SOME PROPERTIES OF SYMBOL ALGEBRAS OF DEGREE THREE

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1 SOME PROPERTIES OF SYMBOL ALGEBRAS OF DEGREE THREE CRISTINA FLAUT and DIANA SAVIN Communicated by the former editorial board In this paper, we study some properties of the matrix representations of the symbol algebras of degree three Moreover, we study some equations with coefficients in these algebras, we define the Fibonacci symbol elements and we obtain some of their properties AMS 2010 Subject Classification: 15A24, 15A06, 16G30, 1R52, 11R37, 11B39 Key words: symbol algebras, matrix representation, quaternion and octonion algebras, Fibonacci numbers 0 PRELIMINARIES Let n be an arbitrary positive integer, let K be a field whose chark does not divide n and contains ω, a primitive n-th root of the unity Let K = K\{0}, a, b K and let S be the algebra over K generated by elements x and y where x n = a, y n = b, yx = ωxy This algebra is called a symbol algebra also known as a power norm residue algebra and it is denoted by J Milnor, in [13], calls it symbol a, b K,ω algebra because of its connection with the K-theory and with the Steinberg symbol For n = 2, we obtain the quaternion algebra For details about Steinberg symbol, the reader is referred to [11] and for other details about quaternions, the reader is referred to [14] and [16] In the paper [2], using the associated trace form for a symbol algebra, the author studied some properties of such objects and gave some conditions for a symbol algebra to be with division or not Starting from these results, we intend to find some examples of division symbol algebras Since such an example is not easy to provide, we try to find first sets of invertible elements in a symbol algebra and study their algebraic properties and structure see, for example, Proposition 44 MATH REPORTS 1666, ,

2 444 Cristina Flaut and Diana Savin 2 In the special cases of quaternion algebras, octonion algebras and symbol algebras, the definition of a division algebra is equivalent to the fact that all its nonzero elements are invertible since these algebras admit a type of norm n that permits composition, ie nab = n a n b, for all elements a, b in such an algebra From this idea, for a symbol algebra with a = b = 1 Theorem 47, we proved that all Fibonacci symbol elements are invertible This set is included in the set of the elements denoted by M from Proposition 44, in which we proved that M is a Z-module We intend to complete this set with other invertible elements such that the obtained set to be an algebra not only a Z-module and we intend to study some properties of this module if there exist some properties similar with the properties of the modules from [18] The study of symbol algebras in general, and of degree three in particular, involves very complicated calculus and, usually, can be hard to find examples for some notions Hence, the present paper is rather technical, which is however unavoidable given the subject The paper is structured in four sections In the first section, some definitions and general properties of these algebras are introduced Since the theory of symbol algebras has many applications in different areas of mathematics and other applied sciences, in sections two, we studied matrix representations of symbol algebras of degree three and, in section three, we used some of these properties to solve some equations with coefficients in these algebras Since symbol algebras generalize the quaternion algebras, starting from some results given in the paper [9], in which the author defined and studied Fibonacci quaternions, we define in a similar manner the Fibonacci symbol elements and we study their properties We computed the formula for the reduced norm of a Fibonacci symbol element Proposition 45 and, using this expression, we find an infinite set of invertible elements Theorem 47 1 INTRODUCTION In the following, we assume that K is a commutative field with chark 2, 3 and A is a finite dimensional algebra over the field K The center CA of an algebra A is the set of all elements c A which commute and associate with all elements x A An algebra A is a simple algebra if A is not a zero algebra and {0} and A are the only ideals of A The algebra A is called central simple if the algebra A F = F K A is simple for every field extension F of K Equivalently, a central simple algebra is a simple algebra with CA = K We remark that each simple algebra is central simple over its center If A is a central simple algebra, then dim A = n = m 2, with m N The degree of the central simple algebra A, denoted by DegA, is DegA = m

3 3 Some properties of symbol algebras of degree three 445 If A is an algebra over the field K, a subfield of the algebra A is a subalgebra L of A such that L is a field The subfield L is called a maximal subfield of the algebra A if there is not a subfield F of A such that L F If the algebra A is a central simple algebra, the subfield L of the algebra A is called a strictly maximally subfield of A if [L : K] = m, where [L : K] is the degree of the extension K L Let L M be a field extension This extension is called a cyclic extension if it is a Galois extension and the Galois group G M/L is a cyclic group A central simple algebra A is called a cyclic algebra if there is L, a strictly maximally subfield of the algebra A, such that L/K is a cyclic extension Proposition 11 [14], Proposition a, p 277 Let K L be a cyclic extension with the Galois cyclic group G = G L/K of order n and generated by the element σ If A is a cyclic algebra and contains L as a strictly maximally subfield, then there is an element x A {0} such that: i A = x j L; 0 j n 1 ii x 1 γx = σ γ, for all γ L; iii x n = a K We will denote a cyclic algebra A with L, σ, a We remark that a symbol algebra is a central simple cyclic algebra of degree n For details about central simple algebras and cyclic algebras, the reader is referred to [14] Definition 12 [14] Let A be an algebra over the field K If K L is a finite field extension and n a natural number, then a K-algebra morphism ϕ : A M n L is called a representation of the algebra A The ϕ characteristic polynomial of the element a A is P ϕ X, a = detxi n ϕ a, the ϕ norm of the element a A is η ϕ a = det ϕ a and the ϕ trace of the element a A is τ ϕ a = tr ϕ y If A is a K central simple algebra such that n = DegA, then the representation ϕ is called a splitting representation of the algebra A Remark 13 i If X,Y M n K, K an arbitrary field, then we know that tr X t = tr X and tr X t Y = tr XY t It results that tr XY X 1 = tr Y ii [14], p 296 If ϕ 1 : A M n L 1, ϕ 2 : A M n L 2 are two splitting representations of the K algebra A, then P ϕ1 X, a = P ϕ2 X, a It results that the ϕ 1 characteristic polynomial is the same with the ϕ 2 characteristic polynomial, ϕ 1 norm is the same with ϕ 2 norm, the ϕ 1 trace is the same with ϕ 2 trace and we will denote them by P X, a instead of P ϕ X, a, η A/K instead of η ϕ1 or simply η and τ instead of τ ϕ, when is no confusion in

