Soft Comput DOI 10.1007/s00500-015-1714-5

FOUNDATIONS

The Belluce-lattice associated with a bounded Hilbert algebra Dumitru Bu¸sneag1 · Dana Piciu1

© Springer-Verlag Berlin Heidelberg 2015

Abstract In this paper, for a bounded Hilbert algebra A we construct the Belluce-lattice associated with A and define the notion of reticulation for A. Keywords Belluce-lattice · Hilbert algebras · Prime spectrum · Maximal spectrum · Reticulation · Bounded distributive lattices

1 Introduction The reticulation was first defined for commutative rings by Simmons (1980) and it was extended by Belluce to noncommutative rings in Belluce (1991). The reticulation of a ring R is a bounded distributive lattice L(R) such that the prime spectrum of R, endowed with the Zariski topology, is homeomorphic to the prime spectrum of L(R), endowed with the Stone topology. By this connection many properties can be transported from R to L(R) and vice versa. Hence, a natural problem is to define a reticulation for others classes of universal algebra. This was done by Belluce for M V -algebras in Belluce (1986), Georgescu for quantales in Georgescu (1995) (which

Communicated by A. Di Nola.

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Dumitru Bu¸sneag [email protected] Dana Piciu [email protected]

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Department of Mathematics, Faculty of Mathematics and Natural Sciences, University of Craiova, 13, Al.I. Cuza st., 200585 Craiova, Romania

constitute a good abstraction of the lattice of congruences for many types of algebraic structures), Leu¸stean for B Lalgebras in Leustean (2003) and Mure¸san for residuated lattices in Mure¸san (2008, 2010). So, generally speaking, the reticulation for an algebra A of types mentioned above is a pair (L A , λ) consisting of a bounded distributive lattice L A and a surjection λ : A → L A so that the function given by the inverse image of λ induces (by reticulation) a homeomorphism of topological spaces between the prime spectrum of L A and that of A. This construction allows many properties to be transferred between L A and A. Using the model of the above structures, in this paper we will define the reticulation of a bounded Hilbert algebra and prove several of its properties. The paper is divided in five sections. In the first two sections, we recall all the preliminary notations and results relative to Hilbert algebra and relevant to the paper. In Sect. 3, we study the prime spectrum Spec(A) and the maximal spectrum Max(A) of a bounded Hilbert algebra A, following a standard method as in Atiyah and Macdonald (1969). It turns out that Max(A) is compact Hausdorff topological space (Theorem 27). In Sect. 4 we construct the Belluce-lattice L A associated with a bounded Hilbert algebra A (Proposition 31) and prove that the topological spaces Max(A) and Max(L A ) are homeomorphic (Theorem 42). In Sect. 5 we define the notion of reticulation of a bounded Hilbert algebra A (Definition 10) and prove the uniqueness of the reticulation (Theorem 43). In Sect. 6 we define the notion of normal bounded Hilbert algebra (using the model of lattices) and prove that if A is a normal bounded Hilbert algebra, then L A is normal lattice (Corollary 51).

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2 Preliminaries Definition 1 (Bu¸sneag 2006; Diego 1966; Monteiro 1980) A Hilbert algebra is an algebra (A, →, 1) of type (2,0) such that the following axioms are fulfilled for every x, y, z ∈ A : (a1 ) x → (y → x) = 1; (a2 ) (x → (y → z)) → ((x → y) → (x → z)) = 1; (a3 ) If x → y = y → x = 1, then x = y. In Diego (1966) it is proved that the system of axioms {a1 , a2 , a3 } is equivalent with the system {a4 , a5 , a6 , a7 }, where: (a4 ) x → x = 1; (a5 ) 1 → x = x; (a6 ) x → (y → z) = (x → y) → (x → z); (a7 ) (x → y) → ((y → x) → x) = (y → x) → ((x → y) → y). For examples of Hilbert algebras, see Bu¸sneag (2006), Diego (1966) and Monteiro (1980). If A is a Hilbert algebra, then the relation x ≤ y iff x → y = 1 is a partial order on A (which will be called the natural ordering on A); with respect to this ordering 1 is the largest element of A. A bounded Hilbert algebra is a Hilbert algebra A with a smallest element 0; in this case for x ∈ A we denote x ∗ = x → 0. A Hilbert algebra is called trivial if A = {1} and nontrivial if A = {1}. For x1 , . . . , xn , x ∈ A(n ≥ 1) we define (x1 , x2 , . . . , xn−1 ; xn ) = xn if n = 1 and x1 → (x2 , . . . , xn−1 ; xn ) if n ≥ 2. In Bu¸sneag (2006), Diego (1966), Monteiro (1980), we have the following rules of calculus for x, y, z ∈ A : (h 1 ) x → 1 = 1; (h 2 ) x ≤ y → x; (h 3 ) x ≤ (x → y) → y; (h 4 ) ((x → y) → y) → y = x → y; (h 5 ) If x ≤ y then z → x ≤ z → y and y → z ≤ x → z; (h 6 ) x → y ≤ (y → z) → (x → z); (h 7 ) x → (y → z) = y → (x → z); (h 8 ) If σ is a permutation of {1, 2, . . . , n} and x1 , . . . , xn , x ∈ A then, (xσ (1) , . . . , xσ (n) ; x) = (x1 , . . . , xn ; x); (h 9 ) x → (x1 , x2 , . . . , xn−1 ; xn ) = (x, x1 , . . . , xn−1 ; xn ) = · · · = (x1 , x2 , . . . , xn−1 , x; xn ); (h 10 ) (x1 , . . . , xn ; x → y) = (x1 , . . . , xn ; x) → (x1 , . . . , xn ; y). If A is a bounded Hilbert algebra, for x, y ∈ A we denote x  y = x ∗ → y and x  y = (x → y ∗ )∗ .

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In a bounded Hilbert algebra A, for x, y, z ∈ A we have the following rules of calculus [see Bu¸sneag (2006) and Bu¸sneag (2011)]: (h 11 ) 0∗ = 1, 1∗ = 0; (h 12 ) x → y ∗ = y → x ∗ ; (h 13 ) x → x ∗ = x ∗ , x ∗ → x = x ∗∗ ; (h 14 ) If x ≤ y then y ∗ ≤ x ∗ ; (h 15 ) x ≤ x ∗∗ , x ∗∗∗ = x ∗ ; (h 16 ) (x → y)∗∗ = x ∗∗ → y ∗∗ = x → y ∗∗ ; (h 17 ) x → y ≤ y ∗ → x ∗ ; (h 18 ) x  0 = 0, x  1 = x ∗∗ , x  x ∗ = 0 and x  x = x ∗∗ ; (h 19 ) x  y = y  x ≤ x ∗∗ , y ∗∗ and x ∗∗  y ∗∗ = x  y; (h 20 ) If x ≤ y then x  y = x ∗∗ and x  z ≤ y  z; (h 21 ) x  (y  z) = (x  y)  z; (h 22 ) x, y ≤ x  y, x  x = x ∗∗ , x  0 = x ∗∗ , x  1 = 1, x  x ∗ = 1; (h 23 ) x  (y → z) = (x  y) → (x  z); (h 24 ) x  (y  z) = (x  y)  z = y  (x  z). From the associativity of  and , for x1 , . . . , xn ∈ A, n ≥ 3, we can define x1  x2  · · ·  xn = x1  (x2  (. . .  (xn−1  xn ) . . .) and x1  x2  . . .  xn = x1  (x2  (. . .  (xn−1  xn ) . . .). Clearly, (h 25 ) x1  x2  · · ·  xn = (x1 → (x2 → (. . . → (xn−1 → xn∗ ) . . .)∗ = (x1 , x2 , . . . , xn−1 ; xn∗ )∗ . Also, ∗ → (h 26 ) x1  x2  . . .  xn = x1∗ → (x2∗ → . . . → (xn−1 ∗ ∗ ∗ xn ) . . .) = (x1 , x2 , . . . , xn−1 ; xn ).

