# Noetherian and Artinian Lattices

Authors: Derya Keskin Tütüncü; Sultan Eylem Toksoy; Rachid Tribak
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## Abstract

It is proved that ifL is a complete modular lattice which is compactly generated, then Rad(L)/0 is Artinian if, and only if for every small element a of L, the sublattice a/0 is Artinian if, and only if L satisfies DCC on small elements.


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## Body

## 1. Introduction

By alattice we mean a partially ordered set (L,≤) such that every pair of elements a, b in L has a greatest lower bound (or a meet) a∧b and a least upper bound (or a join) a∨b; that is, (i)
a∧b≤a, a∧b≤b, and c≤a∧b for all c∈L with c≤a, c≤b,(ii)
a≤a∨b, b≤a∨b, and a∨b≤d for all d∈L with a≤d, b≤d.Note that, for givena, b∈L, a∧b and a∨b are unique, and(1.1)a≤b⟺a=a∧b⟺b=a∨b.
Let (L,≤,∧,∨) (or just L) be any lattice. Given a, b∈L, we set(1.2)a≤′b⟺b≤a.
Then (L,≤′) is a partially ordered set; moreover, for any a, b∈L, a, b have greatest lower bound a∨b and least upper bound a∧b. We call (L,≤′,∨,∧) the opposite lattice of L, and denote it by L∘.Let(L,≤∧,∨) be any lattice. Let a≤b in L. We define(1.3)ba={x∈L:a≤x≤b}.
(Sometimes b frac a is denoted by b/a.)A lattice(L,≤,∧,∨)  has a least element if there exists z∈L such that z≤a(a∈L). In this case, z is uniquely defined and is usually denoted by 0. The lattice Lhas a greatest element if there exists u∈L such that a≤u(a∈L). In this case, u is uniquely defined and is usually denoted by 1. A lattice L is called complete if every subset of L has a meet and a join, and it is called modular if a∧(b∨c)=b∨(a∧c) for all a, b, c in L with b≤a. For more information about lattice theory, refer to [1–3].Throughout this paper(L,≤,∨,∧,0,1) will be a complete modular lattice. An element e∈L is called an essential element if e∧x≠0 for every nonzero element x∈L. An element s∈S is said to be small if s is an essential element of the opposite lattice L∘. Let E(L) denote the set of all essential elements of L. The set of all small elements of L will be denoted by S(L).A set{ci∣i∈I}⊆L is called a direct set if, for all i, j∈I, there exists k∈I with ci∨cj≤ck. The lattice L is said to be upper continuous if, for every direct set {ci∣i∈I} in L and element a∈L, we have a∧(⋁i∈Ici)=⋁i∈I(a∧ci). On the other hand, L is said to be lower continuous if for every inverse set {ci∣i∈I} (i.e., for all i, j in I, there exists k∈I with ck≤ci∧cj) and element a∈L, a∨(⋀i∈Ici)=⋀i∈I(a∨ci). We will call an element f in Lfinitely generated element (or compact element) if whenever f≤∨S, for some direct set S in L, then there exists x∈S such that f≤x. Note that 0 is always a finitely generated element of L. It is known that an element f is finitely generated if and only if for every nonempty subset U of L with f≤∨U there exists a finite subset F of U such that f≤∨F. A lattice L is said to be finitely generated (or compact) if 1 is finitely generated. We call the lattice L  compactly generated if each of its elements is a join of finitely generated elements (see [2]). Note that every compactly generated lattice is upper continuous (see, e.g., [4, Proposition 2.4]). Moreover, it is shown in [4, Exercises 2.7 and 2.9] that for every element a of a compactly generated lattice L, the sublattices a/0 and 1/a are again compactly generated. A lattice L is called a finitely cogenerated (or cocompact) lattice, if for every subset X of L such that ∧X=0 there is a finite subset F of X such that ∧F=0. An element g∈L is said to be finitely cogenerated (or cocompact) if the sublattice g/0 is a finitely cogenerated lattice. If a<b and a≤c<b imply c=a, then we say that a is covered by b (or b  coversa). If 0 is covered by an element a of L, then a is called an atom element of L. A lattice L is said to be semiatomic if 1 is a join of atoms in L (see [4]). The meet of all maximal elements (different from 1) in L is denoted by Rad(L), and it is called the radical of L (see [2]). If L is compactly generated, then Rad(L) is the join of all small elements of L (see [2, Theorem 8]). The join of all atoms of L, denoted by Soc(L), is called the socle of L. The socle of a compactly generated lattice is equal to the meet of all essential elements (see [4, Theorem 5.1]).A non-empty subsetS of L is called an independent set if, for every x∈S and finite subset T={t1,…,tn} of S with x∉T, x∧(t1∨⋯∨tn)=0. We say that a nonzero lattice L has finite uniform(or Goldie)dimension if L contains no infinite independent sets; equivalently, sup{k∣L contains an independent subset of cardinality equal to k}=n<∞. In this case L is said to have uniform (or Goldie) dimension n and this is denoted by u(L). We shall say that L has hollow (or dual Goldie) dimension n, provided the opposite lattice L∘ has uniform dimension n. The lattice L is said to be Artinian (noetherian) if L satisfies the descending (ascending) chain condition on its elements. A lattice L will be called an E-complemented lattice if, for each a∈L, there exists b∈L such that a∧b=0 and a∨b∈E(L).In Section2 we mainly prove that a lattice L is noetherian if and only if L is E-complemented and every essential element of L is finitely generated (Corollary 2.4). In Section 3 we generalize Theorem  5 in [5] to lattice theory (Theorem 3.7).


