CC BY
12
Vladikavkaz Mathematical Journal 2018, Volume 20, Issue 2, P. 86-94
V/J,K 517.5+517.9
DOI 10.23671 /VNC.2018.2.14725
CHARACTERIZATIONS OF FINITE DIMENSIONAL ARCHIMEDEAN VECTOR LATTICES
F. Polat1, M. A. Toumi2 1 Cankiri Karatekin University; 2 University of Carthage
To Professor Anatoly Kusraev with warmest wishes on the occasion of his 65th anniversary
Abstract. In this paper, we give some necessary and sufficient conditions for an Archimedean vector lattice A to be of finite dimension. In this context, we give three characterizations. The first one contains the relation between the vector lattice A to be of finite dimension and its universal completion Au. The
A
two equivalent conditions holds : (a) every maximal modular algebra ideal in Au is relatively uniformly complete or (b) Orth(A, Au) = Z(A, Au) where Orth(A, Au) and Z(A,AU) denote the vector lattice of all orthomorphisms from A to Au and the sublattice consisting of orthomorphisms n with |n(x)| < A|x| (x € A) for some 0 < A e R, respectively. It is well-known that my universally complete vector lattice A is of the form C(X) for some Hausdorff extremally disconnected compact topological space X. The point x e X is railed ct- isolated if the intersection of every sequence of neighborhoods of x is a neighborhood xA be a vector lattice and let Au(= CTC (X)) be its universal completion. Then A is of finite dimension if and only if each element of X is c-isolated. Bresar in [1] raised a question to find new examples of zero product determined algebras. Finally, as an application, we give a positive answer to this question.
Key words: hyper-Archimedean vector lattice, /-algebra, universally complete vector lattice. Mathematical Subject Classification (2000): 47B60, 16E40.
1. Introduction
Throughout the paper, R and C denote real numbers and complex numbers, respectively and let N = {1, 2,...}. For a set A, An denotes the cartesian product A x xA, n € N.
n times
A
algebra, then A is isomorphic to R, C or the quaternion field. For the details, we refer to [2]. For the case of lattice-ordered algebras, Huijsmans [3] proved that an Archimedean lattice-ordered algebra with unit element e > 0 in which every positive element has a positive
R
of Huijsmans for Banach lattice algebras. Later on, these two results are generalized and combined in [5] by using an easy observation as follows:
© 2018 Polat F., Toumi M. A.
Theorem 1. Let A be an Archimedean lattice-ordered algebra with unit element e > 0. Then the following statements are equivalent:
(i) Every positive element has a positive inverse.
(ii) A is a d-algebra and every positive element has a positive inverse.
(iii) A is an f-algebra and each positive element has a positive inverse.
(iv) A is an almost f-algebra and each positive element has an inverse. Af
(vi) A is order and algebra isomorphic to R.
A
||e|| = 1, then A is also isometric to R.
As far as we know, no attention at all has been paid in the literature to the problem when a lattice-ordered algebra is of finite dimension. The aim of this paper is to give a positive answer to this problem. In connection with our problem, Bresar studied the class of finite dimensional spaces which are zero product determined. Recall that an algebra A over a field K is said to be zero product determined, if for every bilinear map f : A x A ^ B, where B is an arbitrary vector space over K, with the property that for all x,y £ A, f (x, y) = 0 whenever xy = 0, is of the form f (x,y) = $(xy) for some linear map $ : A ^ B. This concept was introduced in [1]. The original motivation for this concept was problems on the zero product preserving linear maps. Recently, Bresar [1] proved the following result.
Theorem 2. A finite dimensional algebra is zero product determined if and only if it is generated by idempotents.
Bresar pointed out that the main purpose of the paper [1] was to find new examples of zero product determined algebras, but ultimately it is restricted to an unexpected characterization of finite-dimensional algebras that are generated by idempotents and the initial problem of finding new examples still remains open. Moreover, the problem of finding other classes of algebras for which the characterization of Theorem 1 holds is fully open.
As an application of our study, we will give a new class of non-finite dimensional zero product algebras for which the characterization of the previous theorem holds.
2. Preliminaries
In order to avoid unnecessary repetitions, we assume that all vector lattices under consideration are Archimedean.
