Group law and the Security of elliptic curves on
Volume 2, Issue 5, Page No 104-108, 2017
Author’s Name: Abdelalim Seddika), Chaichaa Abdelhak, Souhail Mohamed
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Laboratory of Topology, Algebra, Geometry and Discrete Mathematics, Department of Mathematical and computer sciences, Faculty of sciences Ain Chock Hassan II University of Casablanca.
a)Author to whom correspondence should be addressed. E-mail: seddikabs@hotmail.com
Adv. Sci. Technol. Eng. Syst. J. 2(5), 104-108 (2017); DOI: 10.25046/aj020517
Keywords: The Discrete Logarithm Problem, Group Low, The Localization of the Ring, The maximal ideal Complexity
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In this paper, we study the elliptic curve , with the localization of the ring where and if , in the maximal ideal . Finally we show that and the execution time to solve the problem of discrete logarithm in is , such that the execution time to solve the problem of discrete logarithm in is . The motivation for this work came from search for new groups with intractable (DLP) discrete logarithm problem is there great importance.
Received: 12 April 2017, Accepted: 04 May 2017, Published Online: 28 December 2017
Introduction
The elliptic curves are a very fashionable subject in mathematics. They are the basis of the demonstration of Fermat’s great theorem by Andrew Wiles, it was proposed for cryptographic use independently by Neal Koblitz[1] and Victor Miller in 1985, claim that elliptic curve cryptography requires much smaller keys than those used in conventional public key cryptosystems, while maintaining an equal level of security. In 2008, Virat introduced the elliptic curves over local ring Fp[] = Fp[X]/(X2)[2], and a proposed a new public key cryptosystem which is a variant of the ElGamal cryptosystem on an elliptic curve, in 2013 Chillali generalized the Virat result for the ring Fp
Chillali and Abdelalim constructed a ring Fp[e1,e2,e3], defined an elliptic curve over Ea,b(Fp[e1,e2,e3]) and they showed that Card(Ea,b(Fp[e1,e2,e3])) ≥ (Card(Ea,b(Fp))− 3)n +Card(Ea,b(Fp)) [4].
In this work we will generalize the construction of Fp[e1,e2,e3] to Fp[e1,..,en], but not a local ring to define a group law in Fp[e1,..,en],we localized the ring Fp[e1,..,en] in a maximal ideal, and we give it a group law and show that Card(Ea,b(Fp[e1,..,en])) ≥ (Card(Ea,b(Fp)) − 3)n + Card(Ea,b(Fp)), then shows the discrete logarithmic complexity is Ω(N) Such that N = Ea,b(Fp) by using the attacks baby step/giant step and ρ−Pollard.
Let p be an odd prime number and n be an integer such that n ≥ 1. we consider the ring An = Fp[e1,…,en] = {a0+a1e1+…+anen/a0,a1,…,an ∈Fp, eiei = ei and eiej = 0 if i , j}. Fp[e1,…,en,] is vector space over Fp with basis (1,e1,…,en). We have
X +Y = (x0 +y0)+(x1 +x1)e1 +…+(xn +yn)en and X.Y = t0 +t1e1 +…+tnen with:
( t0 = x0y0 if i = 0
ti = xiy0 +x0yi +xiyi if i , 0
Proposition 1. Let X = x0 +x1e1 +…+xnen ∈Fp[e1,…,en] then X is invertible if and only if x0 , 0 and xi , −x0 for all i ∈{1,…,n}.
Proof. Let X = x0+x1e1+…+xnen ∈Fp[e1,…,en] a invertible element, there Y = y0 +y1e1 +…+ynen ∈Fp[e1,…,en] such that X.Y = 1, this implies that
( x0y0 = 1 if i = 0
xiy0 +x0yi +xiyi = 0 if i , 0
therefore, x0 , 0 and xi ,−x0 for all i ∈{1,…,n}. In this case:
( y0 = x0−1 if i = 0
yi = −(x0 +xi)−1xix0−1 if i , 0
The other since is evident.
Proposition 2. The ideal P = (e1,…,en) is a maximal
(prime) of a ring Fp[e1,…,en].
Proof. We have Fp[e1,…,en]/P ‘ Fp is a field, there P is maximal.
