Morphisms

Instead of doing a huge amount of theory defining precisely what morphisms are, we will in each case say explicitly what a morphism is.

As usual K is a field. A morphism from the projective line to the projective plane is given by a triple of polynomials G_0, G_1, G_2 homogeneous of the same degree d in two variables Y_0, Y_1. On points the map is given by

(a : b) |——> (G_0(a, b) : G_1(a, b) : G_2(a, b))

This makes sense because if (a : b) = (a' : b') in P^1, then there exists a nonzero scalar λ in K such that a = λ a', b = λ b' and we see that

(G_0(a, b), G_1(a, b), G_2(a, b)) =
(G_0(λa', λb'), G_1(λa', λb'), G_2(λa', λb')) =
(λ^dG_0(a', b'), λ^dG_1(a', b'), λ^dG_2(a', b')) =
λ^d(G_0(a', b'), G_1(a', b'), G_2(a', b'))

which defines the same point in P.

But for this to make sense we need to make sure that G_0, G_1, G_2 do not vanish simultaneously, because (0 : 0 : 0) isn't a point of P^2. Note that if G_0(a, b) = G_1(a, b) = G_2(a, b) = 0, then bY_0 - aY_1 divides G_0, G_1, and G_2. Because our field K is not algebraically closed we may not have enough points. Thus we will require the stronger property that gcd(G_0, G_1, G_2) = 1 in the polynomial ring K[Y_0, Y_1]. Finally, if we multiply all three by the same nonzero scalar, then we obtain the same morphism.

In conclusion: A morphism P^1 —> P^2 is given by a triple of polynomials G_0, G_1. G_2 homogeneous of the same degree d with gcd(G_0, G_1, G_2) = 1. Two morphisms are considered the same if the triples of polynomials differ by a scalar. The integer d is called the degree of the morphism. If d = 0 we say that the morphism is constant.

We say (G_0, G_1, G_2) is a morphism into the curve C : F = 0 if the substitution of polynomials

F(G_0, G_1, G_2) ∈ K[Y_0, Y_1]

is zero as a polynomial (not just zero on all points). In this case we say that it is a morphism onto C if the morphism is not a constant morphism (this only makes sense because we are mapping from a curve to another curve here; in general the definition is more complicated).

Now it turns out that, any morphism P^1 —> P^2 of degree 2 maps into a line or a conic. Two examples

(a : b ) |——→ (a^2, b^2, 0)

maps to the line X_2 = 0 and the map

(a : b) |——→ (a^2, ab, b^2)

maps into the conic X_0X_2 - X_1^2 = 0.

Exercise 16: Let K = Z/pZ. Write a script that takes three polynomials and checks whether they define a morphism from P^1 to P^2. For example, have it return the degree of the morphism or -1 if it is not a morphism. (Question: Can the degree be zero? I think so!)

Exercise 17.a: Prove that a degree two morphism P^1 —> P^2 maps onto either a line or a conic.

Exercise 17.b: Let K = Z/pZ. Write a script that starts with G_0, G_1, G_2 homogeneous of degree 2, checks whether they define a morphism from P^1 —> P^2 and finds the equation of the line or conic that the morphism maps into. 