In the following lines, we recall some definitions and basic facts about vector lattices, lattice-ordered algebras and multilinear maps.
For the unexplained terminology on vector lattices, lattice-ordered algebras and multilinear maps, we refer the reader to [6, 7, 8, 9].
Given a vector lattice A, the set A+ = {a £ A : a ^ 0} is called the positive cone of A. Let a, e £ A+, then e is called a component of a if e A (a — e) = 0. A
the multiplication in A are compatible, that is a,b £ A+ implies ab £ A+, is called lattice-ordered algebra (briefly a i-algebra). An ¿-algebra A is called an f-algebra if A verifies the property that a A b = 0 and c ^ 0 imply ac A b = ca A b = 0. Any /-algebra is automatically
i A f
it follows from a A b = 0 that ab = ba = 0. An i-algebra A is called a d-algebra if A verifies the property that a A b = 0 and c ^ 0 imply ac A bc = ca A cb = 0.
The vector lattice A is called Dedekind complete if for each non-empty subset B of A which is bounded above, sup B exists in A. The vector lattice A is called laterally complete
if every orthogonal system in A has a supremum in A. If A is Dedekind complete and
AA completion Au, this means that there exists a unique (up to a lattice isomorphism) universally complete vector lattice Au such that A can be identified with an order dense sublattice of Au (see [6, Section 8, Exercise 13] for an interesting approach to the existence of the universal completion by using orthomorphisms). A
D has a supremum, then there exists an at most countable subset C of D with sup C = sup D. A Dedekind complete vector lattice with the countable sup property is called super Dedekind complete vector lattice.
Let A be a vector lattice. A subset S of A+ is called an orthogonal system of A if 0 € S and u A v = 0 for each pair (u, v) of distinct elements in S. It follows from Zorn's lemma that every orthogonal system of A is contained in a maximal orthogonal system.
A subset S in a vector lattice E is called solid if it foliows from |u| ^ |v| in E and v € S that u € S. A solid vector subspace of a vector lattice is called an ideal. The ideal P in a vector lattice is prime whenever it follows from inf(a, b) € P that at least one of a € P or b € P
E {u}
by Eu. Clearly, Eu = {v € E : 3 A ^ 0 such that |v| ^ A|u|}.
An order closed ideal in a vector lattice is called a band. A band B of a vector lattice E is said to be a projection band if B © Bd = E where Bd denotes the disjoint complement of B. A vector lattice has the projection property if every band is a projection band.
Let A be a vector lattice and v £ A+. Then the sequence (an)ngN in A is called (v) relatively uniformly convergent to a € A if for every real number e > 0 there exists a natural number ne such that |an — a| ^ ev for all n ^ ne. This will be denoted by an ^ a (v). If an ^ a (v) for some 0 ^ v € A, then the sequence (an)neN is called (relatively) uniformly convergent to a, which will be denoted by an ^ a (r.u). The notion of (v) (relatively) uniformly Cauchy sequence is defined in the obvious way. A vector lattice is called (relatively) uniformly complete
A
A
Let A and B be vector lattices. A multilinear map ^ : An ^ B is said to be positive whenever (a1,..., an) € (A+)n implies ^ (a1,..., an) € B+. A multilinear map ^ is said to be orthosymmetric if for all (a1,..., an) € An such th at ai A aj = 0 for some 1 ^ i, j ^ n implies ^ (a1,..., an) = 0.
3. Main results
We start with some auxiliary results which will be used in the sequel.
Proposition 1. Let A be a vector lattice and n € N. Then the followings are equivalent:
1) A = 11 © I2 © ... © In for some simple order ideals 11,11,..., In in A.
2) A = Rx1 © Rx2 © ... © Rxn for some elements x1, x2,..., xn € A.
< (1) ^ (2) Since Ii is simple for each i = 1, 2 ..., n, and according to [3, Proposition 1], it follows that Ii = Rxi for some elem ents x1, x2,..., xn € A Consequently, A = Rx1 © Rx2 © ... © Rxn for som e n € N and some elem ents x1, x2,..., xn € A.
