Cubic Equations

August 26, 2008 by ellipticcurves

Now assume that f(x,y) is cubic. The previous method no longer works. Indeed, if  we have a point P(x_0, y_0) \in C(K) and draw a line of slope \alpha \in K, then equation

f(x, \alpha (x - x_0) + y_0)=0,

which we need to solve to find the points of intersection of the line with the curve C, will be cubic:

ax^3+bx^2+cx + d=0.

After factoring out x-x_0, we are left with a quadratic equation, whose roots need not lie in K.

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Quadratic Equations

August 26, 2008 by ellipticcurves

Consider a quadratic polynomial f(x,y). Assume, as usual, that it is irreducible, i.e., that the corresponding curve C:f(x,y)=0 is not a union of two lines (or a double line). Then C is a conic section: a circle, ellipsis, parabola, or a hyperbola.

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Linear Equations

August 26, 2008 by ellipticcurves

Let us begin with a curve

C: ax+by = c,

with a,b,c\in K. Without loss of generality, we may assume that b\neq 0. Then the map C\to {\mathbb A}^1 given by (x,y)\mapsto x is a morphism defined over K with an inverse morphism {\mathbb A}^1\to C given by x \mapsto (x, (c-ax)/b). Therefore, every line C in {\mathbb A}^2 that is defined over K is isomorphic to {\mathbb A}^1. In particular, one can establish a bijection between C(L) and {\mathbb A}^1(L) = L for all L/K.

Basic Algebraic Geometry

August 26, 2008 by ellipticcurves

The subject of Algebraic Geometry starts out as a study of solution of systems of polynomial equations in several variables. To make our life easier, we will restrict ourselves to solving just one equation in only two variables:

f(x, y) = 0.

For now, let us denote this equation by X_f. When the coefficients of f belong to a field K, we say that X_f is defined over K.

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Welcome

August 1, 2008 by ellipticcurves

This Fall, I will be teaching a course in Elliptic Curves with an emphasis on computations. I am thinking that posting some notes online would not hurt, hence this blog. Feel free to say hi.

Have a nice semester, everyone.