aa r X i v : . [ m a t h . L O ] A p r On Simple finitely generated simple polyadicfinitely generated simple polyadic
Tarek Sayed AhmedApril 3, 2019
Abstract
We show that simple finitely generated polyadic equality algebrasmay not be generated by a single element. We prove an analogous resultfor quasi-polyadic equality algebras of infinite dimensions. In contrast,we show that any simple finitely generated infinite dimensional polyadicequality algeba is generated by a one element. The two most famous algebraisations of first order logic are Tarski’s cylin-dric algebras and Halmos’ polyadic algebras. It is commonly accepted thatsuch algebras belong to different paradigms showing a lot of contradictorybehaviour. In this paper, we add one such property (that mentioned in theabstract) to this long list establishing this dichotomy.Let α be an ordinal. Let U be a set. Let Γ ⊆ α, E ⊆ α × α , τ : α → α and X ⊆ α U : c (Γ) X = { s ∈ α U : ∃ t ∈ X, t ( i ) = s ( i ) for all i / ∈ Γ } , s τ X = { s ∈ α U : s ◦ τ ∈ X } , d E = { s ∈ α U : s i = s j for ( i, j ) ∈ E } . The operations are called cylindrifications , substitutions , and diagonal ele-ments , respecively. We use simplified notation for some of these operations c i = c ( { i } ) and d ij = d { ( i,j ) } . Let S be the operation of forming subalgebras, P be that of forming products, and H be the operation of forming homomorphicimages. Then(i) RCA α = SP { ( B ( α U ) , c i , d ij ) i,j<α : U is a set } . (ii) RQEA α = SP { ( B ( α U ) , c i , d ij , s [ i,j ] ) i,j<α : U is a set } . Mathematics Subject Classification.
Primary 03G15.
Key words : algebraic logic,quasi-polyadic algebras, polyadic algebra V = α U ( p ) = { s ∈ α U |{ i ∈ α : s i = p i }| < ω } where p ∈ α U , and the operations are the relativization of the abovedefined operations on V . Dc α is the class of dimension complemented cylindric algbras. A ∈ Dc α ,if ∆ x = α for all x ∈ A . Here ∆ x = { i ∈ α : c i x = x } is called the dimensionset of x . It is known that, Dc α for infinite α is a nice proper generalizationof locally finite algebras (the algebraic counterpart of first order logic, cf [5]section 4.3), sharing a lot of its properties [9]. In this case cylindric andpolyadic algebras coincide. It is not hard to show that in this case simplefinitely generated algebras are generated by a single element.Andr´eka and N´emeti have shown that this property for cylindric algebrasdoes not generalize to Cs regα , even when α is finite ≥ We use a fairly diect modification of their construction to prove the quasipolyadicequality case. Let α be an ordinal > Q denotes the set of rational numbers.Let r ∈ Q , then r = α × { r } . α Q ( r ) is the weak space { s ∈ α Q : | ] { i ∈ α : s i = r }| < ω } and V = α Q ( ) . Throughout r ∈ α Q and t ∈ Q .[ t, r ] = { s ∈ V : t = X { r i s i : i < α }} ,P ol = { [ t, r ] : r ∈ α Q ( ) ∪ α Q ( ) } ,P o = { [ t, r ] ∈ P ol : { } ⊆ Rg ( r ) } ∪ { d ij : i < j < α } ,P of = { [ t, ] : t ∈ Q } and A = S g C P ol.
