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So I'm trying to understand the group of points of $y^2=x^3+1$ over $\mathbb{F}_5$ and for some reason I seem to be getting nonsense answers and I'm not sure what I'm doing wrong.

So basically my formulas for this particular elliptic curve are:

$\lambda=\frac{y_1-y_2}{x_1-x_2}$, or $\frac{3x^2}{2y}$ if we're doubling the point.

$x_3=\lambda^2 -x_1-x_2$

$y_3=\lambda x_3+y_1-\lambda x_1$

My points are: $O,(0,1),(0,4),(4,0),(2,3),(2,2)$. Now maybe someone could just clue me in right away as to whether any of these points can be the identity element in disguise, because $(0,1)$ and $(0,4)$ definitely seem to be. But upon adding some of the other points to each other I'm getting that the other three points all have order two, so I was thinking the Klein-4 Group, but then for instance $(4,0)+(2,2)=(2,2)$ which means that $(4,0)$ must be the identity, and I'm just thoroughly confused.

It would seem to me that the only points which could be the identity would be the $(0,1)$ and $(0,4)$ since those have zero in the x-coordinate, but the only way all my computation makes sense is if they're all the identity. However I'm not sure how much of this reasoning carries over to finite fields.

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Have you double checked your calculations? Anyways, the identity of an elliptic curve (in Weierstrass form) is always the point at infinity; none of the points that can be expressed in $(x,y)$ coordinates can be the identity. – Hurkyl Feb 15 '13 at 22:33
Well I just checked again that $2(0,1)=(0,1)$. – Thoth Feb 15 '13 at 22:35
Yah Wikipedia's is a bit different. Mine's coming from Rational Points on Elliptic Curves by Silverman. But no matter what $(4,0)+(2,2)$ equals, the fact that $(0,1)+(0,1)=(0,1)$ is already nonsense. – Thoth Feb 15 '13 at 22:45
up vote 1 down vote accepted

The point $(x_3, y_3)$ is the third points of intersection of your curve with the line through $(x_1, y_1)$ and $(x_2, y_2)$; it is not their sum, but rather minus their sum (three co-linear points $P,Q,R$ satisfy $P+Q+R=O$). With the formulae you have written down, the sum of $(x_1, y_1)$ and $(x_2, y_2)$ is in fact $(x_3, -y_3)$.

As Hurkyl remarks, the identity is the point at infinity on your curve. None of your points could be "the identity in disguise".

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oh CHrist! I forgot about that, thanks. – Thoth Feb 15 '13 at 23:29
@cat No worries, it happens. All the best! – Bruno Joyal Feb 15 '13 at 23:37

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