After the discovering of pairing-based cryptography, de-

velopers and researchers have been studding and developing

new techniques and methods for constructing more efﬁciently

implementation of pairings protocols and algorithms.The ﬁrst

pairing is introduced by Weil Andre in 1948 called Weil

pairing, after that more pairing are appear like tate pairing,

ate pairing and a lot more. The beneﬁce of Elliptic curve

cryptosystems which was discovered by Neal Koblitz [1] and

Victor Miller [2] are to reduce the key sizes of the keys utilize

in public key cryptography. Some works like presented in

[3] interested in signature numeric. The authors in [4] show

that we can use the ﬁnal exponentiation in pairings as one

of the countermeasures against fault attacks. In [5], [6], [7],

[13] Nadia El and others show a study case of working with

elliptic curve with embedding degree 5,9,15 and 27. Also in

[9], [10], [11], [12] researchers show the case of working with

curve with embedding degree 18. In [8] they give a study of

security level of optimal ate pairing.

In the present article, we seek to obtain efﬁcient ways to

pairing computation for curves of embedding degree 54. We

will see how to improve arithmetic operation in curves with

embedding degree 54 by using the tower building technique.

We will give three cases studies that show, when we use a

degree 2 twists, we can handle most operations in Fp2,Fp6

and Fp18 , and when we use a degree 3 twists, we can handle

most operations in Fp3,Fp6,Fp9,Fp18 and Fp27 instead. By

making use of an tower building technique, we also improve

the arithmetic of Fp6,Fp18 and Fp54 in order to get better

results. Finally we will compare these cases to know which

path is the optimal path.

In this paper, we will investigate and examine what will

happens in case of optimal ate pairing with embedding degree

54.

The paper is organized as follow. Section 2 we recall some

background on the main pairing proprieties also ate pairing,

and Miller Algorithm. Section 3 presents our main theorem

in this work. Section 4 will presents the results of our work.

Finally, Section 5 concludes this paper.

In everything that follows, Ewill represent an elliptic curve

with equation

y2=x3+ax+bfor b∈Fqwith qprime number. The symbol

aopt will denote the optimal ate pairing. We shall use, without

explicit mention, the following :

•G1⊂(E(Fq)): additive group of cardinal n∈N∗.

•G2⊂(E(Fqk)): additive group of cardinal n∈N∗.

•G3⊂F∗

qk⊂µn: cyclic multiplicative group of cardinal

n∈N∗.

•µn={u∈¯

Fq|un= 1}.

•P∞: the point at inﬁnity of the elliptic curve.

•k: the embedding degree: the smallest integer such that

rdivides qk−1.

•fs,P : a rational function associated to the point P and

some integer s.

•m,s,i: multiplication, squaring, inversion in ﬁeld Fp.

•M2, S2, I2: multiplication, squaring, inversion in ﬁeld

Fp2.

•M3, S3, I3: multiplication, squaring, inversion in ﬁeld

Fp3.

•M6, S6, I6: multiplication, squaring, inversion in ﬁeld

Fp6.

•M9, S9, I9: multiplication, squaring, inversion in ﬁeld

Fp9.

•M18, S18, I18: multiplication, squaring, inversion in ﬁeld

Fp18

•M27, S27, I27: multiplication, squaring, inversion in ﬁeld

Fp27 .

Tower Building Technique on Elliptic Curve with Embedding

Degree 54

ASSOUJAA ISMAIL, EZZOUAK SIHAM

Sidi Mohammed Ben Abdellah University

Faculty of science Dhar El Mahrez

Department of mathematics, Lab: LASMA

Fez, MOROCCO.

Abstract: Pairing based cryptography is one of the newest security solution that attract a lot of attention, because

we can work with efficient and faster pairing to make the security a lot practical, also the working with extension

finite field of the form 𝔽pk is more useful and secure with k ≥ 12 the implementation become more important.

In this paper, we will presents cases studies of improving pairing arithmetic calculation on curves with

embedding degree 54. We use the tower building technique, and study the case when using a degree 3 twist to

carry out most operations in 𝔽p3 and 𝔽p6 or 𝔽p9 and 𝔽p18 or 𝔽p27, or when using a degree 2 twist to handle most

of the operations in 𝔽p2 and 𝔽p6 and 𝔽p18 .

Key-Words: —Optimal ate pairing, Miller Algorithm, Embedding degree 54, Twist curve

Received: April 22, 2024. Revised: September 13, 2024. Accepted: October 14, 2024. Published: November 14, 2024.

