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Why is it so that $Y^2Z = X^3 + AXZ^2 + BZ^3$ is a non-singular elliptic curve if $4A^3 - 27B^2 \neq 0$? If we check the partial derivatives we get that

$\frac{\partial F}{\partial Z} = Y^2 - 2AXZ - 3BZ^2$

$\frac{\partial F}{\partial X} = -3X^2 - AZ^2$

$\frac{\partial F}{\partial Y} = 2YZ$

What restriction makes $(0,0,0)$ not a valid singular point?

2) I am mainly considering elliptic curves over $\mathbb{Q}$ and would like to end up with elliptic curves on the form $Y^2 = X^3 + AX + B$. I'm a little stuck on continuing my argumentation from the form $Y^2Z = X^3 + AXZ^2 + BZ^3$.

Thanks.

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    $\begingroup$ Notice your first equation is homogeneous in $X,Y,Z$. The equation defines a curve in the projective space (and not in affine 3-space, in which case you will have a surface and not a curve). Projective space has points $(a,b,c)$ with at least one of them non-zero, and only their ratios matter. $\endgroup$ – Mohan Jan 10 '16 at 14:35
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2) The passage you need is simply de-homogeneizing with respect to $Z$, which in algebraic terms mean to set $Z=1$. The geometric interpretation of this is choosing as the line at infinity the one given by $Z=0$, and intersecting your projective curve with the copy of the affine plane given by $\{(X:Y:Z)\in \mathbb P^2\colon Z\neq 0\}$. Notice that here the fact that you are working over $\mathbb Q$ does not play any role, the argument is the same for any field $K$.

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