4 446 Cristina Flaut and Diana Savin 4 notation In this case, the polynomial P X, a is called the characteristic polynomial, the norm η is called the reduced norm of the element a A and τ is called the trace of the element a A Proposition 14 [14], Corollary a, p 296 If A is a central simple algebra over the field K of degree m, ϕ : A M r L is a matrix representation of A, then m r, η ϕ = η r/m, τ ϕ = r/m τ and η ϕ a, τ ϕ a K, for all a A 2 MATRIX REPRESENTATIONS FOR THE SYMBOL ALGEBRAS OF DEGREE THREE Let ω be a cubic root of unity, K be a field such that ω K and S = a, b K,ω be a symbol algebra over the field K generated by elements x and y where 21 x 3 = a, y 3 = b, yx = ωxy, a, b K In [17], the author gave many properties of the left and right matrix representations for the real quaternion algebra Since symbol algebras generalize the quaternion algebras, using some ideas from this paper, in the following, we will study the left and the right matrix representations for the symbol algebras of degree 3 A basis in the algebra S is 22 B = {1, x, x 2, y, y 2, xy, x 2 y 2, x 2 y, xy 2 }, see [2, 4, 11, 14] Let z S, 23 z = c 0 + c 1 x + c 2 x 2 + c 3 y + c 4 y 2 + c 5 xy + c 6 x 2 y 2 + c 7 x 2 y + c 8 xy 2 and Λ z M 9 K be the matrix with the coefficients in K which its columns are the coordinates of the elements {z 1, zx, zx 2, zy, zy 2, zxy, zx 2 y 2, zx 2 y, zxy 2 } in the basis B : Λ z = c 0 ac 2 ac 1 bc 4 bc 3 abω 2 c 6 abω 2 c 5 abωc 8 abωc 7 c 1 c 0 ac 2 bc 8 bc 5 bω 2 c 4 abω 2 c 7 abωc 6 bωc 3 c 2 c 1 c 0 bc 6 bc 7 bω 2 c 8 aω 2 c 3 bωc 4 bωc 5 c 3 aωc 7 aω 2 c 5 c 0 bc 4 ac 2 abωc 8 ac 1 abω 2 c 6 c 4 aω 2 c 6 aωc 8 c 3 c 0 aωc 7 ac 1 aω 2 c 5 ac 2 c 5 ωc 3 aω 2 c 7 c 1 bc 8 c 0 abωc 6 ac 2 bω 2 c 4 c 6 ω 2 c 8 ωc 4 c 7 c 2 ωc 5 c 0 ω 2 c 3 c 1 c 7 ωc 5 ω 2 c 3 c 2 bc 6 c 1 bωc 4 c 0 bω 2 c 8 c 8 ω 2 c 4 aωc 6 c 5 c 1 ωc 3 ac 2 aω 2 c 7 c 0

5 5 Some properties of symbol algebras of degree three 447 Let α ij M 3 K be the matrix with 1 in position i, j and zero in the rest and 0 0 a γ 1 = 1 0 0, β 1 = 0 0 1, β 2 = 1 0 0, β 3 = ω 0, β 4 = ω 0 0 Proposition 21 The map Λ : S M 9 K, z Λ z is a K algebra morphism γ Proof With the above notations, let Λ x = X = 0 α 31 aβ 2 0 β 1 α 13 0 bα 12 ωbβ 4 M 9 K and Λ y = Y = β 3 α 21 0 ω 2 α 23 ωα 33 ω 2 α 12 By straightforward calculations, we obtain: Λ x 2 = X 2, Λ y 2 = Y 2, Λ xy = XY, Λ x 2 y = Λ x 2 Λ y = ΛxΛxy = X 2 Y, Λ xy 2 = Λ x Λ y 2 = ΛxyΛy = XY 2, Λ x 2 y 2 = Λ x 2 Λ y 2 = ΛxΛxy 2 = Λx 2 yλy = X 2 Y 2 Therefore, we have Λ z 1 z 2 = Λ z 1 Λ z 2 It results that Λ is a K algebra morphism The morphism Λ is called the left matrix representation for the algebra S Definition 22 For Z S, we denote by Z = c 0, c 1, c 2, c 3, c 4, c 5, c 6, c 7, c 8 t M 9 1 K the vector representation of the element Z Proposition 23 Let Z, A S, then: i 1 Z = Λ Z 0 ii AZ = Λ A Z Proof ii From i, we obtain that 1 1 AZ = Λ AZ = Λ A Λ Z 0 0, where 0 M 8 1 K is the zero matrix = Λ A Z Remark 24 1 We remark that an element z S is an invertible element in S if and only if det Λ z 0

