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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?

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@jspecter: By ‘in practice’, do you allow computer algebra? – Zhen Lin Dec 30 '11 at 1:10
@Zhen Lin. Why yes. But of course it would be nice to have a technique that's applicable by hand. – 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 $(0,0) \times (0,0)$.

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