Pseudo -MV algebras

Pseudo - MV algebras George Georgescu∗ and Afrodita Iorgulescu∗∗ ∗

Institute of Mathematics,

Calea Grivit¸ei 21, P.O. Box 1-764, Bucharest, Romania E-mail: [email protected] ∗∗

Department of Computer Science, Academy of Economic Studies,

Piat¸a Roman˘ a Nr.6 - R 70167, Oficiul Po¸stal 22, Bucharest, Romania E-mail: [email protected]

Dedicated to the memory of our dear Professor, Grigore C. Moisil Abstract Since MV algebras are categorically equivalent to abelian l-groups with strong unit, we have started from arbitrary l-groups and thus we have obtained the more general notion of pseudo-MV algebra. We followed closely [1] and [9]. Keywords MV-algebra, pseudo-MV algebra, l-group, normal ideal, Boolean center, Boolean sheaf

1

Pseudo-MV algebras. Definition and Properties.

We consider a structure A = (A, ⊕, − , ∼ , 0, 1), where ”⊕ ” is a binary operation, ”− ” and ”∼ ” are unary operations and 0, 1 are constants. We put by definition: def

y · x = (x− ⊕ y − )∼ and we consider that the operation ”·” has priority to the operation ”⊕”, i.e. we shall write simply ”x ⊕ y · z” instead of ”x ⊕ (y · z)”. Definition 1.1 A is a pseudo-MV algebra if the following axioms hold for any x, y, z ∈ A: (A1) x ⊕ (y ⊕ z) = (x ⊕ y) ⊕ z, (A2) x ⊕ 0 = 0 ⊕ x = x, (A3) x ⊕ 1 = 1 ⊕ x = 1, (A4) 1∼ = 0;

1− = 0,

(A5) (x− ⊕ y − )∼ = (x∼ ⊕ y ∼ )− , (A6) x ⊕ x∼ · y = y ⊕ y ∼ · x = x · y − ⊕ y = y · x− ⊕ x, (A7) x · (x− ⊕ y) = (x ⊕ y ∼ ) · y, (A8) (x− )∼ = x. We shall write x= instead of (x− )− and we shall write x≈ instead of (x∼ )∼ .

1

Definition 1.2 For any n ∈ IN, x ∈ A, we put: def def (i) 0x = 0 and (n + 1)x = nx ⊕ x = x ⊕ nx; def

def

(i’) x0 = 1 and xn+1 = xn · x = x · xn . Example 1.3 Let us consider an arbitrary l-group (G, +, −, 0) and let u ∈ G, u ≥ 0. We put by definition: def def def x ⊕ y = (x + y) ∧ u, x− = u − x, x∼ = −x + u. Then we have the following proposition. Proposition 1.4 (A = [0, u], ⊕, − , ∼ , 0, u) is a pseudo-MV algebra. Proof. Note that A is closed under these operations. Also we have: (a) x · y = (y − ⊕ x− )∼ = −[(u − y + u − x) ∧ u] + u = [(x − u + y − u) ∨ (−u)] + u = (x − u + y) ∨ 0 (b) x · y − = (x − u + u − y) ∨ 0 = (x − y) ∨ 0 = (x − y)+ (c) x∼ · y = (−x + u − u + y) ∨ 0 = (−x + y) ∨ 0 = (−x + y)+ . We shall verify now the above axioms (A1) - (A8): (A1) x ⊕ (y ⊕ z) = (x + [(y + z) ∧ u]) ∧ u = (x + y + z) ∧ (z + u) ∧ u = (x + y + z) ∧ u = . . . = (x ⊕ y) ⊕ z (A2) x ⊕ 0 = (x + 0) ∧ u = x (A3) x ⊕ u = (x + u) ∧ u = u (A4) u∼ = −u + u = 0, u− = u − u = 0 − − ∼ (A5) (x ⊕ y ) = (u − x + u − y)∼ = −(u − x + u − y) + u = y − u + x − u + u = y − u + x (x∼ ⊕ y ∼ )− = u − (−x + u − y + u) = u − u + y − u + x = y − u + x (A6) x ⊕ x∼ · y = (x + [(−x + y) ∨ 0]) ∧ u = [(x − x + y) ∨ x] ∧ u = (y ∨ x) ∧ u = x ∨ y x · y − ⊕ y = ([(x − y) ∨ 0] + y) ∧ u = [(x − y + y) ∨ y] ∧ u = x ∨ y (A7) Since x− ⊕ y = (u − x + y) ∧ u, x · (x− ⊕ y) = (x − u + [(u − x + y) ∧ u]) ∨ 0 = [(x − u + u − x + y) ∧ x] ∨ 0 = x ∧ y; On the other side, since x ⊕ y ∼ = (x − y + u) ∧ u, (x ⊕ y ∼ ) · y = ([(x − y + u) ∧ u] − u + y) ∨ 0 = [(x − y + u − u + y) ∧ (u − u + y)] ∨ 0 = x ∧ y (A8) (x− )∼ = −x− + u = −(u − x) + u = x − u + u = x. 2 Remark 1.5 In the pseudo-MV algebra [0, u] we have: x= = u + x − u,

x≈ = −u + x + u.

Open problem 1.6 Is any pseudo-MV algebra obtained like in Example 1.3? Let A be an arbitrary pseudo-MV algebra. Proposition 1.7 The following properties hold: (1) y · x = (x∼ ⊕ y ∼ )− , (2) (x∼ )− = x, (3) 0∼ = 0− = 1, (4) x · 1 = 1 · x = x, (5) x · 0 = 0 · x = 0, (6) x ⊕ x∼ = 1, (7) x · x− = 0,

x− ⊕ x = 1, x∼ · x = 0,

(8) (x ⊕ y)− = y − · x− ,

(x ⊕ y)∼ = y ∼ · x∼ , 2

(9) (x · y)− = y − ⊕ x− ,

(x · y)∼ = y ∼ ⊕ x∼ ,

(10) x ⊕ y = (y − · x− )∼ = (y ∼ · x∼ )− , (11) x∼ · y ⊕ y ∼ = y ∼ · x ⊕ x∼ , (12) x · (x− ⊕ y) = y · (y − ⊕ x), (13) x · (y · z) = (x · y) · z. Proof. (1) follows from (A5) and by definition of ”·”. (2)In (x∼ ⊕ y ∼ )− = (x− ⊕ y − )∼ we make x = 1. Then (1∼ ⊕ y ∼ )− = (1− ⊕ y − )∼ ⇒ (0 ⊕ y ∼ )− = (0 ⊕ y − )∼ ⇒ (y ∼ )− = (y − )∼ = y, by (A8). (3) 0∼ = (1− )∼ = 1, by (A4) and (A8) and 0− = (1∼ )− = 1, by (A4) and (2) (4) x · 1 = (x∼ ⊕ 1∼ )− = (x∼ ⊕ 0)− = (x∼ )− = x, by (1), (A4),(A2),(2); 1 · x = 1 analogously. (5) x · 0 = (x∼ ⊕ 0∼ )− = (x∼ ⊕ 1)− = 1− = 0, by (1), (3), (A3) and (A8). (6) In x ⊕ x∼ · y = y ⊕ y ∼ · x we make x = 1; then y ⊕ y ∼ = 1. In y ⊕ y ∼ = 1 we make y = x− ; then x− ⊕ x = x− ⊕ (x− )∼ = 1. (7) x · x− = ((x− )∼ ⊕ x∼ )− = (x ⊕ x∼ )− = 1− = 0. The same proof for the other. (8) y − · x− = ((x− )∼ ⊕ (y − )∼ )− = (x ⊕ y)− (9) (x · y)− = ((y − ⊕ x− )∼ )− = y − ⊕ x− (10) (y ∼ · x∼ )− = ((x ⊕ y)∼ )− = x ⊕ y (11) In x · y − ⊕ y = y · x− ⊕ x we make the sustitution of x by x∼ and of y by y ∼ ; then x∼ · y ⊕ y ∼ = x∼ · (y − )∼ ⊕ y ∼ = y ∼ · (x∼ )− ⊕ x∼ = y ∼ · x ⊕ x∼ (12) x · (x− ⊕ y) = [(x− ⊕ y)∼ ⊕ x∼ ]− = [y ∼ · x ⊕ x∼ ]− = (x∼ · y ⊕ y ∼ )− = y · (y − ⊕ x) (13) (x · y) · z = (z − ⊕ (x · y)− )∼ = (z − ⊕ y − ⊕ x− )∼ = . . . = x · (y · z). 2 We immediately get the following Corollary 1.8 If A = (A, ⊕, − , ∼ , 0, 1) is a pseudo-MV algebra, then Ad = (A, ·, ∼ , − , 1, 0) is also a pseudo-MV algebra, called ”the dual pseudo-MV algebra” of A. Proposition 1.9 The following properties are equivalent: (a) x− ⊕ y = 1, (b) y ∼ · x = 0, (c) y = x ⊕ x∼ · y, (d) x = x · (x− ⊕ y), (e) ∃a, such that y = x ⊕ a, (f ) x · y − = 0, (g) y ⊕ x∼ = 1. Proof. (a)⇒ (b): x− ⊕ y = 1 ⇒ y ∼ · x = (x− ⊕ y)∼ = 1∼ = 0 (b)⇒ (a):y ∼ · x = 0 ⇒ x− ⊕ y = (y ∼ · x)− = 0− = 1 (b)⇒ (c):y ∼ · x = 0 ⇒ y = y ⊕ y ∼ · x = x ⊕ x∼ · y, by (A6) (c)⇒ (e): obvious (e)⇒ (a): x− ⊕ y = x− ⊕ (x ⊕ a) = x− ⊕ x ⊕ a = 1 ⊕ a = 1 (d)⇒ (b): x = x · (x− ⊕ y) = y · (y − ⊕ x) ⇒ y ∼ · x = y ∼ · y · (y − ⊕ x) = 0 (a)⇒ (d): obvious (a)⇒ (f):x = x · 1 = x · (x− ⊕ y) = (x ⊕ y ∼ ) · y ⇒ x · y − = (x ⊕ y ∼ ) · y · y ∼ = 0 (f)⇒ (a): y = 0 ⊕ y = x · y − ⊕ y = x ⊕ x∼ · y, by (A6), hence x− ⊕ y = x− ⊕ x ⊕ x∼ · y = 1 ⊕ x∼ · y = 1 (f)⇒ (g): x · y − = 0 ⇒ (x · y − )∼ = 1 ⇒ (y − )∼ ⊕ x∼ = 1 ⇒ y ⊕ x∼ = 1 (g)⇒ (f): analogous. 2 We denote by ”x ≤ y” if and only if one of the above properties holds. Then we have the following 3

Proposition 1.10 ” ≤ ” is an order relation on A. Proof. • x ≤ x : x− ⊕ x = 1 • x ≤ y, y ≤ x ⇒ y = x ⊕ x∼ · y = y ⊕ y ∼ · x = x • x ≤ y, y ≤ z ⇒ there exists a, b such that y = x ⊕ a and z = y ⊕ b ; then z = x ⊕ (a ⊕ b) ⇒ x ≤ z. 2 Remark 1.11 0 ≤ x ≤ 1, for any x ∈ A, since 0− ⊕ x = 1 ⊕ x = 1 and x · 1− = x · 0 = 0. Proposition 1.12 The following properties hold: (a) x ≤ y ⇔ y − ≤ x− ⇔ y ∼ ≤ x∼ , (b) x ≤ y ⇒ a ⊕ x ≤ a ⊕ y, x ⊕ a ≤ y ⊕ a, (c) x ≤ y ⇒ a · x ≤ a · y, x · a ≤ y · a, (d) x · y ≤ z ⇔ y ≤ x− ⊕ z ⇔ x ≤ z ⊕ y ∼ , (e) x · y ≤ x, x · y ≤ y, x ≤ x ⊕ y, y ≤ x ⊕ y. Proof. (a) y − ≤ x− ⇔ (x− )∼ · y − = 0 ⇔ x · y − = 0 ⇔ x ≤ y y ≤ x∼ ⇔ (y ∼ )− ⊕ x∼ = 1 ⇔ y ⊕ x∼ = 1 ⇔ x ≤ y (b) x ≤ y ⇒ there exists z, such that y = x ⊕ z ⇒ a ⊕ y = (a ⊕ x) ⊕ z ⇒ a ⊕ x ≤ a ⊕ y x ≤ y ⇒ y ⊕ x∼ = 1 ⇒ y ⊕ a ⊕ (x ⊕ a)∼ = y ⊕ a ⊕ a∼ · x∼ = y ⊕ x∼ ⊕ x≈ · a = 1 ⊕ x≈ · a = 1 ⇒ x ⊕ a ≤ y ⊕ a (c) x ≤ y ⇒ y ∼ ≤ x∼ ⇒ y ∼ ⊕ a∼ ≤ x∼ ⊕ a∼ ⇒ a · x ≤ a · y; the other proof is analogous (d) x · y ≤ z ⇔ (x · y)− ⊕ z = 1 ⇔ y − ⊕ x− ⊕ z = 1 ⇔ y ≤ x− ⊕ z; x · y ≤ z ⇔ z ⊕ (x · y)∼ = 1 ⇔ z ⊕ y ∼ ⊕ x∼ = 1 ⇔ x ≤ z ⊕ y ∼ (e) follows from (b) and (c). 2 ∼

