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?

  • $\begingroup$ Actually, there is no point at infinity in affine coordinates. Elliptic curves are often defined as the points of the affine curve plus an additional symbol , which is then called the "point at infinity". The mapping of that point to (1:1:0) is then by definition only. As you noticed, the point at infinity has the property that it results in adding P and -P. In affine coordinates, this calculation is not possible as the denominator in the addition formula gets zero. $\endgroup$
    – user27950
    Nov 14, 2016 at 8:51

1 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 $t=-1$).

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 $R^* = R \setminus \{0\}$.

  • $\begingroup$ you are not answering why the point at infinity corresponds to (1:1:0) $\endgroup$
    – user27950
    Nov 15, 2016 at 22:00
  • $\begingroup$ @Cryptostasis ? $\endgroup$
    – user94293
    Nov 16, 2016 at 4:29
  • $\begingroup$ You clearly describe, how the point at infinity is mapped to the corresponding Jacobian coordinates. This is all correct and fine. The mapping of the point at infinity to the jacobian coordinate is a definition . In my opinion, the question asks, why this definition is natural. $\endgroup$
    – user27950
    Nov 16, 2016 at 4:49

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