So, (h 27 ) (x1  x2  . . .  xn )∗ = (x1 , x2 , . . . , xn−1 ; xn∗ ). Proposition 1 Let A be a bounded Hilbert algebra and x, y, z ∈ A. Then: (h 28 ) (x  y)∗ = x ∗  y ∗ ; (h 29 ) (x  y)∗ = x ∗  y ∗ ; (h 30 ) If x  z ≤ y, then z ≤ x → y ∗∗ ; (h 31 ) x  (x → y) = x  y ≤ y ∗∗ ; (h 32 ) x  (y  z) ≤ (x  y)  (x  z).

The Belluce-lattice associated…

Proof (h 28 ). We have x ∗  y ∗ = x ∗∗ → y ∗ = y → x ∗∗∗ = y → x ∗ and (x  y)∗ = (x → y ∗ )∗∗ = x → y ∗∗∗ = x → y ∗ , hence (x  y)∗ = x ∗  y ∗ . (h 29 ). We have x ∗  y ∗ = (x ∗ → y ∗∗ )∗ = ((x ∗ → y)∗∗ )∗ = (x ∗ → y)∗∗∗ = (x ∗ → y)∗ = (x  y)∗ . (h 30 ). If x  z ≤ y, then (x → z ∗ )∗ ≤ y, so y ∗ ≤ (x → ∗ z )∗∗ = x → z ∗∗∗ = x → z ∗ , hence x ≤ y ∗ → z ∗ = z → y ∗∗ . (h 31 ). We have x  (x → y) = (x → (x → y)∗ )∗ = ((x → y) → x ∗ )∗ = (x → (y → 0))∗ = (x → y ∗ )∗ = x  y ≤ y ∗∗ . (h 32 ). We have x  (y  z) = (x → (y  z)∗ )∗ = (x → ∗ (y → z)∗ )∗ and (x  y)  (x  z) = (x  y)∗ → (x  z) = (x → y ∗ )∗∗ → (x → z ∗ )∗ = (x → z ∗ ) → (x → y ∗ )∗∗∗ = (x → z ∗ ) → (x → y ∗ )∗ , so h 32 is equivalent with (x → (y ∗ → z)∗ )∗ ≤ (x → z ∗ ) → (x → y ∗ )∗ or x → z ∗ ≤ (x → (y ∗ → z)∗ )∗ → (x → y ∗ )∗ . Since (x → (y ∗ → z)∗ )∗ → (x → y ∗ )∗ = (x → ∗ y ) → (x → (y ∗ → z)∗ )∗∗ = (x → y ∗ ) → (x → (y ∗ → z)∗∗∗ ) = (x → y ∗ ) → (x → (y ∗ → z)∗ ) = x → (y ∗ → (y ∗ → z)∗ ) = x → ((y ∗ → z) → y ∗∗ ) = x → (y ∗ → (z → 0)) = x → (y ∗ → z ∗ ) = y ∗ → (x → z ∗ ), finally, h 32 is equivalent with x → z ∗ ≤ y ∗ → (x → z ∗ ), which is obvious.  Proposition 2 Let A be a bounded Hilbert algebra and x1 , x2 , . . . , xn , a ∈ A (n ≥ 2). If (x1 , x2 , . . . , xn ; a) = 1, then x1  x2  · · ·  xn ≤ a ∗∗ . Proof Mathematical induction relative to n. Consider n = 2 and (x1 , x2 ; a) = 1, that is, x1 → (x2 → a) = 1. From a ≤ a ∗∗ we deduce that 1 = x1 → (x2 → a) ≤ x1 → (x2 → a ∗∗ ), hence x1 → (x2 → a ∗∗ ) = 1, that is, x1 ≤ x2 → a ∗∗ = a ∗ → x2∗ . Then a ∗ ≤ x1 → x2∗ , hence (x1 → x2∗ )∗ ≤ a ∗∗ , that is, x1  x2 ≤ a ∗∗ . Suppose that the assertion is true for n−1 and suppose that (x1 , x2 , . . . , xn ; a) = 1. Since 1 = (x1 , x2 , . . . , xn ; a) = (x1 , x2 , . . . , xn−1 ; xn → a) then x1  x2  · · ·  xn−1 ≤ h

16 (xn → a)∗∗ = xn → a ∗∗ , hence (x1  x2  · · ·  xn−1 ) → (xn → a ∗∗ ) = 1. Then (x1  x2  · · ·  xn−1 )  xn ≤  (a ∗∗ )∗∗ = a ∗∗ , that is, x1  x2  · · ·  xn ≤ a ∗∗ .

We denote by Ds(A) the set of all deductive systems of A. If A is bounded, then a deductive system D is proper iff 0∈ / D. If A is a Hilbert algebra and S ⊆ A is a nonempty set, by [S) we denote the lowest deductive system of A (relative to inclusion) which contains S; [S) will be called the deductive system of A generated by S. If D ∈ Ds(A) and a ∈ A\D, then we denote D(a) = [D ∪ {a}). Proposition 3 (Bu¸sneag 2006) Let A be a Hilbert algebra and a1 , a2 , . . . , an ∈ A. Then [{a1 , a2 , . . . , an }) = {x ∈ A : (a1 , a2 , . . . , an ; x) = 1}. In particular, if a ∈ A, then [{a}) = {x ∈ A : x ≥ a} = [a) ([a) will be called principal). In Tarski (1956) Tarski proves that [S) = ∪{[F) : F ⊆ S, F finite}. Theorem 4 (Bu¸sneag 2006; Diego 1966; Monteiro 1980) Let A be a Hilbert algebra and S ⊆ A is a nonempty set, D ∈ Ds(A) and a ∈ A\D. Then: (i) [S) = {x ∈ A : there are a1 , a2 , . . . , an ∈ S such that (a1 , a2 , . . . , an ; x) = 1}; (ii) D(a) = {x ∈ A : a → x ∈ D}; (iii) (Ds(A), ⊆) is a complete distributive lattice, where for D1 , D2 ∈ Ds(A), D1 ∧ D2 = D1 ∩ D2 and D1 ∨ D2 = [D1 ∪ D2 ) = {x ∈ A : there are a1 , a2 , . . . , an ∈ D1 such that (a1 , a2 , . . . , an ; x) ∈ D2 }. A proper deductive system P of a Hilbert algebra A is said to be irreducible (prime) if it is a meet-irreducible (meetprime) element of the lattice Ds(A). Since (Ds(A), ⊆) is distributive, then the notions of irreducible and prime coincides. We denote by Spec(A) the set of all prime deductive systems of A. Theorem 5 (Bu¸sneag 2006; Celani and Jansana 2012; Diego 1966; Monteiro 1980) Let A be a Hilbert algebra and P ∈ Ds(A), P = A. Then the following are equivalent:

(a8 ) 1 ∈ D; (a9 ) If x, x → y ∈ D, then y ∈ D.

(i) P ∈ Spec(A); (ii) If D1 ∩ D2 ⊆ P with D1 , D2 ∈ Ds(A), then D1 ⊆ P or D2 ⊆ P; (iii) If P = D1 ∩ D2 with D1 , D2 ∈ Ds(A), then P = D1 or P = D2 ; (iv) For every x, y ∈ A\P, there is z ∈ A\P such that x, y ≤ z; (v) For every x, y ∈ A\P, there is z ∈ A\P such that x → z, y → z ∈ P.

It is immediate that {1} and A are deductive systems of A; every deductive system different from A will be called proper.

Theorem 6 (Bu¸sneag 2006; Celani and Jansana 2012; Monteiro 1980) Let A be a Hilbert algebra, D ∈ Ds(A) and a ∈ A\D. Then:

2.1 Deductive systems in a Hilbert algebra Definition 2 If A is a Hilbert algebra, a subset D of A will be called deductive system if

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(i) There is P ∈ Spec(A) such that D ⊆ P and a ∈ / P; (ii) If a  b, then there is P ∈ Spec(A) such that a ∈ P and b ∈ / P; (iii) Every deductive system D is the intersection of the prime deductive system that contain D. A proper deductive system M of a Hilbert algebra A is said to be maximal if it is a maximal element in the lattice Ds(A), that is, if it is not a proper subset of any proper deductive system of A. We denote by Max(A) the set of all maximal deductive systems of A. For x, y ∈ A we denote x y = (x → y) → ((y → x) → x) = y x. Theorem 7 (Bu¸sneag 2006; Celani and Jansana 2012; Diego 1966; Monteiro 1980; Ta¸sc˘au 2007) Let A be a Hilbert algebra and M ∈ Ds(A), M = A. The following are equivalent: (i) M ∈ Max(A); (ii) If x, y ∈ / M, then x → y ∈ M; (iii) If x y ∈ M, then x ∈ M or y ∈ M. Theorem 8 (Bu¸sneag 2006; Dan 2009; Monteiro 1980) Let A be a bounded Hilbert algebra and M ∈ Ds(A), M = A. The following are equivalent: (i) M ∈ Max(A); (ii) If x ∈ / M, then x ∗ ∈ M; (iii) If x  y ∈ M, then x ∈ M or y ∈ M.