Proposition 1.13 (A, ≤) is a lattice in which the join x ∨ y and the meet x ∧ y of any two elements x and y are given by: (a) x ∨ y = x ⊕ x∼ · y = y ⊕ y ∼ · x = x · y − ⊕ y = y · x− ⊕ x, (b) x ∧ y = x · (x− ⊕ y) = y · (y − ⊕ x) = (x ⊕ y ∼ ) · y = (y ⊕ x∼ ) · x. Proof. (a) x− ⊕ x ⊕ x∼ · y = 1 ⇒ x ≤ x ⊕ x∼ · y and similarly y ≤ y ⊕ y ∼ · x. Let x, y ≤ z. We shall prove that y ⊕ y ∼ · x ≤ z. (y ⊕ y ∼ · x)− ⊕ z = (y ∼ · x)− · y − ⊕ z = (x− ⊕ y) · y − ⊕ y ⊕ z ∼ · y = y · (x− ⊕ y)− ⊕ x− ⊕ y ⊕ z ∼ · y = y · (x− ⊕ y)− ⊕ x− ⊕ z ⊕ y ∼ · x = 1. Then use (A6). (b) It is obvious that x · (x− ⊕ y) = (x ⊕ y ∼ ) · y ≤ x, y, by (A7) and Proposition 1.12(e). Let z ≤ x, y; then x− , y − ≤ z − , hence x− ∨ y − ≤ z − . It follows that z = (z − )∼ ≤ (x− ∨ y − )∼ = (x− ⊕ (x− )∼ · y − )∼ = (x− ⊕ x · y − )∼ = (x · y − )∼ · (x− )∼ = ((y − )∼ ⊕ x∼ ) · x = (y ⊕ x∼ ) · x = x · (x− ⊕ y). 2 Remark 1.14 For any x, y ∈ A we have: x · y ≤ x ∧ y ≤ x ∨ y ≤ x ⊕ y. We shall consider that the operations ”⊕” and ”·” have priority to the operations ”∨” and ”∧” (which have the same priority). Let I be an arbitrary set throughout this paper. Proposition 1.15 a ⊕ (∧i∈I bi ) = ∧i∈I (a ⊕ bi ),

(∧i∈I bi ) ⊕ a = ∧i∈I (bi ⊕ a),

whenever the arbitary meets exist. 4

Proof. Obviously, a ⊕ (∧i∈I bi ) ≤ a ⊕ bi , for every i ∈ I. Let now z ≤ a ⊕ bi , i ∈ I, i.e. z ≤ (a∼ )− ⊕ bi , ∼ i ∈ I; it follows that a∼ · z ≤ bi , i ∈ I, by Proposition 1.12(d) ¡V and ¢hence V a · z ≤ ∧i∈I bi ; it follows that ∼ − z ≤ (a ) ⊕ (∧i∈I bi ) = a ⊕ (∧i∈I bi ) . Therefore we get a ⊕ i∈I bi = i∈I (a ⊕ bi ). To prove the second equality, remark first that (∧i∈I bi )⊕a ≤ bi ⊕a, for every i ∈ I. Let now u ≤ bi ⊕a, for every i ∈ I; then u ≤ bi ⊕ (a− )∼ , i ∈ I and hence u · a− ≤ bi , i ∈ I, by Proposition 1.12(d); it follows that u·a− ≤ ∧i∈I bi and hence u ≤ (∧i∈I bi )⊕(a− )∼ = (∧i∈I bi )⊕a. Therefore we get (∧i∈I bi )⊕a = ∧i∈I (bi ⊕a). 2 Proposition 1.16 a · (∨i∈I bi ) = ∨i∈I (a · bi ),

(∨i∈I bi ) · a = ∨i∈I (bi · a),

whenever the arbitrary unions exist. Proof. Obviously, a · bi ≤ a · (∨i∈I bi ), for each i ∈ I. Let a · bi ≤ z, i ∈ I; then bi ≤ a− ⊕ z, i ∈ I, by Proposition 1.12(d) and hence ∨i∈I bi ≤ a− ⊕ z; it follows that a · (∨i∈I bi ) ≤ z. Therefore we get that a · (∨i∈I bi ) = ∨i∈I (a · bi ). To prove the second equality, remark first that bi · a ≤ (∨i∈I bi ) · a, i ∈ I. Let bi · a ≤ u, i ∈ I; then bi ≤ u ⊕ a∼ , i ∈ I, by Proposition 1.12(d) and hence ∨i∈I bi ≤ u ⊕ a∼ ; it follows that (∨i∈I bi ) · a ≤ u and therefore we get (∨i∈I bi ) · a = ∨i∈I (bi · a). 2 Proposition 1.17 For any g, h, k ∈ A we have: (1) g ∧ (h ⊕ k) ≤ (g ∧ h) ⊕ (g ∧ k), (2) g ∨ (h · k) ≥ (g ∨ h) · (g ∨ k). Proof. (1) (g ∧ h) ⊕ (g ∧ k) = [(g ∧ h) ⊕ g] ∧ [(g ∧ h) ⊕ k] = = (g ⊕ g) ∧ (h ⊕ g) ∧ (g ⊕ k) ∧ (h ⊕ k) ≥ g ∧ g ∧ g ∧ (h ⊕ k) = g ∧ (h ⊕ k), by Proposition 1.15. (2) (g ∨ h) · (g ∨ k) = [(g ∨ h) · g] ∨ [(g ∨ h) · k] = g · g ∨ h · g ∨ g · k ∨ h · k ≤ g ∨ g ∨ g ∨ h · k = g ∨ (h · k), by Proposition 1.16. 2 Proposition 1.18

(1) a ∧ (∨i∈I bi ) = ∨i∈I (a ∧ bi ),

(2) a ∨ (∧i∈I bi ) = ∧i∈I (a ∨ bi ), whenever the arbitrary unions and meets exist. Proof. (1) a ∧ (∨i∈I bi ) = (∨i∈I bi ) · ((∨j∈I bj )− ⊕ a) = ∨i∈I [bi · ((∨j∈I bj )− ⊕ a)], by Propositions 1.13(b) and 1.16. But, for any i ∈ I : bi ≤ ∨j∈I bj ⇒ (∨j∈I bj )− ≤ (bi )− ⇒ (∨j∈I bj )− ⊕ a ≤ (bi )− ⊕ a − ⇒ bi · ((∨j∈I bj )− ⊕ a) ≤ bi · (b− i ⊕ a) = a ∧ bi ⇒ ∨i∈I [bi · ((∨j∈I bj ) ⊕ a)] ≤ ∨i∈I (a ∧ bi ). Hence a ∧ (∨i∈I bi ) ≤ ∨i∈I (a ∧ bi ). The other inequality is obvious. (2) a ∨ (∧i∈I bi ) = (∧i∈I bi ) ⊕ (∧j∈I bj )∼ · a = ∧i∈I [bi ⊕ (∧j∈I bj )∼ · a], by Propositions 1.13 (a) and 1.15. ∼ ∼ ∼ But, for any i ∈ I, ∧j∈I bj ≤ bi ⇒ (∧j∈I bj )∼ ≥ b∼ i ⇒ ((∧j∈I bj ) · a ≥ bi · a ⇒ bi ⊕ (∧j∈I bj ) · a ≥ ∼ ∼ bi ⊕ bi · a = a ∨ bi ⇒ ∧i∈I [bi ⊕ (∧j∈I bj ) · a] ≥ ∧i∈I (a ∨ bi ). Hence, a ∨ (∧i∈I bi ) ≥ ∧i∈I (a ∨ bi ). The other inequality is obvious. 2 Corollary 1.19 For every pseudo-MV algebra A = (A, ⊕, − , ∼ , 0, 1), the structure L(A) = (A, ∨, ∧, 0, 1) is a bounded distributive lattice. Proof. By Propositions 1.13 and 1.18 and Remark 1.11. We get immediately that

2

Corollary 1.20 The structure L(A)d = (A, ∧, ∨, 1, 0) is also a bounded distributive lattice, called ”the dual” of the lattice L(A). Proposition 1.21 a ⊕ (∨i∈I bi ) = ∨i∈I (a ⊕ bi ),

(∨i∈I bi ) ⊕ a = ∨i∈I (bi ⊕ a),

whenever the arbitrary unions exist. 5

Proof. Obviously, a ⊕ bi ≤ a ⊕ (∨i∈I bi ), for any i ∈ I. Let a ⊕ bi ≤ u, i ∈ I; remark that a ≤ u. Then for any i ∈ I: a∼ · (a ⊕ bi ) ≤ a∼ · u. But a∼ · (a ⊕ bi ) = a∼ · ((a∼ )− ⊕ bi ) = a∼ ∧ bi , by Proposition 1.13(b). Hence a∼ ∧ bi ≤ a∼ · u. It follows that a∼ ∧ (∨i∈I bWi ) = ∨i∈I (a∼ ∧ bi ) ≤ a∼ · u, by Proposition 1.18(1). Hence, a ⊕ [a∼ ∧ ( i∈I bi )] ≤ a ⊕ a∼ · u = a ∨ u = u, by Proposition 1.13(a) and since a ≤ u; but a ⊕ [a∼ ∧ (∨i∈I bi )] = (a ⊕ a∼ ) ∧ (a ⊕ (∨i∈I bi )) = a ⊕ (∨i∈I bi ), by Proposition 1.15; hence a ⊕ (∨i∈I bi ) ≤ u. Therefore we get that a ⊕ (∨i∈I bi ) = ∨i∈I (a ⊕ bi ). The second equality has a similar proof. 2 Proposition 1.22 a · (∧i∈I bi ) = ∧i∈I (a · bi ),

(∧i∈I bi ) · a = ∧i∈I (bi · a),

whenever the arbitrary meets exist. Proof. Obviously, a · (∧i∈I bi ) ≤ a · bi , for any i ∈ I. Let z ≤ a · bi , i ∈ I; remark that z ≤ a. For any i ∈ I: a− ⊕ a · bi ≥ a− ⊕ z. But a− ⊕ a · bi = a− ⊕ [(a− )∼ · bi ] = a− ∨ bi , i ∈ I, by Proposition 1.13(a). Hence a− ∨ bi ≥ a− ⊕ z, i ∈ I. It follows that a− ∨ (∧i∈I bi ) = ∧i∈I (a− ∨ bi ) ≥ a− ⊕ z, by Proposition 1.18(2). Hence a · [a− ∨ (∧i∈I bi )] ≥ a · (a− ⊕ z) = a ∧ z = z, by Proposition 1.13(b) and since z ≤ a; but a · [a− ∨ (∧i∈I bi )] = a · a− ∨ a · (∧i∈I bi ) = a · (∧i∈I bi ), by Proposition 1.16; hence a · (∧i∈I bi ) ≥ z. Therefore we get that a · (∧i∈I bi ) = ∧i∈I (a · bi ). The second equality has a similar proof. 2 Proposition 1.23

(i) (x ∧ y)− = x− ∨ y − ,

(i’) (x ∧ y)∼ = x∼ ∨ y ∼ ,

(x ∨ y)− = x− ∧ y − ,

(x ∨ y)∼ = x∼ ∧ y ∼ .

Proof. (i) (x ∧ y)− = ((x ⊕ y ∼ ) · y)− = y − ⊕ (x ⊕ y ∼ )− = y − ⊕ y · x− = y − ⊕ (y − )∼ · x− = x− ∨ y − , by Proposition 1.13. Then (x∨y)− = (x⊕x∼ ·y)− = (x∼ ·y)− ·x− = (y − ⊕x)·x− = (y − ⊕(x− )∼ )·x− = x− ∧y − , by Proposition 1.13. Thus (i) holds. (i’) (x ∧ y)∼ = [x · (x− ⊕ y)]∼ = (x− ⊕ y)∼ ⊕ x∼ = y ∼ · x ⊕ x∼ = y ∼ · (x∼ )− ⊕ x∼ = x∼ ∨ y ∼ . The second part of (i’) has a similar proof. 2 Proposition 1.24

(i) x · y − ∧ y · x− = 0,

(ii) y ⊕ x∼ ∨ x ⊕ y ∼ = 1,

x∼ · y ∧ y ∼ · x = 0.

y − ⊕ x ∨ x− ⊕ y = 1.

Proof. (i) x · y − = 0 ∨ x · y − = x · x− ∨ x · y − = x · (x− ∨ y − ). Analogously y · x− = y · (x− ∨ y − ). Then x · y − ∧ y · x− = [x · (x− ∨ y − )] ∧ [y · (x− ∨ y − )] = (x ∧ y) · (x− ∨ y − ) = (x ∧ y) · (x ∧ y)− = 0. The second equality is obtained by replacing x by x∼ and y by y ∼ in the first one. (ii) follows from (i) by applying ∼ and − , respectively. 2 Proposition 1.25

(i) x∼ · y − ∧ y · x = 0,

(ii) x ⊕ y = x ⊕ y · x ⊕ y, (iii) x = (x ⊕ y) · y − ⊕ y · x,

y = y · x ⊕ x∼ · (x ⊕ y).

