Projective coordinates for point at infinity on elliptic curve What is the unique characteristic of the projective coordinates of a point at infinity? I am specifically looking for a characteristic on (short) weierstrass curves. I know that the point at infinity is usually denoted with (0, 1, 0), however, (0, 2, 0) would be the exact same point. And what about the points (x, y, 0) where both x and y are non-zero integers? Such as (1, 1, 0) and (2, 3, 0)? What about x being non-zero and y being zero, like (1, 0, 0)?
What about (0, 0, 0)?
Would checking the z-coordinate be sufficient? That is, is "A point $P$ is the point at infinity if and only if $z_p=0$" true?
 A: In projective coordinates, $[x:y:z]$ denotes the same point as $[tx:ty:tz]$ (if $t\ne 0$), hence indeed $[0:1:0]$ and $[0:2:0]$ denote the same point. It is convenient but not mandatry to take the "simplest" representative, which would be $[0:1:0]$. While $[1:1:0]$, $[2:3:0]$, $[1:0:0]$ etc. are also infinte points they are different points; they all lie on the line at infinity (i.e. $z=0$) of the projective plane (so the answer to your final question is "yes").
To check which of these points at infinity is on your curve, just plug in the coordinates $x,y,0$ into the (homogenuos!) equation of your curve. Since anything containing $z$ disappears, you get a very simple equation.
For example, the affine equation $$y^2=x^3-x$$
becomes
$$ y^2z=x^3-xz^2$$
in homogenuous coordinates. Letting $z=0$, this becomes
$$0=x^3, $$
i.e., $x=0$, so that $[0:1:0]$ is the one and only point at infinitiy on this curve.
A: Just for completeness:
$\mathbb{P}^2$, which is the set of all homogeneous points, is defined to be the set $\{x:y:z\} $ where x,y,z are not all zero and {x:y:z} = {tx:y:tz} for $t \neq 0$.
That is, $[0:0:0]$ is excluded by definition.
There is a reason for this. Supposed  $[0:0:0] \in \mathbb{P}^2 $, and $t$ can be $0$. Then all points $\{x:y:z\}  = t [0:0:0]$. In other words, there is only one point in $\mathbb{P}^2 $, since all points are in fact multiples of $[0:0:0]$. This means $\mathbb{P}^2 $ is actually a single point, which makes it a rather boring algebraic object.
