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**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.