Proof. • (i) By replacing x by x∼ in Proposition 1.24 (i). • (ii) Since x∼ · y − ∧ y − ∧ y · x ≤ x∼ · y − ∧ y · x, it follows that x∼ · y − ∧ (y · x ⊕ y) · y − = x∼ · y − ∧ (y · x ⊕ (y − )∼ ) · y − = x∼ · y − ∧ y − ∧ y · x = 0. Then, by Proposition 1.15, we get [x∼ ∧ (y · x ⊕ y)] · y − = 0, and therefore, by Proposition 1.9, we obtain: x∼ · (x ⊕ y · x ⊕ y · x) = x∼ · ((x∼ )− ⊕ y · x ⊕ y) = x∼ ∧ (y · x ⊕ y) ≤ y. By Proposition 1.12 (d), x ⊕ y · x ⊕ y ≤ (x∼ )− ⊕ y = x ⊕ y. The other inequality is obvious. • (iii) We prove by two equalities. Since x ∧ y − = (x ⊕ (y − )∼ ) · y − = (x ⊕ y) · y − and y − ∨ x = y − ⊕ (y − )∼ · x = y − ⊕ y · x, we get: (x ⊕ y) · y − ⊕ y · x = (x ∧ y − ) ⊕ y · x = (x ⊕ y · x) ∧ (y − ⊕ y · x) = 6

(x ⊕ y · x) ∧ (y − ∨ x) = [(x ⊕ y · x) ∨ y − ] ∨ [(x ⊕ y · x) ∧ x] = [(x ⊕ y · x) ∨ y − ] ∨ x. From x ⊕ y · x ⊕ y ≤ x ⊕ y = x ⊕ (y − )∼ it follows, by Proposition 1.12, (x ⊕ y · x) ∨ y − = (x ⊕ y · x ⊕ (y − )∼ ) · y − = (x ⊕ y · x ⊕ y) · y − ≤ y, hence (x ⊕ y) · y − ⊕ y · x = x. We prove now the second equality. From x ⊕ y · x ⊕ y ≤ x ⊕ y = (x∼ )− ⊕ y it follows x∼ ∧ (y · x ⊕ y) = x∼ · ((x∼ )− ⊕ y · x ⊕ y) = x∼ · (x ⊕ y · x ⊕ y) ≤ y. Consequently, y = [x∼ ∧ (y · x ⊕ y)] ∨ y = [x∼ ∧ (y · x ⊕ y)] ∨ [y ∧ (y · x ⊕ y)] = (x∼ ∨ y) ∧ (y · x ⊕ y) = (y · x ⊕ x∼ ) ∧ (y · x ⊕ y) = y · x ⊕ (x∼ ∧ y) = y · x ⊕ x∼ · (x ⊕ y). Proposition 1.26

2

(i) x ∨ y = x · (x ∧ y)− ⊕ y,

(ii) x ∧ y = 0 ⇒ x ⊕ y = x ∨ y, (iii) x ∧ y = 0 ⇒ x ∧ (y ⊕ z) = x ∧ z. Proof. (i) x · (x ∧ y)− ⊕ y = x · (x− ∨ y − ) ⊕ y = (x · x− ∨ x · y − ) ⊕ y = x · y − ⊕ y = x ∨ y. (ii) follows by (i) (iii) Remark that x ≤ x ∧ (x ⊕ z) = [(x ∧ y) ⊕ x] ∧ (x ⊕ z) = (x ⊕ a) ∧ (y ⊕ x) ∧ (x ⊕ z). Then x ∧ z ≤ x ∧ (y ⊕ z) ≤ (x ⊕ x) ∧ (y ⊕ x) ∧ (x ⊕ z) ∧ (y ⊕ z) = (x ∧ y) ⊕ (x ∧ z) = x ∧ z. Proposition 1.27 (x ⊕ z) · y ≤ x ⊕ z · y,

2

y · (x ⊕ z) ≤ y · x ⊕ z.

Proof. • [(x ⊕ z) · y]− ⊕ x ⊕ z · y = y − ⊕ z − · x− ⊕ x ⊕ z · y = y − ⊕ (x ∨ z − ) ⊕ z · y = (y − ⊕ x ⊕ z · y) ∨ (y − ⊕ z − ⊕ z · y) = (y − ⊕ x ⊕ z · y) ∨ (y − ⊕ z − ⊕ (z − )∼ · y) = (y − ⊕ x ⊕ z · y) ∨ (y − ⊕ (z − ∨ y)) = (y − ⊕ x ⊕ z · y) ∨ (y − ⊕ z − ) ∨ (y − ⊕ y) = 1. • y · x ⊕ z ⊕ (y · (x ⊕ z))∼ = y · x ⊕ z ⊕ (x ⊕ z)∼ ⊕ y ∼ = y · x ⊕ z ⊕ z ∼ · x∼ ⊕ y ∼ = y · x ⊕ (z ∨ x∼ ) ⊕ y ∼ = (y · x ⊕ z ⊕ y ∼ ) ∨ (y · x ⊕ x∼ ⊕ y ∼ ). But y · x ⊕ x∼ ⊕ y ∼ = y · (x− )∼ ⊕ x∼ ⊕ y ∼ = (x∼ ∨ y) ⊕ y ∼ = (x∼ ⊕ y ∼ ) ∨ (y ⊕ y ∼ ) = 1. Then, by Proposition 1.9 (g), we get the second inequality. 2 Proposition 1.28

(a) y ⊕ x = z ⊕ x and x · y = x · z ⇒ y = z,

(b) x ⊕ y = x ⊕ z and y · x = z · x ⇒ y = z, (c) x ⊕ y = y ⇔ x− ⊕ y − = x− , (d) x ⊕ y = x ⇔ x∼ ⊕ y ∼ = y ∼ . Proof. (a) x− ∨ y = x− ⊕ (x− )∼ · y = x− ⊕ x · y = x− ⊕ x · z = . . . = x− ∨ z , x ∧ y = (y ⊕ (x− )∼ ) · x− = (y ⊕ x) · x− = (z ⊕ x) · x− = . . . = x− ∧ z. Since (A, ∨, ∧) is a distributive lattice, it follows that y = z. (b) analogous. (c) x ⊕ y = y ⇒ (x ⊕ (y − )∼ ) · y − = y · y − = 0 ⇒ x ∧ y − = 0 ⇒ x− = x− ⊕ (x ∧ y − ) = (x− ⊕ x) ∧ (x− ⊕ y − ) = x− ⊕ y − . x− ⊕ y − = x− ⇒ x ∧ y − = x · (x− ⊕ y − ) = x · x− = 0 ⇒ y = (x ∧ y − ) ⊕ y = (x ⊕ y) ∧ (y − ⊕ y) = x ⊕ y. (d) x ⊕ y = x ⇒ x∼ ∧ y = x∼ · ((x∼ )− ⊕ y) = x∼ · (x ⊕ y) = x∼ · x = 0 ⇒ y ∼ = (x∼ ∧ y) ⊕ y ∼ = (x∼ ⊕ y ∼ ) ∧ (y ⊕ y ∼ ) = x∼ ⊕ y ∼ . x∼ ⊕ y ∼ = y ∼ ⇒ x∼ ∧ y = (x∼ ⊕ y ∼ ) · y = y ∼ · y = 0 ⇒ x ⊕ (x∼ ∧ y) = x ⇒ (x ⊕ x∼ ) ∧ (x ⊕ y) = x ⇒ x ⊕ y = x. 2 −

Proposition 1.29 Every commutative pseudo-MV algebra A (i.e. ⊕ is commutative) is an MV algebra.

7

Proof. Since x ⊕ y = y ⊕ x for any x, y ∈ A, it follows that x · y = (y − ⊕ x− )∼ = (x− ⊕ y − )∼ = y · x. Hence x− ⊕ x = 1 = x∼ ⊕ x and x · x− = 0 = x · x∼ , by Proposition 1.7(6), (7). Then by Proposition 1.28(a) we get that x− = x∼ , for every x ∈ A. Then (A, ⊕, ·, − , 0, 1) is an MV algebra. 2 Proposition 1.30 Let A be a totally ordered pseudo-MV algebra (a pseudo-MV chain). Then the following properties hold: (i) x ⊕ y < 1 ⇒ y · x = 0, (ii) y ⊕ x = z ⊕ x < 1 ⇒ y = z, (iii) x · y = x · z > 0 ⇒ y = z, (iv) y ⊕ x = x ⇔ x = 1 or y = 1. Proof. (i) x ⊕ y < 1 ⇒ x∼ 6≤ y ⇒ y < x∼ ⇒ y · x = y · (x− )∼ = 0. (ii) y ⊕ x = z ⊕ x < 1 ⇒ x · y = 0 = x · z, by (i). Then, by Proposition 1.28 (a), y = z. (iii) follows immediately by (ii). (iv) y ⊕ x = 0 ⊕ x < 1 ⇒ y = 0, y ⊕ x = x = 1 ⇒ x = 1.

2

Proposition 1.31 In a pseudo-MV algebra the following properties hold: (1) (x ⊕ y)≈ = x≈ ⊕ y ≈ ;

(x · y)≈ = x≈ · y ≈ ,

(2) (x ⊕ y)= = x= ⊕ y = ;

(x · y)= = x= · y = ,

(3) (x= )≈ = x = (x≈ )= , (4) (x− )≈ = x∼ ;

(x∼ )= = x− ,

(5) 0= = 0 = 0≈ , (6) 1= = 1 = 1≈ , (7) x= ⊕ x− = 1 = x∼ ⊕ x≈ , (8) ((x− )− )− ⊕ x− = 1 = x∼ ⊕ ((x∼ )∼ )∼ . Let us put by definition: def def (i) x≈(0) = x and x≈(m+1) = (x≈(m) )≈ , def def (ii) x=(0) = x and x=(m+1) = (x=(m) )= . Then we have: (9) (nx)≈(m) = n(x≈(m) ), (10) (nx)=(m) = n(x=(m) ), (11) (x ∧ y)≈ = x≈ ∧ y ≈ ,

(x ∨ y)≈ = x≈ ∨ y ≈ ,

(12) (x ∧ y)= = x= ∧ y = ,

(x ∨ y)= = x= ∨ y = .

Proof. Routine.

2

Lemma 1.32 Let x, y ∈ A. If x ∧ y = 0, then for each integer n ≥ 1, m ≥ 1: (1) nx ∧ ny = 0, (2) n(x≈ ) ∧ n(y ≈ ) = 0, (3) n(x≈(m) ) ∧ n(y ≈(m) ) = 0, (4) n(x= ) ∧ n(y = ) = 0, (5) n(x=(m) ) ∧ n(y =(m) ) = 0. 8

Proof. (1) If x ∧ y = 0, then x = x ⊕ 0 = x ⊕ (x ∧ y) = (x ⊕ x) ∧ (x ⊕ y) = (2x) ∧ (x ⊕ y) ≥ 2x ∧ y, since x ⊕ y ≥ y. Hence, 2x ∧ y ≤ x. Then (2x ∧ y) ∧ y ≤ x ∧ y = 0, i.e. 2x ∧ y ≤ 0 which implies 2x ∧ y = 0. Similarly, y = 0 ⊕ y = (2x ∧ y) ⊕ y = (2x ⊕ y) ∧ 2y ≥ 2x ∧ 2y, since 2x ⊕ y ≥ 2x. Thus 2x ∧ 2y ≤ y; hence 2x ∧ (2x ∧ 2y) ≤ 2x ∧ y = 0, i.e. 2x ∧ 2y = 0. It follows that 0 = x ∧ y = 2x ∧ 2y = 2(2x) ∧ 2(2y) = 2(2(2x)) ∧ 2(2(2y)) = . . ., i.e. 2n x ∧ 2n y = 0, for each integer n ≥ 1. Since n ≤ 2n , it follows that nx ∧ ny ≤ 2n x ∧ 2n y, which implies nx ∧ ny = 0. (2) By (1), nx ∧ ny = 0 ⇔ (nx ∧ ny)≈ = 0≈ ⇔ n(x≈ ) ∧ n(y ≈ ) = 0. (3) nx ∧ ny = 0 ⇔ (nx ∧ ny)≈ = 0≈ ⇔ . . . ⇔ (nx ∧ ny)≈(m) = 0≈(m) ⇔ (nx)≈(m) ∧ (ny)≈(m) = 0 ⇔ n(x≈(m) ) ∧ n(y ≈(m) ) = 0. (4) has a proof similar to (2) (5) has a proof similar to (3). 2 Lemma 1.33 For every x, y, z ∈ A we have: (i) x · y − ⊕ y · z − ≥ x · z − and (i’) x∼ · y ⊕ y ∼ · z ≥ x∼ · z. Proof. (i) x · y − ⊕ y · z − ≥ (x · y − ⊕ y) · z − = (x ∨ y) · z − = x · z − ∨ y · z − ≥ x · z − , by Propositions 1.27 and 1.16. (i’) x∼ · y ⊕ y ∼ · z ≥ (x∼ · y ⊕ y ∼ ) · z = (x∼ · (y ∼ )− ⊕ y ∼ ) · z = (x∼ ∨ y ∼ ) · z = x∼ · z ∨ y ∼ · z ≥ x∼ · z, by Propositions 1.27 and 1.16. 2 Definition 1.34 If A = (A, ⊕, − , ∼ , 0, 1) is a pseudo-MV algebra, then two distance functions d1 , d2 : A × A −→ A are defined by: def

def

d1 (x, y) = x · y − ⊕ y · x− ,

d2 (x, y) = x∼ · y ⊕ y ∼ · x.

Proposition 1.35 The two distances verify the following properties: (0) d1 (x, y) = x · y − ∨ y · x− , (1) d1 (x, y) = 0 ⇔ x = y,

d2 (x, y) = x∼ · y ∨ y ∼ · x d2 (x, y) = 0 ⇔ x = y

(2) d1 (x, 0) = x = d2 (x, 0) (3) d1 (0, y) = y = d2 (0, y) (4) d1 (x, 1) = x− ,

d2 (x, 1) = x∼

(5) d1 (1, y) = y − ,

d2 (1, y) = y ∼

(6) d1 (x∼ , y ∼ ) = d2 (x, y),

d2 (x∼ , y ∼ ) = (d1 (x, y))≈

(7) d1 (x− , y − ) = (d2 (x, y))= , (8) d1 (x, y) = d1 (y, x),

d2 (x− , y − ) = d1 (x, y)

d2 (x, y) = d2 (y, x)

(9) d1 (x, z) ≤ d1 (x, y) ⊕ d1 (y, z) ⊕ d1 (x, y) (9’) d1 (x, z) ≤ d1 (y, z) ⊕ d1 (x, y) ⊕ d1 (y, z) (10) d2 (x, z) ≤ d2 (x, y) ⊕ d2 (y, z) ⊕ d2 (x, y) (10’) d2 (x, z) ≤ d2 (y, z) ⊕ d2 (x, y) ⊕ d2 (y, z). Proof. (0) By Propositions 1.24 and 1.26. (1) d1 (x, x) = x · x− ⊕ x · x− = 0 ⊕ 0 = 0; d1 (x, y) = 0 ⇔ x · y − ∨ y · x− = 0 ⇒ x · y − = 0 = y · x− ⇔ x ≤ y and y ≤ x ⇒ x = y, by Proposition 1.9. The second part of (1) has a similar proof. (2) d1 (x, 0) = x · 0− ⊕ 0 · x− = x · 1 ⊕ 0 = x and similarly d2 (x, 0) = x. 9

