# Elliptic Curve in Weierstrass Form

My book says that an Elliptic Curve is a curve of the form

$$y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6\text{ over a field.}$$

My question is: Why there are some parts like $a_7xy^2$ missing?

• Not an answer, but if you look, each monomial has “weight” 6, where $y$ has weight 3 and $x$ has weight 2. So the index on the coefficient in a supposed monomial $xy^2$ would have to be $-2$. – Lubin Apr 15 '17 at 2:11

An elliptic curve over a field $F$, in fact, is a projective, smooth curve of genus $1$, with at least one point defined over $F$. It turns out (as Lord Shark discusses) there is a change of variables that brings any elliptic curve to a model of the form you write (which is called a Weierstrass equation). In fact, if the characteristic of $F$ is not $2$ or $3$, then you can bring it to a model of the form $y^2=x^3+Ax+B$, which is called a short Weierstrass form.
For instance, the curve $C$ over $\mathbb{Q}$ given by $x^3+y^3=1$ (the curve $X^3+Y^3=Z^3$ in projective space) is also an elliptic curve (there is at least one point, namely $(1,-1)$, the point $[1,-1,0]$ in projective coordinates), even though it is not given a priori by a Weierstrass form. A change of variables brings $C$ to the equation $$y^2 - 9y = x^3 - 27$$ in Weierstrass form.
The Weierstrass form follows from looking at a "base point" $O$ on the elliptic curve $E$ and considering rational functions on $E$ with no poles outside $O$. Apart from the constant functions all functions do have a pole at $O$. By Riemann-Roch considerations there are no such functions with simple poles, and essentially only one with a double and one with a triple pole. What I mean by this is that if $x$ has a double pole and $y$ a triple pole then the functions in the Riemann-Roch space $L(2O)$ are linear combinations of $1$ and $x$ and those in $L(3O)$ are linear combinations of $1$, $x$ and $y$. By replacing $x$ and $y$ by suitable scalar multiples we can cancel off the sextuple pole of $y^2-x^3$ in order to get it into the space $L(5O)$. This space has basis $1$, $x$, $y$, $x^2$ and $xy$, so $y^2-x^3=a_6+a_4x-a_3x+a_2x^2-a_1xy$ for some constants $a_i$. This is the Weierstrass equation.
Notation: I use $L(kO)$ for the rational functions with no poles outside $O$ and at most a $k$-fold pole at $O$. By Riemann-Roch, $L(kO)$ has dimension $k$ for $k\ge1$.