the complement of the point at infinity). The coordinate on $\mathbb {A}$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 $\mathbb {A}$² defined by the equations

y² + a(x)y + b(x) = 0

f (x) = 0

In other words, let E = $\mathbb {F}$[x]/(f (x)) be the finite extension of the ground field $\mathbb {F}$ 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 y² + $\alpha$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$$\mathbb {P}$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 $\mathbb {F}$[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 = $\mathbb {F}$[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 = $\mathbb {F}$[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 $\mathbb {F}$ suffice), we can find that each closed point has many ``elements''. In fact, let $\mathbb {F}$(P) denote the field E = $\mathbb {F}$[x]/(f (x)) in cases (2) and (4). In case (3) let $\mathbb {F}$(P) be the quadratic extension of E where the irreducible quadratic polynomial y² + $\alpha$y + $\beta$ has its roots. We note that $\mathbb {F}$(P) is a finite extension of the finite field $\mathbb {F}$ 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($\mathbb {F}$(P)) is a finite set of cardinality equal to the degree [$\mathbb {F}$(P) : $\mathbb {F}$]; note that this is deg(f ) in cases (2) and (4) and is 2 deg(f ) in case (3). This number $\mathbb {F}$(P) : P] is called the degree of the closed point P and denoted deg(P).