In this YouTube video, Dan Boneh mentions that if both points are defined on the base field then the pairing is degenerate.

Why is that? And specifically is this true if I use the Weil Pairing?


1 Answer 1


(I write a more extensive answer with the background of all this; the result you are looking for is Theorem 3. A proof would be too long for this site, you can consult standard textbooks.)

Let $E$ be an elliptic curve over $\mathbf{F}_p$ and $P \in E(\mathbf{F}_p)$ be a point of prime order $r \ne p$. So $r$ divides the order of $E(\mathbf{F}_p)$, and suppose further that $r^2$ does not divide the order of $E(\mathbf{F}_p)$. For any extension field $K$ of $\mathbf{F}_p$, we let $E(K)[r]$ be the group of all points $Q \in E(K)$ such that $rQ = \infty$. Note that $P \in E(K)[r]$, so $\langle P\rangle$ is a subgroup of $E(K)[r]$.

Theorem 1. The group $E(\overline{\mathbf{F}_p})[r]$ is isomorphic to $\mathbf{Z}_r \oplus \mathbf{Z}_r$.

Corollary. There is an integer $k > 1$ such that $E(\mathbf{F}_{p^k})[r]$ is isomorphic to $\mathbf{Z}_r \oplus \mathbf{Z}_r$.

In the following, we let $k$ be the smallest such integer (called the embedding degree), $K = \mathbf{F}_{p^k}$, and $E[r] = E(K)[r]$. Thus all the points of $E[r]$ have coordinates in $K$.

Then the mathematician's Weil pairing $e_r$ is a map from $E[r] \oplus E[r]$ to $\mathbf{F}_{p^k}^*$ with the following properties (among others), which makes it different from the cryptographer's pairing.

Bilinearity. For all points $P,Q,R \in E[r]$ we have $$ \begin{align*} e_r(P+Q,R) &= e_r(P,R)\times e_r(Q,R), \text{and} \\ e_r(P,Q+R) &= e_r(P,Q)\times e_r(P,R). \end{align*} $$

Alternacy. For all points $P \in E[r]$, we have $e_r(P,P) = 1$.

Non-degeneracy. For all points $P \in E[r]$, if $e_r(P,Q) = 1$ for all points $Q \in E[r]$, then $P = \infty$.

The property of alternacy of the Weil pairing makes it unsuitable for cryptography, since in cryptography we require $e(P,P) \ne 1$. So the Weil pairing needs to be modified.

Theorem 3. Let $P,Q \in E[r]$. Then $e_r(P,Q) = 1$ if and only if $Q$ is a multiple of $P$.

Corollary 1. A point $Q \in E[r]$ has coordinates in $\mathbf{F}_p$ if and only if it is a multiple of $P$.

Corollary 2. The group of $r$th roots of unity in $K$ is cyclic of order $r$.

In the following, let $Q$ be a point of $E[r]$ which is not a multiple of $P$. This implies that $Q$ (which has coordinates in $\mathbf{F}_{p^k}$) does not have coordinates in $\mathbf{F}_p$. Let $G = \langle P\rangle$, and let $G_T$ be the group of $r$th roots of unity in $K$ (both groups are cyclic of order $r$).

We define the modified Weil pairing $\hat{e}_r$ as a map from $G \times G$ to $G_T$ such that $\hat{e}_r(P,P) = e_r(P,Q)$. Then $\hat{e}_r$ is a suitable cryptographic pairing: its bilinearity follows from the bilinearity of $e_r$, and its non-degeneracy (in the cryptographic sense) follows from Theorem 3 since $Q$ is not a multiple of $P$.

There are many ways to do this in practice, one is to let $p \equiv 2 \pmod 3$ and such that $(p+1)/6$ is prime. In this case the curve $Y^2 = X^3 + 1$ has $p+1$ points (it is supersingular) and we can choose a point of order $r = (p+1)/6$. It can also be shown that in this case, we have $k = 2$. Since $p^2-1$ is a multiple of $3$, there is a primitive $3$rd root of unity in $K = \mathbf{F}_{p^2}$, call it $\zeta$. Then for a point $P = (x,y)$, the point $Q$ is $(\zeta x, y)$.

  • $\begingroup$ Thank you very much!Is that possible to tell me about the proof of Corollary 1? $\endgroup$
    – les2004
    Mar 9, 2016 at 20:43

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