Weil pairing on $E[p]$ is trivial

I'm currently working through Elliptic Curves, L. C. Washington. On page 147 he writes "The Weil pairing is not defined on $E[p]$ (or, if we defined it, it would be trivial since $E[p]$ is cyclic and also since there are no nontrivial $p$th roots of unity in characteristic $p$)."

Here $E$ is an elliptic curve and $E[p]$ the group of $p$-torsion points.

I don't see why this should be the case (assuming that $p$ is prime?). The definition of the Weil pairing doesn't seem to exclude primes at all.

The other interpretation I can think of this is that $p$ is the characteristic of the field that $E$ is over, but this doesn't make sense either since we have the definitions of ordinary and supersingular elliptic curves (when $E[p] \cong \mathbb{Z}_p$ and $E[p] \cong 0$, respectively).

• In those cases $E[p]$ is cyclic. Let $Q$ be a generator. So if $A,B$ are arbitrary elements of $E[p]$ we have $A=mQ$ and $B=nQ$ for some integers $m,n$. Then $$(A,B)=(mQ,nQ)=(Q,Q)^{mn}=1^{mn}=1.$$ – Jyrki Lahtonen Dec 30 '17 at 10:49
• I'm happy with $E[p]$ being cyclic in those cases, but I'm not really understanding why he says that the Weil pairing isn't defined on such an $E[p]$ – Irregular User Dec 30 '17 at 10:51
• And also: $pA=O$ so $$(A,B)^p=(pA,B)=(O,B)=1.$$ Because there are no non-trivial roots of unity of order $p$ we therefore must also have $(A,B)=1$. – Jyrki Lahtonen Dec 30 '17 at 10:51
• I guess the pairing can still be defined , but it is trivial, and therefore not useful for any purpose. I'm rusty though so I may be wrong. – Jyrki Lahtonen Dec 30 '17 at 10:53
• Looking more carefully, I suppose that the proposed trivial map will fail the Weil pairing definition at nondegeneracy (if $e_n$ is the map, then $e_n(S, T) = 1$ for all $T \in E[n]$ need not imply that $S = \infty$, and similarly for when the roles of $S$ and $T$ are swapped). Thanks for the realisation! Would you mind summarising these comments into an answer so that I can accept it? – Irregular User Dec 30 '17 at 10:59