On a generalization of the Gaussian plane to three dimensions

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1 arxiv: v1 [math.cv] 4 Jan 019 On a generalization of the Gaussian plane to three dimensions Ján Haluška, Math. Institute SAS, Slovakia Abstract Each associative division algebra over the real number field of finite dimension n N is isomorphic (1) to R (the field of all real numbers, n = 1), or, () to C (the field of all Gaussian complex numbers, n = ), or, (3) to H (the algebra of all quaternions, n = 4) by the theorem by G. F. Frobenius. In this paper, we study an associative, commutative, distributive division vector algebra T over the real field R such that its factor algebra T modulo an ideal D is a three dimensional factor field, where D is a real line in E 3 and the set of non-trivial zero divisors. A topology is given via an absolute value defined in the paper. The field T/D is a generalization of the structure of the Gaussian complex plane C to three dimensions. Although this paper is technically most of algebraic nature, after building up an appropriate theory, the field T/D may serve for creating a theory generalizing the complex analysis theory to three dimensions. 1 Preliminaries The Frobenius theorem about division algebras over R is commonly known, e.g., [7], p Let us pitch shortly some facts about three kinds of complex numbers rings in the plane. If we denote by ı,j,ε complex units for these types of complex numbers, respectively, for elliptic numbers we have ı = 1, ı = 1), hyperbolic numbers we have j = 1, j = 1), and for parabolic numbers Author s address: Ján Haluška, Mathematical Institute, Slovak Academy of Sciences, Grašák Str. 6, Košice, Slovakia 1

2 we have ε = 0, ε = 1. Together all three types of complex numbers are called the generalized complex numbers. Elliptic complex numbers are simply called as complex numbers. For the review and geometrical imagination about the all types of complex numbers in the plane, cf. [6]. In [4] we studied a complex plane C constructed from three non-collinear points. It is well generally known that the only set of complex numbers is a commutative Archimedean continuous field (with respect to isomorphisms and subsets) for E n, n. This fact has its consequence in the construction of the theory called the complex mathematical analysis. However, in the real world, there is needed a calculus whith at least three (or more) geometrical dimensions. In this paper we show how to solve this problem mathematically in three dimensions. Here are some quotation ([1,, 3, 9]) which study the notion of multiplication in modules more deeply. By the theorem by G. F. Frobenius from 1877, each associative division algebra over the real number field of finite dimension n N is isomorphic (1) tor(the field of all real numbers, n = 1), or, () toc(the field of all Gaussian complex numbers, n = ), or, (3) to H (the algebra of all quaternions, n = 4; in general, quaternions are not commutative). In this paper, after an introducing of the operation of multiplication on E 3, we study an associative, commutative, distributive division vector algebra T over the real field R such that the factor algebra T modulo an ideal D is a three dimensional field, where D is a real line in E 3. The algebra T is a generalization of the Gaussian complex plane C to three dimensions. A topology is given via an absolute value where a triangle inequality neither does. or, does not hold be true in the general. But the structure of T/D can be clearly applied to build up a calculus in three dimensions, similarly as it is in the case of the complex analysis theory. Operation of multiplication in the module T A ring is an commutative additive group, commutative multiplicative semigroup, and the operation of addition is distributive with respect to multiplication. The aim of this section is to equip the standard algebra E 3 over R with an operation of multiplication so that the resulting vector space T becomes a ring. Concerning operation of addition it is known that elements of the space

