In light of the holiday, I would like to air a grievance.

I have no good way to recoordinatize a morphism of varieties as I move between coordinate neighborhoods.

Let me explain what I mean with an example. Consider the affine variety $E_Y$ cut out by the equation

$$Z + Z^2 - (X^3 + ZX^2) = 0 \mbox{ }\mbox{ }\mbox{ }\mbox{ } (1)$$

The projective closure of $E_Y$, which we'll denote by $E,$ is an elliptic curve with identity element $O:=(0,0)$ in $X,Z$-coordinates.

So there is addition morphism $\mu:E\times E \rightarrow E$ and an inverse morphism $-:E \rightarrow E.$ A calulation shows, $-\mu$ is given by

$$X(-\mu) = \frac{(z_1 -z_0)^2 - (x_1 -x_0)(x_1z_0 - x_0z_1)}{(x_1 -x_0)^2 + (x_1 - x_0)(z_1 - z_0)} - x_0 - x_1$$

$$Z(-\mu) = \frac{z_1 - z_0}{x_1 -x_0}\left(\frac{(z_1 -z_0)^2 - (x_1 -x_0)(x_1z_0 - x_0z_1)}{(x_1 -x_0)^2 + (x_1 - x_0)(z_1 - z_0)} - x_0 - x_1\right) + \frac{x_1z_0 - x_0z_1}{x_1 -x_0} $$

on $E_Y \times E_Y$ outside of the locus $(x_1 -x_0)^2 + (x_1 - x_0)(z_1 - z_0) = 0.$

I would like to express the morphism $-\mu$ in terms of regular functions on some open neighborhood of $O \times O,$ but my current method of obtaining such an expression is "to move symbols around" in my expression for $-\mu$ using the relation of the curve, $(1),$ until I obtain a regular expression at $(0,0) \times (0,0).$ This is often a huge waste of time and becomes nearly impossible as the equations defining the variety become more complicated.

So I'm wondering if there is a more methodical way to approach this problem? How does one do this in practice?

  • 1
    $\begingroup$ @jspecter: By ‘in practice’, do you allow computer algebra? $\endgroup$ – Zhen Lin Dec 30 '11 at 1:10
  • $\begingroup$ @Zhen Lin. Why yes. But of course it would be nice to have a technique that's applicable by hand. $\endgroup$ – jspecter Dec 30 '11 at 5:20

I don't know how much this will help you in general but here is how I would deal with your particular example :

Put $dx = x_1 - x_0$ and $dz = z_1 - z_0$, so you have a problem at $dx(dx+dz)=0$. Then write $$X(- \mu) = \frac{dz^2-dx(z_0dx-x_0dz)}{dx(dx+dz)}- x_0-x_1 $$ You can see that $X(- \mu)$ (as well as $Z(- \mu)$) is homogeneous of degree $0$ in $dx,dz$

From your equation $z+z²=x^3+zx²$, with a process analogous to differentiation, you get $dx(x_1^2+x_0x_1+x_0^2+z_0x_0+z_0x_1) = dz(1+z_1+z_0-x_1^2)$

This allows you to replace $dx$ with $1+z_1+z_0-x_1^2$ and $dz$ with $x_1^2+x_0x_1+x_0^2+z_0x_0+z_0x_1$ in your expressions, and you obtain a rational function with denominator $dx(dx+dy) = 1 + \ldots$ for $X$ and $dx^2(dx+dy) = 1 + \ldots$ for $Z$, so they are valid on a neighboorhoud of $(0,0) \times (0,0)$.


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.