Proof We have y → (x  y) = y → (x → y ∗ )∗ = (x → y ∗ ) → y ∗ ∈ D (since by h 3 , x ≤ (x → y ∗ ) → y ∗ ). Since y ∈ D, then x  y ∈ D.  Let A be a Hilbert algebra. Definition 3 (Bu¸sneag 2006; Monteiro 1980; Porta 1981) An element r ∈ A will be called regular if (r → x) → x = x for every x ∈ A. An element d ∈ A will be called dense if d → r = r for every regular element r ∈ A. We denote by R(A) (D(A)) the set of all regular (dense) elements of A. Theorem 13 (Bu¸sneag 2006; Monteiro 1980; Porta 1981) If A is a Hilbert algebra, then R(A) is a Hilbert subalgebra of A and D(A) ∈ Ds(A). Corollary 14 (Bu¸sneag 2006; Monteiro 1980; Porta 1981) If A is a bounded Hilbert algebra, then R(A) = {r ∈ A : r ∗∗ = r } and D(A) = {d ∈ A : d ∗ = 0}. Definition 4 If A is a Hilbert algebra, the radical Rad(A) of A is defined as the intersection of all maximal deductive systems of A. For x, y ∈ A, we denote r (x, y) = ((x → y) → x) → x. Theorem 15 (Monteiro 1980; Porta 1981) Let A be a Hilbert algebra. Then: (i) Rad(A) = [{r (x, y) : x, y ∈ A}); (ii) Rad(A) ⊆ D(A); (iii) If A is a bounded Hilbert algebra, then Rad(A) = D(A) = {x ∈ A : x ∗ = 0}.

Corollary 9 If M ∈ Max(A) and x ∗∗ ∈ M, then x ∈ M . Proof If x ∈ / M, then by Theorem 8, (ii), we deduce that x ∗ ∈ M. From x ∗ , x ∗∗ ∈ M we deduce that 0 ∈ M, hence M = A, a contradiction.  A subset S of a bounded Hilbert algebra A is called closed if x, y ∈ S implies x  y ∈ S. Theorem 10 (Dan 2009) Let A be a bounded Hilbert algebra, D ∈ Ds(A), D = A and S ⊆ A a -closed subset of A such that S ∩ D = ∅. Then there exists M ∈ Max(A) such that D ⊆ M and M ∩ S = ∅. Corollary 11 Let A be a bounded Hilbert algebra and D ∈ Ds(A), D = A. Then there exists M ∈ Max(A) such that D ⊆ M. Proof If D = A, then 0 ∈ / D and apply Theorem 10 with S = {0}.  Lemma 12 If A is a bounded Hilbert algebra, D ∈ Ds(A) and x, y ∈ D, then x  y ∈ D.

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Theorem 16 (Monteiro 1980; Porta 1981) Let A be a Hilbert algebra. If P ∈ Spec(A) and Rad(A) ⊆ P, then P ∈ Max(A). Remark 1 If A is a Hilbert algebra, then Max(A) ⊆ Spec(A). Definition 5 (Monteiro 1980; Porta 1981) A Hilbert algebra A will be called semi-simple (or Tarski algebra) if Rad(A) = {1}. So, A is a Tarski algebra iff (x → y) → x = x, for every x, y ∈ A. Corollary 17 If A is a Tarski algebra, then Max(A) = Spec(A).

3 The topological spaces Spec(A) and Max(A) Let A be a bounded Hilbert algebra. For X ⊆ A, we shall denote D(X ) = {P ∈ Spec(A) : X  P}. For any a ∈ A, D({a}) will be denote by D(a) (that is, D(a) = {P ∈ Spec(A) : a ∈ / P}).

The Belluce-lattice associated…

Proposition 18 Let A be a Hilbert algebra and X, Y ⊆ A. Then: D(∅) = ∅ and D(A) = Spec(A); If X ⊆ Y,then D(X ) ⊆ D(Y ); If F, G ∈ Ds(A), then F ⊆ G iff D(F) ⊆ D(G); D(X ) = D([X )); D(X ) = D(Y ) iff [X ) = [Y ); If {X i : i ∈ I } is a family of subsets of A, then D(∪i∈I X i ) = ∪i∈I D(X i ); (vii) If {Fi : i ∈ I } is a family of deductive systems of A, then D(∨i∈I X i ) = ∪ i ∈ I D(X i ); (viii) D(X ) ∩ D(Y ) = D([X ) ∩ [Y )). (i) (ii) (iii) (iv) (v) (vi)

Proof (i), (ii). Obviously. (iii) ⇒ Apply (ii).  ⇐ If D(F) ⊆ D(G), then for every P ∈ Spec(A), G ⊆ P implies F ⊆ P. Following Theorem 6, (iii), this implies that F ⊆ G. (iv) This is obvious, since every deductive system of A that includes X also includes [X ). (v) Apply (ii)–(iv). (vi) The inclusion ⊇ follows from (ii). If P ∈ D(∪i∈I X i ), then there exists i ∈ I such that X i  P, which is equivalent with P ∈ D(X i ) ⊆ ∪i∈I D(X i ), hence D(∪i∈I X i ) = ∪i∈I D(X i ). (vii) Follows from (iv) and (vi). (viii) Since [X ) ∩ [Y ) ⊆ [X ), [Y ), then by (i) and (iii), D([X ) ∩ [Y )) ⊆ D([X )) ∩ D([Y )) = D(X ) ∩ D(Y ). If P ∈ D(X ) ∩ D(Y ), then P ∈ D([X )) ∩ D([Y )), hence P ∈ Spec(A), [X )  P and [Y )  P. Since P ∈ Spec(A), by Theorem 5, we deduce that [X ) ∩ [Y )  P, hence P ∈ D([X ) ∩ [Y )).  Corollary 19 For any Hilbert algebra A, the family τ A = {D(X ) : X ⊆ A} is a topology on Spec(A), having {D(a) : a ∈ A} as basis. Proof From Proposition 18, (i), (vi) and (vii) we deduce that {D(X ) : X ⊆ A} is a topology on Spec(A). If X ⊆ A, then X = ∪a∈X {a}, hence D(X ) = D(∪a∈X {a}) = ∪a∈X D(a).  Definition 6 The topology τ A from Corollary 19 is called the Zariski topology on Spec(A) and the topological space (Spec(A), τ A ) is called the prime spectrum of A. Remark 2 By Proposition 18, (iv), we deduce that τ A = {D(F) : F ∈ Ds(A)}. Following Remark 1, we can consider on Max(A) the topology induced by Zariski topology. Thus we obtain a topological space called maximal spectrum of A. For any X ⊆ A and a ∈ A, let us define DMax (X ) = D(X ) ∩ Max(A) = {M ∈ Max(A) : X  M} and / M}. DMax (a) = D(a) ∩ Max(A) = {M ∈ Max(A) : a ∈