3 E 3 form an additive group with null Λ def = (0, 0, 0). Let us denote by w def = (1,0,0),u def = (0,1,0),v def = (0,0,1). The set {w,u,v} is a basis of the three dimensional vector space E 3 over real line R. So, every element x E 3 can be written as x def = X w w X u u X v v, where X w,x u,x v R. The sign denotes an usual parallelepiped addition in the vector space E 3 and the sign denotes is its inverse group operation; an inverse element to a E 3 is denoted as ( a), we write also a ( a) = a a = Λ. Definition 1 Let x = (X w,x u,x v ) T and y = (Y w,y u,y v ) T. Then x y = (X w,x u,x v ) (Y w,y u,y v ) def = w(x w Y w X u Y v X v Y u ) u(x w Y u +X u Y w X v Y v ) v(x w Y v +X u Y u +X v Y w ). (1) Let us denote by T the vector space E 3 over R equipped with this operation of multiplication. Remark 1 The operation of multiplication in E 3 can be equivalently introduced via multiplication of basic elements as follows: w w = w,u u = v,v v = u, w u = u w = u,w v = v w = v,u v = v u = w. () Concerning the operations of multiplication and addition, a commutative and distributive laws hold for the set T: Lemma 1 Commutative law. Let x, x T, then x y = y x. Proof. By Definition 1. 3

4 Lemma Distributive law. Let x = (X w,x u,x v ) T, y = (Y w,y u,y v ) T, and z = (Z w,z u,z v ) T. Then x [y z] = {x y} {x z}. Proof. By Definition 1, x [y z] = (X w,x u,x v ) [(Y w,y u,y v ) (Z w,z u,z v )] = w[x w (Y w +Z w ) X u (Y v +Z v ) X v (Y u +Z u )] u[x w (Y u +Z u )+X u (Y w +Z w ) X v (Y v +Z v )] v[x w (Y v +Z v )+X u (Y u +Z u )+X v (Y w +Z w )] = w[x w Y w X u Y v X v Y u ] w[x w Z w X u Z v X v Z u ] u[x w Y u +X u Y w X v Y v ] u[x w Z u +X u Z w X v Z v ] v[x w Y v +X u Y u +X v Y w ] v[x w Z v +X u Z u +X v Z w ] = {w[x w Y w X u Y v X v Y u ] u[x w Y u +X u Y w X v Y v ] v[x w Y v +X u Y u +X v Y w ]} {w[x w Z w X u Z v X v Z u ] u[x w Z u +X u Z w X v Z v ] v[x w Z v +X u Z u +X v Z w ]} = {x y} {x z}. (3) 3 Properties of T We collect some properties of T into a theorem. Theorem 1 The set T is a vector algebra over R. Proof. Let x,y,z T, where x = (X w,x u,x v ),y = (Y w,y u,y v ),z = (Z w,z u,z v ) and X w,x u,x v ; Y w,y u,y v ; Z w,z u,z v R. Vector space. The space T with respect to the operation of addition is identical to the vector space E 3 over R. Commutative monoid. The set T is a multiplicative commutative monoid with the unit w = (1,0,0). The associativity of the set T are proved in the next paragraph. 4

5 Associative law. To prove that [x y] z = x [y z] needs a rather longer technical evalue. Let the coordinates of [x y] = (a w,a u,a v ) be respectively as follows: a w = X w Y w X u Y v X v Y u R, a u = X w Y u +X u Y w X v Y v R, a v = X w Y v +X u Y u +X v Y w R. (4) Similarly, let us denote the coordinates of [z y] = (b w,b u,b v ) respectively as follows: b w = Y w Z w Y u Z v Y v Z u R, We have: b u = Y w Z u +Y u Z w Y v Z v R, b v = Y w Z v +Y u Z u +Y v Z w R. (5) [x y] z = (wa w ua u va v ) (wz w uz u vz v ) By () and the distributivity, then = (w w)a w Z w (v u)a v Z u (u v)a u Z v (u w)a u Z w (w u)a w Z u (v v)a v Z v (v w)a v Z w (u u)a u Z u (w v)a w Z v (6) = w[(x w Y w X u Y v X v Y u )Z w (X w Y v +X v Y w +X u Y u )Z u (X w Y u +X u Y w X v Y v )Z v ] (7) u[(x w Y u +X u Y w X v Y v )Z w +(X w Y w X u Y v X v Y u )Z u (X w Y v +X u Y u +X v Y w )Z v ] (8) 5