(2) ^ (1) This direction is trivial. >
Definition 1. The depth of a vector lattice is the supremum (possibly infinite) of the lengths of maximal orthogonal system.
A
1) A
2) A = Rxi © Rx2 © ... © R xn for som e n £ N and some elern ents x1,x2 ,...,xn £ A.
< (1) ^ (2): Let n £ N be the dept h of ^^d {e1 ,e2,en} be the maximal orthogonal system of length n in A. Since A has the projection property, A = {e1}dd©{e2}dd©... {ei}dd© ... © {en}dd. Let 1 ^ i ^ n be fixed and y £ {ei}dd. Then y = y+ — y- where y+,y- £ {ei}dd. Let fi = n{y+}dd (ei) and f2 = n{y+}d (ei) where n{y+}d are the band projections
on {e1 ,e2,..., en} with ranges {y+}d^^d {y+}d, respectively. It is easy to see that the system {ei, e2,.. .,ei-l,fl,f2 , ei+1 ■)...■) en } is an orthogonal system of length n + 1. Therefore, either f1 = 0 or f2 = 0.
Case 1: If f = 0, then ei = f2 £ {y+}d. Hence {y+}dd c {ei}dd c {y+}d and so y+ = 0. Case 2: If f2 = 0, then ei = f £ {y+}dd .Hence {y-}dd c {ei }dd c {y+}dd c {y-}d and
y- = 0
Then, the band {ei}dd is totally ordered for all 1 ^ i ^ n. By [3, Proposition 1], {ei}dd = Rei for all 1 ^ i ^ ^^en A = Re1 © Re2 © ... © Ren for some n £ N. The implication (2) ^ (1) ^s trivial. >
To clarify next result, we give the following well-known lemma.
Ae there exists a unique multiplication on A such that A is an f -algebra with e as a unit element.
A
spaces E/I, where I is an order ideal in A, are Archimedean.
Several characterizations of hyper-Archimedean vector lattices are known (see, for example, [10], [11, Theorem 37.6, 61.1 and 61.2]). We collect some of them in the following lemma.
A
equivalent conditions holds.
A
A
(iii) The span of the set of all components of u is the principal ideal generated by u for all u £ A+.
A
A
A
A
on bands).
Definition 5. Let A be a vector lattice. An element x of A is said to be super atomic if the oder ideal Ax generated by x is of finite dimensional.
We now give the following result which shows the relation between the dimensions of a vector lattice and its universal completion.
Proposition 3. Let A be a vector lattice and Au be its universal completion. Then A is finite dimension if and only if Au is finite dimension.
< Let A be a finite dimension vector lattice. Since A is finite dimension and the elements of any finite orthogonal system of A are linearly independent, it follows that A has {e1 ,e2,..., en} as a maximal orthogonal system of length n. Let y £ {ei}dd to a fixed index i with 1 ^ i ^ n. Then y = y+ — y- with y+, y- £ {ei}dd .If y+ = ^d y- = 0, it is easy to see that the system {e1, e2,..., ei-1, y+, y-, ei+1,..., en} is an orthogonal system of length n + 1, which
is a contradiction. Hence, y+ = 0 or y- = 0. Then, the band {ei}dd is totally ordered for all 1 ^ i ^ n. By [3, Propositi on 1], {ei}dd = Rei for all 1 ^ i ^ n. Hence, the band generated by eon Au will be equal to Re^ Then Au = Re1 © Re2 © ... © Ren for some n € ^^nce Au is finite dimension. Conversely, if Au is finite dimension, then so is clearly A >
Remark 1. If a vector lattice A is finite dimension then A = Au.
We now have all ingredients to give the first main result of this section. A Au
followings are equivalent: A
(2) A = I1 © I2 © ... © In for some simple order ideals I1; I2,..., In in A, and n € N.
Au Au Au Au Au
A
< (1) ^ (2) This follows from Propositions ^d 2.
(2) ^ (3) Since A = Rx1 © Rx2 © ... © Rxn for some n € N and some elements x1;x2,... ,xn € A, it follows that Au = A Hence, the set of all order ideals of A is finite and
A
(3) ^ (4) This path is trivia.
