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ON REMARKS OF LIFTING PROBLEMS FOR ELLIPTIC CURVES 1 ...

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<strong>ON</strong> <strong>REMARKS</strong> <strong>OF</strong> <strong>LIFTING</strong> <strong>PROBLEMS</strong> <strong>FOR</strong> <strong>ELLIPTIC</strong> <strong>CURVES</strong> 13<br />

Since ˜P , ˜Q satisfy the ɛ-difficult ECDLP, we get<br />

ɛ √ ∣<br />

∑ ∣ ∣ ∣∣ ∣∣ ∑ ∣ ∣∣)<br />

q < max(<br />

xi α i , yi α i<br />

(7)<br />

< r max<br />

i<br />

The following lemma proves the theorem.<br />

(|x i |, |y i |) max(|α i |).<br />

Lemma 4.5. Let E be an elliptic curve defined over K and the points P 1 , · · · , P r<br />

of E(K) are linearly independent. Define the matrix<br />

where<br />

A = (a ij ) 1≤i,j≤r<br />

a ij = ĥ(P i + P j ) − ĥ(P i) − ĥ(P j),<br />

i<br />

(i, j = 1, · · · , r).<br />

If we define<br />

c k = a kk − a k A −1<br />

k aT k (k = 1, · · · , r)<br />

where a k = (a k1 , · · · , a kk−1 , a kk+1 , · · · , a kr ) and A k is the matrix obtained by removing<br />

kth-row and kth-column in A, then, for any integers n 1 , · · · , n r ,<br />

ĥ(n 1 P 1 + · · · + n r P r ) ≥ c max(n1, 2 · · · , nr)<br />

2<br />

where<br />

Especially, in the case of r = 2,<br />

Proof. Note that<br />

c = c(E, P 1 , · · · , P r ) = 1 2 min(c 1, · · · , c r ) > 0.<br />

c = min(ĥ(P 1), ĥ(P 2)) − (ĥ(P 1 + P 2 ) − ĥ(P 1) − ĥ(P 2)) 2<br />

4 max(ĥ(P 1), ĥ(P .<br />

2))<br />

〈P, Q〉 = ĥ(P + Q) − ĥ(P ) − ĥ(Q)<br />

is the positive-definite symmetric bilinear form on E(K)/E(K) tor . Moreover, it<br />

can be extended to<br />

〈 , 〉 : (E(K)/E(K) tor ⊗ R) 2 → R<br />

with taking tensor product on (E(K)/E(K) tor by R [10].<br />

For simplicity, we may assume that |n 1 | = max i (|n i |). Then if we define a ij =<br />

〈P i , P j 〉 and x i = n i /n 1 (i = 1, · · · , r), then<br />

〈n 1 P 1 + · · · + n r P r , n 1 P 1 + · · · + n r P r 〉 = n 2 1(a 11 + 2 ∑ a 1i x i + ∑<br />

a ij x i x j ).<br />

i≥2<br />

i,j≥2<br />

Since 〈, 〉 is the positive-definite symmetric bilinear form, it is easy to show that<br />

f(x 2 , · · · , x r ) = a 11 + 2 ∑ a 1i x i + ∑<br />

a ij x i x j<br />

i≥2<br />

i,j≥2<br />

has the minimum<br />

when<br />

That is,<br />

a 11 − (a 12 , · · · , a 1r )A −1<br />

1 (a 12, · · · , a 1r ) T<br />

(x 2 , · · · , x r ) T = −A −1<br />

1 (a 12, · · · , a 1r ) T .<br />

2ĥ(n 1P 1 + · · · + n r P r ) = 〈n 1 P 1 + · · · + n r P r , n 1 P 1 + · · · + n r P r 〉 ≥ c 1 n 2 1.

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