Proposition 3. Let S = Fp[e1,…,en]− P = {s0 + s1e1 + … + snen/s0 ∈ Fp,s0 , 0}. Then the localized of Fp[e1,…,en] in P is:
AP = {xs00++xs11ee11++……++sxnneenn /x0,x1,…,xn,s0,s1,…,sn ∈Fp,s0 , 0}
= {x0 +x1e1 +…+xnen /x0,x1,…,xn,s1,…,sn ∈Fp}
1+s1e1 +…+snen
AP is a local ring its maximal ideal is
M = {1+x1se11e+1+……++xnsnenen /x1,…,xn,s1,…,sn ∈Fp} and the residual field is K ‘Fp.
Proposition 4. The homomorphism :
π : AP −→ Fp
x1+0+sx11ee11++……++sxnneenn −→ x0
is a surjective homomorphism of rings.
2 The Elliptic Curve over the ring AP
Let n ∈ N∗, A = Fp[e1,…,en], p a prime number p ≥ 5, P = (e1,…,en) and AP the localized of A in P .
Definition 1. An elliptic curve over ring AP is curve that is given by such Weierstrass equation:
Y2Z = X3 +aXZ2 +bZ3
with a,b ∈ AP and 4a3 + 27b2 is invertible on AP , and the reduction over Fp is
Y2Z = X3 +π(a)XZ2 +π(b)Z3
Ea,b(Fp) = {[X : Y : Z] ∈P2(Fp)/Y2Z = X3 +aXZ2 +bZ3}
Ea,b(A) = {[X : Y : Z] ∈P2(A)/Y2Z = X3 +aXZ2 +bZ3} Ea,b(AP ) = {[X : Y : Z] ∈P2(AP )/Y2Z = X3 +aXZ2 +bZ3}
Theorem 1. The mapping | ||
φ : Ea,b(A) [x : y : z] | −→
−→ |
Ea,b(AP ) [1x : y1 : 1z ] |
is injective.
Proof. Let [x : y : z],[x0 : y0 : z0] ∈ Ea,b(A) Suppose that [x : y : z] = [x0 : y0 : z0]
⇒ ∃λ ∈ A such that (x0,y0,z0) = λ(x,y,z)
⇒ (x0,y0,z0) = (λx,λy,λz)
⇒ x0 = λx, y λy and z0 = λz
⇒ and
Then [
we deduce that φ([x : y : z]) = φ([x0 : y0 : z0])
Then φ is well defined.
f : A −→ AP is injective. Note that mapping x −→ 1x
We deduce φ is injective.
Theorem 2. Let a,b ∈Fp, the mapping
ϕ : Ea,b(AP ) −→ Ea,b(A)
[sx1 : sy2 : sz3] −→ [xs2s3 : ys1s3 : zs1s2] is surjective.
Proof. Let [sx1 : y z 0 0 0 (A )
Suppose that [s
Then ∃λ ∈ A∗P such that (sx1 s2 s3 (s1, s2, sz3)
⇒ sx00 = λsx1; sy200 = λsy2; sz3sz3
1
⇒ s1s2s3 sx100 = λs2s3x; s1s2s3 sy200 = λs1s3y; s1s2s3 sz300 = λs1s2z
Then ϕ[sx00 : sy00 : sy00 ] = ϕ[s1s2s3 sx00 : s1s2s3 sy00 : s1s2s3 sz300 ]
Then ϕ is well defined.
Let [x : y : z] ∈ Ea,b(A), we have ϕ[1x : y1 : 1z ] = [x : y : z].
Finally the mapping ϕ is surjective.
Corollary 1. Let a,b ∈Fp, then
Card(Ea,b(AP )) = Card(Ea,b(A))
Proposition 5. A Weierstrass equation is defined a elliptic curve over AP if and only if the reduction ever Fp is a elliptic curve.
Proposition 6. Let a,b ∈ AP such that 4a3+27b2 is invertible on AP .The set Ea,b(AP ) together with a special point O = [0 : 1 : 0], a commutative binary operation denoted by +. It is well known that the binary operation + endows the set Ea,b(AP ) with an abelian group with O as identity element.
Proof. The proof in M.Virat theses[2] page 56, based on [5] page 117 corollary 6.6 and [6] page 63.