6 So, v[(x w Y v +X u Y u +X v Y w )Z w +(X w Y u +X u Y w X v Y v )Z u ] = w[(x w Y w Z w X u Y v Z w X v Y u Z w ) (X w Y v Z u +X v Y w Z u +X u Y u Z u ) +(X w Y w X u Y v X v Y u )Z v ] (9) (X w Y u Z v +X u Y w Z v X v Y v Z v )] (10) u[(x w Y u Z w +X u Y w Z w X v Y v Z w ) +(X w Y w Z u X u Y v Z u X v Y u Z u ) (X w Y v Z v +X u Y u Z v +X v Y w Z v )] (11) v[(x w Y v Z w +X u Y u Z w +X v Y w Z w ) +(X w Y u Z u +X u Y w Z u X v Y v Z u ) +(X w Y w Z v X u Y v Z v X v Y u Z v )] (1) and = w[x w (Y w Z w Y u Z v Y v Z u ) X u (Y w Z v +Y u Z u +Y v Z w ) X v (Y w Z u +Y u Z w Y v Z v )] (13) u[(x w (Y u Z w +Y w Z u Y u Z v ) +X u (Y w Z w Y u Z v Y v Z u ) X v (Y w Z u +Y u Z w +Y v Z v )] (14) v[x w (Y w Z v +Y u Z u +Y v Z w ) +X u (Y w Z u +Y u Z w Y v Z v ) +X v (Y w Z w Y u Z v Y v Z u )] (15) 6

7 = w[x w b w X u b v X v b u ] u[x w b u +X u b w X v b v ] v[x w b v +X u b u +X v b w ] (16) = (wx w ux u vx v ) (wb w ub u vb v ) = x [y z]. Comparing the coordinats of expressions [x y] z and x [y z], they are equal. The associativity of T with respect to the operation is proved. Stretching/contraction of T by scalars. Let α R, α 0, and T α def = {αx = (αx w,αx u,αx v ) x T}. Then T α = αt concerning all introduced structures in T. The real number α R represents a stretching/contraction scale of algebras T and T α. 4 A homomorphism of a plane C T to C Let x = (X w,x u,x v ) T,y = (Y w,y u,y v ) T. Let us to consider a plane C = {(X w,x v ) X w R,X v R}. Let C be equipped with following operations of addition and multiplication: (X w,x v )+(Y w,y v ) def = (X w +Y w,x v +Y v ), (X w,x v ) (Y w,y v ) def = (X w Y w X v Y v,x v Y w +Y v X w ). Definition We say that φ : T C is a (dimensional) reduction of T to C, if φ(x) def = (X w +X u,x v ). 7

8 Theorem Let x = (X w,x u,x v ),y = (Y w,y u,y v ) be two elements in T where X u = 0,Y u = 0. Then φ(x y) = φ(x)+φ(y),φ(x y) = φ(x) φ(y). The plane C is isomorphic to the Gaussian complex plane C where the vector w T corresponds to the real unit 1 and the vector v corresponds to the imaginary unit ı, respectively, and according this homomorphism, in T, resp. in C. v v = w φ ı ı = 1 Proof. Let (X w,x u,x v ) T,(Y w,y u,y v ) T, where X w,x u,x v R, Y w,y u,y v R. Let X u = 0,Y u = 0. Concerning the mapping φ, (a) The mapping {(X w,x u,x v ) T X u = 0} = C C is a set bijection; (b) The morphism T C of operations of addition: T (X w,0,x v ) (Y w,0,y v ) φ (X w +Y w,x v +Y v ) C; (c) The morphism T C of operations of multiplication: T (X w,0,x v ) (Y w,0,y v ) φ w(x w Y w X v Y v ) v(x v Y w +X w Y v ) C. (17) (d) Concerning the operation of multiplication, the morphism φ is derived from the Definition 1 : Consequently, φ[(x w,0,x v ) (Y w,0,y v )] = φ[w(x w Y w ) u( X v Y v ) v(x v Y w +X w Y v )] = w(x w Y w X v Y v ) v(x v Y w +X w Y v ). (18) v v = w φ ı ı = 1 in T, resp. in C. To verify the rest properties of the isomorphism C C we let to the reader. 8