6 448 Cristina Flaut and Diana Savin 6 2 The Λ norm of the element z S is η Λ z = det Λ z = η 3 z and τ Λ z = 9trΛ z Indeed, from Proposition 14, if A = S, K = L, m = 3, r = 9, ϕ = Λ, we obtain the above relation 3 We have that τ Λ z = trλ z = 9c 0 Let z S, z = A + By + Cy 2, where A = c 0 + c 1 x + c 2 x 2, B = c 3 + c 5 x + c 7 x 2, C = c 4 + c 8 x + c 6 x 2 We denote by z ω = A + ωby + ω 2 Cy 2, z ω 2 = A + ω 2 By + ωcy 2 Proposition 25 Let S = 1,1 K,ω Then η Λ z = η Λ z ω = η Λ z ω 2 Proof The left matrix representation for the element z S is Λ z = c 0 c 2 c 1 c 4 c 3 ω 2 c 6 ω 2 c 5 ωc 8 ωc 7 c 1 c 0 c 2 c 8 c 5 ω 2 c 4 ω 2 c 7 ωc 6 ωc 3 c 2 c 1 c 0 c 6 c 7 ω 2 c 8 ω 2 c 3 ωc 4 ωc 5 c 3 ωc 7 ω 2 c 5 c 0 c 4 ac 2 ωc 8 c 1 ω 2 c 6 c 4 ω 2 c 6 ωc 8 c 3 c 0 ωc 7 c 1 ω 2 c 5 c 2 c 5 ωc 3 ω 2 c 7 c 1 c 8 c 0 ωc 6 c 2 ω 2 c 4 c 6 ω 2 c 8 ωc 4 c 7 c 2 ωc 5 c 0 ω 2 c 3 c 1 c 7 ωc 5 ω 2 c 3 c 2 c 6 c 1 ωc 4 c 0 ω 2 c 8 c 8 ω 2 c 4 ωc 6 c 5 c 1 ωc 3 c 2 ω 2 c 7 c 0 and for the element z ω is c 0 c 2 c 1 ω 2 c 4 ωc 3 ωc 6 c 5 c 8 ω 2 c 7 c 1 c 0 c 2 ω 2 c 8 ωc 5 ωc 4 c 7 c 6 ω 2 c 3 c 2 c 1 c 0 ω 2 c 6 ωc 7 ωc 8 c 3 c 4 ω 2 c 5 c 3 ω 2 c 7 c 5 c 0 ω 2 c 4 c 2 c 8 c 1 ωc 6 Λ z ω = c 4 ωc 6 c 8 ωc 3 c 0 ω 2 c 7 c 1 c 5 c 2 c 5 ω 2 c 3 c 7 c 1 ω 2 c 8 c 0 c 6 c 2 ωc 4 c 6 ωc 8 c 4 ωc 7 c 2 ω 2 c 5 c 0 c 3 c 1 c 7 ω 2 c 5 c 3 c 2 ω 2 c 6 c 1 c 4 c 0 ωc 8 c 8 ωc 4 c 6 ωc 5 c 1 ω 2 c 3 c 2 c 7 c 0 Denoting by D rs = d rs ij M 9 K the matrix defined such that d rs kk = 1 for k / {r, s}, d rs rr = d rs ss = 0, d rs rs = d rs sr = 1 and zero in the rest, we have that det D rs = 1 If we multiply a matrix A to the left with D rs, the new matrix is obtained from A by changing the line r with the line s and if we multiply a matrix A to the right with D rs, the new matrix is obtained from A by changing the column r with the column s By straightforward calculations, it results that Λ z ω = D 79 D 48 D 46 D 57 D 21 D 23 Λ z D 12 D 23 D 49 D 79 D 48 D 59, therefore η Λ z = det Λ z = det Λ z ω = η Λ z ω In the same way, we get that η Λ z = det Λ z = det Λ z ω 2 = η Λ z ω 2

7 7 Some properties of symbol algebras of degree three 449 Similar to the matrix Λ z, z S, we define Γ z M 9 K to be the matrix with the coefficients in K of the basis B for the elements {z 1, xz, x 2 z, yz, y 2 z, xyz, x 2 y 2 z, x 2 yz, xy 2 z} in their columns This matrix is Γ z = c 0 ac 2 ac 1 bc 4 bc 3 abω 2 c 6 abω 2 c 5 abωc 8 abωc 7 c 1 c 0 ac 2 bωc 8 bω 2 c 5 bc 4 abωc 7 abω 2 c 6 bc 3 c 2 c 1 c 0 bω 2 c 6 bωc 7 bωc 8 ac 3 bc 4 bω 2 c 5 c 3 ac 7 ac 5 c 0 bc 4 aω 2 c 2 abω 2 c 8 aωc 1 abωc 6 c 4 ac 6 ac 8 c 3 c 0 aω 2 c 7 aω 2 c 1 aωc 5 aωc 2 c 5 c 3 ac 7 ωc 1 bω 2 c 8 c 0 abωc 6 aω 2 c 2 bc 4 c 6 c 8 c 4 ω 2 c 7 ωc 2 ωc 5 c 0 c 3 ω 2 c 1 c 7 c 5 c 3 ω 2 c 2 bωc 6 ωc 1 bc 4 c 0 bω 2 c 8 c 8 c 4 ac 6 ωc 5 ω 2 c 1 c 3 aωc 2 aω 2 c 7 c 0 Let α ij M 3 K be the matrix with 1 in position i, j and zero in the rest and β 5 = 0 0 1, β 6 = ω 0 0, 0 ω β 7 = , β 8 = Proposition 26 With the above notations, we have: i Γ d 1 z 1 + d 2 z 2 = d 1 Γ z 1 +d 2 Γ z 2, for all z 1, z 2 S and d 1, d 1 K ii Γ z 1 z 2 = Γ z 2 Γ z 1, for all z 1, z 2 S Proof Let Γ x = U = Γ y = V = 0 bα 12 ωbβ 8 β 7 α 21 0 α 23 α 33 α 12 γ ωα 31 aωβ 6 0 ωβ 5 ω 2 α 13 M 9 K and By straightforward calculations, we obtain: Γ x 2 = U 2, Γ y 2 = V 2, Γ yx = UV, Γ x 2 y = Γ y Γ x 2 = ΓxyΓx = V U 2, Γ xy 2 = Γ y 2 Γ x = ΓyΓxy = V 2 U, Γ x 2 y 2 = Γ y 2 Γ x 2 = Γxy 2 Γx = ΓyΓx 2 y = V 2 U 2 Therefore, we have that Γ z 1 z 2 = Γ z 2 Γ z 1 and Γ is a K algebra morphism