Proposition 7. Let P = [x : y : z] and P 0 = [x0 : y0 : z0] in Ea,b(AP ), we have
- if πEa,b(AP )(P ) , πEa,b(AP )(P 0) then P + P 0 = [p1 : q1 : r1] with p1 = y2x0z0 − zxy02 − a(zx0 + xz0)(zx0 − xz0)+(2yy0 −
3bzz0)(zx0 −xz0)
q1 = yy0(z0y − zy0) − a(xyz02 − z2x0y0) + (−2azz0 − 3xx0)(x0y −xy0)−3bzz0(z0y −zy0) r1 = (zy0 +z0y)(z0y −zy0)+(3xx0 +azz0)(zx0 −xz0)
- if πEa,b(R)(P ) = πEa,b(R)(P 0) then P + P 0 = [p2 : q2 : r2] with
p2 = (yy0 − bzz0)(x0y + xy0) + (a2zz0 − 2axx0)(zy0 + z0y)−3b(xyz02 +z2x0y0)−a(yzx02 +x2y0z0)
q2 = y2y02 + 3ax2x02 + (−a3 − 9b2)z2z02 − a2(zx0 + xz0)2 −2a2zxz0x0 +(9bxx0 −3abzz0)(zx0 +xz0) r2 = (yy0 + 3bzz0)(zy0 + z0y) + (3xx0 + 2azz0)(x0y + xy0)+a(xyz02 +z2x0y0)
Proof. The proof in M.Virat theses[2] page 57 Proposition 2.1.2. based on formulas I, II and III in the article of H.Lange and W.Ruppert[7].
Corollary 2. The mapping
πE(AP ) : E(AP ) −−→→ [π(x) :Eπ((Fyp) :) π(z)] is a [x : y : z] homomorphism of groups.
Theorem 3. Let a,b ∈Fp. Then
Card(Ea,b(A)) > (Card(Ea,b(Fp))−3)n +Card(Ea,b(Fp))
Proof. The proof for n = 3 exist in article of A.Chilali, S.Abdelalim[4].
For n ∈N∗ We consider the set:
T = {[x : y : z]/y2z = x3 +axz2 +bz3}
= {[x : 0 : z]/0 = x3 +axz2 +bz3} = {[x : 0 : 1]/0 = x3 +ax +b}
then Card(T ) is exactly the number of the solution of
the equation x3 +ax +b = 0 therefore Card(T ) ≤ 3.
Let
G = Ea,n(Fp)−T
= {[x : y : z] ∈P2(Fp)/y2z = x3 +axz2 +bz3 avec y , 0}
= {[x : 1 : z] ∈P2(Fp)/z = x3 +axz2 +bz3}
= {[x : 1 : z] ∈P2(Fp)/[x : 1 : z] ∈ Ea,n(Fp)} therefore Card(G) ≥ Card(Ea,n(Fp))−3 We consider the mapping:
α : Gn −→ Ea,b(A)
([xi : yi : zi])ii==1n −→ [Pni=0xiei : 1 : Pin=0ziei] We have:
n n
(Xxiei)3 = X(xiei)3
i=0 i=0 n n
X 2 = X(ziei)2
( ziei)
i=0 i=0 n n n
X X
a( xiei)( ziei)2 = aX(xiei)(ziei)2
i=0 i=0 i=1
Since [xi : 1 : zi] ∈ Ea,n(Fp) then zi = xi3 +axizi2 +bzi3. It is clear that : ziei = (xiei)2 +a(xiei)(ziei)2 +b(ziei)3, ∀i ∈{1,..,n}
n n n n
Xziei = X(xiei)3 +aX(xiei)(ziei)2 +bX(ziei)3
i=0 i=0 i=0 i=0
n n n n n
X X 3 +a(Xxiei)(Xziei)2 +b(Xziei)3 ziei = ( xiei)
i=0 i=0 i=0 i=0 i=0 we deduce that:
n n
X X
[ xiei : 1 : ziei] ∈ Ea,b(A)
i=0 i=0
α is injective. Then
Ea,n(Fp) ⊆ Ea,b(A) α(Gn) ⊆ Ea,b(A)
and Ea,n(Fp)∩α(Gn) = {[0 : 1 : 0]}
We result that
Card(Ea,b(A)) ≥ Card(α(Gn))+Card(Ea,n(Fp))−1
Since α is injective, then
Card(Gn) = Card(G)n ≥ Card(Ea,n(Fp)−3)n we deduce that:
Card(Ea,b(A)) ≥ Card(Ea,n(Fp)−1)n +Card(Ea,n(Fp))
Corollary 3. Let a and b two elements of Fp. Then
Card(Ea,b(AP )) > (Card(Ea,b(Fp))−3)n +Card(Ea,b(Fp))
Proof. In Corollary 1 Card(Ea,b(AP )) = Card(Ea,b(A)), and in Theorem 3
Card(Ea,b(A)) > (Card(Ea,b(Fp))−3)n +Card(Ea,b(Fp)) we deduce that:
Card(Ea,b(AP )) > (Card(Ea,b(Fp))−3)n +Card(Ea,b(Fp))
3 The discrete logarithm problem: complexity and security
The discrete logarithm problem for G may be stated as: Given g ∈ G and h ∈< g >, find an integer x such that h = gx and ord(g) = q a prime number.