Her we are taking the closure under all polyadic operations. Let
P er α be theset of all transpositions { [ i, j ] : i, j ∈ α } . For a set X , let X S = { s τ x : x ∈ X, τ ∈ P er α } . Then clearly X ⊆ X S and ( X S ) S = X S . Then, it is absolutelystraightforward to show that P ol S = P ol and
P o S = P o and
P of S = P of .Also a n = [ n, ], 1 A = [0 , ], 0 A = [1 , ] P o ∪ P of ⊆ P ol ⊆ ℘ ( V ) = C , { a n : n ∈ ω } ⊆ P of and { , } ⊆ P ol ∼ ( P o ∪ P of ). Let X ⊆ C be arbitary,then X ∗ = { Y { p i : i < n } : n ∈ ω and ( ∀ i < n )( p i ∈ X or − p i ∈ X ) } ,X ∗∗ = { X { π i : i < n } : n ∈ ω and ( ∀ i < n ) π i ∈ X ∗ } ,G ( X ) = ( X S ∪ P o ) ∗∗ . Note the definition of G ( X ) is different than that in [2], it is in fact bigger. Inwhat follows, we show that the proof in [2] survives introducing substitutionscoresponding to transpositions. 2 emma 1.1. ( ∀ Y ⊆ G ( P ol ))( G ( Y ) ⊆ C Proof. [2] Lemma 1. The two technical lemmas 1.1 in and 1.2 in [2] are thesame. G ( Y ) is closed all the cylindric operations is done exactly like the prooffor the CA case. We need to check that G ( Y ) is closed under substitutions.Let p ∈ G ( Y ). Then p ∈ ( Y S ∪ P o ) ∗∗ . Since Y S and P o are closed undersubstitutions, and substitutions are Boolean endomorphisms, then we are done.
Lemma 1.2. A is simple, and every subalgebra of A is simple Proof.
Like the proof of claim 1.1 p. 868 in [2].
Lemma 1.3. (i) nat : P of ∗∗ → G ( P of ) /P oz is an isomorphism ofBoolean algebras, that is P oz ∩ P of ∗∗ = { } (ii) P of ∗∗ is an atomic Boolean algebra, and the set of its atoms is P of .(iii) Let Y ⊆ P of . Then G ( Y ) ∩ P of = Y S and G ( Y ) ∩ P of ∗∗ = ( Y S ) ∗∗ . Proof.
We need to check only the last statement, which is slightly differentfrom the corresponding one in [2], namely that formulated in lemma 2 (iii)p. 868 of [2], for we are allowing substitutions. Let Y ⊆ P ol . First weshow that G ( Y ) ∩ P of ∗∗ ⊆ ( Y S ) ∗∗ . Let d ∈ G ( Y ) ∩ P of ∗∗ . Then Y S ∪ P generates d in G ( P of ). Consider the factor algebra G ( P of ) /P oz . Then Y S ∪ P generates d in G ( P of ). Then
Y /P oz generates d/P oz in G ( P of ) /P oz . But nat : P of ∗∗ → G ( P of ) /P oz is an isomorphism. Since Y S ∪ { d } ⊆ P of ∗∗ ,this implies that Y S generates d in P of ∗∗ , i.e d ∈ Y S . We have proved that G ( Y ) ∩ P of ∗∗ = Y S . Let y ∈ G ( Y ) ∩ P of . Then y ∈ ( Y S ) ∗∗ . But P of is theset of atoms of
P of ∗∗ . Since | P of | ≥ ω , no set of generators generates newatoms in P of ∗∗ . Hence y ∈ Y S We have seen that G ( Y ) ∩ P of = Y S . Definition 1.4.
Let
N eg = G (0) ∼ P oz . Claim 4 . (i) P oz ⊆ G (0) (ii) N eg is an ultrafilter of the Boolean algebra ( G (0) , ∩ , − ) . N eg is the filtergenerated by {− p : p ∈ P o } .(iii) ( ∀ g ∈ P of ∗∗ ∼ { } )( ∀ σ ∈ G (0)[ g · σ ∈ P oz ⇔ σ ∈ P oz ] . (iv) ( ∀ Y ⊆ P of )( ∀ g ∈ G ( Y ))( ∃ ρ ∈ G (0)( ∃ n ∈ ω )( ∃ v ∈ n N eg )( ∃ y ∈ n Y ) ρ + P i ∈ n v i y i ∈ { g, − g } . roof. [2] Lemma 3. We need to check the last item only. Let Y ⊆ P of and g ∈ G ( Y ) = ( Y ∪ P o ) ∗∗ . Then there is a finite W ⊆ Y S such that g ∈ G ( W )and ( ∀ W ⊆ W )( g / ∈ G ( W ). Let n = | W | and { y i : i ∈ n } . Then every y i iseither an element of Y or a substitution corresponding to a transposition of anelement in Y and they