## 2. Noetherian Lattices

The following lemma was given us by Patrick F. Smith from his unpublished notes.Lemma 2.1.
LetL be a lattice. Consider the following statements.(i)
L is noetherian.(ii)
L has finite uniform dimension.(iii)
L is E-complemented. Then (i)⇒(ii)⇒(iii).Proof.
(i)⇒(ii) Suppose L is noetherian but that L does not have finite uniform dimension. Then there exists an infinite independent set of nonzero elements xn(n∈ℕ). Consider the ascending chain x1≤x1∨x2≤⋯ in L. Because L is noetherian, there exists a positive integer n such that x1∨⋯∨xn=x1∨⋯∨xn∨xn+1. This implies that xn+1≤(x1∨⋯∨xn)∧xn+1=0, a contradiction. Therefore L has finite uniform dimension.
(ii)⇒(iii) Let a∈L. If a∈E(L), we are done. If a∉E(L), then there exists 0≠b1∈L such that a∧b1=0. If a∨b1∈E(L), we are done. Otherwise, there exists 0≠b2∈L such that (a∨b1)∧b2=0. Repeating this argument we produce an independent set {a,b1,b2,…}. Thus this process must stop, so there exists k∈ℕ such that a∧(b1∨⋯∨bk)=0 and a∨(b1∨⋯∨bk)∈E(L).Remark 2.2.
Note that iff is a finitely generated element of a lattice L, then for every non-empty set U with f=∨U there exists a finite subset F of U such that f=∨F.Proposition 2.3.
LetL be a lattice such that x is finitely generated for every x∈E(L). Then the following are equivalent.(i)
L is noetherian.(ii)
L has finite uniform dimension.(iii)
L is E-complemented.Proof.
We only need to prove(iii)⇒(i) by Lemma 2.1. Let a be a nonzero element in L. By (iii), there exists an element b of L such that a∧b=0 and a∨b∈E(L). By hypothesis, a∨b is finitely generated. Let a=∨S for a nonempty set S in L. Then a∨b=∨(S∪{b}). Since a∨b is finitely generated, a∨b=∨F∨b for a finite subset F of S. Since L is modular, we have a=∨F. Therefore every element in L is finitely generated. Hence L is noetherian by [4, Proposition 2.3].Corollary 2.4.
A latticeL is noetherian if and only if L is E-complemented and every essential element of L is finitely generated.Lemma 2.5.
Every upper continuous latticeL is E-complemented.Proof.
Leta∈L. Let S={b∈L∣a∧b=0}. Clearly, 0∈S. Let {ci∣i∈I} be a chain in S and let c=⋁i∈Ici. Then a∧c=a∧(⋁i∈Ici)=⋁i∈I(a∧ci)=0. By Zorn’s lemma, S contains a maximal member u. Then a∧u=0. Suppose that (a∨u)∧x=0 for some x∈L. Then a∧(u∨x)=0, and hence u∨x∈S. Since u≤u∨x, we have u=u∨x and x≤u. Thus x=(a∨u)∧x=0. It follows that a∨u∈E(L). Therefore L is E-complemented.Corollary 2.6.