Corollary 20 The family τ A,Max(A) = {DMax (X ) : X ⊆ A} is the family of open sets of the maximal spectrum of A and the family {DMax (a) : a ∈ A} is a basis for the topology τ A,Max(A) of Max(A). For any X ⊆ A we define V (X ) = {P ∈ Spec(A) : X ⊆ P}, that is, V (X ) = Spec(A)\D(X ). For any a ∈ A, V ({a}) will be denote by V (a) (that is, V (a) = {P ∈ Spec(A) : a ∈ P} = Spec(A)\D(a)). Proposition 21 Let A be a Hilbert algebra and X, Y ⊆ A. Then: (i) V (0) = ∅ and V (∅) = V ({1}) = Spec(A); (ii) If X ⊆ Y , then V (Y ) ⊆ V (X ); (iii) V (X ) = ∅ iff [X ) = A; (iv) V (X ) = Spec(A) iff X = ∅ or X = {1}; (v) V (X ) = V ([X )); (vi) V (X ) = V (Y ) iff [X ) = [Y ); (vii) If F, G ∈ Ds(A), then V (F) = V (G) iff F = G; (viii) V (X ) ∪ V (Y ) = V ([X ) ∩ [Y )). (ix) If {X i : i ∈ I } is a family of subsets of A, then V (∪i∈I X i ) = ∩i∈I V (X i ). Proof (i), (ii). Obviously. (iii) ⇒ Suppose that [X ) = A. By Theorem 6, (i), there is P ∈ Spec(A) such that [X ) ⊆ P. Since X ⊆ [X ), then X ⊆ P, so P ∈ Spec(X ), hence V (X ) = ∅, a contradiction.  ⇐ If V (X )  = ∅, then there is P ∈ Spec(A) such that X ⊆ P. Then A = [X ) ⊆ P, hence A = P, a contradiction. (iv) ⇐ By (i).  ⇒ Suppose that [X )  = ∅ and X  = {1}. Then there is a ∈ X, a = 1. By Theorem 6, (i), there exists P ∈ Spec(A) such that a ∈ / P. Thus, X  P, so P ∈ / V (X ), that is, V (X ) = Spec(A). (v) This is obvious, since every deductive system of A that includes X, also includes [X ). (vi) ⇐ If [X ) = [Y ), then by (v), V (X ) = V ([X )) = V ([Y )) = V (Y ).  ⇒ If [X ) = A, then by (iii), V ([X )) = ∅. Then V ([Y )) = ∅, hence [Y ) = A = [X ). Suppose that [X ), [Y ) are proper deductive systems of A. Applying twice Theorem 6, (iii) and (v), it follows that [X ) = ∩{P ∈ Spec(A) : P ∈ V ([X ))} = ∩{P ∈ Spec(A) : P ∈ V (X )} = ∩{P ∈ Spec(A) : P ∈ V (Y )} = ∩{P ∈ Spec(A) : P ∈ V ([Y ))} = [Y ). (vii) Follows from (vi), since F = [F) and G = [G). (viii) Since [X )∩[Y ) ⊆ [X ), by (ii) and (v) we deduce that V (X ) = V ([X )) ⊆ V ([X )∩[Y )) and V (Y ) ⊆ V ([X )∩[Y )), hence V (X ) ∪ V (Y ) ⊆ V ([X ) ∩ [Y )). If P ∈ V ([X ) ∩ [Y )), then P ∈ Spec(A) and [X ) ∩ [Y ) ⊆ P. By Theorem 5, (ii), we deduce [X ) ⊆ P or [Y ) ⊆ P, hence Since P ∈ V ([X )) ∪ V ([Y )) = V (X ) ∪ V (Y ). Thus, V ([X ) ∩ [Y )) ⊆ V (X ) ∪ V (Y ). (ix) Obviously, from (i). 

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Corollary 22 Let A be a Hilbert algebra and a, b ∈ A. Then:

(vi) V (a) ∪ V (b) ⊆ V (a  b); (vii) If A is bounded, then V (a) ⊆ D(a ∗ ).

(i) If a ≤ b, then D(b) ⊆ D(a); (ii) D(a) = ∅ iff a = 1; (iii) D(a) = Spec(A) iff [a) = A.

Proof (i)–(vi) Follows from Corollary 22, (i)–(vi). (vii) If P ∈ V (a), then a ∈ P. If a ∗ ∈ P, then 0 ∈ P, / P, hence, P = A, a contradiction. Thus, we must have a ∗ ∈  that is, P ∈ D(a ∗ ), hence V (a) ⊆ D(a ∗ ).

If A is bounded, then: (iv) (v) (vi) (vii)

D(a) = Spec(A) iff a = 0; D(a ∗∗ ) ∪ D(b∗∗ ) = D(a  b); D(a) ∩ D(b) ⊇ D(a  b); D(a) = D(b) iff a = b.

Proof (i) If P ∈ D(b), then b ∈ / P. Clearly, a ∈ / P (since if a ∈ P, from a ≤ b we deduce that b ∈ P), hence, P ∈ D(a), that is, D(b) ⊆ D(a). (ii)  ⇐ Clearly, if a = 1, then D(1) = ∅ (since 1 ∈ P for every P ∈ Ds(A)).  ⇒ Suppose D(a) = ∅. If a  = 1,by Theorem 6, (i), there is P ∈ Spec(A) such that a ∈ / P, hence D(a) = ∅, a contradiction. (iii)  ⇒ If [a) = A, then by Theorem 6, we deduce that there is P ∈ Spec(A) such that [a) ⊆ P. Then a ∈ P, hence P∈ / D(a) = Spec(A), a contradiction.  ⇐ Suppose [a) = A. Clearly, D(a) ⊆ Spec(A). If P ∈ Spec(A), then a ∈ / P (if a ∈ P then A = [a) ⊆ P, hence A = P, a contradiction), hence P ∈ D(a), that is, D(a) = Spec(A). (iv) By (iii), D(a) = Spec(A) iff [a) = A iff 0 ∈ [a) iff a ≤ 0 iff a = 0. (v) Since a  b ≤ a ∗∗ , b∗∗ , by (i), we deduce that D(a ∗∗ ), D(b∗∗ ) ⊆ D(a  b), hence D(a ∗∗ ) ∪ D(b∗∗ ) ⊆ D(a  b). / P Consider P ∈ D(a  b), hence a  b ∈ / P. Then a ∗∗ ∈ ∗∗ ∗∗ / P since if by contrary a ∈ P and b∗∗ ∈ P, by or b ∈ Lemma 12 and h 19 , a ∗∗  b∗∗ = a  b ∈ P, a contradiction. So P ∈ D(a ∗∗ ) ∪ D(b∗∗ ), hence D(a  b) ⊆ D(a ∗∗ ) ∪ D(b∗∗ ), that is, D(a ∗∗ ) ∪ D(b∗∗ ) = D(a  b). (vi) Since a, b ≤ a  b by (i), we deduce that D(a  b) ⊆ D(a), D(b), hence D(a  b) ⊆ D(a) ∩ D(b). (vii) By Proposition 18, (v), D(a) = D(b) iff [a) = [b) iff a = b.  Corollary 23 Let A be a Hilbert algebra and a, b ∈ A. Then: (i) (ii) (iii) (iv) (v)

If a ≤ b then V (a) ⊆ V (b); V (a) = ∅ iff [a) = A; If A is bounded, then V (a) = ∅ iff a = 0; V (a) = Spec(A) iff a = 1; V (a ∗∗ ) ∩ V (b∗∗ ) = V (a  b);

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In the sequel, let A be a Hilbert algebra and for a ∈ A, we define DMax (a) = D(a) ∩ Max(A) = {M ∈ Max(A) : a ∈ / M}, VMax (a) = V (a) ∩ Max(A) = {M ∈ Max(A) : a ∈ M}. Clearly, DMax (a) = Max(A)\VMax (a). From Bu¸sneag (2006) and Dan (2009) and Corollary 22 we deduce: Proposition 24 Let A be a bounded Hilbert algebra and a, b ∈ A. Then: (i) (ii) (iii) (iv) (v) (vi)

VMax (0) = ∅ and VMax (1) = Max(A); If a ≤ b, then VMax (a) ⊆ VMax (b); VMax (a ∗∗ ) = VMax (a); VMax (a) ∪ VMax (b) = VMax (a  b); VMax (a) ∩ VMax (b) = VMax (a  b); VMax (a) = VMax (b) iff a ∗∗ = b∗∗ iff a ∗ = b∗ .