are all distinct. Since y i = y j = ⇒ y i ∩ y j = 0, because Y S ⊆ P of , and the latter is an antichain. Since G ( W ) = ( W S ∪ P o ) ∗∗ , wehave that g = X i ∈ n y i · σ i + σ n · Y i ∈ n − y i for some { σ i : i ∈ n } ⊆ G (0) = P o ∗∗ . Then − g = X i ∈ n y i · − σ i + − σ n · Y {− y i : i ∈ n } . Assume σ n ∈ P oz . Then σ n . Q {− y i : i ∈ n } ∈ P oz and( { σ i : i ∈ n } ∩ P oz = 0 = ⇒ { y i · σ i : i ∈ n } ∩ P oz = 0)by W S ⊆ P of . Then ( ∀ W ⊂ W )( g / ∈ G ( W ) and P oz ⊆ G (0) imply { σ i : i ∈ n } ⊆ N eg so we get the desired form. If σ n / ∈ P oz , then − σ n ∈ P oz and wecan work analogously with − g .Let A n = S g C { a i : i < n } . That is A n is the quasi-polyadic equality subalgebra of C , genertaed by the a i ’s i < n . Recall that A n is simple. Then we have our first main Theorem: Theorem 2 . A n cannot be generated by n elements, but can be generatedby n + 1 elements Proof. [2] Claim 1.2 p. 871. Fix n , and let Y = { a i : i < n } . First we showthat ( ∀ X ⊆ A n )[ | X | ≤ n = ⇒ S g C X = A n ]Suppose to the contrary that S g C X = A n for some X = { g i : i < n } . Then A n ⊆ G ( Y ) and A n ⊆ G ( X ). Let { g i : i ∈ n } = X and let j < n. Then weamy suppose that( ∃ I j ⊆ n )( ∃ ρ j ∈ G (0))( ∃ v j : I j → N eg ) g j = ρ j + X { v ji a i : i ∈ I j } . Let these I j ρ j and v j be fixed for every j ∈ n . By Y ⊆ G ( X ) ⊆ G ( { a i : i ∈ S { I j : j inn } we have that S { I j : j ∈ n } = 2 n . Then ∃ k, h ∈ n ( ∀ j ∈ n )[[ k ∈ I j iff h ∈ I j } by the same reasoning in [2] p. 872. Let those k and h be fixed. By a k ∈ G ( X )we have that a k = P B ′ for some finite B ′ ⊆ ( X S ∪ P o ) ∗ . Let B = [ B ′ ∩ ( X ∪ o ) ∗∗ ]. Then, like the proof in [2] p. 872, it can be shown that B ⊆ G (0).Thus B ′ ⊆ G (0) since the latter is closed under substitutions corresponding totranspositions. Thus a k ∈ G (0) which is a contradiction. The rest of the proofis essentially the same as that in [2] p.872.Finally, we should mention that the above proof survives if we replace Q by any field of characteristic 0. However the proof does not work when Q isreplaced by a finite field.Now we turn to the polyadic paradigm. PA α denotes the class of polyadicalgebras of dimension α , while PEA α denotes the class of polyadic equalityalgebras of dimension α . From now on α will be only infinite. We follow [5]for the abstract axiomatixation of such algebras. Definition 4 . Let A ∈ PA α . Let J ⊆ α , an element a ∈ A is independentof J if c ( J ) a = a . J supports a if a is independent of α ∼ J . We write ∆ a forthe least J that supports a ; ∆ a is called the dimension of a . Definition 5 . Let J ⊆ β and A = h A, + , · , − , , , c (Γ) , s τ i Γ ⊆ β,τ ∈ β β be a PA β .Let N r J B = { a ∈ A : c ( β ∼ J ) a = a } . Then Nr J B = h N r J B , + , · , − , c (Γ) , s ′ τ i Γ ⊆ J,τ ∈ α α where s ′ τ = s ¯ τ . The structure Nr J B is an algebra, called the J compression of B . When J = α , α an ordinal, then Nr α B ∈ PA α and is called the neat α reduct of B and its elements are called α -dimensional. Lemma . (1) Let A ∈ PEA α . Then for all β > α there exists B ∈ PEA β such that A ⊆ Nr α B . Furthermore, every element of B is of the form s B σ a where a ∈ A , σ ∈ β β and σ ↾ α is one to one.(2) If A is a generates B and A is simple, then so is B Proof.