8 450 Cristina Flaut and Diana Savin 8 Proposition 27 Let Z, A S, then: i 1 Z = Γ Z 0 ii ZA = Γ A Z iii Λ A Γ B = Γ B Λ A Proposition 28 Let S = Theorem 29 Let S =, where 0 M 8 1 K is the zero matrix a,b K,ω 1,1 K,ω M 9 = 1, 1 a x, 1 a x2, 1 b y, 1 b y2, 1 ab xy, 1 ab x2 y 2, 1 ab x2 y, 1 ab xy2,, therefore η Γ z = η Γ z ω = η Γ z ω 2 be a symbol algebra of degree three and N 9 = 1, x 2, x, y 2, y, x 2 y 2, xy, xy 2, x 2 y t, M 10 = 1, 1 a x2, 1 a x, 1 b y2, 1 b y, 1 ab x2 y 2, 1 ab xy, 1 ab xy2, 1 ab x2 y, N 10 = 1, x, x 2, y, y 2, xy, x 2 y 2, x 2 y, xy 2 t The following relation is true M 9 Λ z N 9 = M 10 Γ t z N 10 = 3z, z S 3 SOME EQUATIONS WITH COEFFICIENTS IN A SYMBOL ALGEBRA OF DEGREE THREE Using some properties of left and right matrix representations found in the above section, we solve some equations with coefficients in a symbol algebra of degree three Let S be an associative algebra of degree three For z S, let P X, z be the characteristic polynomial for the element a 31 P X, z = X 3 τ z X 2 + π z X η z 1, where τ is a linear form, π is a quadratic form and η a cubic form Proposition 31 [1] With the above notations, denoting by z = z 2 τ z z + π z 1, for an associative algebra of degree three, we have: i π z = τ z ii 2π z = τ z 2 τ z 2 iii τ zw = τ wz iv z = η z z v zw = w z vi π zw = π wz In the following, we will solve some equations with coefficients in the symbol algebra of degree three S = For each element Z S relation a,b K,ω 31 holds First, we remark that if the element Z S has η Z 0, then Z is an invertible element Indeed, from 31, we have that ZZ = η Z, therefore Z 1 = Z ηz

9 9 Some properties of symbol algebras of degree three 451 We consider the following equations: 32 AZ = ZA 33 AZ = ZB 34 AZ ZA = C 35 AZ ZB = C, with A, B, C S Proposition 32 i Equation 32 has non-zero solutions in the algebra S ii If equation 33 has nonzero solutions Z in the algebra S such that η Z 0, then τ A = τ B and η A = η B iii If equation 34 has solution, then this solution is not unique iv If Λ A ΓB is an invertible matrix, then equation 35 has a unique solution Proof i Using vector representation, we have that AZ = ZA It results that Λ A Z = Γ A Z, therefore Λ A Γ A Z = 0 The matrix ΛA Γ A has the determinant equal with zero first column is zero, then equation 32 has non-zero solutions ii We obtain η AZ = η BZ and η A = η B since Z is an invertible element in S Using representation Λ, we get Λ A Λ Z = Λ Z Λ B From Remark 13 i, it results that Λ A = Λ Z Λ B Λ Z 1 Therefore τ A = tr Λ A = tr Λ Z Λ B Λ Z 1 = tr Λ B = τ B K iii Using the vector representation, we have Λ A Γ A Z = C and, since the matrix ΛA Γ A has the determinant equal with zero first column is zero, if equation 34 has solution, then this solution is not unique iv Using vector representation, we have Λ A Γ B Z = C Proposition 33 Let A = a 0 + a 1 x + a 2 x 2 + a 3 y + a 4 y 2 + a 5 xy + a 6 x 2 y 2 + a 7 x 2 y + a 8 xy 2 S, B = b 0 + b 1 x + b 2 x 2 + b 3 y + b 4 y 2 + b 5 xy + b 6 x 2 y 2 + b 7 x 2 y + b 8 xy 2 S, A 0 = A a 0 0 and B 0 = B b 0 0 If a 0 = b 0, A 0 B 0, η A 0 = η B 0 = 0 and π A 0 = π B 0 0, then all solutions of equation 33 are in the K-algebra A A, B, the subalgebra of S generated by the elements A and B, and have the form λ 1 X 1 + λ 2 X 2, where λ 1, λ 2 K, X 1 = A 0 + B 0 and X 2 = π A 0 A 0 B 0 Proof First, we verify that X 1 and X 2 are solutions of the equation 33 Now, we prove that X 1 and X 2 are linearly independent elements If α 1 X 1 + α 2 X 2 = 0, it results that α 2 π A 0 = 0, therefore α 2 = 0 We obtain

10 452 Cristina Flaut and Diana Savin 10 that α 1 = 0 Obviously, each element of the form λ 1 X 1 + λ 2 X 2, where λ 1, λ 2 K is a solution of the equation 33 and since π A 0 0 we have that A A, B = A X 1, X 2 Since each solution of the equation 33 belongs to algebra A A, B, it results that all solutions have the form λ 1 X 1 +λ 2 X 2, where λ 1, λ 2 K 4 FIBONACCI SYMBOL ELEMENTS In this section, we will introduce the Fibonacci symbol elements and we will compute the reduced norm of such an element This relation helps us to find an infinite set of invertible elements First of all, we recall [3], [7] and give some properties of Fibonacci numbers, properties which will be used in our proofs Fibonacci numbers are the following sequence of numbers 0, 1, 1, 2, 3, 5, 8, 13, 21,, with the nth term given by the formula: f n = f n 1 + f n 2, n 2, where f 0 = 0, f 1 = 1 The expression for the nth term is 41 f n = 1 5 [α n β n ], where α = and β = Remark 41 Let f n n 0 be the Fibonacci sequence f 0 = 0, f 1 = 1, f n+2 = f n+1 + f n, n N Then f n + f = 2f n+2, n N f n + f n+4 = 3f n+2, n N Remark 42 Let f n n 0 be the Fibonacci sequence f 0 = 0, f 1 = 1, f n+2 = f n+1 + f n, n N Let ω be a primitive root of unity of order 3, let K = Q ω be the cyclotomic field and let S = be the symbol algebra a,b K,ω of degree 3 Thus, S has a K- basis { x i y j 0 i, j < 3 } such that x 3 = a K, y 3 = b K, yx = ωxy Let z S, z = n i,j=1 x i y j c ij The reduced norm of z is ηz=a 2 c 3 20+bc 3 21+b 2 c bc 20 c 21 c 22 +a c bc 3 11+b 2 c bc 10 c 11 c a c 00 c 10 c 20 +bc 01 c 11 c 21 +b 2 c 02 c 12 c 22-3abω c 00 c 12 c 21 +c 01 c 10 c 22 +c 02 c 11 c 20 -