Baby-Step/Giant-Step Method: The idea behind the Baby-Step/Giant-Step method is a standard divideand-conquer approach found in many areas of computer science. We write
x = x0 +x1dqc
Now, since 0 ≤ x ≤ q, we have that 0 ≤ x0,x1 ≤ dqc We first compute the Baby-Steps gi ←− gi,f or0 ≤ i ≤dqc
The pairs (gi,i) are stored in a table so that one can easily search for items indexed by the first entry in the pair. This can be accomplished by sorting the table on the first entry, or more efficiently by the use of hash tables. To compute and store the Baby-Steps clearly requires O(dqc) time and a similar amount of storage.
We now compute the Giant-Steps hj ←− h.g−jdqc for 0 ≤ j ≤dqc , and try to find a match in the table of BabySteps, i.e. we try to find a value gi such that gi = hj. If such a match occurs we have x0 = i and x1 = j.
since, if gi = hj , we have gi = h.g−jdqc
i.e gi+jdqc = h
Notice that the time to compute the Giant-Steps is at√ most O( q).
Hence, the overall time and space complexity of the√ Baby-Step/Giant-Step method is O( q)[8].
Pollard-Type Methods: We define a partition G = S0 ∪ S1 ∪S2 and the function
(hy,a,a+b mod(q)) if
f (y,a,b) = (y2,2a mod(q),2b mod(q)) if (gy,a+1 mod(q),b) if |
y ∈ S0
y ∈ S1 y ∈ S2 |
Note that if y = gahb, Then, taking (z,k,l) = (y,a,b), z = gkhl.
Then, it is possible to iterate f until finding two identical results: (y,a,b) = (z,k,l) and
gahb = y = z = gkhl ⇒ gl−k = hl−b
As gx = h, we have
ga−k = gx(l−b) ⇒ x(l −b) = (a−k) mod(q)
if l −b Is invertible modulo q, We find the solution x = (a−k)(l −b)−1 mod(q)
√
The time and space complexity of the method is O( q).
Let Card(Ea,b(AP )) = M and Card(Ea,b)(Fp) = N n ≥ 3, and N ≥ 7[8].
Theorem 4.√ The time for solving DLP in Ea,b(AP ) is Ω(N),
and O( N) for solving DLP in Ea,b(Fp).
Proof.√ Its clear that for solving DLP in Ea,b(Fp) is
O(√N). We have the time for solving DLP in Ea,b(AP ) is
O( M). And: M ≥ (N −3)n +N
⇒ M√ ≥ (N −3)3 +N ≥ N2 because n ≥ 3 and N ≥ 7
⇒ √M ≥ N
⇒ M = Ω(N)
Then the time complexity for solving DLP in√ Ea,b(AP ) is
Ω(N) Instead of O( N) for Ea,b(Fp).
Example:
Let n = 3, p = 5, a = 1 and b = 1.
E1,1(F5) = {[0 : 1 : 0],[0 : 1 : 1],[0 : 4 : 1],[2 : 1 : 1],[2 : 4 : 1],[3 : 1 : 1],[3 : 4 : 1],[4 : 2 : 1],[4 : 3 : 1]}
We have Card(E1,1(F5)) = 9. Then
Card(E1,1(AP )) ≥ (E1,1(F5) − 3)3 + Card(E1,1(F5)) = 225 The following table gives the difficulty of calculating in A = F5[e1,e2,e3], and in E1,1(A) as a function of the F5 addition and the multiplication.
