y^{2} + a(x)y + b(x) |
= | 0 | |

f (x) |
= | 0 |

In other words, let

- The point at infinity on
*T*. - An irreducible factor
*f*(*x*) of the discriminant*a*(*x*)^{-}4*b*(*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*= [*x*]/(*f*(*x*)). - 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*). - 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*= [*x*]/(*f*(*x*)).

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
*y*^{2} + *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*).