0
$\begingroup$

If I have $E: y^2 = 2x^3 + 3$, and I want to compute all the points over $\mathbb{F}_5$

Do I simply just plug in 0-4 for y, and 0-4 for $2x^3 +3$ and then all the ones that are equal are the only points? So in this example

$$(1,0), (2,2), (2,3), (4,1), (4,4)$$

How do I find the points at infinity?

thanks

$\endgroup$
  • $\begingroup$ It's one way to do it. An elliptic curve in Weierstrass form has one point at infinity. $\endgroup$ – Lord Shark the Unknown Mar 8 '18 at 20:06
1
$\begingroup$

By definition, the group $E(\mathbb{F}_5)$ is given by $$ E(\mathbb{F}_5)=\{(x,y)\mid y^2=2x^3+3\}\cup \{\mathcal{O}\}, $$ where $\mathcal{O}$ is the point "at infinity". This extra point satisfies $P+\mathcal{O}=\mathcal{O}+P=P$ for the point addition on $E$. One can check the size of such groups by the Hasse-inequality $$ |p+1-\# E(\mathbb{F}_p)|\le 2\sqrt{p}. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.