4 1. Warming up to Enumerative Geometry
Slightly modify our description of polynomials to include
infinity.
In any of the cases (1) (a), (b), and (c), there are now exactly
two solutions. The situation is familiar from high school algebra. If
the coefficients of f(x) are real, then in case (a) there are two real
roots, in case (b) there are two complex conjugate roots, and in case
(c) there is one real root with multiplicity 2.
To handle cases (2), we need to include infinity. We do that by
introducing the complex projective line.
Definition 1.1. The complex projective line CP1 (or just P1) is
the set of all ordered pairs of complex numbers {(x, y) C2 | (x, y) =
(0, 0)} where we identify pairs (x, y) and (x , y ) if one is a scalar
multiple of the other: (x, y) = λ(x , y ) for some λ C∗, where C∗ is
the set of nonzero complex numbers.
So for example (1, 2), (3, 6), and (2 + 3i, 4 + 6i) all represent the
same point of
P1.
This construction is an example of the quotient of a set by an
equivalence relation. See Exercise 2.
The idea is that
P1
can be thought of as the union of the set
of complex numbers C and a single point “at infinity”. To see this,
consider the following subset U0
P1:
U0 = {(x0, x1)
P1
| x0 = 0}.
Then U0 is in one-to-one correspondence with C via the map
(1) φ0 : U0 C : (x0, x1)
x1
x0
.
Note that φ0 is well defined on U0. First of all, x0 is not 0 so the
division makes sense. Secondly, if (x0, x1) represents the same point of
P1 as (x0, x1), then x0 = λx0 and x1 = λx1 for some nonzero λ C.
Thus φ0((x0, x1)) = x1/x0 = (λx1)/(λx0) = x1/x0 = φ((x0, x1)) and
φ is well defined as claimed. The inverse map is given by
ψ0 : C U0, z (1, z).
The complement of U0 is the set of all points of
P1
of the form
(0, x1). But since (0, x1) = x1(0, 1), all of these points coincide with
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