11 11 Some properties of symbol algebras of degree three abω 2 c 00 c 11 c 22 +c 02 c 10 c 21 +c 01 c 12 c 20 +c 3 00+bc 3 01+b 2 c bc 00 c 01 c 02 See [14], p 299 We define the n th Fibonacci symbol element to be the element F n = f n 1 + f n+1 x + f n+2 x 2 + f y+ 43 +f n+4 xy + f n+5 x 2 y + f n+6 y 2 + f n+7 xy 2 + f n+8 x 2 y 2 In [8] AF Horadam generalized the Fibonacci numbers, giving by: h n = h n 1 + h n 2, n N, n 2, h 0 = p, h 1 = q, where p, q are arbitrary integers From [8], relation 7, we have that these numbers satisfy the equality h n+1 = pf n + qf n+1, n N In the following will be easy to use instead of h n+1 the notation h p,q n+1 It is obvious that h p,q n + h p,q n = h p+p,q+q n, for n a positive integers number and p, q, p, q integers numbers See [4] Remark 43 Let f n n 0 be the Fibonacci sequence f 0 = 0, f 1 = 1, f n+2 = f n+1 + f n, n N Then i fn 2 + fn 1 2 = f 1, n N ; ii fn+1 2 f n 1 2 = f, n N ; iii f 2 = 2f n f n+1 2 f n, 2 n N; iv fn 2 f n 1 f n+1 = 1 n 1, n N ; v f = fn 2 + 2f n f n 1 We define the n th generalized Fibonacci symbol element to be the element H p,q n = h p,q n 1 + h p,q n+1 x + hp,q n+2 x2 + h p,q y+ 44 +h p,q n+4 xy + hp,q n+5 x2 y + h p,q n+6 y2 + h p,q n+7 xy2 + h p,q n+8 x2 y 2 What algebraic structure has the set of Fibonacci symbol elements or the set of generalized Fibonacci symbol? The answer will be found in the following proposition Proposition 44 Let M the set { n } M = H p i,q i n i n N, p i, q i Z, i = 1, n {0} is a Z module i=1

12 454 Cristina Flaut and Diana Savin 12 Proof The proof is immediate if we remark that for n 1, n 2 N, p 1, p 2, α 1, α 2 Z, we have: α 1 h p 1,q 1 n 1 + α 2 h p 2,q 2 n 2 = h α 1p 1,α 1 q 1 n 1 + h α 2p 2,α 2 q 2 n 2 Therefore, we obtain that M is a Z submodule of the symbol algebra S In the following, we will compute the reduced norm for the nth Fibonacci symbol element Proposition 45 Let F n be the nth Fibonacci symbol element Let ω be a primitive root of order 3 of unity and let K = Q ω be the cyclotomic field Then the norm of F n is ηf n = 4a 2 h 211,14 +a[f n+2 ωh 30766, h 22358, f 2 n ωh 45013, h 33683,27523 h 84,135 2fn 2 + 8h 8,3 +f h 12,20 fn 2 + ωh 4368,1453 +h 14128, n ωh 1472, h 12982,24138 Proof In this proof, we denote with E x, y, z = x 3 + y 3 + z 3 3xyz We obtain: ηf n = ηf n ω + ω 2 f n 3 + fn fn+2++f 3 n+8 3 = a 2 f n fn fn+8-3f 3 n+2 f n+5 f n+8 +a f n fn fn+7-3f 3 n+1 f f n+7 +a f n 3 + fn fn+2-3f 3 n f n+1 f n+2 +a f 3 + fn fn+5-3f 3 f n+4 f n+5 +a f n fn fn+8-3f 3 n+6 f n+7 f n+8 +a ω 3 fn 3 + fn fn+7-3ωf 3 n f n+5 f n+7 +a f n f 3 + ω 3 fn+8-3ωf 3 n+1 f f n+8 +a f n fn ω 3 fn+6-3ωf 3 n+2 f n+4 f n+6 +a f n 3 + fn ω 6 fn+8-3ω 3 2 f n f n+4 f n+8 +a ω 6 fn fn fn+6-3ω 3 2 f n+1 f n+5 f n+6 +a f n f 3 + ω 6 fn+7 3 3ω 2 f n+2 f f n+7 + fn 3 + f 3 + fn+6-3f 3 n f f n+6 = a 2 E f n+2, f n+5, f n+8 + a E f n+1, f n+4, f n+7 +a E f n, f n+1, f n+2 + a E f, f n+4, f n+5 +a E f n+6, f n+7, f n+8 + a E ωf n, f n+5, f n+7 +a E f n+1, f, ωf n+8 + a E f n+2, f n+4, ωf n+6 - ]

13 13 Some properties of symbol algebras of degree three 455 +a E f n, f n+4, ω 2 f n+8 + a E ω 2 f n+1, f n+5, f n+6 +a E f n+2, f, ω 2 f n+7 + E fn, f, f n+6 Now, we compute E f n+2, f n+5, f n+8 E f n+2, f n+5, f n+8 = = 1 [ 2 f n+2 + f n+5 + f n+8 f n+5 -f n f n+8 -f n f n+8 -f n+2 2] Using Remark 41, Remark 42, Remark 43 iii and the recurrence of the Fibonacci sequence, we obtain: E f n+2, f n+5, f n+8 = = 1 [ 2 2f n+4+f n+8 f n+4 +f n f n+7 +f n f n+1 +2f n+4 +f n+7 2] = = 1 [ 2 f n+4 + 3f n+6 2f 2 + 2f n f + 2f n+6 2] = = 4 11f n f 135f 2 n f 2 n 51f 2 n 1 Then, we have: 46 E f n+2, f n+5, f n+8 =4 11f n+2 +14f 135f 2 n+1+82f 2 n-51f 2 n 1 Using Remark 43 i, ii and the definition of the generalized Fibonacci sequence it is easy to compute that: E f n+2, f n+5, f n+8 =4 11f n+2 +14f 135f +84f 1-2fn 2 = =4 11f n f h 84,135 2fn 2 =4h 11,14 h 84,135 2fn 2 Therefore, we obtain 47 E f n+2, f n+5, f n+8 = 4h 11,14 h 84,135 2fn 2 Replacing n n 1 in relation 46 and using Remark 43, iii and the recurrence of the Fibonacci sequence, we obtain: 48 E f n+1, f n+4, f n+7 =4 3f n+2 +11f 51f 2 n+1+33f 2 n-20f 2 n 1 Now, we calculate E f n, f n+1, f n+2 E f n, f n+1, f n+2 = 1 [ 2 f n+f n+1 +f n+2 f n+1 -f n 2 + f n+2 -f n f n+2 -f n 2] = So, we obtain =f n+2 f 2 n 1 + f 2 n + f 2 n+1 49 E f n, f n+1, f n+2 = f n+2 f 2 n+1 + f 2 n + f 2 n 1