1. Introduction

2. Mathematical Background

PROOF

DOI: 10.37394/232020.2024.4.8

Assoujaa Ismail, Ezzouak Siham

E-ISSN: 2732-9941

Volume 4, 2024

•M54, S54, I54: multiplication, squaring, inversion in ﬁeld

Fp54

Remark 1: In this paper, our main objective is to identify

the optimal path with the lowest cost. Although the cost of

multiplication remains the same in each path we choose, we

aim to determine the path with the minimum cost of squaring

or inversion.

Proposition 1:

We investigate these cases by following the process outlined

below:

1) Transform the elliptic curve with embedding degree k

using the variable change (x, y)→(xu2/d, yu3/d)

2) Choose an appropriate irreducible polynomial for tower

building

3) Construct the twisted isomorphic rational point

4) Determine the cost of multiplication, squaring, and in-

version in the corresponding ﬁeld.

TWIST OF AN ELLIPTIC CURVE

Deﬁnition 1: (Twist of an elliptic curve) [6]

Let E and E’ be two elliptic curves deﬁned over Fq, for q, a

power of a prime number p. Then, the curve E’ is a twist of

degree d of E if we can deﬁne an isomorphism Ψdover Fqd

from E’ into E and such that d is minimal:

Ψd:E0(Fq)→E(Fqd).

Theorem 1: [6] Let E be an elliptic curve deﬁned by the

short Weiestrass equation y2=x3+ax +bover an extension

Fqof a ﬁnite ﬁeld Fp, for p a prime number, k a positive

integer such that q=pk. According to the value of k, the

potential degrees for a twist are d =2, 3, 4 or 6 (in this paper,

we are intersted with the case of d=2 and 3).

•d= 2, Let v∈Fpk/2such that the polynomial X2−vis

irreducible in Fpk/2. The equation of the curve E0deﬁned on

Fpk/2is E0:vy2=x3+ax +b. The morphism Ψ2is deﬁned

by:

Ψ2:E0(Fpk/2)−→ E(Fpk)

(x, y)−→ (x, yv1/2)

•d= 3, the curve Eadmits a twist of degree 3 if and only

a= 0. Let v∈Fpk/d be such that the polynomial X3−vis

irreducible in Fpk/d . The equation of E0is then y2=x3+b

The morphism is:

Ψ3:E0(Fpk/3)−→ E(Fpk)

(x, y)−→ (xv1/3;yv1/2)

Cost calculation:

We use the cost of operation in Quadratic and cubic twisted

curve to calculate the cost of operation in the ﬁeld with

embedding degree 2i.3with the tower building technique for

every path.

•Cost of operation in Quadratic twisted curve:

We already know that the cost of multiplication, squaring and

inversion in the quadratic ﬁeld Fp2are:

M2= 3m, S2= 2m, I2= 4m+irespectively ( [17]).

•Cost of operation in Cubic twisted curve:

We already know that the cost of multiplication, squaring and

inversion in in the cubic twisted ﬁeld Fp3are:

M3= 6m, S3= 5s, I3= 9m+ 2s+irespectively ( [17]).

Vector representation point:

In order to construct a vector representation point in Fpk,

we generally need the following set forms a basis of Fpk

over Fp,Bk={1, u, u2, ..., uk−1}, which is known as

polynomial basis. An arbitrary element A in Fpkis written

as A=a0+a1u+a2u2+... +ak−1uk−1. The vector

representation of A is vA= (a0, a1, a2, ..., ak−1).

We use the vector representation point of Quadratic and cubic

twisted curve to know the vector representation point of

operation in the ﬁeld with embedding degree 2i.3with the

tower building technique for every path.

Vector representation point in Quadratic twisted curve:

We have Eis y2=x3+ax +b.

Let u∈Fpsuch that the polynomial x2−uis irreducible over

Fp.

The equation of E0is uy2=x3+ax +b.

So to map E(Fp)to E0(Fp),we have:

E(Fp)→E0(Fp)

(x, y)→(x1, y1)=(x, yu1/2)

Using ψ2(x, y)=(x, yu1/2)to map E0(Fp)to E(Fp2)

E0(Fp)→E(Fp2)

(x, y)→(x, yu1/2)

Hence, to map E(Fp)to E(Fp2), we have:

E(Fp)→E(Fp2)

(x, y)→(x1, y1)=(x, yu)

•Let map P to P1:

Let P= (x, y) = (a, b)and P1= (x1, y1) = (a1, b1)B2,

where x1, y1, a1, b1∈Fp2.