9 5 σ-conjugation, homomorphism µ : T σ Definition 3 If x = wx w ux u vx v T, then we define a σ-conjugate element x of the element x as following: where X w,x u,x v R. x def = wx w ux v vx u T, Note that the vectors δ def = u v and w are not perpendicular: the scalar product < u v,w >=< (0,1, 1),(1,0,0) >= 0. Theorem 3 If then x = wx w ux u vx v T, x x = w[x w+x u+x v] [u v]{x w X u +X u X v X v X w } = Aw Bδ, where A = X w +X u +X v, B = X wx u +X u X v X v X w, and X w,x u,x v R. Proof. By Definition 1, x x = [wx w ux u vx v ] [wx w ux v vx u ] = w(x w +X u +X v ) [u v](x wx u +X u X v X v X w ). A triple (δ,λ,w) of non-collinear different points defines a plane in E 3, let us denote by σ a complex plane containing the basic vectors w Λ ("real unit") and δ Λ ("imaginary unit"). The plane σ consist of elements x x T. There are exactly three possibilities for σ to be a complex plane, cf. 1. Preliminary; they are elliptic (Gaussian), hyperbolic, and parabolic types of the operations of multiplication (the operation of addition is the same for the all three kinds of complex planes). For our case the hyperbolic complex numbers will be apt. 9

10 Theorem 4 Let the plane σ be equipped with the operation of multiplication derived from the operation on T (in other words, the operation in σ is induced with the operation in T). The operation is a hyperbolic complex multiplication on a plane σ, the real unit is w and the imaginary unit is, respectively, i.e., = w, where = w 3 δ 3 = 1 3 [w u v]. Proof. A question of the a length of an unit "imaginary" element will be definitely solved after introducing of a notion of the absolute value, where is a value on T, an analogue of the absolute value notion in the hyperbolic complex plane. It will be also true that = 1, cf. Theorem 6, (vii). Preliminary we show that σ is a hyperbolic complex plane, we prove that Indeed, ( w = = 3 δ ) 3 ( w = 3 w ) 3 where = w. ( w 3 δ 3 ) ( 3 δ 3 δ ) ( w 3 δ ) 3 = w (δ δ) 4 9 δ = w (( δ) w) 4 δ = w, (19) 9 δ δ = (u v) (u v) = (u u) [( v) ( v)] ( u ( v)) = (v u) w = ( δ) w. To prove the rest properties of the hyperbolic complex plane is trivial and therefore omitted. Lemma 3 The vectors w, σ are not collinear and also they are not orthogonal. 10

11 Proof. It is a trivial fact that vectors w, are not collinear. To show that they are also not orthogonal, the scalar product of these vectors is non-zero: < w, >= w, w 3 δ 3 = The following lemma is useful. ( 1 (1,0,0), 3, ) 3, = Lemma 4 If x = (X w,x u,x v ) T 3 and y = (Y w,y u,y v ) T 3, then (x y) = x y. Proof. Let us denote x = (X w,x u,x v ) = (X w, X v, X u ) and y = (Y w,y u,y v) = (Y w, Y v, Y u ), respectively. We have: x y = w(x wy w X uy v X vy u) u(x wy u+x uy w X vy v) v(x wy v+x uy u+x vy w) = w(x w Y w X v Y u X u Y v ) u( X w Y v X v Y w X u Y u ) = (x y). v( X w Y u +X v Y v X u Y w ) (0) 6 Numbers A and B Let us make the following regular linear transformation (shift) in the plane σ: where Then (w,δ) (w, ), = w 3 δ 3,δ = 3 w. x x = ( Xw +X u +X v X ) wx u +X u X v X w X v w 3(X wx u +X u X v X w X v ) = Aw B, (1) 11