(3) d1 (0, y) = 0 · y − ⊕ y · 0− = 0 ⊕ y · 1 = y and similarly the rest. (4) d1 (x, 1) = x · 1− ⊕ 1 · x− = x · 0 ⊕ x− = x− and similarly the other. (5)d1 (1, y) = 1 · y − ⊕ y · 1− = y − ⊕ y · 0 = y − and similarly the other. (6) d1 (x∼ , y ∼ ) = x∼ · (y ∼ )− ⊕ y ∼ · (x∼ )− = x∼ · y ⊕ y ∼ · x = d2 (x, y), d2 (x∼ , y ∼ ) = x≈ ·y ∼ ⊕y ≈ ·x∼ = x≈ ·(y − )≈ ⊕y ≈ ·(x− )≈ = (x·y − )≈ ⊕(y·x− )≈ = (x·y − ⊕y·x− )≈ = (d1 (x, y))≈ . (7) d1 (x− , y − ) = x− · y = ⊕ y − · x= = (x∼ )= · y = ⊕ (y ∼ )= · x= = (x∼ · y)= ⊕ (y ∼ · x)= = (x∼ · y ⊕ y ∼ · x)= = (d2 (x, y))= . (8) follows by (0) and by commutativity of ∨. (9) d1 (x, y) ⊕ d1 (y, z) ⊕ d1 (x, y) = [(x · y − ∨ y · x− ) ⊕ (y · z − ∨ z · y − )] ⊕ (x · y − ∨ y · x− ) = [(x · y − ⊕ y · z − ) ∨ (y · x− ⊕ y · z − ) ∨ (x · y − ⊕ z · y − ) ∨ (y · x− ⊕ z · y − )] ⊕ (x · y − ∨ y · x− ) ≥ [[x · z − ∨ (y · x− ⊕ y · z − ) ∨ (x · y − ⊕ z · y − ) ∨ (y · x− ⊕ z · y − )] ⊕ x · y − ] ∨ ∨ [[x · z − ∨ (y · x− ⊕ y · z − ) ∨ (x · y − ⊕ z · y − ) ∨ (y · x− ⊕ z · y − )] ⊕ y · x− ] = = (x · z − ⊕ x · y − ) ∨ (y · x− ⊕ y · z − ⊕ x · y − ) ∨ (x · y − ⊕ z · y − ⊕ x · y − ) ∨ (y · x− ⊕ z · y − ⊕ x · y − ) ∨ ∨ (x · z − ) ∨ (y · x− ⊕ y · z − ⊕ y · x− ) ∨ (x · y − ⊕ z · y − ⊕ y · x− ) ∨ (y · x− ⊕ z · y − ⊕ y · x− ) ≥ ≥ (x · z − ⊕ x · y − ) ∨ (x · z − ⊕ y · x− ) ∨ ∨ (x · y − ⊕ z · y − ⊕ y · x− ) ∨ (y · x− ⊕ z · y − ⊕ y · x− ) ≥ ≥ [x · z − ⊕ (x · y − ∨ y · x− )] ∨ (x · y − ⊕ z · x− ) ∨ (y · x− ⊕ z · x− ) = = [x · z − ⊕ d1 (x, y)] ∨ [(x · y − ∨ y · x− ) ⊕ z · x− ] = = [x · z − ⊕ d1 (x, y)] ∨ [d1 (x, y) ⊕ z · x− ] ≥ ≥ x · z − ∨ z · x− = d1 (x, z), by Lemma 1.33. (9’) d1 (y, z) ⊕ d1 (x, y) ⊕ d1 (y, z) = = [(y · z − ∨ z · y − ) ⊕ (x · y − ∨ y · x− )] ⊕ (y · z − ∨ z · y − ) = ((y · z − ⊕ x · y − ) ∨ (z · y − ⊕ x · y − ) ∨ (y · z − ⊕ y · x− ) ∨ (z · y − ⊕ y · x− )) ⊕ (y · z − ∨ z · y − ) = = (y · z − ⊕ x · y − ⊕ y · z − ) ∨ (z · y − ⊕ x · y − ⊕ y · z − ) ∨ (y · z − ⊕ y · x− ⊕ y · z − ) ∨ (z · y − ⊕ y · x− ⊕ y · z − ) ∨ ∨ (y · z − ⊕ x · y − ⊕ z · y − ) ∨ (z · y − ⊕ x · y − ⊕ z · y − ) ∨ (y · z − ⊕ y · x− ⊕ z · y − ) ∨ (z · y − ⊕ y · x− ⊕ z · y − ) ≥ ≥ [(y · z − ⊕ x · z − ) ∨ (z · y − ⊕ x · z − )] ∨ [(z · x− ⊕ y · z − ) ∨ (z · x− ⊕ z · y − )] = = [(y · z − ∨ z · y − ) ⊕ x · z − ] ∨ [z · x− ⊕ (y · z − ⊕ z · y − )] = = [d1 (y, z) ⊕ x · z − ] ∨ [z · x− ⊕ d1 (y, z)] ≥ ≥ x · z − ∨ z · x− = d1 (x, z), by Lemma 1.33. (10) Similar to (9). (10’) Similar to (9’). 2

2

Homomorphisms and ideals

Definition 2.1 Let A and B be pseudo-MV algebras. A function h : A −→ B is a homomorphism iff it satisfies the following conditions, for each x, y ∈ A: (H1) h(0) = 0 (H2) h(x ⊕ y) = h(x) ⊕ h(y) (H3) h(x− ) = (h(x))− (H4) h(x∼ ) = (h(x))∼ . Remark 2.2 It follows that we also have : (H1’) h(1) = 1, since 1 = x ⊕ x∼ in A implies h(1) = h(x ⊕ x∼ ) = h(x) ⊕ h(x∼ ) = h(x) ⊕ (h(x))∼ = 1 in B.

10

(H2’) h(x · y) = h(x) · h(y), since h(x · y) = h((y ∼ ⊕ x∼ )− ) = (h(y ∼ ⊕ x∼ ))− = (h(y ∼ ) ⊕ h(x∼ ))− = ((h(y))∼ ⊕ (h(x)∼ )− = ((h(x) · h(y))∼ )− = h(x) · h(y). (H5) h(x ∨ y) = h(x) ∨ h(y), h(x ∧ y) = h(x) ∧ h(y), i.e. h is a lattice homomorphism, by Proposition 1.13. Following current usage , if h is one-one we shall equivalently say that h is an injective homomorphism, or an embedding. If the homomorphism h : A −→ B is onto B we say that h is surjective. By an isomorphism we shall mean a surjective one-one homomorphism. The kernel of a homomorphism h : A −→ B is the set def Ker(h) = h−1 (0) = {x ∈ A | h(x) = 0}. Definition 2.3 An ideal of a pseudo-MV algebra A is a subset H of A satisfying the following conditions: (I1) 0 ∈ H (I2) If x, y ∈ H then x ⊕ y ∈ H and y ⊕ x ∈ H (I3) If x ∈ H, y ∈ A and y ≤ x, then y ∈ H. Remarks 2.4 (1) We also have in an ideal H the following property: (I2’) If x, y ∈ H, then x · y, y · x ∈ H, since x · y, y · x ≤ x, y. (2) If H is an ideal of the pseudo-MV algebra A, then H is an ideal of the lattice (A, ∨, ∧), since x ∨ y ≤ x ⊕ y. We denote by I(A) the set of ideals of A. The intersection of any family of ideals of A is still an ideal of A. For every subset W ⊆ A, the smallest ideal of A which contains W , i.e. the intersection of all ideals D ⊇ W , is said to be the ideal generated by W, and will be denoted (W ]. The proof of the next lemma is immediate, and will be omitted. Lemma 2.5 Let W be a subset of a pseudo-MV algebra A. If W = ∅, then (W ] = {0}. If W 6= ∅, then (W ] = {x ∈ A | x ≤ w1 ⊕ w2 ⊕ . . . ⊕ wn ,

for some w1 , w2 , . . . , wn ∈ W }

In particular, for each element z of a pseudo-MV algebra A, the ideal (z] = ({z}] is called the principal ideal generated by z, and we have (z] = {x ∈ A | x ≤ nz

for some n ≥ 1}.

(1)

Note that (0] = {0} and (1] = A. Further, for every ideal J of a pseudo-MV algebra A and each z ∈ A we have (J ∪ {z}] = {x ∈ A | x ≤ ⊕m i=1 (ai ⊕ ni z), for some ai ∈ J and for some integers m ≥ 1, ni ≥ 0}, i.e. x ≤ (a1 ⊕ n1 z) ⊕ (a2 ⊕ n2 z) ⊕ . . . ⊕ (am ⊕ nm z). Proposition 2.6 If g, k ∈ A, then (g ∧ k] = (g] ∩ (k].

11

(2)

Proof. Obviously, g ∈ (g] and k ∈ (k]. Since g ∧ k ≤ g, k we get g ∧ k ∈ (g], g ∧ k ∈ (k]; then g ∧ k ∈ (g] ∩ (k], which is an ideal. Then (g ∧ k] ⊆ (g] ∩ (k], since (g ∧ k] = ∩{D ∈ I(A) | D 3 g ∧ k}. Conversely, suppose that h ∈ (g] ∩ (k]; then h ≤ ng and h ≤ mk, for some n, m ≥ 1. It follows that h ≤ ng ∧ mk ≤ n(g ∧ mk) ≤ nm(g ∧ k), by Proposition 1.17; thus h ∈ (g ∧ k]. 2 An ideal H of A is proper iff H 6= A. An ideal H of A is called maximal iff it is proper and no proper ideal of A strictly contains H, i.e. for each ideal J 6= H, if H ⊆ J then J = A. Let H be an ideal of A and let define on A two binary relations by: def

x ≡L(H) y ⇔ d1 (x, y) ∈ H,

(3)

def

x ≡R(H) y ⇔ d2 (x, y) ∈ H,

(4)

where L comes from ”Left”, R comes from ”Right”. Lemma 2.7 The relations ≡L(H) and ≡R(H) are equivalence relations on A. Proof. The relation ≡L(H) is reflexive: x ≡L(H) x ⇔ d1 (x, x) ∈ H ⇔ 0 ∈ H which is true. It is symmetric: x ≡L(H) y ⇔ d1 (x, y) ∈ H ⇔ d1 (y, x) ∈ H ⇔ y ≡L(H) x, by Proposition 1.35(8). It is transitive: (x ≡L(H) y and y ≡L(H) z) ⇔ (d1 (x, y) ∈ H and d1 (y, z) ∈ H) ⇒ (d1 (x, y) ⊕ d1 (y, z) ⊕ d1 (x, y) ∈ H), by (I2); then, by Proposition 1.35(9), d1 (x, z) ∈ H ⇔ x ≡L(H) z. The proof that ≡R(H) is an equivalence relation is similar. 2 Lemma 2.8 Moreover, we have that: H = {x ∈ A | x ≡L(H) 0} and H = {x ∈ A | x ≡R(H) 0}. Proof. x = d1 (x, 0) ∈ H ⇔ x ≡L(H) 0 and x = d2 (x, 0) ∈ H ⇔ x ≡R(H) 0, by Proposition 1.35(2). 2 The relations ≡L(H) and ≡R(H) being equivalence relations, we can consider the quotient sets A/ ≡L(H) and A/ ≡R(H) . Un element of A/ ≡L(H) is denoted by x/ ≡L(H) and is called a left class of x. Un element of A/ ≡R(H) is denoted by x/ ≡R(H) and is called a right class of x. We define now on the set of left classes, A/ ≡L(H) , a binary relation ” ≤L(H) ” by: x/ ≡L(H)

≤L(H)

y/ ≡L(H)

def

⇔

x · y − ∈ H.

(5)

We define also on the set of right classes, A/ ≡R(H) , a binary relation ” ≤R(H) ” by: x/ ≡R(H)

≤R(H)

y/ ≡R(H)

def

⇔

y ∼ · x ∈ H.

(6)

Then we have the following Lemma 2.9 The relations ≤L(H) and ≤R(H) are partial order relations on the respective sets. Proof. We shall prove that ≤L(H) is a partial order relation on A/ ≡L(H) : • reflexivity: x/ ≡L(H) ≤L(H) x/ ≡L(H) ⇔ x · x− = 0 ∈ H, which is true. • antisymmetry: (x/ ≡L(H) ≤L(H) y/ ≡ L(H) and y/ ≡L(H) ≤L(H) x/ ≡ L(H) ) ⇔ (x · y − ∈ H and y · x− ∈ H ) ⇒ d1 (x, y) = x · y − ⊕ y · x− ∈ H ⇔ x ≡L(H) y ⇔ x/ ≡L(H) = y/ ≡L(H) . • transitivity: (x/ ≡L(H) ≤L(H) y/ ≡L(H) and y/ ≡L(H) ≤L(H) z/ ≡L(H) ) ⇔ (x · y − ∈ H and y·z − ∈ H) ⇒ x·z − ∈ H, since x·z − ≤ x·y − ⊕y·z − , by Lemma 1.33. Hence, x/ ≡L(H) ≤L(H) z/ ≡L(H) and the proof is over. The proof that ≤R(H) is a partial order relation is similar. 2 For any ideal H of A, we can define a map φ : A/ ≡L(H) −→ A/ ≡R(H) by def

φ(x/ ≡L(H) ) = (x− )/ ≡R(H) .

