If $\mathbb{F}_q$ is a finite field and the elliptic curve E is defined over finite field $\mathbb{F}_p$ such that ECDLP is hard in E($\mathbb{F}_p$), where $q, p$ are prime and $q \ll p$. Let T $\in E(\mathbb{F}_p)$ with $mT= \mathcal{O}$. My question is in finite field $\mathbb{F}_q, 1=q+1$. It follows that $T=(q+1)T$. Does it imply $q$ have to be divided by $m$? Is it possible that $q$ isn't divided by $m$ with other condition while $1=q+1, T=(q+1)T$? Thank you~
8
-
1$\begingroup$ the error is that $E(\mathbb{F}_p)$ is a $\mathbb{Z}$-module not a $\mathbb{F}_p$-vector space. Curves for which $p$ divides the order of $E(\mathbb{F}_p)$ are sometimes called anomalous. Pick your any elliptic curves and do some computations and you will see your error. (This should be closed.) $\endgroup$– Chris WuthrichCommented Dec 30, 2010 at 10:29
-
$\begingroup$ $E(\mathbb{F}_p)$ consists of all the rational points on E over $\mathbb{F}_p$. Excuse me, but could you tell me why $E(\mathbb{F}_p)$ is a $\mathbb{Z}$ module or isn't a $\mathbb{F}_p$ vector space? $\endgroup$– athenaCommented Dec 30, 2010 at 12:39
-
$\begingroup$ for any $n\in\mathbb{Z}$, we define $n\cdot P=P+P+\cdots +P$ using the addition law $+$ on the curve. There is absolutely no reason that $p\cdot P = O$, just because the coordinates of the points are in $\FF_p$. As I said, take $y^2 + y = x^3 + 1$ over $\FF_2$ it has $3$ rational points, so $2\cdot P = -P$. You should read the basics of elliptic curve in a book like Silverman;s "The arithmetic of elliptic curves". $\endgroup$– Chris WuthrichCommented Dec 30, 2010 at 13:34
-
$\begingroup$ The role of $q$ is unclear. What does it have to do with $E,p,m$? $\endgroup$– Felipe VolochCommented Dec 30, 2010 at 13:57
-
$\begingroup$ You are right, maybe I misread the question and $q$ just stands for the order of the group, in which case $T = (q+1)T$ is obvious. But then $m=q$. $\endgroup$– Chris WuthrichCommented Dec 30, 2010 at 14:25
|
Show 3 more comments