In projective coordinates point-at-infinity can be identified with z=0. How to identify the point-at-infinity in affine coordinate.
Whether x=0 and y=0 can be considered as point-at-infinity in affine coordinate?
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2$\begingroup$ Does crypto.stackexchange.com/questions/6156/… help, by any chance? $\endgroup$– ThomasFeb 25, 2013 at 5:58
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$\begingroup$ @Thomas: I went through the link you posted. My question is not related to encoding of the point-at-infinity. In our implementation after EC operation such as EC point addition we return the point "x=0, y=0" when we encounter point-at-infinity. So I wanted to check is it safe in doing this? Is there a remote possibility of obtaining a valid point with x=0,y=0 when working with standard EC curves? $\endgroup$– AndyFeb 26, 2013 at 4:15
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$\begingroup$ Well, clearly (0,0) is a valid point for the curve y^2=x^3+x which is in the standard reduced Weierstrass form. If you want a simple encoding that can be unambiguous, why not use (p,0) where p is the cardinality of the field of definition ? $\endgroup$– minarJul 3, 2013 at 22:27
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From a mathematical point of view this is not possible. I would say almost by definition because the point at infinity does not lie in the affine part. To be more precise:
Let $\mathbb P^2$ is the two dimensional projective space with coordinates $(x:y:z)$ and $\mathbb A^2 \subset \mathbb P^2$ the affine part where $z \neq 0$ then one can see $X = x/z$ and $Y = y/z$ as coordinates on $\mathbb A^2$. Then $\mathbb P^2 \setminus \mathbb A^2$ is exactly the part where $z=0$ by definition.
Now if $E$ is the curve given by $y^2z = x^3 + axz^2 +bz^3$, then there is exactly one point of $E$ in $\mathbb P^2 \setminus \mathbb A^2$. Indeed by setting $z=0$ one obtains $0 = x^3$ hence $x=0$ showing that (x:y:z) = (0:1:0) is the only point of $E$ not in $\mathbb A^2$. This point (0:1:0) is the point at infinity and does not lie in $\mathbb A^2$ hence you cannot represent (0:1:0) in affine coordinates.