Corollary 25 If A is a bounded Hilbert algebra and a, b ∈ A. Then: (i) (ii) (iii) (iv) (v) (vi)

DMax (0) = Max(A) and DMax (1) = ∅; If a ≤ b then DMax (b) ⊆ DMax (a); DMax (a ∗∗ ) = DMax (a); DMax (a) ∩ DMax (b) = DMax (a  b); DMax (a) ∪ DMax (b) = DMax (a  b); DMax (a) = DMax (b) iff a ∗∗ = b∗∗ iff a ∗ = b∗ .

Proposition 26 Let A be a bounded Hilbert algebra. Then for any a ∈ A, DMax (a) is a compact set in Max(A). Proof It is enough to prove that any cover of D(a) with basic open sets contains a finite cover of DMax (a). Let DMax (a) = ∪i∈I DMax (ai ) = DMax (∪i∈I {ai }) [by Proposition 18, (vi)]. By Proposition 18, (v), we get that [a) = [{ai : i ∈ I }), so, a ∈ [{ai : i ∈ I }). By Proposition 3, there are i 1 , . . . , i n ∈ I such that (ai1 , ai2 , . . . , ain ; a) = 1. We shall prove that DMax (a) = DMax (aii ) ∪ . . . ∪ DMax (ain ) = DMax (aii  · · ·  ain ) [see Corollary 23, (v)]. We have that aii  · · ·  ain = (ai1 , ai2 , . . . , ain−1 ; ai∗n )∗ (see h 25 ). From (ai1 , ai2 , . . . , ain ; a) = 1 and Proposition 2 we deduce that aii  · · ·  ain ≤ a ∗∗ .

The Belluce-lattice associated…

By Corollary 23, (ii), (iii), (v), we deduce that DMax (a) = DMax (a ∗∗ ) ⊆ DMax (aii  . . .  ain ) = DMax (aii ) ∪ . . . ∪ DMax (ain ). The other inclusion is obvious, since DMax (aii ) ∪ . . . ∪  DMax (ain ) ⊆ ∪i∈I DMax (ai ) = DMax (a). Theorem 27 If A is a bounded Hilbert algebra, then Max(A) is a compact Hausdorff topological space. Proof Since Max(A) = DMax (0), by Proposition 26 we deduce that Max(A) is compact. Let M, N ∈ Max(A) with M = N . Then M  N or N  M, hence there are x ∈ M\N or y ∈ N \M. / M, hence From x ∈ M and x ∈ / N we deduce that x ∗ ∈ ∗ M ∈ DMax (x ) and N ∈ DMax (x). Moreover, by Corollary 23, (iv) we deduce that DMax (x)∩ DMax (x ∗ ) = DMax (x  x ∗ ) = DMax (1) = ∅. Similarly if y ∈ N \M. Hence Max(A) is Hausdorff. 

4 The Belluce-lattice associated with a bounded Hilbert algebra For the basic definitions, notations and properties relative to lattices used in this paper see Balbes and Dwinger (1974). Definition 7 Let L be a lattice. A nonempty subset F of L is called filter of L iff it satisfies the following conditions: • (a10 ) If x, y ∈ F,then x ∧ y ∈ F; • (a11 ) If x ∈ F, y ∈ L ,and x ≤ y, then y ∈ F. The set of all filters of L is denoted by F(L). A filter F of L is said to be proper if F = L . Proposition 28 (Balbes and Dwinger 1974) If L is a distributive lattice, then (F(L), ⊆) is a distributive lattice. Proposition 29 (Balbes and Dwinger 1974) Let L be a distributive lattice and P ∈ F(L), P = L . The following conditions are equivalent: (i) P is a meet-prime element in F(L) (that is, if F1 , F2 ∈ F(L) and F1 ∩ F2 ⊆ P, then F1 ⊆ P or F2 ⊆ P); (ii) P is a meet-irreducible element in F(L) (that is, if F1 , F2 ∈ F(L) and F1 ∩ F2 = P, then F1 = P or F2 = P); (iii) If x, y ∈ L and x ∨ y ∈ P, then x ∈ P or y ∈ P. Definition 8 Let L be a distributive lattice. A proper filter P of L will be called prime if it verify one of the equivalent conditions of Proposition 29.

We denote by Spec(L) the set of all prime filters of L. Spec(L) is called the prime spectrum of L . As in the case of Hilbert algebras, if L is a lattice and X ⊆ L we shall denote D(X ) = {P ∈ Spec(L) : X  P}. If a ∈ L ,by D(a) we will denote D({a}), that is, D(a) = / P}. {P ∈ Spec(L) : a ∈ It is well known that the family {D(X ) : X ⊆ L} is the topology on Spec(L) (called Stone topology) where the family {D(a) : a ∈ L} is a basis for the Stone topology. We recall (Balbes and Dwinger 1974) that a proper filter M of a lattice L is called maximal if it is a maximal element of the set of all proper filters of L . The set of all maximal filters of L is called the maximal spectrum of L and is denoted by Max(L). Clearly, Max(L) ⊆ Spec(L). If L is a lattice, for any X ⊆ L we shall denote DMax (X ) = {M ∈ Max(L) : X  M}. If a ∈ L , DMax ({a}) will be / denoted by DMax (a), so, DMax (a) = {M ∈ Max(L) : a ∈ M}. If L is a distributive lattice, since Max(L) ⊆ Spec(L), then the family {DMax (X ) : X ⊆ L} is a topology on Max(L) having {DMax (a) : a ∈ L} as a basis. Let A be a bounded Hilbert algebra. For any a, b ∈ A define a ≡ b iff for any M ∈ Max(A),(a ∈ / M iff b ∈ / M) iff for any M ∈ Max(A), (a ∈ M iff b ∈ M). Remark 3 By Proposition 24, (vi) and Corollary 25, (vi),we deduce that for a, b ∈ A, a ≡ b iff VMax (a) = VMax (b) iff DMax (a) = DMax (b) iff a ∗∗ = b∗∗ iff a ∗ = b∗ . Proposition 30 The relation ≡ is a congruence relation on A with respect to  and . Proof It is obvious that ≡ is an equivalence relation on A. Let x, y, z, t ∈ A such that x ≡ y and z ≡ t. We shall prove that x  z ≡ y  t and x  z ≡ y  t . Let M ∈ Max(A). If x  z ∈ M, since x  z ≤ x ∗∗ , z ∗∗ (by h 19 ), then x ∗∗ , z ∗∗ ∈ M, hence x, z ∈ M (by Corollary 9). Since x ≡ y and z ≡ t we deduce that y, t ∈ M. By Lemma 12 we deduce that y  t ∈ M. If x  z ∈ M, by Theorem 8, (iii), x ∈ M or z ∈ M. Since x ≡ y and z ≡ t we deduce that y, t ∈ M, hence y  t ∈ M.  For a ∈ A let us denote by [a] the congruence class of a and by L A = A/ ≡= {[a] : a ∈ A} the quotient set. We also denote by p A : A → L A the canonical surjection defined by p A (a) = [a], for every a ∈ A. On L A the relation [a] ≤ [b] iff for every M ∈ Max(A), a ∈ M implies b ∈ M is an order on A. Remark 4 If a ≤ b, then [a] ≤ [b] [see Proposition 32, (ii)] and [a] ≤ [b] iff [a  b] = [a] iff [a  b] = [b].