14 456 Cristina Flaut and Diana Savin 14 Replacing n n + 3 in relation 49, using Remark 43 iii and the recurrence of the Fibonacci sequence we obtain: 410 E f, f n+4, f n+5 = f n+2 + 2f 23f 2 n f 2 n 9f 2 n 1 Replacing n n + 3 in relation 410, using Remark 43 iii and the recurrence of the Fibonacci sequence we obtain: 411 E f n+6, f n+7, f n+8 = 3f n+2 +4f 635f 2 n+1+387f 2 n-239f 2 n 1 Adding equalities , after straightforward calculation, we have: E f n+1, f n+4, f n+7 +E f n, f n+1, f n+2 +E f, f n+4, f n+5 +E f n+6, f n+7, f n+8 =f n f 2 n f 2 n-965f 2 n 1 +f 4790f 2 n f 2 n-1854f 2 n 1 Using Remark 43 i, ii, it is easy to compute that: E f n+1, f n+4, f n+7 +E f n, f n+1, f n+2 +E f, f n+4, f n+5 +E f n+6, f n+7, f n+8 = f n f f +27f 2 n +f 2936f f 1 +94f 2 n = 412 = f n+2 h 965, f h 1854, fn 2 h 67,1 Then, we obtained that: n+4 E f n+1, f n+4, f n+7 +E f n, f n+1, f n+2 +E f, f n+4, f n+5 +E f n+6, f n+7, f n+8 = 413 = f n+2 h 965, f h 1854, fn 2 h 67,1 n+4 Replacing n by n 1 in relation 48, using Remark 43 iii and the recurrence of the Fibonacci sequence we obtain: 414 E f n, f, f n+6 =8 8f n+2 +3f 20f 2 n+1+11f 2 n 8f 2 n 1 Using Remark 43 i,ii, the definition of the generalized Fibonacci sequence, we have: E f n, f, f n+6 =8 8f n+2 +3f [20 fn+1 2 fn f 2 n 1 +fn 2 ] f 2 n = =8h 8,3 8 8f n+2 +3f 20f +12f 1 fn 2 Therefore, 415 E f n, f, f n+6 = 8h 8,3 h 12,20 fn 2 Now, we compute E ωf n, f n+5, f n+7 E ωf n, f n+5, f n+7 = 1 2 ωf n+f n+5 +f n+7 + f n+7 -ωf n 2] h 12,20 fn 2 [ f n+7 -f n f n+5 -ωf n 2

15 15 Some properties of symbol algebras of degree three 457 By repeatedly using of Fibonacci sequence recurrence and Remark 43 iii, we obtain: E ωf n, f n+5, f n+7 = 1 2 [2 2+ω f n+2+ 7 ω f ] [104f 2 n+1+65f 2 n 40f 2 n 1+ 8 ωf n+1 +ω 3f n 1 2 In the same way, we have: + 11 ωf n+1 +ω 8f n 1 2 ] E ωf n, f n+5, f n+7 = 1 2 [2 2+ω f n ω f ] [-40ω+287 f 2 n ω+31 f 2 n 1+ 64ω+285 f 2 n-4 16ω-55-1 n] Now, we calculate E f n+1, f, ωf n+8 E f n+1, f, ωf n+8 = 1 2 f n+1 + f + ωf n+8 [ f f n ωf n+8 f 2 + ωf n+8 f n+1 2] By repeatedly using the Fibonacci sequence recurrence and Remark 43 iii, it is easy to compute that: 417 E f n+1, f, ωf n+8 =[5ω 1 f n ω 1 f ] [ 2 696ω +578f 2 n+1 182ω+169f 2 n ω+441 f 2 n+ 970ω n ] Now, we calculate E f n+2, f n+4, ωf n+6 E f n+2, f n+4, ωf n+6 = = 1 [ 2 f n+2+f n+4 +ωf n+6 f n+4 -f n ωf n+6 -f n ωf n+6 -f n+2 2] 418 In the same way, we obtain: E f n+2, f n+4, ωf n+6 = 1 2 [2 ω+1 f n+2+ 3ω+1 f ] [520ω +309 f 2 n+1+ 80ω+47 f 2 n ω 21 f 2 n+8 14ω+3 1 n ] Adding relations 416, 417, 418 and making some calculus, it results: E ωf n, f n+5, f n+7 + E f n+1, f, ωf n+8 + E f n+2, f n+4, ωf n+6 = = f n+2 [10138ω+6127 f 2 n ω+5231 f 2 n+ 301ω+1132 f 2 n 1 1 n 1235ω+5315]+f [4103ω f 2 n ω+8566 f 2 n ω+1771 f 2 n 1 1 n 11396ω+16279]

