# Elliptic curves: Why over the complex numbers?

I am a math undergrad, so much of the literature on elliptic curves escapes me. I'm trying to understand why one considers elliptic curves over the complex numbers. Specifically, this part of the Wikipedia article is very technical and hard to understand for me. Silverman and Tate say the following:

In general, two cubic curves meet in nine points. To make that statement correct, one should first of all use the projective plane, which has extra points at infinity. Secondly, one should introduce multiplicities of intersections, counting points of tangency for example as intersections of multiplicity greater than one. And finally, one must allow complex numbers for coordinates. [emphasis mine]

I understand the two first musts, but why must we allow complex numbers for coordinates? An explanation in language understandable by an undergrad would be appreciated. Thanks.

• Here's an analogy: A polynomial of degree 3 has 3 roots, counting multiplicities. But only if we count roots in $\mathbb C$ as well as $\mathbb R$. – TonyK Jan 27 '14 at 9:46
• Silverman--Tate's statement about 9 intersection points requires coordinates in some algebraically closed field (e.g. the complex numbers). But as the Wikipedia article goes on to discuss (in the section after the one you linked to) there are many interesting questions about elliptic curves over other fields, like the rational numbers. – user64687 Jan 27 '14 at 10:11

In general, Bezout's theorem asserts that two curves in the projective plane over an algebraically closed field intersect in a number of points wich equals the product of their degrees, counting the points with multiplicity and assuming that the two curves have no components in common. You cannot relax hypothesis of this theorem in any way. If you consider the affine plane instead of the projective one, you could not be able to "see" points that lie at infinity, e.g. take two parallel lines in the affine plane over any field. If the field is not algebraically closed the curves could have intersection points which "do not appear" because of the fact that over fields which are not algebraically closed there exist polynomials in one variable of degree $>1$ which are irreducible. In fact, finding points of intersection of two algebraic curves roughly means finding zeroes of polynomials. For example, consider the conic given by $x^2+y^2-z^2=0$ and the line $z=0$. If you consider them as curves over $\mathbb R$, you won't find any intersection point, because the equation $x^2+y^2=0$ has as the only solution over $\mathbb R$ the pair $(0,0)$ which does not give you a point in the projective plane since $(0\colon 0\colon 0)$ is not allowed. But over $\mathbb C$, you can find the two intersection points: they are $(1\colon i\colon 0)$ and $(1\colon -i\colon 0)$.