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Proposition 31 Relative to the above order on L A , the algebra (L A , ∧, ∨, [0], [1]) is a bounded distributive lattice, where for a, b ∈ A, [a]∧[b] = [ab] and [a]∨[b] = [ab]. Proof Let a, b ∈ A. Clearly, [a  b] ≤ [a], [b]. Consider c ∈ A such that [c] ≤ [a], [b]. To prove [c] ≤ [ab] consider M ∈ Max(A) such that c ∈ M. Then, by the definition we deduce that a, b ∈ M, hence a  b ∈ M, by Lemma 12. So, [a] ∧ [b] = [a  b]. Clearly, [a], [b] ≤ [a  b]. Consider now c ∈ A such that [a], [b] ≤ [c]. To prove [a  b] ≤ [c] consider M ∈ Max(A) such that a  b ∈ M. By Theorem 8, (iii), we deduce that a ∈ M or b ∈ M. In both cases we deduce that c ∈ M, hence [a] ∨ [b] = [a  b] . Clearly, for a ∈ A, [0] ≤ [a] ≤ [1], so (L A , ∧, ∨, [0], [1]) is a bounded lattice. To prove the distributivity of L A , it is suffice to prove that for every a, b, c ∈ A, [a] ∧ ([b] ∨ [c]) ≤ ([a] ∧ [b]) ∨ ([a] ∧ [c]) equivalent with [a  (b  c)] ≤ [(a  b)  (a  c)],  which is true by h 32 and Remark 4. Definition 9 The bounded distributive lattice L A will be called the Belluce-lattice associated with A. Remark 5 For the case of M V -algebras see Belluce (1986), quantales see Georgescu (1995), B L-algebras see Leustean (2003) and residuated lattices see Bu¸sneag and Piciu (2012), Mure¸san (2008) and Mure¸san (2010). Proposition 32 Let A be a bounded Hilbert algebra and a, b ∈ A. Then: (i) (ii) (iii) (iv) (v) (vi) (vii)

[a] ≤ [b] iff DMax (b) ⊆ DMax (a); If a ≤ b, then [a] ≤ [b]; [a] = [b] iff a ∗∗ = b∗∗ iff a ∗ = b∗ ; [a] = [a ∗∗ ]; [a] = [0] iff a = 0; [a] = [1] iff a ∗ = 0; If e ∈ B(A), then [e] ≤ [a] iff e ≤ a ∗∗ .

Proof (i) We have [a] ≤ [b] iff [a] ∧ [b] = [a] iff [a  b] = [a] iff DMax (a) = D Max (a  b) = DMax (a) ∪ DMax (b) iff DMax (b) ⊆ DMax (a). (h )

20 [a ∗∗ ] = [a], (ii) If a ≤ b, then [a] ∧ [b] = [a  b] = by Proposition 24, (iii), hence [a] ≤ [b]. (iii) Follows from Remark 3. (iv) Follows from (iii) and h 15 . (v) By (iii), [a] = [0] iff a ∗∗ = 0∗∗ = 0 iff a = 0, see h 15 . (vi) By (iii), [a] = [1] iff a ∗ = 1∗ = 0. (vii) Suppose e ∈ B(A) [hence e∗∗ = e, see Bu¸sneag (2006)] and [e] ≤ [a]. Then [e] ∧ [a] = [e], hence [e  a] = [e]. Then (e  a)∗∗ = e∗∗ = e. But (e  a)∗∗ = (e → a ∗ )∗∗∗ = (e → a ∗ )∗ = e  a, so e = e  a ≤ a ∗∗ , by h 19 . If e ≤ a ∗∗ , by (ii) we deduce that [e] ≤ [a ∗∗ ] = [a]. 

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We recall (Bu¸sneag 2006) that if A and B are Hilbert algebras, then f : A → B is a morphism of bounded Hilbert algebras if f (x → y) = f (x) → f (y), for every x, y ∈ A. If A and B are bounded Hilbert algebras, we ask that f (0) = 0. We denote by Hi (Hi ) the category of Hilbert (bounded Hilbert) algebras and by Ld(0, 1) the category of bounded distributive lattices. Remark 6 If f : A → B is a morphism in Hi , then for every x, y ∈ A, f (x ∗ ) = ( f (x))∗ , f (x  y) = f (x)  f (y) and f (x  y) = f (x)  f (y). Proposition 33 Let f : A → B a morphism of bounded Hilbert algebras. (i) If F ∈ Ds(B), then f −1 (F) ∈ Ds(A); if F is proper then f −1 (F) is proper; (ii) If M ∈ Max(B), then f −1 (M) ∈ Max(A). Proof (i) Consider F ∈ Ds(B). Since f (1) = 1 we deduce that 1 ∈ f −1 (F). If x, y ∈ A such that x, x → y ∈ f −1 (F), then f (x), f (x → y) = f (x) → f (y) ∈ F. Since F ∈ Ds(B) we deduce that f (y) ∈ F, hence y ∈ f −1 (F), that is, f −1 (F) ∈ Ds(A). If F = B, then 0 ∈ / F. If f −1 (F) = A, then 0 ∈ f −1 (F), hence 0 = f (0) ∈ F, a contradiction. (ii) Suppose that M ∈ Max(B). Then M = B and by (i) we deduce that f −1 (M) = A. To prove f −1 (M) ∈ Max(A) consider x ∈ A\ f −1 (M). / M. By Theorem 8, Then x ∈ / f −1 (M), hence f (x) ∈ ∗ ∗ ∗ f (x ) = ( f (x)) ∈ M, hence x ∈ f −1 (M), that is,  f −1 (M) ∈ Max(A). Lemma 34 Let A, B be two bounded Hilbert algebras and f : A → B a morphism of bounded Hilbert algebras. If a, b ∈ A and DMax (a) = DMax (b), then DMax ( f (a)) = DMax ( f (b)). Proof Let P ∈ Max(B); by Proposition 33, (ii), f −1 (M) ∈ / M Max(B). It follows that M ∈ DMax ( f (a)) iff f (a) ∈ iff a ∈ / f −1 (M) iff f −1 (M) ∈ DMax (a) iff f −1 (M) ∈ / f −1 (M) iff f (b) ∈ / M iff M ∈ DMax ( f (b)). DMax (b) iff b ∈ Hence DMax ( f (a)) = DMax ( f (b)). Let f : A → B a morphism of bounded Hilbert algebras and let us define R( f ) : L A → L B by R( f )([a]) = [ f (a)], for every a ∈ A. By Lemma 34 and Remark 6 we deduce that α( f ) is welldefined.  Proposition 35 Let f : A → B a morphism in Hi . Then R( f ) : L A → L B , R( f )([a]) = [ f (a)], for every a ∈ A, is a morphism in Ld(0, 1).

The Belluce-lattice associated…

Proof Clearly, R( f )(0) = R( f )([0]) = [ f (0)] = [0] = 0 and R( f )(1) = R( f )([1]) = [ f (1)] = [1] = 1. If x, y ∈ A, then R( f )([x] ∧ [y]) = R( f )([x  y]) = [ f (x  y)] = [ f (x)  f (y)] (by Remark 6)= [ f (x)] ∧ [ f (y)] = R( f )([x]) ∧ R( f )([y]) and R( f )([x] ∨ [y]) = R( f )([x  y]) = [ f (x  y)] = [ f (x) f (y)] = [ f (x)]∨[ f (y)] = R( f )([x])∨R( f )([y]).  If for every A ∈ Ob(Hi ) we denote R(A) = L A , then we have defined a functor R : Hi → Ld(0, 1) called the reticulation functor. Proposition 36 Let A, B ∈ Hi . (i) If f : A → B is a morphism in Hi , then R( f ) : L A → L B (defined in Proposition 35) is the unique morphism in Ld(0, 1) such that the diagram f

A −→ B ↓ pB ↓ pA R( f )

L A −→ L B is commutative, i.e., p B ◦ f = R( f ) ◦ p A ; (ii) The functor R : Hi → Ld(0, 1) preserves injective and surjective morphisms. Proof (i) Following Proposition 35, R( f ) is defined for a ∈ A, by R( f )([a]) = [ f (a)]. Since [a] = p A (a) and [ f (a)] = p B ( f (a)) we deduce that R( f )( p A (a)) = p B ( f (a)), so (R( f ) ◦ p A )(a) = ( p B ◦ f )(a), that is, p B ◦ f = R( f ) ◦ p A . (ii) Suppose f is injective and consider a, b ∈ A such that R( f )([a]) = R( f )([b]). Then [ f (a)] = [ f (b)], hence ( f (a))∗ = ( f (b))∗ [by Proposition 32, (iii)], that is, f (a ∗ ) = f (b∗ ). Since f is supposed injective, then a ∗ = b∗ , hence [a] = [b], that is, R( f ) is injective. Suppose now that f is surjective and consider b ∈ A. Then there exists a ∈ A such that b = f (a), so, R( f )([a]) = [ f (a)] = [b], that is, R( f ) is surjective.  For a set S we denote by P(S) = {X ⊆ S}. For a bounded Hilbert algebra A, we consider the map p ∗A : P(L A ) → P(L B ), p ∗A (S) = p −1 A (S) = {a ∈ A : p A (a) = [a] ∈ S}, for every S ⊆ L A . Remark 7 Since p A is a surjective map, then is well known that p ∗A is one-to-one and p A ( p ∗A (S)) = S, for every S ⊆ L A. In the following by A we denote a bounded Hilbert algebra. Lemma 37 If F ∈ F(L A ), then p ∗A (F) ∈ Ds(A).