For a subset X of an algebra C , we write Ig A X , for the ideal generatedby X . We show that if I is an ideal in B , then the ideal generated by I ∩ A in B coincides with I . We have A = Nr α B . Only one incusion is non-trivial.Let x ∈ I . The c B (∆ x ∼ α ) x ∈ A , hence in I ∩ A . Since x ≤ c B (∆ x ∼ α ) x , we have x ∈ Ig B ( I ∩ A ). So it follows that if I is a ideal in B , and A is simple, so that A ∩ I = { } , then I = { } . Theorem . Every simple
PEA α generated by finitely many elements isgenerated by a single element Proof.
Let A ∈ PEA α be simple, and assume that it is generated by twoelemens x and y . Let β be an ordinal such that β contains at least two distinct5lements k, l not ∈ α , and let B be a dilation of A so that that A = Nr α B , and A generates B , so that B is, in fact, generated by x and y . Let k, l ∈ β ∼ α .Let b = x · d kl + y · d kl . Then we claim that B = S g B b . First not that b · d kl = x · d kl and z − · d kl = y · − d kl . Then c k ( z · d kl ) = c k ( x · d kl ) = x · c k d kl = x, and c k ( z · − d kl ) = c k ( y · − d kl ) = y · c k − d kl = y · c · − d = y. Hence B is generated by b . Since B is a dilation of A , then there exists σ ∈ β β , σ is one to one on α and a ∈ A such that s B σ a = b . We claim that a generates A .Let Ω = β ∼ α . It suffices to show that if x ∈ B , c B (Ω) x = x, then x ∈ S g A { a } .We first treat several cases separately. Call a term τ positive if it is builtup of succesive applications of substitutions and cylindrifications and has nooccurences of complementation.We start studying some positive terms applied b , and show that when theoutput is in A , then it is actually in S g A { a } . We proceed by induction on the number of unary operations. We start by n = 1.(1) Let x = s B σ ′ b , where ∆ x ⊆ α . Then x = s B σ ′ s B σ a . Let µ = σ ◦ σ ′ . Choose τ ∈ β β , such that τ ◦ µ ( α ) ⊆ α , and τ ↾ α ⊆ Id.
Then since ∆ x ⊆ α , we get x = s B µ a = s B τ ◦ µ a = s B ( τ ◦ µ ) ↾ α ) a = s A τ ◦ µ a ∈ S g A { a } . (2) Assume that x = c (∆) b where ∆ ⊆ β . Then x = c B (∆) s B σ a .Choose µ and τ permutations of β such that ( µ ◦ τ ) | α ⊆ Id , µ ◦ τ ◦ σ ( α ) ⊆ α and µ ◦ τ (∆) ⊆ α . Then c B (∆) s A σ a = s Id c B (∆) s B σ a = s A µ ◦ τ ) ↾ α c B (∆) s B σ a = s B µ s τ c B (∆) s A σ a = s µ c Γ s τ s σ a = c ∆ ′ s µ s τ s σ a = c (∆ ′ ) s µ ◦ τ ◦ σ a = c (∆ ′ ) s µ ◦ τ ◦ σ a Now we have shown that if we have a term τ of the form f ◦ f ◦ . . . f n where the f i ’s are either cylindrifications or substitutionsand τ ( b ) ∈ A , then τ ( b ) ∈ S g A ( a ).Now we consider the Boolean join + and complementation. We start bycomplementation applied to unary terms. If τ is a unary term consisting ofonly substitutions, then − τ ( b ) = τ ( − b ) since the substitutions are Boolean6ndomorphisms, and we are done. Now let c ∂ (∆) be the dual of c (∆) defined by − c ∆ − If complementation is applied to a unary term containg cylindrifications,then by noting that − c (Γ) a = c ∂ (Γ) − x , and the polyadic axioms P11, and P12involving the interaction of substitutions and cylindrifications, hold for theduals of cylindrifications, we are done in this case, too.Now we consider the Boolean join. Let x ∈ A and assume that x = x + x .Let Γ = β ∼ α . Then c (Γ) x = c (Γ) x + c (Γ) x , hence c (Γ x i = x i , so that x i ∈ A for i = 1 , . Now x i ∈ B so we can assume that there exists unary terms(possibly) with negations, such that τ and τ such that x i = τ i ( b ) . But then x i = τ ′ i ( a ) and we are done. References [1] Amer, M., Sayed Ahmed, T.
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