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2 +4ab+3b2 generate, create qr bidimensional barcode none for .net projects Visual Basic and Visual C# and hence .net framework QR-Code e(R, R) = 1 for all points R E[5] except R = O. An example of the opposite situation is given by E : y 2 = x3 + 4 over F997 (this example was provided by Kim Nguyen).

The points P = (0, 2) and Q = (747, 776) generate the 9 points of order 3 in E(F997 ). One can compute that e(P, P ) = 1, e(Q, Q) = 1 and e(P, Q) = e(Q, P ) 1 = 304. Hence we have e(R, R) = 1 for all R E[3].


Di QRCode for .NET stortion Maps. The typical case in cryptographic applications is P E(Fq ) and k > 1.

It is very useful to be able to consider self-pairings in this case. A valuable technique introduced by Verheul [335], which applies to both the Tate and Weil pairings, is to use a non-rational endomorphism. Lemma IX.

14. Let P E(Fq ) have prime order r and suppose k > 1. Suppose that E(Fqk ) has no points of order r2 .

Let be an endomorphism of E. If (P ) E(Fq ), then e(P, (P )) = 1. Proof.

Since is an endomorphism it follows that the order of (P ) is r or 1. The latter case does not occur since (P ) E(Fq ). Hence {P, (P )} is a basis for the r-torsion on E.

Since there is no r2 -torsion it follows that (P ) rE(Fqk ). Finally, since e(P, P ) = 1 it follows that e(P, (P )) = 1. A non-rational endomorphism which maps a point from E(Fq ) to E(Fqk ) as in the lemma is called a distortion map.

For a given supersingular curve (see De nition IX.18) there is at least one nice choice (indeed, Theorem 5 of [336] proves that distortion maps always exist). In Section IX.

13 we give examples of convenient distortion maps for supersingular curves. However, the following result from [335] implies that distortion maps do not exist for ordinary (i.e.

, non-supersingular; see Section IX.10) curves. Theorem IX.

15. Let E be an elliptic curve over Fq which has a distortion map. Then E is supersingular.

Proof. Suppose is an endomorphism which maps some point P E(Fq ) to a point (P ) E(Fq ). Let be the q-power Frobenius map, which is an endomorphism of E.

Then (P ) = P and ( (P )) = (P ) and so = in End(E). Hence, End(E) is non-commutative and E is supersingular according to De nition IX.18.


The Trace Map. An alternative to using a non-rational endomorphism (which works for all elliptic curves) is the following. Let P = (x, y) E(Fqk ) and de ne the trace map.

Tr(P ) =. Gal(Fqk /Fq ). (P ) =. (xq , y q ). where the sum is elliptic curve point addition. The trace map is a group homomorphism and Tr(P ) E(Fq ). Under conditions like those of Lemma IX.

14 it follows that if P E(Fqk ) has prime order r, P E(Fq ) and Tr(P ) = O, then e(Tr(P ), P ) = 1. We stress that the trace map transforms points which are de ned over a large eld into points de ned over a small eld, whereas distortion maps go the other way around. For further details see Boneh, Lynn and Shacham [42].

The trace map enables mapping into a speci c cyclic subgroup of E(Fqk ) of order r (called the trace zero subgroup T ) as follows. If P is a randomly. IX.7. NON-DEGENERACY, SELF-PAIRINGS AND DISTORTION MAPS chosen ele ment of E(Fqk ) of order r, then P = [k]P Tr(P ) is easily seen to satisfy Tr(P ) = O. Furthermore, if P E(Fq ) and if r is coprime to k, then P = O. This idea is used in [42].

A more e cient way to map into the trace zero subgroup using twists is given in [18]. The following degeneracy result was shown to me by Dan Boneh. Lemma IX.

16. Let E be an elliptic curve over Fq . Let r be a prime such that r divides #E(Fq ) and such that the subgroup of r elements has embedding degree k > 1.

Assume that r does not divide either k or (q 1). Let Tr be the trace map with respect to Fqk /Fq as above and the q-power Frobenius. Let T = {P E(Fqk )[r] : Tr(P ) = O}.

Then T = {P E(Fqk )[r] : (P ) = [q]P } and for all P, Q T we have e(P, Q) = 1. Proof. If P E(Fq )[r], then Tr(P ) = [k]P .

Hence T E(Fq )[r] = {O}. On the other hand, it is easy to see that T is a subgroup of E(Fqk )[r]. Hence #T = r and T is cyclic.

Let be the q-power Frobenius map, which acts on both Fqk and on E(Fqk ). It is known that has eigenvalues 1 and q on E(Fqk ). Let {P1 , P2 } be a basis for E(Fqk )[r] such that (P1 ) = P1 and (P2 ) = [q]P2 .

Then Tr(P2 ) = [1 + q + + q k 1 ]P2 = [(q k 1)/(q 1)]P2 = O and it follows that P2 is the generator of T . To complete the proof it su ces to prove that e(P2 , P2 ) = 1 and we do this by showing e(P2 , P2 ) Fq . Consider e(P2 , P2 )q = (e(P2 , P2 )) = e( (P2 ), (P2 )) = e([q]P2 , [q]P2 ) = e(P2 , P2 )q .

Acting by 1 gives e(P2 , P2 ) = e(P2 , P2 )q , which implies e(P2 , P2 ) Fq . IX.7.

5. Symmetry. Another issue for the Tate pairing is to consider the relationship between P, Q n and Q, P n , for points P, Q E(K)[n].

Up to nth powers we have P, Q. = Q, P n en (P, Q).. Since the Weil pairing is non-degenerate, we often have P, Q n = Q, P n up to nth powers. However, if distortion or trace maps are being used, then the pairing is usually restricted to a single cyclic subgroup. In this case Q = [m]P for some m and so e(Q, (P )) = e([m]P, (P )) = e(P, [m] (P )) = e(P, (Q)).

In other words, the pairing is symmetric when restricted to a cyclic subgroup..
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