(7)

Since if x ≡L(H) y , i.e. d1 (x, y) ∈ H, then x− ≡R(H) y − , because d2 (x− , y − ) = d1 (x, y) ∈ H, it follows that the map φ is well defined. 12

Proposition 2.10 The map φ defined above is a bijection between the sets A/ ≡L(H) and A/ ≡R(H) . Proof. • φ is injective: let φ(x/ ≡L(H) ) = φ(y/ ≡L(H) ), i.e. (x− )/ ≡L(H) = (y − )/ ≡L(H) ; hence x ≡R(H) y − , i.e.d2 (x− , y − ) ∈ H; but d2 (x− , y − ) = d1 (x, y); it follows that d1 (x, y) ∈ H; thus x ≡L(H) y and consequently x/ ≡L(H) = y/ ≡L(H) . • φ is surjective: indeed, let x/ ≡R(H) be an element of A/ ≡R(H) ; then there exists x∼ / ≡L(H) ∈ A/ ≡L(H) such that φ(x∼ / ≡L(H) ) = (x∼ )− / ≡R(H) = x/ ≡R(H) . 2 We shall now examine in some detail the set I(A) of ideals of a given pseudo-MV algebra A. We need a little terminology from lattice theory. A lattice is complete if each subset has least W upperWbound and greatest lower bound. A complete lattice is Brouwerian if it satisfies the identity: a∧ i∈I bi = i∈I (a∧bi ). −

Proposition 2.11 Let A be a pseudo-MV algebra. Then (I(A), ∩, ∨) is a complete Brouwerian lattice, where ∩ is the intersection of sets and the join ∨ of an arbitrary collection of ideals is the ideal generated by the ∪ (as sets) of these ideals, the order relation being the inclusion of sets, ⊆. Proof. Routine. 2 Since I(A) is a Brouwerian lattice, it is pseudo-complemented, that is, for each C ∈ I(A), there is a unique maximal ideal C 0 ∈ I(A) such that C ∩ C 0 = {0}, where C 0 = ∨{D ∈ I(A) | D ∩ C = {0}}. We call the ideal C satisfying the equation C = C 00 a polar ideal of A. Remark 2.12 {0} and A are polar ideals. Proposition 2.13 Suppose X is any subset of A. Then Y = {a ∈ A | a ∧ x = 0, for any x ∈ X} is a polar ideal of A. Furthermore, if X is an ideal of A, then Y is precisely X 0 . Proof. • Y is an ideal: (I1): 0 ∈ Y since 0 ∧ x = 0, for all x ∈ X. (I2): Let a, b ∈ Y , i.e. a ∧ x = 0 and b ∧ x = 0. Then, by Proposition 1.17, (a ⊕ b) ∧ x = x ∧ (a ⊕ b) ≤ (x ∧ a) ⊕ (x ∧ b) = 0 ⊕ 0 = 0, hence (a ⊕ b) ∧ x = 0, i.e. a ⊕ b ∈ Y . (I3): If a ∈ Y , i.e. a ∧ x = 0, and b ≤ a, i.e. b ∧ a = b, then b ∧ x = (b ∧ a) ∧ x = b ∧ (a ∧ x) = b ∧ 0 = 0, hence b ∈ Y . • To prove that Y is a polar ideal, we must prove that Y 00 ⊆ Y . Let 0 6= a ∈ Y 00 ; since Y 00 ∩ Y 0 = {0}, by definition of Y 00 , we get that a 6∈ Y 0 ; then, since Y 0 ∩ Y = {0}, it follows that a ∈ Y and thus Y 00 ⊆ Y . Since we also have Y ⊆ Y 00 , it follows that Y = Y 00 and thus Y is a polar ideal. • If X is an ideal of A, we must prove that Y = X 0 . First we shall prove that Y ∩ X = {0}. Indeed, if a ∈ Y ∩ X, i.e. a ∈ Y and a ∈ X, then a ∧ x = 0 for each x ∈ X. In particular for a ∈ X we get a = a ∧ a = 0, thus Y ∩ X = {0}. Since X 0 = ∨{D ∈ I(A) | D ∩ X = {0}} ⊇ D, it follows that Y ⊆ X 0 . Now we shall prove that X 0 ⊆ Y . Indeed, let a ∈ X 0 ; since a ∧ x ≤ a and X 0 ∈ I(A), it follows that a ∧ x ∈ X 0 , for any x ∈ X. But we also have that a ∧ x ≤ x ∈ X and since X is an ideal, we get that a ∧ x ∈ X. Hence, a ∧ x ∈ X 0 ∩ X. But X 0 ∩ X = {0}, by the definition of X 0 . It follows that a ∧ x = 0, i.e. a ∈ Y . Thus X 0 ⊆ Y and therefore X 0 = Y . 2 We shall understand by g 0 and g 00 the polar ideals g 0 and g 00 . The polar ideals g 0 and g 00 are called principal polars (g ∈ A): g 0 = {g}0 = {a ∈ A | a ∧ g = 0} and g 00 = (g 0 )0 = {a ∈ A | a ∧ x = 0, for any x ∈ g 0 } and g 0 ∩ g 00 = {0}. We now need V another definition from lattice theory; an element x of a lattice is meet-irreducible if whenever x = i∈I xi , then x = xi , for some i ∈ I. An element x is finitely meet-irreducible if x satisfies the former when I is finite. We shall now examine the meet-irreducible elements of the lattice I(A). 13

Theorem 2.14 Let P ∈ I(A). Then P is meet-irreducible in I(A) iff P is maximal in I(A) with respect to not containing some g ∈ A. Proof. Like in MV algebras case. 2 We call an ideal with the properties of Theorem 2.14 regular and denote the set of them by Γ(A). If P ∈ Γ(A), then P ∗ is the cover of P and if g ∈ P ∗ \ P , then P is a value of g. Zorn’s lemma gives us the following Proposition 2.15 Every element of a pseudo-MV algebra has at least one value. Consequently, each ideal can be obtained as the intersection of regular ideals. A regular ideal H has the further property that its set of left classes (A/ ≡L(H) ) is totally ordered with respect to ≤L(H) . The ideals with the property that its set of left classes is totally ordered are called prime ideals. They are characterized in the next theorem. Remark 2.16 For any ideal H of A, the set of left classes is totally ordered by ≤L(H) if and only if the − set of right classses is totally ordered by ≤R(H) . Indeed, x/ ≡ ¡ L(H) ≤L(H) ¢ y/ ≡L(H) ¡ ⇐⇒ x · y¢ ∈ H ⇐⇒ − ∼ − − − (x ) · y ∈ H ⇐⇒ (x )/ ≡R(H) ≥R(H) (y )/ ≡R(H) ⇐⇒ φ x/ ≡L(H) ≥R(H) φ y/ ≡L(H) , by (5), (4), (6), (7); then use Proposition 2.10. Theorem 2.17 For P ∈ I(A), the following are equivalent: (a) P is prime (that is A/ ≡L(P ) or, equivalently, A/ ≡R(P ) is totally ordered), (b) {C ∈ I(A) | C ⊇ P } is totally ordered under inclusion, (c) P is finitely meet-irreducible in I(A), (d) If a ∧ b ∈ P then a ∈ P or b ∈ P , (e) If a ∧ b = 0 then a ∈ P or b ∈ P , (f ) For any a, b ∈ A,

a · b− ∈ P or b · a− ∈ P ,

(g) For any a, b ∈ A,

a∼ · b ∈ P or b∼ · a ∈ P .

Consequently, each regular ideal is prime. Proof. •(a)⇒ (b): Suppose that B and C are incomparable ideals containing P , i.e. B ⊇ P, C ⊇ P such that B 6⊆ C and C 6⊆ B; hence there exists b ∈ B such that b 6∈ C and there exists c ∈ C such that c 6∈ B. Let us consider the left classes b/ ≡L(P ) and a/ ≡L(P ) ; we have then, by (a) that : b/ ≡L(P ) ≤L(P ) c/ ≡L(P ) or c/ ≡L(P ) ≤L(P ) b/ ≡L(P ) ⇔ b · c− ∈ P or c · b− ∈ P . − − It follows that b · c ⊕ c = b ∨ c ∈ C or c · b ⊕ b = c ∨ b ∈ B, hence b ∈ C or c ∈ B, by (I3); contradiction. • (b)⇒ (c): If B ∩ C = P , then P ⊆ B, C. By (b) we have either C ⊆ B or B ⊆ C, say C ⊆ B; then P = B ∩ C = C, i.e. P = C. Hence (c) holds. • (c) ⇒ (d): (P ∨ (a]) ∩ (P ∨ (b]) = P ∨ ((a] ∩ (b]) = P ∨ (a ∧ b] = P , by Proposition 2.6 and since (a ∧ b] ⊆ P , by (I2) and (I3). By (c) it follows that P = P ∨ (a] or P = P ∨ (b] and thus a ∈ P or b ∈ P . • (d) ⇒ (e): If a ∧ b = 0 ∈ P , then by (d) it follows that a ∈ P or b ∈ P . • (e) ⇒ (a): Let x/ ≡L(P ) , y/ ≡L(P ) ∈ A/ ≡L(P ) ; since x · y − ∧ y · x− = 0, it follows by (e) that x · y − ∈ P or y · x− ∈ P which means, by definition of ≡L(P ) , that x/ ≡L(P ) ≤L(P ) y/ ≡L(P ) or y/ ≡L(P ) ≤L(P ) x/ ≡L(P ) , i.e. A/ ≡L(P ) is totally ordered by ≤L(P ) . • (e) ⇒ (f): Since a · b− ∧ b · a− = 0 for any a, b ∈ A, by Proposition 1.24. • (f) ⇒ (d): Suppose that a ∧ b ∈ P and that, for instance, a · b− ∈ P ; hence a · b− ⊕ (a ∧ b) ∈ P . But a ≤ (a · b− ⊕ a) ∧ (a ∨ b) = (a · b− ⊕ a) ∧ (a · b− ⊕ b) = a · b− ⊕ (a ∧ b), by Proposition 1.15. We get that a ∈ P. • (e) ⇒ (g): By Proposition 1.24. • (g) ⇒ (d): Suppose that a ∧ b ∈ P and that , for instance, a∼ · b ∈ P . Since b ≤ (a ∨ b) ∧ (b ⊕ a∼ · b) = (a ⊕ a∼ · b) ∧ (b ⊕ a∼ · b) = (a ∧ b) ⊕ a∼ · b ∈ P , it follows that b ∈ P . • That each regular ideal is prime follows immediately from (c). 2 14

Corollary 2.18 If P, Q ∈ I(A),

P ⊆ Q and P is prime, then Q is prime.

Proof. By Theorem 2.17 (f).

2

Corollary 2.19 If P is a prime ideal and a, b ∈ P , then a2 ∈ P or b2 ∈ P . Proof. By Theorem 2.17 (f), we have a·b− ∈ P or b·a− ∈ P , for any a, b ∈ A. Suppose that a·b− ∈ P , hence a · b ⊕ a · b− ∈ P . By using Propositions 1.16 anf 1.27, we get: a2 ∨ a · b = a · (a ∨ b) = a · (a · b− ⊕ b) ≤ a · b ⊕ a · b− , hence a2 ∈ P . 2 Notice that if A is a totally ordered pseudo-MV algebra, then {0} is prime ( since A/ ≡L({0}) = A is totally ordered): x/ ≡L({0}) = {y ∈ A | d1 (x, y) = 0} = {x} and thus, by Theorem 2.17(b), {C ∈ I(A) | C ⊇ {0}} = {C ∈ I(A) | C 3 0} = I(A) is totally ordered under inclusion (i.e. I(A) is a chain). From Theorem 2.17(b) it is apparent that the intersection of a chain of primes is still a prime. Since each prime may evidently be placed in a maximal chain of primes, it follows that each prime contains a (not necessarily unique) minimal prime. With minimal primes comes an important connection between the notions of prime and polar, as the next theorem will reveal. It is apparent from (e) of Theorem 2.17 that if P is prime, then the set F = A \ P satisfies the property that a ∧ b > 0 for all a, b ∈ F . A subset F of A \ {0} maximal with respect to this property is called an ultrafilter. We can now caracterize minimal primes. Theorem 2.20 For P ∈ I(A), the following are equivalent: (a) P is a minimal prime, (b) A \ P is an ultrafilter, (c) P = ∪{g 0 | g 6∈ P }, (d) P is prime, and for all h ∈ P, h0 6⊆ P . Proof. • (a) ⇒ (b): We have already observed that for all a, b ∈ A \ P, a ∧ b > 0. By Zorn’s lemma we may obtain an ultrafilter F which contains A \ P . We shall prove that F = A \ P . For each g ∈ F , we have that g 0 ⊆ A \ F ⊆ P. (8) Let a ∈ g 0 ; then a ∈ A and a 6∈ F (since if a ∈ F, g ∈ F, a ∧ g > 0). Hence a ∈ P (if a 6∈ P, a ∈ A, then a ∈ A \ P ⊆ F , i.e. a ∈ F ). But then Q = ∪{g 0 | g ∈ F } ⊆ P . We shall prove that Q is a prime ideal. (I1): 0 ∈ g 0 , hence 0 ∈ Q. (I2): If a, b ∈ Q, i.e. a ∈ g 0 , b ∈ h0 with g, h ∈ F , then by Proposition 1.17, we get g ∧ h ∧ (a ⊕ b) ≤ (g ∧ h ∧ a) ⊕ (g ∧ h ∧ b) = ((a ∧ g) ∧ h) ⊕ ((b ∧ h) ∧ g) = 0 ⊕ 0 = 0, therefore g ∧ h ∧ (a ⊕ b) = 0, i.e. a ⊕ b ∈ (g ∧ h)0 . Since g ∧ h ∈ F (g ∧ h = (g ∧ h) ∧ f > 0 and h ∈ F ; then apply (I3) to F), it follows that a ⊕ b ∈ Q. (I3) If a ∈ Q, i.e. a ∈ g 0 , g ∈ F , and b ≤ a, then b ∈ g 0 , since g 0 is an ideal; hence b ∈ Q. Thus Q is an ideal. To prove that Q is prime, by Theorem 2.17(e), suppose that c ∧ d = 0 and c 6∈ Q, i.e. c ∧ g > 0, ∀g ∈ F . But since F is an ultrafilter, c ∈ F and so d ∈ c0 ⊆ Q, i.e. d ∈ Q. Thus Q is prime, Q ⊆ P . Since P is a minimal prime, P = Q. Hence A \ F = P , by (10), i.e. F = A \ P and thus (b) holds. • (b) ⇒ (c): The proof above shows that if A \ P is an ultrafilter, then for each g ∈ A \ P , g 0 ⊆ P . Then ∪{g 0 | g 6∈ P } is a prime ideal contained in P . To prove that we also have P ⊆ ∪{g 0 | g 6∈ P } we shall prove equivalently that h 6∈ ∪{g 0 | g 6∈ P } implies h 6∈ P . Indeed, if 0 < h 6∈ ∪{g 0 | g 6∈ P }, then h ∧ g > 0 for all g ∈ A \ P ; but A \ P is an ultrafilter; it follows that h ∈ A \ P , i.e. h 6∈ P . Thus, P = ∪{g 0 | g 6∈ P }. 15