(4) ^ (6) Let (Bn)neN be an increasing sequence of bands in Au. Then (Bn)neN is a dec-
Au A
exists n0 € N such that Bn = B^ for all n ^ n0. Consequently, Bn = Bn0 to all n ^ n0.
Au
(6) ^ (7) Let S = {ei : i € I} be a maximal orthogonal system in Au. Then e = sup {ei :
i € I} is a weak order unit of Au. Sinee Au is super Dedekind complete, it follows that there exists an at most countable subset T = {en : n € N} of S such that e = sup {en : n € N}. Let (Bn)neN be an increasing sequence of bands in Au where Bn = {\/ 1^i^nei}d^. Since Au is super Notherien, it follows thae there exi sts n0 € N such th at Bn = Bn0 to a 11 n ^ n0. Consequently, {\/ 1^i^n0ei}dd = {V 1^i^nei}dd to all n ^ n0. Therefore, en = 0 to all n > n0 and then S is a finite set. Let S = {/jn, jn € Jn} be an orthogonal system of Au such that en = sup {/jn, jn € Jn} to all 1 ^ n ^ n0. Since the set S' = {/jn, jn € Jn, 1 ^ n ^ n0} is a maximal orthogonal system of Au, it follows that S' is a finite set. Using the same argument with each /jn, we deduce that there exist m0 € N and a maximal orthogonal system K = {kn, 1 ^ n ^ m0} of Au with e = sup {kn, 1 ^ n ^ m0} such that K will be the finest orthogonal system meaning that we cannot use the decomposition process another time. Consequently, AU = AU1 © AU2 © ... © AUmo .Let 1 ^ n ^ n0 and let 0 ^ x ^ kn.
Let Hkn = {y € AUn, y = Y11<i<m aihi where ai € R and hi is a component of kn}. It is not hard to prove that He is a hyper-Archimedean vector sublattice of Au. Since K is a finest orthogonal system, it follows that the set of all components of kn is {0, kn}.
By Freudenthal Spectral Theorem, there exists xm = ^ 1<i<m aihi = h xn ^ x
(r. u). Therefore, there exists an € R such that x — ankn- Hence, Hkn = AUn for all 1 ^ n ^ n0.
Since any relatively uniformly complete hyper-Archimedean vector lattice is of finite dimension (see [11, Theorems 37.6, 61.4], [12, Theorem 3]), and AU is relatively uniformly complete, it follows that AU is of finite dimension. Therefore, e is super atomic.
(7) ^ (8) Let e be a super atomic weak order unit of Au. Then AU is of finite dimension. Since x A ne ^ x (r.u) to all 0 ^ x £ Au, it follows that Au is of finite dimension.
(8) & (1) This equivalence is trivial.
(1) ^ (5) Sinee Au is of finite dimension, the set of all order ideals of Au is finite and then Au satisfies the ascending chain condition on ideals and so we are done. >
A E A
n : E ^ A is said to be a positive orthomorphism if n (x) A y = 0 whenever x A y = 0 for each x,y £ E. An orthomorphism is the difference of two positive orthomorphisms. Orth(E, A) will denote the vector lattice of all orthomorphisms from E to A Z (E, A) will denote the sublattice of Orth(E, A) consisting of those n for which there is a non-negative real A with
—\x ^ n (x) ^ Ax
for all x £ E +. Let A be a lattice-ordered algebra. M(A) denotes the set of all maximal two-sided algebra ideals. We consider a subset m(A) of M(A) consisting of relatively uniformly closed ideals.
All prerequisites are made for the second main result of this section.
A Au
f
A
Au
(3) Orth(A, Au) = Z(A, Au).
< (1) ^ (2) Sinee A is of finite dimension, then Au is of finite dimension so that Au becomes a Banach lattice. It is well-known that any maximal modular algebra ideal of a commutative Banach algebra is closed. So we are done.
(2) ^ (3) Since any maximal modular algebra ideal of Au is relatively uniformly closed and by using the main result of [13], we deduce that Orth(Au) = Z(Au). Moreover, it is not hard to prove that Orth(Au) = Orth(A, Au) Mid Z(A, Au) = Z(Au).
