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Bezout’s Theorem

By ellipticcurves

We are coming across this theorem so often, that I think it deserves a post of its own. We keep wishing that the number of intersection of two curves defined by equation of degrees m and n, respectively, be mn, but we constantly hit snags. In this post, we try to iron out things out all the way.

For simplicity, we will refer to the degree of the equation defining the curve as the degree of the curve, and denote it by {\rm deg}(C).

One of the problems is that some curves might have infinitely many points of intersection, but that happens precisely when they have a component in common. We rule that possibility out in out conditions.

The other problem is that there is some inhomogeneity to the affine plane. We fix this by working in the projective plane.

Finally, just like with solving polynomial equations in one variable, we should be looking for points with coordinates in an algebraically closed field and counting them with appropriate multiplicity. The latter actually requires a definition. 

First let us look at the one variable case. Let f(x) \in K[x] be a polynomial with a root x=0. Let K[[x]] denote the ring of formal power series in x. How can you figure out the multiplicity of x=0 (the number of times x divides f(x)) using these two. Sounds like a stupid question, so I will just give the answer. It equals the dimension over K of the vector space K[[x]]/(f), where (f) denotes the ideal of K[[x]] generated by (f). Notice that if we write f(x) = x^a g(x) so that g(x) does not have a root at x=0, then g is invertible in K[[x]], and therefore (f) = (x^a). Finally, the basis K[[x]]/(x^a) is formed by 1, x, \ldots, x^{a-1}, while any higher power of x is congruent to zero modulo x^a. Therefore, {\rm dim}_K K[[x]]/(f) = a, as expected.

Through an appropriate change of variables, we can move any point to the origin of the affine plane. We defined the intersection multiplicity i(P; C, D) of two projective curves C:F(X, Y, Z)=0 and D: G(X, Y, Z)=0 at a point P(0:0:1) by analogy:

i(P; C, D) := K[[X, Y]] / (f, g).

Here, f(x,y) := F(x, y, 1) and g(x,y) := G(x, y, 1).

Bezout’s Theorem. Let C and D be two curves in {\mathbb P^2} that do not have a common component. Then

\sum_{P \in C \cap D} i(P; C, D) = {\rm deg}(C) {\rm deg}(D).

Notice that we do not specify the field over which we are searching for points of intersection of C and D. This means we are looking for them over all fields in the universe. One doesn’t have to look that far however. All point of intersection are defined over the algebraic closure \overline{K} of the common field K of definition of these curves.

This entry was posted on September 3, 2008 at 5:20 pm and is filed under Uncategorized. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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