• (c) ⇒ (d): If c ∧ d ∈ P and c 6∈ P , i.e. c ∧ d ∧ g = 0, for some g 6∈ P and c ∧ h > 0, ∀h 6∈ P , then c ∧ d ∧ g = (c ∧ g) ∧ d = 0 and c ∧ g > 0. But c ∧ g 6∈ P , since c 6∈ P, g 6∈ P (indeed, (c ∧ g) ∧ h = c ∧ (g ∧ h) > 0 ∧ 0 = 0, ∀h 6∈ P ). Hence, (c ∧ g) ∧ d = 0, i.e. d ∈ (c ∧ g)0 ⊆ P , i.e. d ∈ P . Thus P is prime. The rest is obvious. • (d) ⇒ (a): Suppose Q ∈ I(A) and Q ⊂ P . Choose h ∈ P \ Q. Then h0 6⊆ P , i.e. ∃g 6∈ P, g ∧ h = 0. But then Q cannot be prime. 2 Proposition 2.21 Let A be a pseudo-MV algebra, J an ideal of A and a ∈ A \ J. Then there is a prime ideal P of A such that J ⊆ P and a 6∈ P . Proof. A routine application of Zorn’s Lemma shows that there is an ideal H of A which is maximal with respect to the property that J ⊆ H and a 6∈ H. We shall prove that H is a prime ideal. Let x, y ∈ A and suppose that H is not prime, i.e. x·y − 6∈ H and y·x− 6∈ H, by Theorem 2.17(e) (absurdum hypothesis). − 1 Then the ideal generated by H ∪ {x · y − } must contain the element a. By (2), a ≤ ⊕m i=1 (si ⊕ pi (x · y )), for m2 − some si ∈ H and some integers pi ≥ 0 and m1 ≥ 1. Similarly, a ≤ ⊕j=1 (tj ⊕ qj (y · x )), for some tj ∈ H and some integers qj ≥ 0 and m2 ≥ 1. Let s = s1 ⊕ s2 ⊕ . . . ⊕ sm1 and t = t1 ⊕ t2 ⊕ . . . ⊕ tm2 ; then s, t ∈ H. − − 1 Let p = maxi=1,m1 {pi } and q = maxj=1,m2 {qj }. Then a ≤ ⊕m i=1 (s ⊕ p(x · y )) = m1 (s ⊕ p(x · y )) and m2 − − a ≤ ⊕j=1 (t ⊕ q(y · x )) = m2 (t ⊕ q(y · x )). Let now u = s ⊕ t and n = max{p, q}. Then u ∈ H and a ≤ m1 (u ⊕ n(x · y − )) and a ≤ m2 (u ⊕ n(y · x− )). Hence, by Propositions 1.13 and 1.17, together with Propositions 1.15, 1.24 and Lemma 1.32, we have a ≤ m1 (u ⊕ n(x · y − )) ∧ m2 (u ⊕ n(y · x− )) ≤ m2 [m1 (u ⊕ n(x · y − )) ∧ (u ⊕ n(y · x− ))] ≤ m2 m1 [(u ⊕ n(x · y − )) ∧ (u ⊕ n(y · x− ))] = m2 m1 [u ⊕ (n(x · y − ) ∧ n(y · x− )] = m2 m1 [u ⊕ 0] = m2 m1 u ∈ H; hence a ∈ H, a contradiction. 2

3

Normal ideals and congruences

Definition 3.1 An ideal H of a pseudo-MV algebra A is normal iff it satisfies the condition: (I4) for each x, y ∈ A, y · x− ∈ H ⇔ x∼ · y ∈ H. Lemma 3.2 Let H be a normal ideal. Then: (i) the axiom (I4) is equivalent with (I5), where: (I5) for each x ∈ A,

H ⊕ x = x ⊕ H,

i.e. (1): for each h ∈ H, there exists h0 ∈ H, such that h ⊕ x = x ⊕ h0 and (2): for each h0 ∈ H, there exists h ∈ H, such that x ⊕ h0 = h ⊕ x. (ii) the axiom (I4) implies (I6), where: (I6) h ∈ H ⇔ h= ∈ H and h ∈ H ⇔ h≈ ∈ H.

Proof. • (I4) =⇒ (I5): Let x ∈ A and h ∈ H. We put then y = h ⊕ x and we notice that x ≤ y. Then y · x− ⊕ x = y ∨ x = y = x ⊕ x∼ · y. Hence h ⊕ x = y = y · x− ⊕ x = x ⊕ x∼ · y. If y · x− ∈ H, then by (I4) we get that h0 = x∼ · y ∈ H and we get that there exists h0 ∈ H such that h ⊕ x = x ⊕ h0 ; thus (1) holds. (2) has a similar proof. Thus (I5) holds. (I5) =⇒ (I4) (A. Dvurecenskij): Assume that y · x− ∈ H; then putting h1 = y · x− , we have: x ∨ y = y · x− ⊕ x = h1 ⊕ x = x ⊕ x∼ · y and there exists h2 ∈ H such that x ∨ y = x ⊕ h2 , by (I5). Then x∼ · y ≤ x∼ · (x ∨ y) = x∼ · (x ⊕ h2 ) = x∼ ∧ h2 ≤ h2 ∈ H. It follows that x∼ · y ∈ H. Similarly, if we assume that x∼ · y ∈ H, we obtain that y · x− ∈ H. Thus (I4) holds. Hence (i) holds. • (I4) =⇒ (I6): For y = 1, (I4) gives: x− ∈ H ⇔ x∼ ∈ H. Take then x = x∼ and x = x− ; we get that x ∈ H ⇔ x≈ ∈ H and x ∈ H ⇔ x= ∈ H, i.e. (ii) holds. 2 16

Remark 3.3 (I4) is equivalent to y/ ≡L(H)

≤L(H)

x/ ≡L(H)

⇔

y/ ≡R(H)

≤R(H)

x/ ≡R(H) ,

by definition. Lemma 3.4 (A. Dvurecenskij) Let H be a normal ideal of A and z ∈ A. Then (2) becomes (H ∪ {z}] = {x ∈ A | x ≤ h ⊕ nz, for some h ∈ H and n ∈ IN},

(9)

Proof. By (2) we have that (H ∪ {z}] = {x ∈ A | x ≤ (h1 ⊕ n1 z) ⊕ (h2 ⊕ n2 z) ⊕ . . . ⊕ (hm ⊕ nm z)}. If m = 1, then we get x ≤ h1 ⊕ n1 z. If m = 2, then x ≤ (h1 ⊕ n1 z) ⊕ (h2 ⊕ n2 z) = h1 ⊕ (n1 z ⊕ h2 ) ⊕ n2 z = h1 ⊕ (h02 ⊕ n1 z) ⊕ n2 z = (h1 ⊕ h02 ) ⊕ (n1 z ⊕ n2 z) = h12 ⊕ n12 z, by (I5), with h12 = h1 ⊕ h02 ∈ H and n12 = n1 + n2 ∈ IN. By induction we get that x ≤ h12...m ⊕ n12...m z, with h12...m = h12...m−1 ⊕ h0m ∈ H and n12...m = n1 + n2 + . . . + nm ∈ IN. 2 The next proposition generalizes a well known property of maximal ideals in boolean algebras and M V - algebras. The idea that J must be a normal ideal to be able to prove one of the implications by using the above lemma belongs to A. Dvurecenskij. Proposition 3.5 For any proper normal ideal J of a pseudo-MV algebra A, the following conditions are equivalent: (i) J is a maximal ideal of A, (ii) For each z ∈ A, z 6∈ J iff [(nz)∼ ∈ J or (mz)− ∈ J, for some integers n, m ≥ 1]. Proof. (i) ⇒(ii): Suppose that J is a maximal ideal of A. • If z 6∈ J, then ({z} ∪ J] = (J ∪ {z}] = A 3 1 and by (9) 1 = nz ⊕ a = b ⊕ mz for some integers n, m ≥ 1 and a, b ∈ J. Hence, 1 = ((nz)∼ )− ⊕ a = b ⊕ ((mz)− )∼ , i.e. (nz)∼ ≤ a or (mz)− ≤ b by Proposition 1.9(a),(g). Then we get that (nz)∼ ∈ J or (mz)− ∈ J, by (I3). • If z ∈ J, then nz ∈ J, by (I2); since J is proper, i.e. 1 6∈ J, it follows that (nz)∼ 6∈ J and (nz)− 6∈ J. (ii) ⇒ (i): Let K be an ideal of A such that J ⊂ K, i.e. J ⊆ K and J 6= K. For every z ∈ K \ J, we have by (ii) that (nz)∼ ∈ J for some integer n ≥ 1 or (mz)− ∈ J for some integer m ≥ 1, hence (nz)∼ ∈ K or (mz)− ∈ K. But, since z ∈ K we get that nz, mz ∈ K, by (I2). Hence nz ⊕ (nz)∼ = 1 ∈ K or (mz)− ⊕ mz = 1 ∈ K, i.e. 1 ∈ K and thus K = A. 2 Definition 3.6 A congruence on A is an equivalence relation ≡ on A satisfying the following conditions: (C1) if x ≡ y and a ≡ b, then (x ⊕ a) ≡ (y ⊕ b) and (a ⊕ x) ≡ (b ⊕ y), (C2) if x ≡ y, then x− ≡ y − and x∼ ≡ y ∼ . Remark 3.7 A congruence on A satisfies also the following condition: (C1’) if x ≡ y and a ≡ b, then (x · a) ≡ (y · b) and (a · x) ≡ (b · y), since x · a = (a− ⊕ x− )∼ , by Proposition 1.7. Theorem 3.8 If H is a normal ideal, then (i) ≡L(H) ⇔ ≡R(H) ; let ≡H denote one of them. (ii) The binary relation ≡H is a congruence on A and we have: H = {x ∈ A | x ≡H 0} = 0/ ≡H . 17

Proof. (i) If x ≡L(H) y, i.e. d1 (x, y) ∈ H, then x · y − ∈ H and y · x− ∈ H, by (I3). Hence y ∼ · x ∈ H and x∼ · y ∈ H, by (I4). Then d2 (x, y) ∈ H, by (I2); thus x ≡R(H) y. The converse implication has a similar proof. Thus (i) holds. (ii) Obviously, ≡H is an equivalence relation. To prove (C1), let x ≡H y and a ≡H b, i.e. d1 (x, y) ∈ H and d1 (a, b) ∈ H. Then we get that x · y − ∈ H, y · x− ∈ H, a · b− ∈ H, b · a− ∈ H, by (I3). But, x · y − ∈ H and a · b− ∈ H give, by (I5), since x ≤ x ⊕ x∼ · y = x · y − ⊕ y and a ≤ a ⊕ a∼ · b = a · b− ⊕ b, that x ⊕ a ≤ x · y − ⊕ y ⊕ a · b− ⊕ b = x · y − ⊕ (y ⊕ a · b− ) ⊕ b = x · y − ⊕ (h ⊕ y) ⊕ b, with h ∈ H. But x ⊕ a ≤ (x · y − ⊕ h) ⊕ (y ⊕ b) ⇔ (x ⊕ a) · (y ⊕ b)− ≤ x · y − ⊕ h ∈ H, by Proposition 1.12(d). It follows by (I3) that (x ⊕ a) · (y ⊕ b)− ∈ H. Similarly, y · x− ∈ H and b · a− ∈ H give that (y ⊕ b) · (x ⊕ a)− ∈ H. Then d1 (x ⊕ a, y ⊕ b) = (x ⊕ a) · (y ⊕ b)− ⊕ (y ⊕ b) · (x ⊕ a)− ∈ H, by (I2); hence (x ⊕ a) ≡H (y ⊕ b). Similarly, (a ⊕ x) ≡H (b ⊕ y). Thus, (C1) holds. To prove (C2), let x ≡H y, i.e. d1 (x, y) ∈ H. Then d2 (x, y) ∈ H, by (i). It follows that (d2 (x, y))= ∈ H, by (I6) and since (d2 (x, y))= = d1 (x− , y − ), by Proposition 1.35(7), it follows that d1 (x− , y − ) ∈ H, i.e. x− ≡H y − . Similarly, since d1 (x, y) ∈ H, it follows that (d1 (x, y))≈ ∈ H, by (I6). But (d1 (x, y))≈ = d2 (x∼ , y ∼ ), by Proposition 1.35(6). Hence d2 (x∼ , y ∼ ) ∈ H, hence d1 (x∼ , y ∼ ) ∈ H, by (i), i.e. x∼ ≡H y ∼ and thus (C2) holds. Hence, ≡H is a congruence relation. H = {x ∈ A | x ≡H 0} = 0/ ≡H by Lemma 2.8. 2 Conversely, we have the following Theorem 3.9

(i) If ≡ is a congruence on A, then 0/ ≡ = {x ∈ A | x ≡ 0} is a normal ideal of A.