(3) ^ (1) Since Orth(Au) = Orth(A, Au), Z(A,Au) = Z(Au) and Orth(Au) = Z(A,Au), it follows that Orth(Au) = Z(Au) = Au. Consequently, Au becomes a Banach lattice and it is well-known that any Banach universally complete vector lattice is of finite dimension. Hence, A is of finite dimension and we are done. >
It is well-known that any universally complete vector lattice A is of the form C^ (X) for some Hausdorff extremally disconnected compact topological space X (i. e. the closure of every open set of X is also open). The symbol C^ (X) denotes the collection of all continuous functions f : X ^ [—to, for which the open set dom f = {x £ X : —to < f (x) < is dense in X. It is well-known that C^ (X) can be equipped with a unital f-algebra multiplication. Moreover, C^(X) is a Dedekind complete f-algebra with e := as a unit element. The orthomorphisms in C^(X) are the pointwise multiplications, so Orth(C ~(X)) = C ~(X).
In order to reach our aim, we need the following;
X
x £ X
xx
(2) x £ dom f for all f £ C™(X).
(3) If f £ C~(X) and f (x) = 0, then f = 0 in some neighborhood of x.
Definition 6. The point x is called ^-isolated (or a P-point, or bounded) whenever x
enjoys any of the properties in Theorem 5.
Theorem 6 [14, 15]. The maximal algebra ideals of C^(X) for x € X are of the form (C~(X))x := {/ € C~(X) : / = 0 in some neighborhood of x}.
We now have gathered all ingredient for the third result of this section.
Theorem 7. Let A be a vector lattice and let Au (= C^ (X)) be universal completion /A A
(2) Each element of X is a-isolated.
< (1) ^ (2) Since A is of finite dimension, then Au is of finite dimension; therefore Au becomes a Banach unital /-algebra and it is well-known that any maximal algebra ideal
6
form (C^(X))x for some x € X and since (C^(X))x (C^(X))x is relatively uniformly closed, xa
(1) ^ (2) Since any point of X is a-isolated, it follows that any maximal algebra ideal of Au is relatively uniformly closed (see [14, 15]). By the main result of [13], we deduce that Orth(AU) = Z(Au). Since Orth(AU) = Z(Au) = AU,AU becomes a Banach lattice and it is well-known that any Banach universally complete vector lattice is of finite dimension. Hence, A is of finite dimension and we are done. >
Next, we will give a new class of non-finite dimension zero product algebras for which the characterization of Theorem 2 holds.
Theorem 8. Let A be an Archimedean unital /-algebra. Then A is zero product deter-A
< For the proof of "only if" part, we will use the same argument as in [16, Theorem 8]. Assume that every bilinear map / : A x A ^ B, where B is an arbitrary vector space over K, with the property that for all x, y € A, /(x, y) = 0 whenever xy = 0, is of the form /(x, y) = $ (xy) for some linear map $ : A ^ B. Suppose, per contra, that A is not hyper-Archimedean. It follows that there exists a prime ideal I which is not maximal. Hence the quotient A/I is linearly ordered space that is not isomorphic to R (see [11, Theorem 27.3 and 33.2]). Let x, y in A/I that are linearly independent. Hence x + y, x are linearly independent. By using Zorn's Lemma, it is not hard to prove that x, y are contained in a Hamel basis H\ and x + y, x are contained in a Hamel basis H2 such that H1 = H2. Then there exist two linear maps / : A/1 R such that / (x) = 1 and f(y) = -1 and g : A/I R such that g(x + y) = 1 and g(y) = —1. Let the bilinear map f : A x A -> 1 be defined by a,b) = f(a)g(b), for all a, b e A. Let a, b e A such that ab = 0. Since I is a prime ideal, it follows that a £ I or b G I. Hence a = 0 or b = 0. Consequently, f(a) = 0 or g(b) = 0. Therefore a,b) = 0. Hence, ^ is orthosymmetric. Whereas, ^f(x,y) = f(x)g(y) / f(y)g(x) = ^(y,x). That is ^ is not symmetric. Hence ^ is not of the form $(xy) for some linear map $ : A ^ R, which is a contradiction.
