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 (x 0 , x 1 ), then x0 = λx 0 and x1 = λx 1 for some nonzero λ ∈ C. Thus φ0((x0, x1)) = x1/x0 = (λx 1 )/(λx 0 ) = x 1 /x 0 = φ((x 0 , x 1 )) 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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