(ii) x ≡ y iff d1 (x, y) ≡ 0 or, equivalently, x ≡ y iff d2 (x, y) ≡ 0. Proof. Let us put H = 0/ ≡. • 0 ≡ 0 by reflexivity of ≡, hence 0 ∈ H and (I1) holds. • Let x ∈ H and y ∈ H, i.e. x ≡ 0 and y ≡ 0. Then by (C1), x ⊕ y ≡ 0 ⊕ 0 = 0, i.e. x ⊕ y ∈ H. Thus (I2) holds. • Let y ∈ H and x ∈ A with x ≤ y. Then y ≡ 0 and x = x · (x− ⊕ y), by Proposition 1.9(d). Since x− ≡ x− , then by (C1) we get: (x− ⊕ y) ≡ x− ⊕ 0, i.e. x− ⊕ y ≡ x− . Then, since x ≡ x and by (C1’) we get: x = [x · (x− ⊕ y)] ≡ x · x− = 0, i.e. x ≡ 0; then x ∈ H and thus (I3) holds and H is an ideal. • Let y · x− ∈ H, i.e. y · x− ≡ 0. Then y ∨ x = y · x− ⊕ x ≡ 0 ⊕ x = x. But y ∨ x = x ∨ y = x ⊕ x∼ · y, hence x ⊕ x∼ · y ≡ x. It follows that (y ∼ ⊕ x≈ · y)∼ = (x ⊕ x∼ · y)∼ ≡ x∼ , by (C2). Then x∼ · y ≡ (x ⊕ x∼ · y)∼ · y = (y ∼ ⊕ x≈ ) · (x∼ · y). But (y ∼ ⊕ x≈ )− = (x≈ )− · y = x∼ · y. Hence x∼ · y ≡ (y ∼ ⊕ x≈ ) · [(y ∼ ⊕ x≈ )− ] = 0, i.e. x∼ · y ∈ H. Similarly, x∼ · y ∈ H implies y · x− ∈ H and thus (I4) holds. Hence H is a normal ideal and thus (i) holds. To prove (ii), let us consider that x ≡ y. Then x · y − ≡ y · y − = 0 and y · x− ≡ x · x− = 0. Then d1 (x, y) = x · y − ⊕ y · x− ≡ 0 ⊕ 0 = 0. Conversely, let us consider that d1 (x, y) ≡ 0 ⇔ d1 (x, y) ∈ 0/ ≡ and 0/ ≡ is an ideal of A, by (i). Then x · y − , y · x− ∈ 0/ ≡ also, since x · y − , y · x− ≤ d1 (x, y). Hence x · y − ≡ 0 and y · x− ≡ 0. It follows that x ∨ y = x · y − ⊕ y ≡ 0 ⊕ y = y and y ∨ x = y · x− ⊕ x ≡ 0 ⊕ x = x. But x ∨ y = y ∨ x, hence by transitivity of ≡ we get that x ≡ y. Therefore x ≡ y ⇔ d1 (x, y) ≡ 0. The proof of x ≡ y ⇔ d2 (x, y) ≡ 0 is similar. 2 Theorems 3.8 and 3.9 lead to the following Corollary 3.10 There is a bijection between the set of normal ideals and the set of congruences of a pseudo-MV algebra. Given a normal ideal H of the pseudo-MV algebra A = (A, ⊕, − , ∼ , 0, 1), we shall denote the equivalence class of x ∈ A with respect to ≡H by x/H and the quotient set A/ ≡H by A/H for short. Since ≡H is a congruence of A, by Theorem 3.8, by defining in the set A/H the operations: def x/H ⊕ y/H = (x ⊕ y)/H,

def (x/H)− = (x− )/H, 18

def (x/H)∼ = (x∼ )/H,

the system A/H = (A/H, ⊕, − , ∼ , 0/H = H, 1/H) becomes a pseudo-MV algebra, called the quotient algebra of A by the normal ideal H. Moreover, the correspondence x ; x/H defines a homomorphism pH from A onto the quotient algebra A/H, which is called the natural homomorphism from A onto A/H. Note that Ker(pH ) = H, since x ∈ Ker(pH ) ⇔ x/H = pH (x) = 0/H = H ⇔ x ≡H 0 ⇔ d1 (x, 0) ∈ H and d2 (x, 0) ∈ H ⇔ x ∈ H. not

Remarks 3.11 (1) If H is a normal ideal, then ≡L(H) ⇔ ≡R(H) ( ⇔ ≡H ), by Theorem 3.8(i) and then y/ ≡H ≤L(H) x/ ≡H ⇐⇒ y/ ≡H ≤R(H) x/ ≡H , by Remark 3.3. Let denote by ”≤H ” one of the partial relations ≤L(H) , ≤R(H) . Hence ≤H is a partial order relation on A/H. (2) If H is a normal ideal of the pseudo-MV algebra A, then: x ≤ y in A implies x/H ≤H y/H in A/H. Indeed, if x ≤ y in A, then x ≤ y ⇔ x ∨ y = y ⇒ (x ∨ y)/H = y/H ⇔ (y · x− ⊕ x)/H = y/H ⇔ y/H · (x− )/H ⊕ x/H = y/H ⇔ x/H ∨ y/H = y/H ⇔ x/H ≤H y/H. It follows that if A is a pseudo-MV chain, then A/H is a pseudo-MV chain. Note that A/H is a pseudo-MV chain iff H is a normal prime ideal of A. (3) If A is a pseudo-MV chain, then the set of normal ideals of A is totally ordered by inclusion. Indeed, if H, J were normal ideals of A such that H 6⊆ J and J 6⊆ H, then there would be elements a, b ∈ A, such that a ∈ J \ H and b ∈ H \ J, whence a 6≤ b and b 6≤ a, by (I3), which is impossible. Definition 3.12 We shall say that a pseudo-MV algebra A is representable if it can be represented as (i.e. it is ) a subdirect product of totally ordered pseudo-MV algebras. Proposition 3.13 The following are equivalent for a pseudo-MV algebra A: (a) A is representable, (b) there exists a set S of normal prime ideals such that ∩S = {0}, (c) all polars are normal ideals, (d) all minimal prime ideals are normal. Proof. (a) ⇔ (b) QObvious. (a) ⇒ (c) Let A ⊆ i∈I Ai be a subdirect product and P be a polar ideal of A. For any x ∈ A we denote by supp(x) = {i ∈ I | xi 6= 0}. Since every Ai is totally ordered, we shall have, for any x, a ∈ A: x ∧ a = 0 ⇐⇒ supp(x) ∩ supp(a) = ∅. Since x · y − = 0 iff x ≤ y iff y ∼ · x = 0, it follows that supp(x · y − ) = supp(y ∼ · x), hence x · y − ∈ P iff y ∼ · x ∈ P , which proves that P isSnormal. (c) ⇒ (d) This follows from M = {x0 | x 6∈ M }, for any minimal prime ideal M , by Theorem 2.20. (d) ⇒ (a) Since the intersection of all minimal prime ideals is {0}. 2 Remark 3.14 If A is representable, then for all a, b ∈ A, 2(a ∧ b) = 2a ∧ 2b. Indeed, the equality holds in totally ordered pseudo-MV algebras: if a ≤ b, then a ∧ b = a and 2a = a ⊕ a ≤ a ⊕ b ≤ b ⊕ b = 2b; hence 2(a ∧ b) = 2a = 2a ∧ 2b. Hence the equality holds in A. Open problem 3.15 If for all a, b ∈ A, 2(a ∧ b) = 2a ∧ 2b, then is A representable?

19

4

Boolean center and Pierce representation of a pseudo-MV algebra

In this section we shall study the Boolean center B(A) of a pseudo-MV algebra A and with any pseudoMV algebra we shall associate a Boolean sheaf (Pierce sheaf) whose stalks are directly indecomposable. The background for sheaf theory is [16], Ch.5. If L is a bounded distributive lattice, then B(L) will be the Boolean algebra of complemented elements in L. For a pseudo-MV algebra A = (A, ⊕, − , ∼ , 0, 1) we shall denote by B(A) the Boolean algebra associated with the bounded distributive lattice (A, ∨, ∧, 0, 1), where ∨, ∧ are those defined in Proposition 1.13. Elements of B(A) are called the boolean elements of A. We choose to note by ”x− ” the complement of x ∈ B(A). We shall characterize the elements of B(A) in terms of pseudo-MV algebra operations. Let A be a pseudo-MV algebra. Lemma 4.1 For every x ∈ A, x ⊕ x = x iff x · x = x. Proof. By Proposition 1.28 (c), x ⊕ x = x iff x− ⊕ x− = x− iff (x− ⊕ x− )∼ = (x− )∼ iff x · x = x.

2

Proposition 4.2 For every x ∈ A, the following conditions are equivalent: (1) x ⊕ x = x, (2) x ∧ x− = 0, (3) x ∨ x− = 1, (4) x ∧ x∼ = 0, (5) x ∨ x∼ = 1, (6) x ∈ B(A). Proof. (1) =⇒ (2): x ⊕ x = x ⇒ x ∧ x− = (x ⊕ (x− )∼ ) · x− = (x ⊕ x) · x− = x · x− = 0. (2) =⇒ (5): x ∧ x− = 0 ⇒ x ∨ x∼ = x∼ ∨ (x− )∼ = (x ∧ x− )∼ = 1. (5) =⇒ (1): x ∨ x∼ = 1 ⇒ x = (x ∨ x∼ ) · x = x · x ∨ x∼ · x = x · x, by Proposition 1.16. Then apply Lemma 4.1. Hence, (1) ⇔ (2) ⇔ (5). Similarly, (1) ⇔ (3) ⇔ (4). Hence, (1), (2), (3), (4) and (5) are equivalent. Remark that the equivalent conditions (2) and (3) state that x− is a complement of x, thus in particular (2) =⇒ (6) and that the equivalent conditions (4) and (5) state that x∼ is also a complement of x. Finally, (6) =⇒ (2): Assume there exists y ∈ A, x ∧ y = 0 and x ∨ y = 1. Thus x ∨ y = 1 ⇒ x− = (x ∨ y) · x− = x · x− ∨ y · x− = y · x− ⇒ x− ≤ y, x ∧ y = 0 ⇒ x− = x− ⊕ (x ∧ y) · x− = (x− ⊕ x) ∧ (x− ⊕ y) = x− ⊕ y ⇒ y ≤ x− . This yields y = x− , so x ∧ x− = 0. 2 By Proposition 4.2 it follows that for every x ∈ B(A), x− = x∼ . Proposition 4.3 If x ∈ B(A) and y ∈ A, then (1) x ⊕ y = x ∨ y = y ⊕ x, (2) y · x = x ∧ y = x · y. Proof. • We have that x ∨ y ≤ x ⊕ y and (x ⊕ y) · (x ∨ y)− = (x ⊕ y) · x− ∧ (x ⊕ y) · y − = (x ⊕ y) · x− ∧ (x ⊕ (y − )∼ ) · y − = (x ⊕ y) · x− ∧ x ∧ y − ≤ x− ∧ x = 0, by Propositions 1.23, 1.22, 1.13 and 4.2, hence x ⊕ y ≤ x ∨ y; thus x ⊕ y = x ∨ y. Similarly, we have that x ∨ y ≤ y ⊕ x and (x ∨ y)∼ · (y ⊕ x) = x∼ · (y ⊕ x) ∧ y ∼ · (y ⊕ x) = x∼ · (y ⊕ x) ∧ y ∼ · 20

[(y ∼ )− ⊕ x] = x∼ · (y ⊕ x) ∧ y ∼ ∧ x ≤ x∼ ∧ x = 0, hence y ⊕ x ≤ x ∨ y; thus y ⊕ x = x ∨ y. Thus (1) holds. • y · x = (x− ⊕ y − )∼ = (x− ∨ y − )∼ = (y − ⊕ x− )∼ , by the previous (1), since x− ∈ B(A); then we get immediately (2), since (x− ∨ y − )∼ = x ∧ y and (y − ⊕ x− )∼ = x · y. 2 It is routine to prove the two following Corollary 4.4 B(A) is a subalgebra of the pseudo-MV algebra A. A subalgebra C of A is a Boolean algebra iff C ⊆ B(A). Corollary 4.5 A pseudo-MV algebra A is a Boolean algebra if and only if the operation ⊕ is idempotent, i.e. the equation x ⊕ x = x is satisfied in A. Proposition 4.6 For x ∈ A the following are equivalent: (1) there is a natural number n ≥ 1 such that nx ∈ B(A), (2) there is a natural number n ≥ 1 such that x− ∨ nx = 1, (3) there is a natural number n ≥ 1 such that x∼ ∨ nx = 1, (4) there is a natural number n ≥ 1 such that nx = (n + 1)x. Proof. (1) =⇒ (4): If nx ∈ B(A), then 2nx = nx; then nx = (n+1)x, since nx ≤ (n+1)x ≤ 2nx = nx. (4) =⇒ (2): If nx = (n + 1)x, then x− ∨ nx = x− · (nx)− ⊕ nx = [(n + 1)x]− ⊕ nx = (nx)− ⊕ nx = 1. (2) =⇒ (1): Assume that x− ∨ nx = 1; hence [(n + 1)x]− ⊕ nx = (nx ⊕ x)− ⊕ nx = x− · (nx)− ⊕ nx = x− ∨ nx = 1 ⇒ (n + 1)x ≤ nx ⇒ nx = (n + 1)x ⇒ nx = nx ⊕ nx ⇒ nx ∈ B(A). Hence (1) ⇔ (4) ⇔ (2). (4) =⇒ (3): If nx = (n + 1)x, then x∼ ∨ nx = nx ⊕ (nx)∼ · x∼ = nx ⊕ (x ⊕ nx)∼ = nx ⊕ [(n + 1)x]∼ = nx ⊕ (nx)∼ = 1. (3) =⇒ (1): Assume that x∼ ∨ nx = 1; hence nx ⊕ [(n + 1)x]∼ = nx ⊕ [x ⊕ nx]∼ = nx ⊕ (nx)∼ · x∼ = x∼ ∨ nx = 1; then (n + 1)x ≤ nx ⇒ nx = (n + 1)x ⇒ nx = nx ⊕ nx ⇒ nx ∈ B(A). Hence (1) ⇔ (4) ⇔ (3). 2 Remarks 4.7 (1) nx ∈ B(A) ⇔ (nx)− = (nx)∼ ∈ B(A) ⇔ (x− )n = (x∼ )n ∈ B(A). (2) x− ∨ nx = 1 ⇔ (x− ∨ nx)∼ = 1∼ ⇔ x ∧ (x∼ )n = 0. (3) x∼ ∨ nx = 1 ⇔ (x∼ ∨ nx)− = 1− ⇔ x ∧ (x− )n = 0. (4) nx = (n+1)x ⇔ (nx)∼ = [(n+1)x]∼ ⇔ (nx)− = [(n+1)x]− ⇔ (x∼ )n = (x∼ )n+1 ⇔ (x− )n = (x− )n+1 . It follows that the ”dual” of Proposition 4.6 also holds: Proposition 4.8 For x ∈ A the following are equivalent: (1’) there is a natural number n ≥ 1 such that xn ∈ B(A), (2’) there is a natural number n ≥ 1 such that x− ∧ xn = 0, (3’) there is a natural number n ≥ 1 such that x∼ ∧ xn = 0, (4’) there is a natural number n ≥ 1 such that xn = xn+1 . Remark 4.9 A direct proof of Proposition 4.8 would begin like this: (1’) =⇒ (4’): If xn ∈ B(A), then x2n = xn ·xn = xn ; then xn = xn+1 , since xn = x2n ≤ xn+1 = xn ·x ≤ xn (n ≥ 1 implies 2n = n + n ≥ n + 1) and so on. Corollary 4.10 For x ∈ A, the following conditions are equivalent: (1) there is a natural number n ≥ 1 such that nx ∈ B(A); (1’) there is a natural number n ≥ 1 such that xn ∈ B(A).