12 where and A = (X w +X u +X v) X wx u +X u X v X w X v B = 3(X wx u +X u X v X w X v ). Since σ is a hyperbolic complex plane and w w = w, = w. Let us consider the equation which implies (Aw B ) (Aw B ) = Λ [(A w w) (B )] = Λ. Since w w = w and = w, this equality implies equivalently: A w B w = (A+B) (A B)w = Λ. 7 Expression A+B, zero divisors in T Lemma 5 Let x = (X w,x u,x v ) T. Then A+B = (X w +X u ) + (X u +X v ) + (X w X v ). Proof. A+B = (X w +X u +X v)+(x u X v +X u X w X w X v ) = (X w +X u ) + (X u +X v ) + (X w X v ). () Theorem 5 Let x = (X w,x u,x v ) T, X w,x u,x v R. The point Γ T is a (non-trivial) zero divisor if and only if there exists γ R,γ 0, such that Γ = (γ, γ,γ). Proof. We have: x = (X w,x u,x v ) and x = (X w, X v, X u ). Then it is trivial to verify that the point Γ = x = (γ, γ,γ) T is a zero divisor for every γ R, satisfies the definition of zero divisor for x = (γ, γ,γ) 1

13 T,X R. And vice versa, i.e., the element x is a zero divisor with Γ = x. Let us to prove the opposite direction, i.e., the uniqueness of the zero divisors of this form. The system of equations X w +X u = 0 X u +X v = 0 X w X v = 0 has solutions X w = γ X u = γ X v = γ, where γ R. This is a line, the diagonal of the fourth octant of the space E 3. The zero divisors appear for every γ R. The linear transform u = Xw +X u v = Xu +X v w = Xw X v is not regular, because = However, the quantity A+B = A+B saves the information about an element x T, A+B = u +v +w = A+B. If we suppose that (X w,x u,x v ) is not a zero divisor, i.e., if there does not exist γ R such that (X w,x u,x v ) = (γ, γ,γ), then u +v +w > 0. And vice-versa, if this inequality holds, then the element (X w,x u,x v ) is not a zero divisor. 13

14 Definition 4 Denote by D def = {Γ T Γ = (γ, γ,γ),γ R}. We will call the set D to be the set of zero divisors of T. Example 1 If (X w,x u,x v ) = (γ + ε, γ,γ), where γ,ε R,ε 0, then (u,v,w) = (ε,0,ε) and u +v +w = ε > 0. 8 Absolute value Definition 5 Let x = (X w,x u,x v ) T. An absolute value of the element x is a non-negative real number x : x [0, + ) such that where x = A+B, A = X w +X u +X v, B = X wx u +X u X v X v X w. The absolute value has following properties: Theorem 6 Let x = (X w,x u,x v ),y = (Y w,y u,y v ) T; γ R, Γ D. Then (i) Γ = 0; (ii) γx = γ x, (iii) x 0; (iv) x y = x y ; (v) Γ x = x ; (vi) w = u = v = 1; (vii) = 1; (viii) (0, γ,γ) = (γ, γ,0) = (γ,0,γ) = (γ,0,0) = (0, γ,0) = (0,0,γ) = γ 0; 14

15 (ix) x = x x ; (x) x+y x + y. Proof. (i) There exists an γ R such that Γ = (γ, γ, γ) is a zero divisor. Then Γ = 0 from Definition 5 the definition of the notion of the absolute value. (ii) This statement trivially follows from the definition and (). (iii) The value A+B is a sum of quadrants of real numbers, therefore x = A+B 0. (iv) By Lemma, Definition 4, Theorem 7(ii), x y w = (x y) (x y) = (x y) (x y ) Hence x y = x y ; = (x x ) (y y ) = ( x y )w. (v) x Γ = x is implied from Γ = (γ, γ,γ) and expression of A+B via quadrants. (vi) w = u = v = 1 can be evaluated directly from definition; (vii) we have = 1 3 [w u ( v)] = 1 3 [(1,0,0) (0,1,0) (0,0, 1)] = ( 1 3, 3, 3 (1 ) = [ + 3 ( ) + 3 = A+B ), ( ) ] [ ( ) ] =