Proof Since p A (1) = [1] ∈ F, then 1 ∈ p ∗A (F). Consider x, y ∈ A such that x, x → y ∈ p ∗A (F). Then [x], [x → y] ∈ F, hence [x] ∧ [x → y] = [x  (x → y)] ∈ F. By h 31 , x (x → y) ≤ y ∗∗ , hence [x (x → y)] ≤ [y ∗∗ ] = [y], by Corollary 9. So, [y] ∈ F, hence y ∈ p ∗A (F), that is, p ∗A (F) ∈ Ds(A).  Lemma 38 If M ∈ Max(L A ), then p ∗A (M) ∈ Max(A). Proof By Lemma 37, p ∗A (M) ∈ Ds(A). Since p ∗A is one-toone, from M = L A we deduce that p ∗A (M) = p ∗A (L A ) = A, hence p ∗A (M) is proper. To prove that p ∗A (M) ∈ Max(A), consider x, y ∈ A such that x  y ∈ p ∗A (M). Then [x]∨[y] = [x  y] = p A (x  y) ∈ M. Since M ∈ Max(L A ) and Max(L A ) ⊆ Spec(L A ) we deduce that M ∈ Spec(L A ). By Theorem 27, (iii), we deduce that [x] ∈ M or [y] ∈ M, hence x ∈ p ∗A (M) or y ∈ p ∗A (M),  that is, p ∗A (M) ∈ Max(A), by Theorem 8, (iii). Lemma 39 If M ∈ Max(A), then p A (M) ∈ Max(L A ). Proof We have M = A, that is, there is a ∈ A\M. If p A (M) = L A , then p ∗A ( p A (M)) = p ∗A (L A ) = A. Then a ∈ p ∗A ( p A (M)), hence p A (a) = [a] ∈ p A (M), hence there exists b ∈ M such that [a] = [b]. Since a ≡ b and b ∈ M, then a ∈ M, a contradiction. So, M = A implies p A (M) = L A . To prove p A (M) ∈ F(L A ), consider α, β ∈ p A (M), that is, α = [a], β = [b] with a, b ∈ M. Then α∧β = [a]∧[b] = [a  b]. By Lemma 12 we deduce that a  b ∈ M, hence α ∧ β ∈ p A (M). Consider now α ∈ p A (M) and β ∈ L A such that α ≤ β. Then α = [a] with a ∈ M and β = [b] with b ∈ A. Since α ≤ β, then β = β ∨ α = [a] ∨ [b] = [a  b], hence b ≡ (a  b). From a ≤ a  b and a ∈ M we deduce that a  b ∈ M. Since b ≡ (a  b) and a  b ∈ M we deduce that b ∈ M, hence b = [b] ∈ p A (M) . So p A (M) ∈ F(L A ). To prove p A (M) ∈ Max(L A ), consider F ∈ F(L A ) such that p A (M) ⊆ F. Then p ∗A ( p A (M)) ⊆ p ∗A (F). Since M ⊆ p ∗A ( p A (M)), then M ⊆ p ∗A (F). Since p ∗A (F) ∈ Ds(a) (by Lemma 37) and M ∈ Max(A), then M = p ∗A (F) or p ∗A (F) = A. If p ∗A (F) = A, then p A ( p ∗A (F)) = p A (A) = L A , hence F = L A , by Remark 7. If M = p ∗A (F), then p A (M) = p A ( p ∗A (F)) = F. So,  p A (M) ∈ Max(L A ). If A is a bounded Hilbert algebra and L A is the reticulation lattice of A, using Lemmas 38 and 39 we can define two maps u A : Max(A) → Max(L A ), u A (M) = p A (M), for every M ∈ Max(A) and v A : Max(L A ) → Max(A), v A (M) = p ∗A (M), for every M ∈ Max(L A ).

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Proposition 40 If A is a bounded Hilbert algebra the, u A ◦ v A = 1Max(L A ) and v A ◦ u A = 1Max(A) .

of Definition 10). Then there exists a unique isomorphism of bounded lattices f : L 1 → L 2 such that f ◦ λ1 = λ2 .

Proof If M ∈ Max(A), then (v A ◦u A )(M) = v A (u A (M)) = p ∗A ( p A (M)) = p −1 A ( p A (M)). Clearly, M ⊆ p −1 A ( p A (M)). For the reverse inclusion, ( p consider a ∈ p −1 A (M)). Then p A (a) ∈ p A (M), hence A [a] = p A (a) = p A (b) = [b] with b ∈ M. So, a ≡ b and b ∈ M, hence a ∈ M. Thus −1 p −1 A ( p A (M)) ⊆ M, hence p A ( p A (M)) = M, that is, v A ◦ u A = 1Max(A) . If M ∈ Max(L A ), then (u A ◦ v A )(M) = u A (v A (M)) = p A ( p −1 A (M)) = M, by Remark 7, that is, u A ◦ v A =  1Max(L A ) .

Proof If we have two isomorphisms of bounded lattices f, g : L 1 → L 2 such that f ◦ λ1 = g ◦ λ1 = λ2 and b ∈ L 1 , then there exists a ∈ A such that b = λ1 (a). Then f (λ1 (a)) = λ2 (a), hence f (b) = λ2 (a) and g(b) = g(λ1 (a)) = λ2 (a) = f (b), that is f = g. Consider now x ∈ L 1 and a ∈ A such that x = λ1 (a). We define f (x) = λ2 (a). If a, b ∈ A such that x = λ1 (a) = λ1 (b), then by R4 , λ1 (a) = λ1 (b) iff a ∗∗ = b∗∗ iff λ2 (a) = λ2 (b). The direct implication proves that f is well-defined and the converse implication proves that f is injective. The surjectivity of λ2 implies that f is surjective, so f is bijective. For proving that f is a bounded lattice isomorphism it remains to show that f commutes with the operations ∧ and ∨. We have f (0) = f (λ1 (0)) = λ2 (0) = 0, f (1) = f (λ1 (1)) = λ2 (1) = 1. Now let x, y ∈ L 1 . Since λ is surjective, there are a, b ∈ A such that x = λ1 (a) and y = λ1 (b). Applying R2 and R3 we obtain the following equalities: f (x ∧ y) = f (λ1 (a) ∧ λ1 (b)) = f (λ1 (a  b)) = λ2 (a  b) = λ2 (a)∧λ2 (b) = f (x)∧ f (y) and analogous f (x ∨y) = f (x) ∨ f (y). 

Corollary 41 If A is a bounded Hilbert algebra, then u A : Max(A) → Max(L A ) is a bijective map and its inverse is vA. Theorem 42 If A is a bounded Hilbert algebra, then u A : Max(A) → Max(L A ) is a homeomorphism between the topological spaces Max(A) and Max(L A ). Proof By Corollary 41 we deduce that u A is bijective. In order to obtain that u A is a homeomorphism, we shall prove that u A is continuous and open. Let a ∈ A. Then u −1 A (DMax ([a])) = {M ∈ Max(A) : u A (M) ∈ DMax ([a])} = {M ∈ Max(A) : p A (M) ∈ / p A (M)} = {M ∈ DMax ([a])} = {M ∈ Max(A) : [a] ∈ Max(A) : a ∈ / M} = DMax (a). Hence u A is continuous, by Corollary 20. We have u A (DMax (a)) = { p A (M) : M ∈ Max(A), M ∈ / M} = { p A (M) : DMax (a)} = { p A (M) : M ∈ Max(A), a ∈ / M ∈ Max(A), [a] ∈ / p A (M)} = {N ∈ Spec(L A ) : [a] ∈ N } = DMax ([a]).  We have got also that u A is open.