21

We then get the following important result. Corollary 4.11 Let A be a pseudo-MV algebra. If any x ∈ A fulfils the equivalent conditions from Propositions 4.6 and 4.8, then we have that x− = x∼ . Proof. For any x ∈ A, we get by Propositions 4.6, 4.8 and 4.3, that there is a natural number n ≥ 1, such that: x− ⊕ nx = x∼ ⊕ nx and nx · x− = nx · x∼ . Then, by Proposition 1.28(a), we get that x− = x∼ . 2 Open problem 4.12 Let A be a pseudo-MV algebra with the property that the two unary operations, ”− ” and ”∼ ”, coincide. Is then ”⊕” commutative, i.e. is then A an MV algebra? If P is a Boolean ideal of B(A), then (P ], the ideal of A generated by P , is given by (P ] = {x ∈ A | there exists e ∈ P, x ≤ e}. Lemma 4.13 (P ] is a normal ideal of A. Proof. If x, y ∈ A and e ∈ P , then, by Proposition 4.3, we get: x · y − ≤ e ⇔ x ≤ e ⊕ (y − )∼ = e ⊕ y = y ⊕ e = (y ∼ )− ⊕ e ⇔ y ∼ · x ≤ e, therefore x · y − ∈ (P ] iff y ∼ · x ∈ (P ]. 2 Remark 4.14 If e ∈ B(A), then (e] = {x ∈ A | x ≤ ne = e, for some n ∈ IN}, by Proposition 4.2, and (e] ( the principal ideal generated by e) is a normal ideal of A, by Lemma 4.13. ∗∗∗ As in MV - algebras case, a sheaf F of pseudo-MV algebras , defined on a Boolean space X, will be called Boolean sheaf (of pseudo-MV algebras). We shall denote by F (X) the pseudo-MV algebra of global sections of the Boolean sheaf F defined on X. For any x ∈ X, we shall denote by Fx the stalk of F in x. If a ∈ F (X) and x ∈ X, then we write ax for the image of a in Fx (the germ of a at x). Generally, if a, b ∈ F (X), then k a = b k= {x ∈ X | ax = bx } is an open set of X. If k a = b k is a clopen set for all a, b ∈ F(X), then we shall say that F is a Hausdorff Boolean sheaf. The Hausdorff Boolean sheaves correspond to Boolean products [5]. Let X = XA be the prime spectrum of B(A) (i.e. X is the set of all prime ideals of B(A)) endowed with the Stone topology; hence X is a Boolean space. The sets V (e) = {P ∈ X | e 6∈ P },

e ∈ B(A),

form a basis for this topology. • For any element e ∈ B(A), denote by ≡e the congruence of A associated with the normal ideal (e− ]: x ≡e y Lemma 4.15 For any x, y ∈ A,

⇔

d1 (x, y) ≤ e−

x ≡e y

iff

⇔

d2 (x, y) ≤ e− .

x · e = y · e.

Proof. By using Propositions 1.35(0), 1.12(d) and (a), 4.2 and Lemma 4.1, one gets the following equivalences: x ≡e y ⇔ x · y − ⊕ y · x− = x · y − ∨ y · x− ≤ e− ⇔ (x · y − ≤ e− and y · x− ≤ e− ) ⇔ (y − ≤ x− ⊕ e− = (e · x)− and x− ≤ y − ⊕ e− = (e · y)− ) ⇔ (e · x ≤ y and e · y ≤ x) ⇔ (x · e ≤ y and y · e ≤ x) ⇔ x · e = y · e. 2 Denote T (V (e)) = Ae = A/ ≡e and x/e = x/ ≡e , for any e ∈ B(A) and x ∈ A. If e, f ∈ B(A) and V (e) ⊆ V (f ), then e ≤ f so one can define a morphism of pseudo-MV algebras : ρf e : Af −→ Ae by putting ρf e (x/f ) = x/e, for any x ∈ A. Similarly to [10], by using Proposition 4.3 and Lemma 4.15, one can prove that the assignment: V (e) 7−→ T (V (e)),

V (e) ⊆ V (f ) 7−→ ρf e : T (V (f )) −→ T (V (e)) 22

can be extended to a Boolean sheaf T of pseudo-MV algebras. T will be called the Pierce sheaf of A. This construction was done, in various forms, for rings (see eg. [16], [17]), for distributive lattices ([7], [12]), for MV-algebras ([18], [19], [11], [10]) and for other algebraic structures. We remark that A is isomorphic to the pseudo-MV algebra of global sections of the Pierce sheaf T ( the proof is like in the case of commutative rings (see [16], p.183) ). A natural problem is to characterize some classes of pseudo-MV algebras in terms of the stalks of Pierce sheaf. • For any Boolean ideal P of B(A), consider the congruence ≡P of A associated with the normal ideal (P ] of A: x ≡P y ⇔ d1 (x, y) ∈ (P ] ⇔ d2 (x, y) ∈ (P ], hence x ≡P y

⇔

there is e ∈ P,

d1 (x, y) ≤ e

⇔

there is e ∈ P,

d2 (x, y) ≤ e.

If P ∈ X, then one can prove that AP = A/ ≡P is the stalk of the Pierce sheaf T at P . For x ∈ A, let us denote xP = x/ ≡P . It is easy to prove that a pseudo-MV algebra A is directly indecomposable iff B(A) = {0, 1} (the proof is like in MV algebras case; see [9], Theorem 6.4.7). Proposition 4.16 For any P ∈ X, the stalk AP is directly indecomposable. Proof. • Consider xP ∈ B(AP ); hence xP ⊕ xP = xP , i.e. d1 (x ⊕ x, x) = (x ⊕ x) · x− ≤ e, for some e ∈ P . It follows that (x ⊕ x) · (e ⊕ x)− = (x ⊕ x) · x− · e− = 0, by Proposition 1.9(f); hence x ⊕ x ≤ e ⊕ x. Then, x ⊕ e ≤ (x ⊕ e) ⊕ (x ⊕ e) = x ⊕ (e ⊕ x) ⊕ e = x ⊕ (x ⊕ e) ⊕ e = (x ⊕ x) ⊕ (e ⊕ e) = (x ⊕ x) ⊕ e ≤ (e ⊕ x) ⊕ e = (x ⊕ e) ⊕ e = x ⊕ (e ⊕ e) = x ⊕ e, by Proposition 4.3(1). Thus (x ⊕ e) ⊕ (x ⊕ e) = x ⊕ e and therefore x ⊕ e ∈ B(A), by Proposition 4.2. • Similarly, x− ⊕ f ∈ B(A), for some f ∈ P . • Let a = e ⊕ f ∈ P ; remark that a = f ⊕ e also, by Proposition 4.3(1). Then x ⊕ a ∈ B(A) and x− ⊕ a ∈ B(A). But x ∧ x− = (x ⊕ (x− )∼ ) · x− = (x ⊕ x) · x− ≤ e ≤ a, by Proposition 1.13. Hence (x∧x− )⊕a = (x⊕a)∧(x− ⊕a) ≤ a⊕a = a ∈ P , by Proposition 1.15; it follows that (x⊕a)∧(x− ⊕a) ∈ P , since P is an ideal of B(A). But P is a prime ideal of B(A), therefore x ⊕ a ∈ P or x− ⊕ a ∈ P ; it follows that x ∈ P or x− ∈ P , i.e. d1 (x, 0) = x ≤ x ∈ P or d1 (x− , 0) = x− ≤ x− ∈ P ; hence x ≡P 0 or x− ≡P 0, i.e. xP = 0P or x− 2 P = 0P . Then B(AP ) = {0P , 1P }, hence AP is directly indecomposable. Let (L, ∨, ∧, 0, 1) be a bounded distributive lattice. For a ∈ L let us denote a0 = {x ∈ L | x ∧ a = 0},

a# = {x ∈ L | x ∨ a = 1},

(a] = {x ∈ L | x ≤ a},

[a) = {x ∈ L | a ≤ x}.

L is a Stone algebra if for any a ∈ L, there is b ∈ B(L) such that a0 = (b]. L is a double Stone algebra if it is a Stone algebra and if the dual of the lattice L is also a Stone algebra. A Stone algebra can also be defined as a structure (L, ∨, ∧, ∗ , 0, 1), where (L, ∨, ∧, 0, 1) is a bounded distributive lattice and ”∗ ” is an unary operation such that 0∗ = 1, x ∧ x∗ = 0 and (x ∧ y)∗ = x∗ ∨ y ∗ , for any x, y ∈ L. Lemma 4.17 If A is a pseudo-MV algebra, then (A, ∨, ∧, 0, 1) is a Stone algebra iff (A, ∨, ∧, 0, 1) is a double Stone algebra. Lemma 4.18 If A is a pseudo-MV chain, then (A, ∨, ∧, 0, 1) is a Stone algebra. Proof. Consider a ∈ A. If a = 0, then a0 = (1]; if a 6= 0, then a0 = (0]. Lemma 4.19 If A is a pseudo-MV algebra, then the following are equivalent: (1) A is directly indecomposable and (A, ∨, ∧, 0, 1) is a Stone algebra, (2) A is a pseudo-MV chain. 23

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Proof. (1) =⇒ (2): We shall prove that in A the following holds x ∧ y = 0 ⇒ x = 0 or y = 0.

(10)

Assume x ∧ y = 0; then x ∈ y 0 . But A is a Stone algebra, hence y 0 = (e], for some e ∈ B(A) = {0, 1}. If e = 0, then y 0 = {0}, so x = 0; if e = 1, then y 0 = A, so y = 1 ∧ y = 0. Thus, (10) is proved. Let us consider a, b ∈ A; we have a · b− ∧ b · a− = 0, by Proposition 1.24(i). Then, by (10), a · b− = 0 or b · a− = 0, i.e. a ≤ b or b ≤ a, by Proposition 1.9(f). Hence A is a pseudo-MV chain. (2) =⇒ (1): By Lemma 4.18, A is a Stone algebra. It is then obvious that B(A) = {0, 1}, hence A is directly indecomposable. 2 Lemma 4.20 Let A be a pseudo-MV algebra such that (A, ∨, ∧, 0, 1) is a Stone algebra. If P is a prime ideal of B(A), then (P ] is a normal prime ideal of A and AP is a pseudo-MV chain. Proof. One remarks that any quotient of a Stone algebra is a Stone algebra, so the Lemma follows by Proposition 4.16 and Lemma 4.19. 2 Proposition 4.21 Assume that (A, ∨, ∧, 0, 1) is a Stone algebra. Then T is a Hausdorff Boolean sheaf whose stalks are pseudo-MV chains. Proof. For a ∈ A let us denote by λ(a) the unique element of B(A) such that a0 = (λ(a)]. Then for all x ∈ A, x ∧ a = 0 iff x ≤ λ(a). For any a ∈ A we shall prove that {P ∈ X | aP = 0P } = V (λ(a)). Indeed, if aP = 0P , then a ≤ e for some e ∈ P , so a ∧ e− = a · e− = 0, by Proposition 4.3. Then e− ≤ λ(a), so (λ(a))− ≤ e. This yields (λ(a))− ∈ P , so λ(a) 6∈ P , i.e. P ∈ V (λ(a)). Conversely, if λ(a) 6∈ P , then (λ(a))− ∈ P , so a ∈ (P ], because a ≤ (λ(a))− . If a, b ∈ A, then {P ∈ X | aP = bP } = {P ∈ X | d1 (a, b)P = 0P } = V (λ(d1 (a, b))). This last term is a clopen set of X, so T is a Hausdorff Boolean sheaf. By Lemma 4.20, the stalks AP , P ∈ X of T are pseudo-MV chains. 2 The next theorem generalizes a well-known result proved by Torrens for MV algebras ([18], [19]). Theorem 4.22 If A is a pseudo-MV algebra, then the following are equivalent: (1) (A, ∨, ∧, 0, 1) is a Stone algebra, (2) (A, ∨, ∧, 0, 1) is a double Stone algebra, (3) The stalks of the Pierce sheaf T of A are pseudo-MV chains, (4) A is isomorphic to the pseudo-MV algebra of global sections of a Hausdorff Boolean sheaf whose stalks are pseudo-MV chains. Proof. (1) ⇐⇒ (2), by Lemma 4.17. (1) =⇒ (3), by Lemma 4.20. (3) =⇒ (4), obvious. (4) =⇒ (1): Let us consider a Hausdorff Boolean sheaf F of pseudo-MV algebras defined on a Boolean space X such that every stalk Fx (x ∈ X) is a pseudo-MV chain. We shall prove that the pseudo-MV algebra F (X) of global sections of F has a subjacent structure of Stone algebra. Consider a ∈ F (X); then U = {x ∈ X | ax = 0x } is a clopen, because F is a Hausdorff Boolean sheaf. X being a Boolean space, one can find a∗ ∈ F (X) such that a∗ |U = 1 |U and a∗ |X−U = 0 |X−U . In this way, one obtains an unary operation ∗ on F (X) and (F (X), ∨, ∧, ∗ , 0, 1) is a Stone algebra. We shall verify only the equality (a ∧ b)∗ = a∗ ∨ b∗ , for all a, b ∈ F (X). For any x ∈ X, the stalk Fx is a pseudo-MV chain; thus the following equivalences hold: (a ∧ b)∗x = 1x ⇐⇒ ax ∧ bx = (a ∧ b)x = 0x ⇐⇒ (ax = 0x or bx = 0x ) ⇐⇒ (a∗x = 1x or b∗x = 1x ) ⇐⇒ a∗x ∨ b∗x = 1x . Thus the desired equality follows. 2 Acknowledgements The authors acknowledge partial support by Grant C.N.C.S.I.S. 101/1999 of the Ministry of National Education.

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