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I am trying to understand the Frey-Rück attack and found different ways of a possible implementation. Since I am not yet very familiar with the Tate-Lichtenbaum pairing and the theory of divisors I wanted to ask which one of the different realizations of the FR reductions is the most efficient or ''prettiest'' in your opinion.

In the following let $E/\mathbb{F}_q$ be an elliptic curve and $P$ a point of prime order $l$. Let $Q=n P$ with $n\in\mathbb{Z}$. And $\tau(*,*)$ is the modified Tate-Lichtenbaum pairing.

The FR-Attack as presented in ''The Tate Pairing and the Discrete Logarithm Applied to Elliptic Curve Cryptosystems'':

Suppose that $l^k$ is the exact $l$-power dividing $\#E(\mathbb{F}_q)$ with $k>1$. (The case $k=1$ is ''easy'' since we have in this case that $\tau_l(P,P)$ is a primitive $l$th root of unity.)
Suppose $P'\in E(\mathbb{F}_q)$ is any point of order $l^k$. In this case $\tau_l(P,P')$ is a primitive $l$th root of unity. (*)
Now calculate $\tau_l(P,P')$ and $\tau_l(Q,P')$ with the divisors $D_P=(P)-(\infty)$ , $D_Q=(Q)-(\infty)$ and $D_{T'}=(2P')-(P')$. Here $2P'$ or $P'$ can never be $iP$ or $jQ$, so we can calculate the pairings without any problems and get as above the DLP $\tau_{l}(Q,P') = \tau_l(P,P')^n$ in $\mathbb{F}_q^*$.

My first question is why (*) holds. Why is $\tau_l(P,P')$ a primitive $l$th root of unity if $P'$ has order $l^k$?
And why can $2P'$ and $P'$ never be $iP$ or $jQ$? Because of their order?

And in Steven Galbraith ''Supersingular Curves in Cryptography'' he does not specify this proposition, but rather picks random points $P'\in E(\mathbb{F}_q)$ until $\tau_l(P,P')$ is a primitive $l$th root of unity. I'm asking myself which way is more efficient? Since in Frey's version it seems more costly to look for this specific point $P'$ of order $l^k$ rather than trying random $P'$.

In the paper of Ryuichi Harasawa et al. ''Compairing the MOV and FR Reductions in Elliptic Curve Cryptography'' they work with two random points to calculate the DLP:

  1. Determine the embedding degree $m$, i.e. the smallest integer such that $n|q^m-1$. Define $k:=\mathbb{F}_{q^m}$.

  2. Pick $S,T\in E(k)$ randomly (not equal to $P,Q$ or the point at infinity $\mathscr{O}$).

  3. Compute two rational functions $f_P$ and $f_Q$ such that $div(f_P) = l(P)-l(\infty)$ and $div(f_Q) = l(Q)-l(\infty)$.

  4. Compute $\alpha = \big(\frac{f_Q(S)}{f_Q(T)}\big)^{\frac{q^m-1}{n}}$ If $\alpha=1$ return to Step 2.

  5. Define $\beta := \big(\frac{f_P(S)}{f_P(T)}\big)^{\frac{q^m-1}{l}}$

  6. Solve the DLP $\beta = \alpha^n$ in $k^*$

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I can answer your very first question on why $\tau_\ell(P,P')$ is an $\ell$th root of unity. If I'm not mistaken, they are assuming in the paper that $E(\mathbb F_q)[\ell]$ (the $\ell$-torsion points in $E(\mathbb F_q)$) is a cyclic group.

(In general, if $\ell$ and $q$ are coprime, $E[\ell]\cong (\mathbb Z/\ell\mathbb Z)^2$ has order $\ell^2$, and all its proper subgroups have order $\ell$.)

Therefore if $P$ has order $\ell$ and $E(\mathbb F_q)[\ell]$ is cyclic, then $P$ is a generator of this group. This is important, because of the domain of the Tate-Lichtenbaum pairing: $$\tau_\ell\colon E(\mathbb F_q)[\ell] \times E(\mathbb F_q)/\ell E(\mathbb F_q)\to \mu_\ell.$$

(this is also why every output of $\tau_\ell$ is an $\ell$th root of unity. To get an $\ell^k$th root, we'd have to use $\tau_{\ell^k}$.)

So assume $\tau_\ell(P,P')=1$ (if not, it is automatically a primitive $\ell$th root of unity). Then for every $u=1,\dots,\ell$, $\tau_\ell(uP,P')=\tau_\ell(P,P')^u=1$. But the non-degeneracy of $\tau_\ell$ implies $P'\in\ell E(\mathbb F_q)$, say $P'=\ell P_1$, $P_1\in E(\mathbb F_q)$. It follows that $\ell^{k+1}P_1=\ell^k P'=\infty$. This is a contradiction, because $\ell^{k+1}$ cannot be the order of a point (because it does not divide $\#E(\mathbb F_q)$), and then $\ell^kP_1=\ell^{k-1}P'=\infty$! Therefore $\tau_\ell(P,P')\neq 1$.

Finally, $P'$ can never be $iP$ or $jQ$ (because the first has order $\ell^k$, and the others have orders dividing $\ell$). The paper specifies that $\ell\neq 2$, and then $2P'$ also has order $\ell^k$, and the same argument holds.

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