5 The reticulation of a bounded Hilbert algebra

Remark 8 If A is a bounded Hilbert algebra, then the pair (L A , p A ) from Sect. 4 (see Proposition 31) is a reticulation of A in the sense of Definition 10. We recall that in Sect. 3, for a ∈ A we have defined VMax (a) = {M ∈ Max(A) : a ∈ M}. Consider S A = {VMax (a) : a ∈ A} ⊆ P(Max(A)). Following Corollary 25 we deduce that S A is a distributive sublattice of the lattice (P(Max(A)), ⊆). Corollary 44 The pair (S A , VMax ) is a reticulation of A (in the sense of Definition 10). Following Theorem 43 we deduce that the lattices L A and S A are isomorphic.

In this section, by A we denote a bounded Hilbert algebra. Definition 10 A reticulation of A is a pair (L , λ), where L is a bounded distributive lattice and λ : A → L is a surjective map that satisfies the following conditions for every x, y ∈ A: (R1 ) λ(0) = 0 and λ(1) = 1; (R2 ) λ(x  y) = λ(x) ∧ λ(y); (R3 ) λ(x  y) = λ(x) ∨ λ(y); (R4 ) λ(x) = λ(y) iff x ∗∗ = y ∗∗ iff x ∗ = y ∗ . Theorem 43 Let A be a bounded Hilbert algebra and (L 1 , λ1 ), (L 2 , λ2 ) be two reticulations on A (in the sense

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6 Normal Hilbert algebras Let us remind (Cornish 1972) that a bounded distributive lattice is called normal if any prime filter of L is contained in a unique maximal filter of L . Normal lattices were introduced by Wallman (1938) as an abstraction of the lattice of closed sets of a normal topological spaces. We recall the following result: Theorem 45 (Cornish 1972) Let L be a bounded distributive lattice. Then the following are equivalent:

The Belluce-lattice associated…

(i) L is normal; (ii) For every a, b ∈ L , a ∧ b = 0 implies there exist u, v ∈ L such that a ∧ u = b ∧ v = 0 and u ∨ v = 1.

(ii) If x, y ∈ A such that x  y = 0, then there exist z, t ∈ A such that x  z = y  t = 0 and z  t = 1. Then

Using the model of distributive lattices and residuated lattices [see Leustean (2009)], we define the notion of normal (bounded) Hilbert algebra in the following way: Definition 11 A bounded Hilbert algebra A will be called normal if every prime deductive system of A is contained in a unique maximal deductive system. We recall some notions of ideals for posets. Let (P, ≤) a poset. A subset S ⊆ P is a down-set provided that for every a, b ∈ P, if a ≤ b and b ∈ S, then a ∈ S. A non-empty set I ⊆ P is said to be ideal of P if it is up-directed down-set, that is, a down-set P such that for every a, b ∈ I, there exists c ∈ I such that a, b ≤ c. Theorem 46 (Mure¸san 2008) Let A be a bounded Hilbert algebra, D ∈ Ds(A), I an non-empty ideal of (A, ≤) such that D ∩ I = ∅. Then there exists P ∈ Spec(A) such that D ⊆ P and P ∩ I = ∅. Lemma 47 Let A be a bounded Hilbert algebra and a, b, x, y ∈ A. Then (h 33 ) If a ≤ x and b ≤ y then a  b ≤ x  y. Proof From a ≤ x we deduce x ∗ ≤ a ∗ , so a ∗ → b ≤ x ∗ → b. From b ≤ y we deduce x ∗ → b ≤ x ∗ → y, hence  a ∗ → b ≤ x ∗ → y, that is, a  b ≤ x  y. Let A be a bounded Hilbert algebra and x, y ∈ A such that x  y = 0. Consider I x,y = {a ∈ A : a ≤ b  c with b  x = c  y = 0}. Lemma 48 Let A be a bounded Hilbert algebra. If x, y ∈ A such that x  y = 0, then I x,y is an ideal in (A, ≤). Proof Since 0 ≤ y x and y x = x  y = 0, then 0 ∈ I x,y . Clearly, if α ≤ β and β ∈ I x,y , then α ∈ I x,y . Consider now α, β ∈ I x,y and want be prove that α  β ∈ I x,y . We have α ≤ b  c with b  x = c  y = 0 and β ≤ d  e with d  x = e  y = 0. By Lemma 47, we deduce that α  β ≤ (b  c)  (d  e) = (b  d)  (c  e) (by h 24 ). By h 32 , we have: (b  d)  x ≤ (b  x)  (d  x) = 0, hence (b  d)  x = 0, and similarly (c  e)  y = 0, so  α  β ∈ I x,y , that is, I x,y is an ideal in (A, ≤). Theorem 49 Let A be a bounded Hilbert algebra. Consider the following assertions: (i) A is normal;

(1) (i) ⇒ (ii); (2) If A is a Tarski algebra, then (ii) ⇒ (i), hence (i) ⇔ (ii). Proof (1) Consider x, y ∈ A such that x  y = 0 and I x,y = {α ∈ A : α ≤ a  b with a  x = b  y = 0}, which, by Lemma 48, is an ideal in (A, ≤). Clearly, y  x ∈ I x,y . Since x, y ≤ y  x, we deduce that x, y ∈ I x,y . To prove (ii) should be proved that 1 ∈ I x,y . Suppose by contrary that 1 ∈ / I x,y . If in Theorem 46 we consider D = {1}, since D ∩ I x,y = ∅, then there exists P ∈ Spec(A) such that P ∩ I x,y = ∅. / P. If P(x) = A, then Since x, y ∈ I x,y , then x, y ∈ 0 ∈ P(x), hence x ∗ ∈ P. Since x  x ∗ = 0, then x ∗ = x ∗  0 ∈ I x,y , hence x ∗ ∈ P ∩ I x,y = ∅, contradiction. Then P(x) = A and similarly, P(y) = A. Following Corollary 11, there are M1 , M2 ∈ Max(A) such that P(x) ⊆ M1 and P(y) ⊆ M2 . By hypothesis, M1 = M2 = M. Since x, y ∈ M, then 0 = x  y ∈ M (by Lemma 12), hence M = A, a contradiction. (2) Suppose A is a Tarski algebra and suppose by contrary that there exists P ∈ Spec(A) such that P ⊆ M1 ∩ M2 and M1 = M2 .Then there exist x ∈ M1 \M2 or y ∈ M2 \M1 . / M2 . Since Suppose that there exists x ∈ M1 such that x ∈ M2 ∈ Max(A), by Theorem 8, (ii), we deduce that x ∗ ∈ M2 . Since x  x ∗ = 0, by hypothesis there exist z, t ∈ A such that x  z = x ∗  t = 0 and z  t = 1. Since A is a Tarski algebra, by Corollary 17, P ∈ Max(A). Since z  t = 1, by Theorem 8, (iii), we deduce that z ∈ P or t ∈ P. If z ∈ P, then z ∈ M1 . Since x ∈ M1 , by Lemma 12, we deduce that 0 = x  z ∈ M1 , a contradiction. If t ∈ P, then t ∈ M2 and analogously we deduce that  0 = x ∗  t ∈ M2 , a contradiction. Corollary 50 If A is a Tarski algebra then A is normal iff for every x, y ∈ A such that x  y = 0, there exist z, t ∈ A such that x  z = y  t = 0 and z  t = 1. Corollary 51 Let A be a bounded Hilbert algebra. If A is normal, then L A is normal lattice. Proof Consider x, y ∈ A such that [x] ∧ [y] = 0. Then [x  y] = [0], so x  y = 0. Since A is normal Hilbert algebra, then by Theorem 49 we deduce that there exist z, t ∈ A such that x  z = y  t = 0 and z  t = 1. Then [x  z] = [y  t] = [0] = 0 and [z  t] = [1] = 1, hence [x] ∧ [z] = [y] ∧ [t] = 0 and [z] ∨ [t] = 1, that is, L A is a normal lattice. 

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D. Bu¸sneag, D. Piciu Conflict of interest conflict of interest.

This section is to certify that we have no potential

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