1
$\begingroup$

Suppose $P\in E(\mathbb{F}_q)$ and $R=dP$.

In the MOV attack, we compute $\alpha=e(P,T)$ and $\beta=e(R,T)$ and try to solve the discrete logarithm problem for $\alpha$ and $\beta$ in the finite field $\mathbb{F}_{q^k}$ where $k$ is the embedding degree.

But if $E(\mathbb{F}_q)$ is cyclic (which it can be even in the supersingular case), $\alpha,\beta$ are doomed to be $1$.

So what am I missing here?

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ Why are they doomed to be 1? You would have $\alpha=\beta=1_{T}$ iff. $P=R=G_{1}$ and $T=G_2$, where $G_1$ and $G_2$ are the generators of the first group and second group respectively, which I suppose, is not the case. In the target group you have that $\beta=d\alpha$. Thus we have reduced the discrete logarithm problem on the group of points on an elliptic curve to the discrete logarithm on finite fields, where subexponential attacks are known. $\endgroup$
    – István András Seres
    Commented May 9, 2020 at 7:54

1 Answer 1

Reset to default
1
$\begingroup$

For the purposes of the MOV attack you choose $T$ to be a point in $E(\mathbb{F}_{q^k})$.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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