**Quadric and cubic surfaces**

Often when you hear the term “quadric surface” or “cubic surface” this refers to a hypersurface S : F = 0 in **P**^3 defined by a homogeneous polynomial F ∈ K[X_0, X_1, X_2, X_3] of degree 2 or 3. As usual we say that S : F = 0 and S' : F' = 0 define the same surface if F = λ F' for some nonzero λ ∈ K. We say S is *irreducible* if F is irreducible. We say that S is *smooth*
or *nonsingular* if the system of equations

F = F_0 = F_1 = F_2 = F_3 = 0

in X_0, X_1, X_2, X_3 has no nonzero solution over the algebraic closure of K. Here F_i is the partial derivative of F with respect to X_i.

I'm going to discuss a way to construct *rational curves on S : F = 0*, i.e., morphisms

φ = (G_0, G_1, G_2, G_3) : **P**^1 —→ **P**^3

which map into S, i.e., such that F(G_0, G_1, G_2, G_3) = 0 as a polynomial. My discussion of
this will **not** be exhaustive and it will not even be a good way of doing this (I think).

Given a quadric or cubic surface S : F = 0 then we can consider C : F(Y_0, Y_1, Y_2, 0) = 0
which is a conic or a cubic in the **P**^2 with coordinates Y_0, Y_1, Y_2. We can more
generally take 4 linear forms L_0, L_1, L_2, L_3 in Y_0, Y_1, Y_2 and consider the conic
or cubic

C : F(L_0, L_1, L_2, L_3) = 0

in **P**^2. If L_0, L_1, L_2, L_3 span the degree 1 part of K[Y_0, Y_1, Y_2] we say that this
is a *plane section* of S. Namely, the condition on the span means exactly that

ψ = (L_0, L_1, L_2, L_3) : **P**^2 —> **P**^3

is a morphism. We have seen in our earlier work that if S is a quadric surface, then the curve C can be parametrized or contains a line, i.e., there is a nonconstant map

χ = (H_0, H_1, H_2) : **P**^1 —> **P**^2

into C where H_0, H_1, H_2 ∈ K[Z_0, Z_1] are homogeneous of degree 2. Combining this with ψ we obtain

ψ o χ : **P**^1 —> **P**^3

whose image is contained in S. What does this mean? It just means that

ψ o χ = (L_0(H_0, H_1, H_2), …)

in other words, you substitute H_i for Y_i. If S is a cubic surface, then you can try to do the same thing by finding L_i such that F(L_0, L_1, L_2, L_3) is singular and using the parametrization of singular cubic curves.

**Exercise 24:**
Use the ideas above to prove (from scratch) that a cubic surface over an algebraically closed field has at least one rational curve on it.

It turns out that both quadrics and cubics always have lines over the **algebraic closure**.
A *line on S* is just a morphism

φ = (G_0, G_1, G_2, G_3) : **P**^1 —→ **P**^3

of degree **1** which map into S. It is a marvelous classical fact that a nonsingular cubic
has exactly 27 lines over an algebraically closed field.

**Exercise 25:**
Work as a group to give a guess as to the possible numbers of lines nonsingular quadric
surface over Z/pZ has. (It turns out not every quadric surface has the same number of
lines.)

**Exercise 26:**
(Too hard probably/just a curiosity.) Can you find a cubic surface over Z/pZ which has 27 lines over Z/pZ?

**Exercise 27:**
The methods above give low degree rational curves. Can you think of methods for finding
higher degree rational curves on the cubic surface S : X_0^3 + X_1^3 + X_2^3 + X_3^3 = 0
over Z/2Z? This is relevant for our project as we later need to do the same thing for
the degree 5 Fermat equation in **P**^5. Thus I suggest you really try this: first think of
algorithms! Don't implement algorithms that take a lot of steps.

Continue reading why rational curves?. Back to the start page.