Is there an economic way to check if an eliptic curve point in jacobian coordiate is the same as a compressed point On an elliptic curve such as $y^2 = x^3 + b$, we define $a$ compressed point $P = (x, y)$ by it's coordinate $x$ and the parity of the $y$ coordinate. $y$ can be computed using $y = \pm\sqrt{x^3 + b}$ and choosing the right value of $y$ using the parity information.

We also define a point in Jacobian coordinate such as $P = (a, b, z)$ with $x = a / z^2$ and $y = b / z^3$ .

I would like to know if there is a known way to check if 2 points in these two set of coordinates are the same point that is cheap. Computing if the $x$ coordinates are the same is easy by checking if $x * z^2 = a$ .

To check if the $y$ coordinate are the same, the best I can come up with is to compute $y = a * z^{-3}$ and check if it has the same parity as the compressed point. However, it requires an expensive computation of the multiplicative inverse of $z$ to get $z^{-1}$. This seems rather overkill as it compute way more information than required. Only one bit is needed.

I was wondering if there is a known way to check if thee 2 points are the same that do not involve expensive computations such as the finding multiplicative inverse or a quadratic residue to decompress the point.
Computing the $z^{-1}$ is very cheap. We can do this by Euclidean algorithm, and in logarithmic time. But you can set $z=1$ in projective coordination. This make computations easy($O(1)$ since $z^{-1}=1$), and with this representation you can see the equality of points evidently.


  Note that, instead of computing $y=b*z^{-3}$ you can check  $y*z^3=b$. In this state, you don't need to finding multiplicative inverse.


Also in powerful programs such as MAGMA, $z$ is equal to $1$, by default.