LetL be an upper continuous lattice. Then L is noetherian if and only if every essential element in L is finitely generated.Lemma 2.7 (see [4, Lemmas 7.3 and 7.5]).
LetL be a lattice and k a positive integer. Then(i)
ift∈S(L), then s∈S(L) for every s≤t;(ii)
ifs1,s2,…,sk∈S(L), then s1∨s2∨⋯∨sk∈S(L).As an easy observation of Lemma2.7, we can give the following two results.Proposition 2.8 (see cf. [5, Proposition 2]).
LetL be a compactly generated lattice. Then Rad(L)/0 is noetherian if and only if L satisfies ACC on small elements.Proof.
(⇒) By [2, Theorem 8].
(⇐) By assumption, L contains a maximal small element x. Since x is small in L, x≤Rad(L). Suppose that x≠Rad(L). Then there exists a small element s of L such that s∉x/0. On the other hand, s∨x is a small element of L by Lemma 2.7(ii). By the maximality of x, we have s∨x=x. This gives s∈x/0, a contradiction. Thus x=Rad(L). By Lemma 2.7(i), Rad(L)/0⊆S(L). Consequently, Rad(L)/0 is noetherian.Proposition 2.9 (see cf. [5, Proposition 3]).
LetL be a compactly generated lattice. Then the following are equivalent.(i)
Rad(L)/0 has finite uniform dimension.(ii)
There exists a positive integerk such that for every small element s of L we have u(s/0)≤k.(iii)
L does not contain an infinite independent set of nonzero small elements.Proof.
(i)⇒(ii) Let s be a small element of L. By [2, Theorem 8], s≤Rad(L). Since u(s/0)≤u(Rad(L)/0), s/0 has finite uniform dimension. The rest is clear.
(ii)⇒(iii) Let {s1,s2,…} be an infinite independent set of nonzero small elements of L. By Lemma 2.7(ii), s1∨s2∨⋯∨sk+1∈S(L), and u((s1∨s2∨⋯∨sk+1)/0)≥k+1, a contradiction.
(iii)⇒(i) Suppose that Rad(L)/0 does not have finite uniform dimension. Then there exists an infinite independent set of nonzero elements {x1,x2,…} of Rad(L)/0. Let i≥1. Since Rad(L)/0 is compactly generated, there exists a nonzero finitely generated element ki of Rad(L)/0 such that ki≤xi. So by Lemma 2.7, ki∈S(L). Therefore {k1,k2,…} is an infinite independent set of nonzero small elements of L, a contradiction. Thus Rad(L)/0 has finite uniform dimension.


