Point at infinity for Jacobian coordinates

Question

Why is the point at infinity ∞ in affine space corresponding to (1 : 1 : 0) for Jacobian coordinates? How can this be shown? Is ist just to add the negative and see that the outcome is 1:1:0 or is there some "higher magic" to verify that a priori?

Answer 1

Jacobian coordinates are obtained by letting $x = X/Z^2$ and $y = Y/Z^3$. The usual Weierstrass equation $y^2 = x^3 + a x+ b$ then, by multiplying both sides by $Z^6$, becomes $$ Y^2 = X^3 + aXZ^4 + bZ^6 $$

An affine point $P = (x,y)$ corresponds to the Jacobian point $P = (t^2x:t^3y:t)$ for any $t \neq 0$. Observe that letting $P = (X:Y:Z)$, one gets $X/Z^2 = (t^2x)/t^2 = x$ and $Y/Z^3 = (t^3y)/t^3=y$; that is, the affine coordinates of $P$, as expected.

The point at infinity $\mathcal{O}$ corresponds to $Z = 0$. Plugging $Z=0$ in the above equation yields $Y^2 = X^3$. The point at infinity has therefore for representation $(t^2:t^3:0)$ for any $t \neq 0$ ---you can check that $(t^3)^2 = (t^2)^3$. For example, taking $t=1$ yields $\mathcal{O} = (1:1:0)$, which is the usual representation chosen to represent the point at infinity in Jacobian coordinates. But any $t \neq 0$ is valid; $(1:-1:0)$ is also a valid representation for the point at infinity in Jacobian coordinates (this corresponds to $t=-1$).

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Note. The case $t=0$ is not valid because $(0:0:0)$ is not. This supposes that the elliptic curve is defined over a field. Over a ring $R$, $t$ should be chosen in $R^*$. Remark that if $R$ is a field then $R^* = R \setminus \{0\}$.