16 458 Cristina Flaut and Diana Savin 16 Now, we compute E f n, f n+4, ω 2 f n+8 E f n, f n+4, ω 2 f n+8 = = 1 fn + f n+4 +ω 2 f n+8 [f n+4 -f n 2 + ω 2 2 f n+8 f n+4 + ω 2 ] 2 f n+8 f n 2 Using Remark 42, the recurrence of the Fibonacci sequence we obtain: f n +f n+4 =3f n+2 ; f n+4 f n = 3f n+1 +f n ; f n+8 = 5f n+2 +8f Using Remark 43 iv it is easy to compute that: E f n, f n+4, ω 2 f n+8 = [ 5ω + 8 fn+2 +8 ω+1 f ] [1460ω+325 f 2 n ω+42 f 2 n ω+127 f 2 n- 1030ω n ] Similarly, it is easy to compute that: E f n+2, f, ω 2 f n+7 = [3ω + 2 fn+2 + 5ω + 4 f ] [546ω+112 f 2 n+1+ 80ω+17 f 2 n 1 418ω+87 f 2 n 418ω+87 1 n ] Now, we compute E f n+5, f n+6, ω 2 f n+1 E f n+5, f n+6, ω 2 1 f n+1 = fn+5 + f n+6 + ω 2 f n+1 2 [f n+6 f n f n+5 ω 2 2 f n+1 + fn+6 ω 2 ] 2 f n+1 Using the recurrence of the Fibonacci sequence, we have: f n+5 +f n+6 +ω 2 f n+1 = ω+4 f n+2 + ω+4 f ; f n+6 f n+5 =3f n+1 +2f n, f n+5 ω 2 f n+1 = ω+6 f n+1 +3f n ; f n+6 ω 2 f n+1 = ω+9 f n+1 +4f n Using Remark 43 iv it is easy to compute that: 422 E f n+5, f n+6, ω 2 f n+1 = [ω + 4 fn+2 + ω + 4 f ] [23ω+151 f 2 n+1+19f 2 n 1 6ω+107 f 2 n 6ω n] Adding equalities , we have: E f n, f n+4, ω 2 f n+8 +E ω 2 f n+1, f n+5, f n+6 +E fn+2, f, ω 2 f n+7 = f n+2 [17785ω f 2 n ω+7667 f 2 n+ 2542ω+1658 f 2 n ω n ]+f [ 2650ω 6171 f 2 n ω+7859 f 2 n ω f 2 n ω n ] Adding ,423, it results:

17 17 Some properties of symbol algebras of degree three E f n+1, f n+4, f n+7 + E f n, f n+1, f n+2 + E f, f n+4, f n+5 + E f n+6, f n+7, f n+8 + E ωf n, f n+5, f n+7 + E f n+1, f, ωf n+8 + E f n+2, f n+4, ωf n+6 + E f n, f n+4, ω 2 f n+8 + E ω 2 f n+1, f n+5, f n+6 + E fn+2, f, ω 2 f n+7 = f n f n f 2 n 2 965fn 1 +f 4790f 2 n f 2 n fn f n+2 [27923ω f 2 n ω f 2 n ω f 2 n ω n ]+f [1453ω f 2 n ω f 2 n ω f 2 n ω n ] Using Remark 43 i, ii and the definition of the generalized Fibonacci numbers, it is easy to compute that: E f n+1, f n+4, f n+7 +E f n, f n+1, f n+2 +E f, f n+4, f n+5 f 2 n +E f n+6, f n+7, f n+8 + E ωf n, f n+5, f n+7 +E f n+1, f, ωf n+8 +E f n+2, f n+4, ωf n+6 + E f n, f n+4, ω 2 f n+8 +E ω 2 f n+1, f n+5, f n+6 +E fn+2, f, ω 2 f n+7 = f n+2 ωh 30766, h 22358, f ωh 45013, h 33683, n+1 Therefore, we obtain: ωh 4368,1453 +h 14128,12163 ωh 1472, h 12982, E f n+1, f n+4, f n+7 +E f n, f n+1, f n+2 +E f, f n+4, f n+5 +E f n+6, f n+7, f n+8 + E ωf n, f n+5, f n+7 + E f n+1, f, ωf n+8 + E f n+2, f n+4, ωf n+6 + E f n, f n+4, ω 2 f n+8 + E ω 2 f n+1, f n+5, f n+6 + E fn+2, f, ω 2 f n+7 = f n+2 ωh 30766, h 22358, f fn 2 ωh 45013, h 33683, n+1 Adding equalities 47, 415, 425, we have: ηf n =4a 2 h 211,14 h 84,135-2fn 2 +8h 8,3 +a[f n+2 ωh 30766, h 22358,20533 ωh 4368,1453 +h 14128,12163 ωh 1472, h 12982,24138 h 12,20 fn 2 +

18 460 Cristina Flaut and Diana Savin 18 f ωh 4368,1453 +h 14128,12163 fn 2 1 n ωh 1472, h 12982,24138 ωh 45013, h 33683,27523 ] Corollary 46 Let F n be the n th Fibonacci symbol element Let ω be a primitive root of order 3 of unity and let K = Q ω be the cyclotomic field Then, the norm of F n is ηf n =f n+2 ωh 30766, h 26822, f ωh 4368, h 19120,20203 fn 2 ωh 45013, h 33835, n+1 ωh 1472, h 12982,24138 Proof From the relation 47, we know that E f n+2, f n+5, f n+8 = 4 11f n f Therefore, we obtain 426 Ef n+2, f n+5, f n+8 =f n+2 h 3696,5940 From the relation 415, we know that 88f 2 n E f n, f, f n+6 = 8 8f n+2 + 3f h 84,135 2fn 2 +f h 4704,7560 h 12,20 fn 2 112f 2 n Then, we obtain 427 E f n, f, f n+6 = f n+2 h 768, fn 2 + f h 288,480 24fn 2 Adding equalities 426 and 427 it is easy to compute that: =f n+2 h 3696,5940 +h 768,1280 It results E f n+2, f n+5, f n+8 +E f n, f, f n+6 = +f h 4704,7560 +h 288,480 =f n+2 h 4464,7220 +f h 4992,8040 f 2 nh 152,136 ηf n =f n+2 h 4464,7220 +f h 4992,8040 fnh 2 152,136 +f n+2 ωh 30766, h 22358, f ωh 4368,1453 +h 14128,12163 fn 2 ωh 45013, n ωh 1472, h 12982,24138 f 2 n 152f n f +h 33683,27523 Therefore ηf n =f n+2 ωh 30766, h 26822, f ωh 4368,1453 +h 19120,20203 fn 2 ωh 45013, h 33835, n+1 ωh 1472, h 12982,24138

