Cubics and more generally plane curves
Let K be a field. Hopefully it should be clear how to think about cubic, quartic, quitic, and higher degree curves in P^2 at this point. But just to be sure, let's define a degree d curve
C : F = 0
in P^2 as determined by a homogeneous polynomial F of degree d in variables X_0, X_1, X_2 with coefficients in K. Two such curves C and C' are the same if and only if F and F' differ by a multiplying by a nonzero scalar in K.
We will often write C(K) to denote the set of points of C.
We say C is irreducible if and only if F is an irreducible polynomial. We say C is geometrically irreducible if F remains irreducible over a separable algebraic closure of K.
Given a pair of curves C and C' of degree d and d' we can ask the following two questions:
Both questions have a naive answer in terms of points, but because we may not have enough points over K we won't usually think of this as the correct answer. The first question has the following natural answer:
The answer to the second question is known as Bezout's theorem. It says that if C and C' have no components in commen (which means that gcd(F, F') = 1 by definition), then the number of points of the intersection C ∩ C' counted with multiplicity is dd'. Even just defining all the terms in this sentence takes a lot of work, but computationally, in examples, it is often very clear why this is so.
Exercise 18: Find an example of curves where C(K) ⊂ C'(K) but not C ⊂ C'.
Exercise 19: For K = Z/pZ write a script checking whether C ⊂ C'.
Exercise 20: Find a pair of conics intersecting in exactly 4 points.
Back to cubics. It turns out that it is usually impossible to parametrize a cubic curve. More generally the same thing is true for higher degree curves. However, if a cubic is irreducible and singular then it can be parametrized.
We say a curve C : F = 0 is singular if there is point (a : b : c) of P^2 with
F(a, b, c) = F_0(a, b, c) = F_1(a, b, c) = F_2(a, b, c) = 0
where F_0, F_1, F_2 are the partial derivatives of F with respect to X_0, X_1, X_2. The problem with this definition (as usual) is that K may not be algebraically closed. In other words, we have to look whether there are solutions to those equations in P^2(K') for all finite extensions K' of K. That is a bit cumbersome and in fact, it is somewhat difficult to come up with an algebraic criterion (in terms of polynomials) that is equivalent to this geometric property. We'll try to avoid it…
Exercise 21: For K = Z/pZ write a script that finds the P ∈ C(K) which are singular.
Exercise 22: Show that a geometrically irreducible cubic curve C with a singular P ∈ C(K) can be parametrized provided the characteristic of K is not 3. If you like, assume that K is algebraically closed (and then it also works in characteristic 3).
Exercise 23: For K = Z/pZ write a script that takes a geometrically irreducible cubic and a singular P ∈ C(K) and finds a parametrization of C.
Continue reading about rational plane curves. Back to the start page.