Order of a point on an Elliptic Curve

Question

I am currently struggling with the determination the order of a point on an elliptic curve.
We had to do the following exercise:

$C = V(y^2+x^3-1)$ and $P = (0,1)$. Now Wikipedia told me that I can calculate the sum of two point with the following formulas:

Let $P=(x_P,y_P), Q=(x_Q,y_Q)$. Then $S=P+Q=(X_S,y_S)$
with $x_S=m^2-x_P-x_Q$ and $y_S=-y_P+s(x_P-x_S)$, where $m = \frac{y_P-y_Q}{x_P-x_Q}$.
So then I get $P+P = (0,-1)= -P$ so that $P+P+P=\infty$.

Now my question is, how this formula works. Because it does not depend anyhow on the formula of the elliptic curve right? So for any elliptic curve my point $P=(0,1)$ has the order $3$? This seems very weird to me.

(I also don't see where the formula for the addition of two points comes from.)

I would be very happy if someone could explain me how this works, and if you know good literature about elliptic curves.

Best, Luca

Answer 1

Instead of following blindly Wikipedia's formulas, it is best to understand how to calculate $P+Q$, or $P+P$, given an elliptic curve. Let us assume for simplicity that the curve is given by $E:y^2=x^3+Ax+B$, and $P,Q\in E$.

• In order to find $P+Q$, first find the equation of the line $L$ through $P$ and $Q$, find the third point $R$ of intersection of $L$ and $E$. Then $P+Q+R=\mathcal{O}$, so that $R=-(P+Q)$. In this case $-(x_0,y_0)=(x_0,-y_0)$, so we can find $P+Q$ as $-R$.

• In order to find $P+P$, first find the tangent line $L$ to $E$ at $P$, and find the third point $R$ of intersection of $L$ and $E$. Then, $2P+R=\mathcal{O}$, and so $2P=-R$.

In your case, $E: y^2=x^3+1$ and $P=(0,1)$. The tangent line is $y=1$, and it turns out that this line has a triple point of intersection with $E$, thus $R=P$. In particular, $2P=-R=-P$, and so $3P=\mathcal{O}$.

Another example with the same curve $E$: let $P=(2,3)$ and $Q=(0,1)$. The line through $P$ and $Q$ is $y=x+1$, and the third point of intersection is $R=(-1,0)$. Since $R=-R$, we obtain $P+Q=(-1,0)$.

Finally, here is a picture of $P=(2,3)$, $2P$, $3P$, $4P$, and $5P$, which shows that $6P=\mathcal{O}$.