P1has a special vector representation with 2 Fpelements

for each x1and y1coordinates. We have B2= (1, u),ψ2:

E0(Fp)→E(Fp2),

ψ2(x, y)=(x1, y1)=(x, yu),(see [9]) we have:

P→P1

E(Fp)→E(Fp2)

(x, y)→(x1, y1)=(x, yu)=(a, bu)B2

P1= (x1, y1)=(x, yu)=(a, bu)B2= ((a, 0),(0, b))

•Let remap P1to P: obtained easily by just placing a and

b in the correct basis position.

P1→P

E(Fp2)→E(Fp)

(x1, y1)→(x, y) = (a, b)

P= (x, y)=(a, b)

So we can easly map and remap between P and P1.

Vector representation point in Cubic twisted curve:

PROOF

DOI: 10.37394/232020.2024.4.8

Assoujaa Ismail, Ezzouak Siham

E-ISSN: 2732-9941

Volume 4, 2024

The curve Eadmits a twist of degree 3 if and only if a= 0

i,e y2=x3+b.

Let u∈Fpsuch that the polynomial x3−uis irreducible over

Fp.

The equation of E0is y2=x3+b/u.

So to map E(Fp)to E0(Fp),we have:

E(Fp)→E0(Fp)

(x, y)→(x1, y1)=(xu1/3, yu1/2)

Using ψ3(x, y)=(xu2/3, yu1/2)to map E0(Fp)to E(Fp3)

E0(Fp)→E(Fp3)

(x, y)→(xu2/3, yu1/2)

Hence, to map E(Fp)to E(Fp3), we have:

E(Fp)→E(Fp3)

(x, y)→(x1, y1) = (xu, yu)

•Let map P to P1:

Let P= (x, y) = (a, b)and P1= (x1, y1) = (a1, b1)B3,

where x1, y1, a1, b1∈Fp3.

P1has a special vector representation with 3 Fpelements for

each x1and y1coordinates.

We have B3= (1, u, u2),ψ3:E0(Fp)→E(Fp3),

ψ3(x, y)=(x1, y1)=(xu, yu),(see [9]) we have:

P→P1

E(Fp)→E(Fp3)

(x, y)→(x1, y1)=(xu, yu)=(au, bu)B3

P1= (x1, y1)=(xu, yu)=(au, bu)B3= ((0, a, 0),(0, b, 0))

•Let remap P1to P: obtained easily by just placing a and b

in the correct basis position

P1→P

E(Fp3)→E(Fp)

(x1, y1)→(x, y) = (a, b)

P= (x, y)=(a, b)

So we can easly map and remap between P and P1.

Corollary 1: :

We can do an extension for the above vector representation,

we have:

E(Fpk/2)→E(Fpk)

(x, y)→(x, yu)

and,

E(Fpk/3)→E(Fpk)

(x, y)→(xu, yu)

The ﬁgure below show all path possible for building an

elliptic curve with embedding degree 54

There is four path possible to building this curve

Fp−→ Fp2−→ Fp6−→ Fp18 −→ Fp54

Fp−→ Fp3−→ Fp6−→ Fp18 −→ Fp54

Fp−→ Fp3−→ Fp9−→ Fp18 −→ Fp54

Fp−→ Fp3−→ Fp9−→ Fp27 −→ Fp54

Exploring the ﬁrst path

Fp−→ Fp2−→ Fp6−→ Fp18 −→ Fp54

The appropriate choices of irreducible polynomial deﬁned

by:

Fp2=Fp[u]/(u2−β),with βa non-square and u2= 2

Fp6=Fp2[v]/(v3−u),with va non-cube and v3= 21/2

Fp18 =Fp6[t]/(t3−v),with ta non-cube and t3= 21/6

Fp54 =Fp18 [w]/(w3−t),with wa non-cube and w3= 21/18

P4(x4, y4) = ((a, 0, ..., 0),(0, ..., 0, b)) with x4, y4∈Fp54

P000 (x000 , y000 ) = ((a, 0, ..., 0),(0, ..., 0, b)) with x000 , y000 ∈Fp18

P00 (x00 , y00 ) = ((a, 0,0,0,0,0),(0,0,0,0,0, b)) with x00 , y00 ∈Fp6

P0(x0, y0) = ((a, 0),(0, b)) with x0, y0∈Fp2

P(x, y) = (a, b)with x, y ∈Fp

3. Tower Building Technique for Elliptic

Curve with Embedding Degree 54

PROOF

DOI: 10.37394/232020.2024.4.8

Assoujaa Ismail, Ezzouak Siham

E-ISSN: 2732-9941

Volume 4, 2024

The cost of multiplication, squaring and inversion in in the

54th twisted ﬁeld Fp54 are:

M54 = (M18)Fp3= (M6)Fp3)Fp3= ((M2)Fp3)Fp3)Fp3

= ((3m)Fp3)Fp3)Fp3= ((3M3)Fp3)Fp3= ((18m)Fp3)Fp3

= (18M3)Fp3= (108m)Fp3= 108M3= 648m,

S54 = (S18)Fp3= (S6)Fp3)Fp3= ((S2)Fp3)Fp3)Fp3

= ((2m)Fp3)Fp3)Fp3= ((2M3)Fp3)Fp3= ((12m)Fp3)Fp3

= (12M3)Fp3= (72m)Fp3= 72M3= 432m,

I54 = (I18)Fp3= (I6)Fp3)Fp3= ((I2)Fp3)Fp3)Fp3

= ((4m+i)Fp3)Fp3)Fp3= ((4M3+I3)Fp3)Fp3

= ((33m+ 2s+i)Fp3)Fp3= (33M3+ 2S3+I3)Fp3

= (207m+ 12s+i)Fp3= 207M3+ 12S3+I3

= 1251m+ 62s+i,

Exploring the second path

Fp−→ Fp3−→ Fp6−→ Fp18 −→ Fp54

The appropriate choices of irreducible polynomial deﬁned

by:

Fp3=Fp[u]/(u3−β),with βa non-cube and u3= 2

Fp6=Fp3[v]/(v2−u),with va non-square and v2= 21/3

Fp18 =Fp6[t]/(t3−v),with ta non-cube and t3= 21/6

Fp54 =Fp18 [w]/(w3−t),with wa non-cube and w3= 21/18

P4(x4, y4) = ((a, 0, ..., 0),(0, ..., 0, b)) with x4, y4∈Fp54

P000 (x000 , y000 ) = ((a, 0, ...0),(0, ..., 0, b)) with x000 , y000 ∈Fp18

P00 (x00 , y00 ) = ((a, 0,0,0,0,0),(0,0,0,0,0, b)) x00 , y00 ∈Fp6

P0(x0, y0) = ((a, 0,0),(0,0, b)) with x0, y0∈Fp3

P(x, y)=(a, b)with x, y ∈Fp

The cost of multiplication, squaring and inversion in in the

54th twisted ﬁeld Fp54 are:

M54 = (M18)Fp3= (M6)Fp3)Fp3= ((M3)Fp2)Fp3)Fp3

= ((6m)Fp2)Fp3)Fp3= ((6M2)Fp3)Fp3= ((18m)Fp3)Fp3

= (18M3)Fp3= (108m)Fp3= 108M3= 648m,

S54 = (S18)Fp3= (S6)Fp3)Fp3= ((S3)Fp2)Fp3)Fp3= ((5s)Fp2)Fp3)Fp3

= ((5S2)Fp3)Fp3= ((10m)Fp3)Fp3= (10M3)Fp3= (60m)Fp3

= 60M3= 360m,

I54 = (I18)Fp3= (I6)Fp3)Fp3= ((I3)Fp2)Fp3)Fp3

= ((9m+ 2s+i)Fp2)Fp3)Fp3= ((9M2+ 2S2+I2)Fp3)Fp3

= ((35m+i)Fp3)Fp3= (35M3+I3)Fp3= (219m+ 2s+i)Fp3

= 219M3+ 2S3+I3= 1323m+ 12s+i,

Exploring the third path

Fp−→ Fp3−→ Fp9−→ Fp18 −→ Fp54

The appropriate choices of irreducible polynomial deﬁned

by:

Fp3=Fp[u]/(u3−β),with βa non-cube and u3= 2

Fp9=Fp3[v]/(v3−u),with va non-cube and v3= 21/3

Fp18 =Fp9[t]/(t2−v),with ta non-square and t2= 21/9

Fp54 =Fp18 [w]/(w3−t),with wa non-square and w3= 21/18

P4(x4, y4) = ((a, 0, ..., 0),(0, ..., 0, b)) with x4, y4∈Fp54

P000 (x000 , y000 ) = ((a, 0, ..., 0),(0, ..., 0, b)) with x000 , y000 ∈Fp18

P00 (x00 , y00 ) = ((a, 0, ..., 0),(0, , ..., 0, b)) with x00 , y00 ∈Fp9

P0(x0, y0) = ((a, 0,0),(0,0, b)) with x0, y0∈Fp3

P(x, y) = (a, b)with x, y ∈Fp

The cost of multiplication, squaring and inversion in in the

36th twisted ﬁeld Fp36 are:

M54 = (M18)Fp3= (M9)Fp2)Fp3= ((M3)Fp3)Fp2)Fp3

= ((6m)Fp3)Fp2)Fp3= ((6M3)Fp2)Fp3= ((36m)Fp2)Fp3

= (36M2)Fp3= (108m)Fp3= 108M3= 648m,

S54 = (S18)Fp3= (S9)Fp2)Fp3= ((S3)Fp3)Fp2)Fp3= ((5s)Fp3)Fp2)Fp3

= ((5S3)Fp2)Fp3= ((25s)Fp2)Fp3

= (25S2)Fp3= (50m)Fp3= 50M3= 300m,

PROOF

DOI: 10.37394/232020.2024.4.8

Assoujaa Ismail, Ezzouak Siham

E-ISSN: 2732-9941

Volume 4, 2024

I54 = (I18)Fp3= (I9)Fp2)Fp3= ((I3)Fp3)Fp2)Fp3

= ((9m+ 2s+i)Fp3)Fp2)Fp3= ((9M3+ 2S3+I3)Fp2)Fp3

= ((63m+ 12s+i)Fp2)Fp3= (63M2+ 12S2+I2)Fp3

= (227m+i)Fp3= 227M3+I3= 1371m+ 2s+i,

Exploring the forth path

Fp−→ Fp3−→ Fp9−→ Fp27 −→ Fp54

The appropriate choices of irreducible polynomial deﬁned

by:

Fp3=Fp[u]/(u3−β),with βa non-cube and u3= 2

Fp9=Fp3[v]/(v3−u),with va non-cube and v3= 21/3

Fp27 =Fp9[t]/(t3−v),with ta non-cube and t3= 21/9

Fp54 =Fp27 [w]/(w2−t),with wa non-square and w2= 21/27

P4(x4, y4) = ((a, 0, ..., 0),(0, ..., 0, b)) with x4, y4∈Fp54

P000 (x000 , y000 ) = ((a, 0, ..., 0),(0, ..., 0, b)) with x000 , y000 ∈Fp27

P00 (x00 , y00 ) = ((a, 0, ..., 0),(0, ..., 0, b)) with x00 , y00 ∈Fp9

P0(x0, y0) = ((a, 0,0),(0,0, b)) with x0, y0∈Fp3

P(x, y) = (a, b)with x, y ∈Fp

The cost of multiplication, squaring and inversion in in the

54th twisted ﬁeld Fp54 are:

M54 = (M27)Fp2= (M9)Fp3)Fp2= ((M3)Fp3)Fp3)Fp2

= ((6m)Fp3)Fp3)Fp2= ((6M3)Fp3)Fp2= ((36m)Fp3)Fp2

= (36M3)Fp2= (216m)Fp2= 216M2= 648m,

S54 = (S27)Fp2= (S9)Fp3)Fp2= ((S3)Fp3)Fp3)Fp2

= ((5s)Fp3)Fp3)Fp2= ((5S3)Fp3)Fp2= ((25s)Fp3)Fp2

= (25S3)Fp2= (125s)Fp2= 125S2= 250m,

I54 = (I27)Fp2= (I9)Fp3)Fp2= ((I3)Fp3)Fp3)Fp2

= ((9m+ 2s+ +i)Fp3)Fp3)Fp2= ((9M3+ 2S3+I3)Fp3)Fp2

= ((63m+ 12s+i)Fp3)Fp2= (63M3+ 12S3+I3)Fp2

= (387m+ 62s+i)Fp2= 387M2+ 62S2+I2

= 1289m+i,

TABLE I

COST OF OPERATIONS IN EACH THE TOWER FIELDS

Path O Cost

1M54 648m

S54 432m

I54 1251m+62s+i

2M54 648m

S54 360m

I54 1323m+12s+i

3M54 648m

S54 300m

I54 1371m+2s+i

4M54 648m

S54 250m

I54 1289m+i

The table above give the overall cost of operations in each

the tower ﬁelds.

We found that the cost of multiplication and squaring is the

same for any path chosen, however the cost of inversion

change on the path, so we can see that the minimal cost for

inversion is 1289m+i.

In this paper, we give some methods for tower building of

extension of ﬁnite ﬁeld of embedding degree 54. We show

that there are four efﬁcients paths for constructions of these

extensions of degree 54. We show that by using a degree 2 or 3

twist we handle to perform most of the operations in F6

p,Fp9,

Fp18 ,Fp27 and Fp54 . By using this tower building technique,

we also improve the arithmetic of Fp54 , in order to get better

results of calculate the cost of their multiplication, squaring

and inversion.

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5. Conclusion

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PROOF

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Assoujaa Ismail, Ezzouak Siham

E-ISSN: 2732-9941

Volume 4, 2024

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PROOF

DOI: 10.37394/232020.2024.4.8

Assoujaa Ismail, Ezzouak Siham

E-ISSN: 2732-9941

Volume 4, 2024