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

• 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