## 3. Artinian Lattices

Lemma 3.1.
LetL be a compactly generated semiatomic lattice. Then the following are equivalent.(i)
L is finitely generated.(ii)
L is finitely cogenerated.(iii)
1 is a finite independent join of atoms.(iv)
L is Artinian.Proof.
(i)⇔(ii)⇔(iii) By [4, Theorem 11.1].
(iv)⇒(ii) By [4, Proposition 11.2].
(iii)⇒(iv) Note that if a is an atom in L, then a/0 is Artinian. Assume that 1=a1∨a2∨⋯∨an such that the join is independent and each ai is atom in L. Since each ai/0 is Artinian, (a1∨a2∨⋯∨an)/0 is Artinian, and hence L is Artinian.Lemma 3.2.
LetL be a compactly generated lattice which satisfies DCC on small elements. If f is a finitely generated element of Rad(L)/0, then f/0 is Artinian.Proof.
Letf be a finitely generated element of Rad(L)/0. Then f≤Rad(L)=⋁I{si∣si∈S(L)} implies that f≤⋁F{si∣si∈S(L)}for some finite subset F of I. By Lemma 2.7, f∈S(L). By assumption and Lemma 2.7(i), f/0 is Artinian.Lemma 3.3.
LetL be a compactly generated lattice which satisfies DCC on small elements. Then, for every k<Rad(L), Soc(Rad(L)/k) is an essential element of Rad(L)/k.Proof.
Letk<Rad(L), and let Soc(Rad(L)/k)=t. Let k≤h≤Rad(L) such that t∧h=k. Assume that k<h. Since Rad(L)/0 is compactly generated, there exists a nonzero finitely generated element x in Rad(L)/0 such that x≤h but x∉k/0. By Lemma 3.2, x/0 is Artinian. Then x/(x∧k)≅(k∨x)/k implies that (k∨x)/k is a nonzero Artinian sublattice. By [4, Proposition 1.4], (k∨x)/k has an atom element p′. Note that k<p′≤x∨k≤h. Since p′ is atom in Rad(L)/k, we have p′≤t. Thus k<p′≤t∧h. This contradicts the fact that t∧h=k. Therefore k=h and t∈E(Rad(L)/k). This completes the proof.Lemma 3.4.
Leta be an element of a compactly generated lattice L. If a is a finitely generated element of a/0, then a is a finitely generated element of L.Proof.
SinceL is compactly generated, a=∨U where U is a set of finitely generated elements in L. Since a is a finitely generated element of a/0, a=⋁(1≤i≤n)ai for some elements ai(1≤i≤n) of U. Therefore a is a finitely generated element of L.Lemma 3.5.
LetL be a compactly generated lattice which satisfies DCC on small elements. Suppose that the set
(3.1)Ω={ai∣0≤ai≤Rad(L)andRad(L)aiisnotfinitelycogenerated}
is nonempty. Then:(1)
the setΩ has a minimal member p which is a small element of L;(2)
ifSoc(Rad(L)/p)=s, then s is not a finitely generated element of Rad(L)/p and s is a small element of L.Proof.
(1) LetΓ be any chain in Ω. Let c=⋀ci∈Γci. If c∉Ω, then Rad(L)/c is finitely cogenerated. Therefore c=ci for some ci∈Γ, a contradiction. By Zorn’s Lemma, Ω has a minimal member p. Let Soc(Rad(L)/p)=s. By Lemma 3.3, s∈E(Rad(L)/p). Thus s is not a finitely generated element of Rad(L)/p by [4, Theorem 11.2]. Let q∈L with 1=p∨q. Then s=s∧1=s∧(p∨q)=p∨(s∧q). It follows that s/p=[p∨(s∧q)]/p≅(s∧q)/(p∧q). Suppose that p∧q≠p. Then Rad(L)/(p∧q) is finitely cogenerated. Let Soc(Rad(L)/(p∧q))=α. Then α is finitely generated in Rad(L)/(p∧q) by [4, Theorem 11.2]. Therefore α/(p∧q) is Artinian by Lemma 3.1. Since Rad(L)/p is a sublattice of Rad(L)/(p∧q), we have s≤α. Thus s∧q≤α≤Rad(L). Since α/(p∧q) is Artinian, (s∧q)/(p∧q) is also Artinian by [4, Proposition 1.5]. This implies that s/p is Artinian, and hence s is a finitely generated element of s/p by Lemma 3.1. Since Rad(L)/p is compactly generated, s is a finitely generated element of Rad(L)/p (see Lemma 3.4), a contradiction. So p∧q=p and hence q∨p=q=1. This gives p∈S(L).