Then "if" part remains. We will use the same argument as in [17, Theorem 1]. Let ^ : A x A ^ B, where B is an arbitrary vector sp ace over K, with the property that for all x, y € A *(x, y) = 0 whenever xy = 0.
Let x,y € A. It follows that x = ^™=1 ai ei and y = /?'> ^ere e^d / are
components of e = |x| + |y|. Then
1<i<n, 1<j<m
Let ed be the disjoint complement of e^^nce e = eit + ed where eit A ed = 0. Then
fj = fj A e = fj A {ei + ed) = {fj A ei) + f A ed). Since fj A ed) A ei = 0, then (fj A ed) ei = 0. Consequently,
H(ef) = ^, {fj A ei) + {f A ed)) = H{el,f3 A ei). ei = ei A e = ei A fj + fjd = fj A ei + ei A fjd .
H(ef) = ^,fj A ei) = H{{f A ei) + {ei A fd),fj A ei). ei A fjd A fj A ei = 0, ei A fjd fj A ei = 0
*(ei, fj) = H (fj A ei, fj A e^ = , ei) By using the same argument, we prove that
*(ei, fj) = Hei, fj A e^ = H (fj A ei, fj A e^ = , ei). Therefore, in view of equality (1), we have
H(x,y) = H(y,x). (2)
Let u be the unit of A and let z £ A. Then the following bilinear map Hz : A x A ^ B defined by Hz (x,y) = H(xz,y), to all x,y £ A, satisfies the property that H(x,y) = 0 whenever xy = 0. Therefore, by using the argument as for H, we deduce that Hz (x,y) = Hz (y,x) = Hx (z,y) = Hx (y,z), to all x,y £ A. In particular if z = u, it follows that H (x, y) = Hx (y, e) = H (xy, e). Consequently, H is not of the form $ (xy) , where $ : A ^ B is defined by $ (x) = H (x, e), to all x £ A and we are done. >
f
gebra is of finite dimension (see [11, Theorems 37.6, 61.4], [12, Theorem 3]). Consequently, we have the following characterization.
Af
ties are equivalent. A A A
References
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Received July 17, 2017 Faruk Polat
Department of Mathematics, Faculty of Science, Cankiri Karatekin University, Uluyazi Kampusu, Cankiri 18100, Turkey E-mail: faruk.polat@gmail.com
Mohamed Ali Toumi Department of Mathematics,
Faculty of Science of Bizerte, University of Carthage Zarzouna, Bizerte 7021, Tunisia E-mail: MohamedAli. Toumi@fsb. rnu. tn
Владикавказский математический журнал 2018, Том 20, Выпуск 2, С. 86-94
ХАРАКТЕРИЗАЦИЯ КОНЕЧНОМЕРНЫХ АРХИМЕДОВЫХ ВЕКТОРНЫХ РЕШЕТОК
Полат Ф.. Toy ми М. А.
Аннотация. Статья посвящена условиям конечномерности архимедовых векторных решеток. Найдены три новые характеризации таких решеток. Первая описывает конечномерность векторной решетки A на языке ее универсального пополнения Au. Вторая утверждает, что векторная решетка конечномерна в том и только в том случае, когда выполнено одно из следующих двух условия: (а) всякий максимальный модулярный алгебраический идеал в Au равномерно полон; (б) Orth(A, Au) = Z(A, Au), где Orth(A, Au) векторная решетка мех ортоморфизмов из A в Au, a Z(A, Au) — подрешетка, состоящая из ортоморфизмов п, удовлетворяющих условию |n(x)| < A|x| (x £ A) при некотором положительном A £ R. Хорошо известно, что всякая универсально полная векторная решетка представляется в виде CTO(X) для некоторого экстремально несвязного компакта X. Точку x £ X называют ст-нзолпрованной, если пересечение любой последовательности окрестностей точки x является окрестностью точки x. Третья характеризация состоит в том, что векторная решетка A с универсальным расширением Au = CTO(X) конечномерна тогда и только тогда, когда каждая точка в X ст-нзолпрована. В качестве приложения получен положительный ответ на вопрос
Брезара о существовании новых примеров алгебр, определяемых нулевыми произведениями.
f