16 (viii) By the item (v) and Definition 5; (ix) x = x x = A+B. (x) If x = y = Λ, then the triangle inequality holds trivially. If one of coordinates of x and y is non-zero, then apply Theorem 6(v). Without loss of generality, let it be the first coordinate. This way we may suppose, without loss of generality, that both x, y are of the following form: x = (0,X u,x v ), y = (0,Y u,y v ), respectively. By the definition, x = (Xw +Xu +Xv)+(X w X u +X u X v X v X w ) = (Xu ) ( ) Xv Xu +Xv +X u X v = + + Analogously, which implies ( )( Xu Xv ) = X u +X v. (3) y = Y u +Y v, x y = (X u +Y u )+(X v +Y v ), x+y x + y. 9 Expression A B The homogeneous system of linear equations X w X u = 0 X u X v = 0 X w +X v = 0 has only a trivial solution (X w,x u,x v ) = (0,0,0). 16

17 Lemma 6 The linear transform, U = Xw X u V = Xu X v W = Xw +X v inversely, X w = W+U+V X u = W U+V X v = W U V is regular and A B = U +V +W = A B = (X w X u ) + (X u X v ) + (X w +X v ). Proof. The system is regular, since: = 0 and A B = (X w +X u +X v)+(x w X v X u X v X u X w ) = (X w X u ) + (X u X v ) + (X w +X v ) = U +V +W = A B. Remark So, the expression A B is only a source of trivial zero divisor Λ of T but it will be practically useful when divide numerically, cf. the next section. 10 Division in T Definition 6 Let x = (X w,x u,x v ) T,x / D. We say that an element y T is an -inverse to x if x y = y x = w; we write this y = w x T x w. The proof of the following lemma is trivial and therefore omitted. 17

18 Lemma 7 Let x,y,z T, x / D, y / D,λ R,λ 0. Then (i) w w = w = w w T, (ii) z w T = z, (iii) z λx T = 1 λ z x T, (iv) x z x y T = z y T. Lemma 8 Let x = (X w,x u,x v ) T, x / D. An -inverse y to x is expressed as follows: y = x (Aw B ) A B, where A B R and A = (Xw +Xu +Xv) X wx u +X u X v X w X v, Proof. By Theorem 4, = w. For any arbitrary x T, x / D, we have: B = 3(X wx u +X u X v X w X v ). (4) y = x w = w x T = w x x (Aw B ) x x T = (Aw B ) (Aw B ) T x (Aw B ) = A (w w) AB(w ) AB(w ) B ( ) T = = x (Aw B ) (A B )w T = x (Aw B ) w x (Aw B ) A B = A B Indeed, x y = x x (Aw B ) A B = (x x ) (Aw B ) (x x ) (Aw B ) T = w. 18 = x (Aw B ) (A+B)(A B). (5) = (x x ) (Aw B ) A B

19 11 The factor ring T/D is a field In this section we will discuss in detail zero divisors of the ring T. Theorem 7 The set of zero divisors D is a (sub-) ring of the ring T. More precisely, if Γ,Θ D, then (i) Γ Θ D is an Abel group, (ii) Γ Θ D is a multiplicative semi-group, (iii) the operation is distributive with respect to the operation in D. Proof. The first item is trivial. Concerning the second item, the multiplications of zero divisors: if Γ = (γ, γ,γ) for some γ R and Θ = (θ, θ,θ), then Γ Θ = (γθ +γθ+γθ, γθ γθ γθ,γθ+γθ +γθ) = (3γθ, 3γθ,3γθ) D. Since D is a prime line, the rest item (iii) of the assertion is trivial. Lemma 9 The set D is an ideal in the ring T. Proof. Verify the properties of the ideal. (i) If Γ,Θ D, then Γ Θ = (γ, γ,γ) (θ, θ,θ) D for some γ,θ R. (ii) Let Γ = (γ, γ,γ) D where γ R. Let x = (X w,x u,x v ) T where X w,x u,x v R. Then Γ x D. Indeed, Γ x = (γ, γ,γ) (X w,x u,x v ) = (X w γ X u γ +X v γ, X w γ +X u γ X v γ,x w γ X u γ +X v γ) = (ε, ε,ε) D, where ε = X w γ X u γ +X v γ R. Thus, D is an ideal. 19