(2) Note thats is not a finitely generated element of Rad(L)/0 as we prove in (1). Let υ∈L such that 1=s∨υ. Note that s/p is a semiatomic lattice. Then s/[p∨(s∧υ)] is also semiatomic by [4, Corollary 6.3]. Therefore,
(3.2)1p∨υ=s∨υp∨υ=[s∨(p∨υ)]p∨υ≅s[s∧(p∨υ)]=s[p∨(s∧υ)].
This implies that 1/(p∨υ) is semiatomic. Suppose that 1≠p∨υ. By [4, Lemma 6.12], there exists a maximal element w of 1/(p∨υ). Clearly, w is a maximal element of L and υ≤w. Thus 1=s∨υ≤s∨w. But s≤Rad(L)≤w. Then w=1, a contradiction. It follows that 1=p∨υ. Since p∈S(L), we have υ=1. Thus s∈S(L).Remark 3.6.
By dualizing [6, Theorem 3.4], we have the fact that if L is upper continuous and a/0 is Artinian for every small element a of L, then ∨S(L)/0 is Artinian. Therefore for compactly generated lattices (ii)⇒(i) in Theorem 3.7 holds, but our aim is to give a proof in a different way. We should call attention to the fact that ∨S(L) need not to be the radical of any upper continuous lattice L.Theorem 3.7 (see cf. [5, Theorem 5]).
LetL be a compactly generated lattice. Then the following are equivalent.(i)
Rad(L)/0 is Artinian.(ii)
For every small elementa of L the sublattice a/0 is Artinian.(iii)
L satisfies DCC on small elements.Proof.
(i)⇒(ii) Clear by [2, Theorem 8].
(ii)⇒(iii) This is immediate.
(iii)⇒(i) Suppose that Rad(L)/0 is not Artinian. By [4, Proposition 11.2], there exists an element g in L with g≤Rad(L) such that Rad(L)/g is not finitely cogenerated. By Lemma 3.5, the set
(3.3)Ω={ai∣0≤ai≤Rad(L)andRad(L)aiisnotfinitelycogenerated}
has a minimal member p such that Soc(Rad(L)/p)=s∈S(L) and s is not a finitely generated element of Rad(L)/p. By (iii) and Lemma 2.7(i), s/0 is Artinian. By Lemma 3.1, s/0 is finitely generated. Therefore s is a finitely generated element of Rad(L)/p by Lemma 3.4. This is a contradiction. Therefore Rad(L)/0 is Artinian.Corollary 3.8.
LetL be a compactly generated lattice. If 1/s is finitely cogenerated for every small element s of L, then Rad(L)/0 is Artinian.Proof.
Consider the descending chain(3.4)x1≥x2≥⋯
of small elements of L. Put x=⋀i≥1xi. Thus x is small in L. By assumption, 1/x is finitely cogenerated. So there exists an integer n such that x=⋀i=1nxi=xn. Hence L has DCC on small elements. By Theorem 3.7, Rad(L)/0 is Artinian.Leta and b be elements of L. Then b is called a supplement of a in L if b is minimal with respect to a∨b=1. Equivalently, b is a supplement of a if and only if a∨b=1 and a∧b∈S(a/0) (see [4, Proposition 12.1]). The lattice L is said to be supplemented if every element a of L has a supplement in L.The following result may be proved in much the same way as [5, Lemma 6], and 1/Rad(L) is a semiatomic lattice by [4, Proposition 12.3] already.Lemma 3.9.
LetL be a compactly generated supplemented lattice with DCC on supplement elements. Then 1/Rad(L) is a finitely generated semiatomic lattice.By using Theorem3.7 and Lemma 3.9, we get the following theorem.Theorem 3.10.
LetL be a compactly generated lattice. Then L is Artinian if and only if L is supplemented and L satisfies DCC on supplement elements and small elements.Proof.
The necessity is clear. Conversely, suppose thatL is a supplemented lattice which satisfies DCC on supplement elements and small elements. By Theorem 3.7, Rad(L)/0 is Artinian, and by Lemmas 3.1 and 3.9, 1/Rad(L) is Artinian. Thus L is Artinian.



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