# Intersection of projective curves

In general, how do we find the intersection of projective curves?

For example suppose I have $V_{1}(x^{2}+y^{2}-2z^{2}),V_{2}(x^{2}+y^{2}-z^{2})$ and I want to find $V_{1} \cap V_{2}$ viewed as subsets of $\mathbb{P}^{2}$ over an algebraically closed field $K$.

Well the first two equations imply that $z=0$. If $y \neq 0$ then $x^{2}+y^{2}=0$, from here what to do? wouldn't the answer depend on $K$?

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1. Yes, you're exactly right in this example. (And yes, it does depend what field we work over).

2. In general, the subject of finding algorithms to do this is called elimination theory. It's a classical subject that, while it lost a lot of traction last century, has found some resurgence in popularity lately due to its utility. The modern approach uses the theory of Grobner bases, which you can find a lot of stuff about online. I am a non-expert so will say no more - basically I'm just telling you to google the buzzwords.

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Small examples like this can usually be done by messing around. For example, you have already realized that $x^2+y^2=0$. Similarly, $z^2=0$. There are a few cases for what happens, based on whether or not $K$ has a square root of $-1$ and whether or not the characteristic is $2$. By the way, notice that when you went from $z^2=0$ to $z=0$, you were implicitly passing to the radical of the ideal. I've been an algebraic geometer long enough that I don't do that without noticing it.

If we want to do significantly larger examples, Bernd Sturmfels' book Solving Systems of Polynomial Equations is very good.

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But there's another point of view that approaches "solving a system of equations" differently. Under this point of view, if you try to solve a system of $n$ equations in $m$ unknowns, the first thing you could ask for is what is the dimension of the solution space. In general, the solution space will have dimension $\geq m - n$, but equality might not hold. And if the dimension is zero, then there are only finitely many solutions, so one might ask to count the solutions. To make the theory work out better, it is advantageous to assign multiplicities to the solutions. This is the perspective of Intersection Theory, which has as its starting point Bezout's Theorem. Two good introductions to this theory are Chapter IV of Shafarevich's Basic Algebraic Geometry I and Appendix C of Hartshorne's Algebraic Geometry.
Edit: For your specific example, you can factor $x^2 + y^2 = (x + iy)(x - iy)$ if your field $K$ has a square root of $-1$; otherwise, you can't solve $x^2 + y^2 = 0$ (in projective space).
Dear Michael, user 10 can always solve $x^2+y^2=0$ because $K$ is explicitly assumed algebraically closed. –  Georges Elencwajg Apr 27 '12 at 7:14