20 Lemma 10 Let Γ = (γ, γ,γ) D where γ R. Then Γ = (γ, γ,γ) and Γ Γ D. Proof. For the arbirtary x T, x = (X w,x u,x v ) = (X w, X v, X u ) and hence Γ = (γ, γ,γ) = (γ, γ,γ). Also, Γ Γ = (γ, γ,γ) (γ, γ,γ) = (3γ, 3γ,3γ) D. Lemma 11 Let x = (X w,x u,x v ) T where X w,x u,x v R. Then where x x = (A,B, B) = (A+B)w ( B,B, B) = (A+B)w Γ, A = X w +X u +X v,b = X wx u +X v X u X w X v, and Γ = ( B,B, B) D. Proof. A trivial verification by definition of the multiplication. We can factorize the set T with this set D. In the following lemma are written equivalence classes applied to the ideal D together with corresponding operations of representatives. These classes of equivalence form the factor ring T modulo D, also equivalently denoted as T/D. There exists a map p from T to a ring T/D defined via: p(x) def = x D. It is a natural homomorphism ring which is surjective. Its kernel is exactly the ideal D. Since D is an ideal of non-zero divisors, the algebra T/D over R is a (three-dimensional) factor field. Lemma 1 Let x, y T, y / D. Then (i) (x D) (y D) = (x y) D, (ii) (x D) (y D) = (x y) D, 0

21 (iii) (x D) (y D) = (x y) D, (iv) x D y D T = x y T D; (iv) A basis of the vector space T/D over R, which is a field, is a set of three classes of equivalences. {w D,u D,v D} Proof. The assertions are only a collection of the generally known facts about factor rings applied to the situation in the present paper. 1 Possible applications 1.1 Application to mathematical analysis Although this paper is technically mostly of algebraic nature, after building up an appropriate theory, the field T/D may serve for creating a theory generalizing the complex analysis theory to three dimensions. 1. Applications to theory of colour It is commonly known that we may satisfactorily-well decompose the white light into three colours (red, blue, green). In [4] we constructed a plane with 3 dependent vector axes which is isomorphic to the Gaussian complex plane having in mind this white light decomposition. The complex plane is used because the eye retina is flat. The theory in this paper 1 can be mentioned as an equally sided axonometric projection of our three dimensional ring T to a Gaussian complex plane. In this paper, if we would apply the mathematics to the theory of colours. Then the three basic colour decomposition can be dealt as generated from a point source as a spacial thee-dimensional case directly, not as a twodimensional projection of the three-dimensional light to a plane as it is done, e.g., in [4]. 1 This idea was explicitly generalized in the paper [5] where we mentioned functions instead of mumbers and a semiring instead of R which situation better models the biological human retina and vision. 1

22 References [1] Anderson, F. W. Fuller, K. R.: Rings and Categories of Modules, Graduate Texts in Mathematics, Vol. 13, nd Ed., Springer-Verlag, Berlin Heidelberg New York, 199. [] El-Bast, Z. A. Smith, P. F.: Multiplication modules, Comm. Algebra 16(1988), [3] Faith C.: Algebra: rings, modules and Categories, I., Springer Verlag, Berlin Heidelberg New York, [4] Gregor, T. Haluška, J.: Lexicographical ordering and field operations in the complex plane. Stud. Mat. 41(014), [5] Haluška, J.: On fields inspired with a polar HSV-RGB theory of colour.in: Hejduk, J. Kowalczyk, S. Pawlak, R. Turowska, M. M.: Modern Real Analysis, Wydawnictwo Universsytetu Lodzkiego, Lodz 015, pp ISBN [6] Harkin, A.A. Harkin, J. B.: Geometry of general complex numbers. Mathematics magazine, 77(004), [7] Nechaev, V. I.: Number systems (in Russian), Prosveshchenie, Moscow, [8] Rudin, W.: Functional analysis, second ed., Mc Graw-Hill, Hamburg 1973, 1991, pp. 44. [9] Werner, S.: Topological Fields, Mathematical Studies 157, Notas Mathematica (16), Ed.: Leopoldo Nachbin, Nord-Holland Amsterdam N.Y. Oxford Tokyo pp. 563.