2
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ 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 $-2$. $\endgroup$ – Lubin Apr 15 '17 at 2:11
5
$\begingroup$

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.

If you are interested in how one finds such changes of variables, this is briefly explained in Silverman and Tate's "Rational Points on Elliptic Curves".

$\endgroup$
6
$\begingroup$

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 $k\ge1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.