19 19 Some properties of symbol algebras of degree three 461 In conclusion, even indices of the top of generalized Fibonacci numbers are very large, the expressions of the norm ηf n from Proposition 45 and from Corollary 46 are much shorter than the formula founded in [14] and, in addition, the powers of Fibonacci numbers in these expressions are 1 or 2 Let S = 1,1 Q,ω be a symbol algebra of degree 3 Using the norm form given in the Corollary 46, we obtain: η F n =ω f n+2 h 30766, f h 4368,1453 fnh , n+1 h 1472, f n+2 h 26822, f h 19120,20203 fnh , n+1 h 12982,24138 If η F n 0, we know that the element F n is invertible From relation η F n 0, it results 428 f n+2 h 30766, f h 4368,1453 f 2 nh 45013, n+1 h 1472, or 429 f n+2 h 26822, f h 19120,20203 f 2 nh 33835, n+1 h 12982, Using relation 41 and Remark 43, i and v, we obtain that f n+2 h 26822, f h 19120,20203 fnh , n+1 h 12982,24138 = 26822f n+2 f f n+2 f f f f f 33835f 2 nf n f 2 nf + 1 n f n n f = 26822f n+2 f 2 n f n+2 f 2 n f n+2 f 2 n f n 1 f n f n f f 2 n f f 2 n f f 33835f 2 nf n f 2 nf + 1 n f n n f = 26822f n+2 f 2 n n f n f n+2 f 2 n f n+2 f 2 n 27659f 2 nf f n 1 f n f n n f f f 2 n f f 33835f 2 nf n f f 2 n 1 We remark that 26822f n+2 f 2 n 1 + 1n f n+2 > 0 We have 26822f n+2 fn f n+2 fn fnf 2 = 54575f n+2 fn fnf 2 0, since 54575f n+2 > 27659f is equivalently with f n+2 f > , 506,which is true for all n 1For n = 0, we have 54575f n+2fn fnf 2 = 0 Obviously, f n 1 f n f n n f = 55506f n 1 f n f n n f > 0, since f f n+2 < , 29

20 462 Cristina Flaut and Diana Savin 20 And, finally, 19120f fn f f 33835fnf 2 n+2 > 0, since f > fn 2 and = > From the above, we have that the term f n+2 h 26822, f h 19120,20203 fnh , n+1 h 12982,24138 > 0 for all n N We just proved the following result: Theorem 47 Let S = 1,1 Q,ω be a symbol algebra of degree 3 All symbol Fibonacci elements are invertible elements 5 CONCLUSIONS In this paper, we studied some properties of the matrix representation of symbol algebras of degree 3 Using some properties of left matrix representations, we solved equations with coefficients in these algebras The study of symbol algebras of degree three involves very complicated calculus and, usually, can be hard to find examples for some notions We introduced the Fibonacci symbol elements, we gave an easier expression of reduced norm of Fibonacci symbol elements, and, from this formula, we found examples of many invertible elements, namely all Fibonacci symbol elements are invertible Using some ideas from this paper, we expect to obtain some interesting new results in a further research and we hope that this kind of sets of invertible elements will help us to provide, in the future, examples of symbol division algebra of degree three or of degree greater than three This idea will be developped in our next researchs Acknowledgments The authors would like to thank the referee for his/her patience and suggestions, which helped us improve this paper Since the second author had the first ideas of this paper during her visit to the International Centre of Theoretical Physics ICTP in Trieste in August 2011, she wishes to express her thanks to the Department of Mathematics of ICTP for hospitality and support REFERENCES [1] JR Faulkner, Finding octonion algebras in associative algebras Proc Amer Math Soc , [2] R Flatley, Trace forms of Symbol Algebras Algebra Colloq , [3] C Flaut and V Shpakivskyi, On Generalized Fibonacci Quaternions and Fibonacci- Narayana Quaternions Adv Appl Clifford Algebr , [4] B Gezer and O Bizim, Cubes in elliptic divisibility sequences, Math Rep , 1, [5] P Gille and T Szamuely, Central Simple Algebras and Galois Cohomology Cambridge University Press, 2006 [6] M Hazewinkel, Handbook of Algebra Vol 2, North Holland, Amsterdam, 2000

21 21 Some properties of symbol algebras of degree three 463 [7] S Halici, On some inequalities and Hankel matrice involving Pell, Pell-Lucas numbers Math Rep , 1, 1 10 [8] AF Horadam, A Generalized Fibonacci Sequence Amer Math Monthly , [9] AF Horadam, Complex Fibonacci Numbers and Fibonacci Quaternions Amer Math Monthly , [10] L Kaplansky, Infinite-dimensional quadratic forms admitting composition Proc Amer Math , [11] TY Lam, Introduction To Quadratic Forms Over Fields Graduate Studies in Mathematics 67, American Mathematical Society, Providence RI, 2005 [12] JS Milne, Class field theory [13] J Milnor, Introduction to Algebraic K-Theory Annals of Mathematics Studies, Princeton Univ Press, 1971 [14] RS Pierce, Associative Algebras Springer Verlag, New York, Heidelberg, Berlin, 1982 [15] D Savin, C Flaut and C Ciobanu, Some properties of the symbol algebras Carpathian J Math , 2, [16] M Tărnăuceanu, A characterization of the quaternion group An Ştiinţ Univ Ovidius Constanţa Ser Mat , [17] Y Tian, Matrix reprezentations of octonions and their applications Adv Appl Clifford Algebra , [18] E Türkmen, Pad Supplemented modules An Ştiinţ Univ Ovidius Constanţa Ser Mat , Received 2 December 2012 Ovidius University, Faculty of Mathematics and Computer Science, Bd Mamaia 124, , Constanţa, Romania cflaut@univ-ovidiusro, cristina flaut@yahoocom Ovidius University, Faculty of Mathematics and Computer Science, Bd Mamaia 124, , Constanţa, Romania savindiana@univ-ovidiusro, dianet72@yahoocom