Operations | + in F5 | × in F5 |
+ in F5[e1,e2,e3] | 3 | 0 |
× in F5[e1,e2,e3] | 6 | 10 |
+ in E1,1(F5[e1,e2,e3]) | 582 | 870 |
+ in E1,1(F5) | 6 | 4 |
Note that the computational difficulty in
E1,1(F5[e1,e2,e3]) is much more difficult than in E1,1(F5).
4 The fundamental algorithms
4.1 Algorithms in A
Algorithm 1 sum_1(p)
Input:(X = (x0,x1,…,xn),Y = (y0,y1,…,yn));
Output:(Z = X +Y); for i = 0 to n do: zi = xi +yi;
end for; return Z = (z0,z1,…,zn); end;
Algorithm 2 prod_1(p)
Input:(X = (x0,x1,…,xn),Y = (y0,y1,…,yn)); Output:(Z = X.Y); z0 = x0.y0; for i = 1 to n do:
zi = x0yi +xiyi +xiy0;
en for; return Z = (z0,z1,…,zn); end;
Algorithm 3 if _invertible_1(p)
Input:(X = (x0,x1,…,xn)) Output:(true,false)
for i = 1 to n do if (x0 +xi == 0 mod(p) or x0 = 0) return false;
end if
end for; return true; end;
Algorithm 4 invertible_1(p)
Input:(X = (x0,x1,…,xn)); Output:(Z = X−1); if (if _invertible_1(X) == f alse);
print(“X is not invertible); return null;
end if;
;
for i = 1 to n do
zi ; end for; return Z = (z0,z1,…,zn); end;
Algorithm 5 projection_1(p)
Input:(X = (x0,x1,…,xn)); Output: x0; return x0; end;
4.2 Algorithms in AP
Algorithm 6 prod_2(p)
Input: (X = ((x0,x1,…,xn),(1,s1,…,sn)),
Y = ((y0,y1,…,yn),(1,t1,…,tn)));
Output: XY;
- = pod_1((x0,x1,…,xn),(y0,y1,…,yn));
- = pod_1((1,s1,…,sn),(1,t1,…,tn));
Z = (A,B); return Z; end;
Algorithm 7 sum_2(p)
Input: (X = ((x0,x1,…,xn),(1,s1,…,sn)),
Y = ((y0,y1,…,yn),(1,t1,…,tn)));
Output: X +Y;
- = sum_1(prod_1((x0,x1,…,xn),(1,t1,…,tn)), prod_1((y0,y1,…,yn),(1,s1,…,sn)));
- = pud_1((1,s1,…,sn),(1,t1,…,tn));
Z = (A,B); return Z; end;
Algorithm 8 projection_2(p)
Input:(X = ((x0,x1,…,xn),(1,s1,…,sn)));
Output: x0; return x0; end;
4.3 Algorithms in Ea,b(AP )
Algorithm 9 projection_3(p)
Input:(X,Y,Z); Output: (π(X),π(Y),π(Z)); return (projection_2(X),projection_2(Y), projection_2(Z)) ;
end;
In this algorithm the + and . in AP for simplicity
Algorithm 10 sum_3(p)
Input: (X = (x,y,z),Y = (x0,y0,z0)); Output: X +Y = (p,q,r);
if projection_3(X) = prejection_3(Y) then;
p = y2x0z0 − zxy02 − a(zx0 + xz0)(zx0 − xz0) + (2yy0 −
3bzz0)(zx0 −xz0); q = yy0(z0y−zy0)−a(xyz02−z2x0y0)+(−2azz0 −3xx0)(x0y−
xy0)−3bzz0(z0y −zy0);
r = (zy0 +z0y)(z0y −zy0)+(3xx0 +azz0)(zx0 −xz0);
else
p = (yy0 − bzz0)(x0y + xy0) + (a2zz0 − 2axx0)(zy0 + z0y) − 3b(xyz02 +z2x0y0)−a(yzx02 +x2y0z0) ; q = y2y02 + 3ax2x02 + (−a3 − 9b2)z2z02 − a2(zx0 + xz0)2 − 2a2zxz0x0 +(9bxx0 −3abzz0)(zx0 +xz0); r = (yy0 + 3bzz0)(zy0 + z0y) + (3xx0 + 2azz0)(x0y + xy0) +
a(xyz02 +z2x0y0); end if; return (p,q,r); fin ;
5 Conclusion
In this work we study the elliptic curves over a special ring, and we construct a new groups with intractable discrete logarithm problem is therefore of great importance, This allows to define more secure cryptographic cryptosystems.
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