Noetherian and Artinian Lattices

Abstract

It is proved that if L 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.

1. Introduction

By a lattice 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 given 𝑎, 𝑏∈𝐿, 𝑎∧𝑏 and 𝑎∨𝑏 are unique, and𝑎≤𝑏⟺𝑎=𝑎∧𝑏⟺𝑏=𝑎∨𝑏. (1.1) Let (𝐿,≤,∧,∨) (or just 𝐿) be any lattice. Given 𝑎, 𝑏∈𝐿, we set𝑎≤^′𝑏⟺𝑏≤𝑎. (1.2) 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 define𝑏𝑎={𝑥∈𝐿∶𝑎≤𝑥≤𝑏}. (1.3) (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 {𝑐_𝑖∣𝑖∈𝐼}⊆𝐿 is called a direct set if, for all 𝑖, 𝑗∈𝐼, there exists 𝑘∈𝐼 with 𝑐_𝑖∨𝑐_𝑗≤𝑐_𝑘. The lattice 𝐿 is said to be upper continuous if, for every direct set {𝑐_𝑖∣𝑖∈𝐼} in 𝐿 and element 𝑎∈𝐿, we have 𝑎∧(⋁_𝑖∈𝐼𝑐_𝑖)=⋁_𝑖∈𝐼(𝑎∧𝑐_𝑖). On the other hand, 𝐿 is said to be lower continuous if for every inverse set {𝑐_𝑖∣𝑖∈𝐼} (i.e., for all 𝑖, 𝑗 in 𝐼, there exists 𝑘∈𝐼 with 𝑐_𝑘≤𝑐_𝑖∧𝑐_𝑗) and element 𝑎∈𝐿, 𝑎∨(⋀_𝑖∈𝐼𝑐_𝑖)=⋀_𝑖∈𝐼(𝑎∨𝑐_𝑖). 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  covers 𝑎). 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 subset 𝑆 of 𝐿 is called an independent set if, for every 𝑥∈𝑆 and finite subset 𝑇={𝑡₁,…,𝑡_𝑛} of 𝑆 with 𝑥∉𝑇, 𝑥∧(𝑡₁∨⋯∨𝑡_𝑛)=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 Section 2 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. Let 𝐿 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 𝑥_𝑛(𝑛∈ℕ). Consider the ascending chain 𝑥₁≤𝑥₁∨𝑥₂≤⋯ in 𝐿. Because 𝐿 is noetherian, there exists a positive integer 𝑛 such that 𝑥₁∨⋯∨𝑥_𝑛=𝑥₁∨⋯∨𝑥_𝑛∨𝑥_𝑛+1. This implies that 𝑥_𝑛+1≤(𝑥₁∨⋯∨𝑥_𝑛)∧𝑥_𝑛+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∨⋯∨𝑏_𝑘)=0 and 𝑎∨(𝑏₁∨⋯∨𝑏_𝑘)∈𝐸(𝐿).

Remark 2.2. Note that if 𝑓 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. Let 𝐿 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 lattice 𝐿 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 lattice 𝐿 is 𝐸-complemented.

Proof. Let 𝑎∈𝐿. Let 𝑆={𝑏∈𝐿∣𝑎∧𝑏=0}. Clearly, 0∈𝑆. Let {𝑐_𝑖∣𝑖∈𝐼} be a chain in 𝑆 and let 𝑐=⋁_𝑖∈𝐼𝑐_𝑖. Then 𝑎∧𝑐=𝑎∧(⋁_𝑖∈𝐼𝑐_𝑖)=⋁_𝑖∈𝐼(𝑎∧𝑐_𝑖)=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. Let 𝐿 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]). Let 𝐿 be a lattice and 𝑘 a positive integer. Then(i)if 𝑡∈𝑆(𝐿), then 𝑠∈𝑆(𝐿) for every 𝑠≤𝑡;(ii)if 𝑠₁,𝑠₂,…,𝑠_𝑘∈𝑆(𝐿), then 𝑠₁∨𝑠₂∨⋯∨𝑠_𝑘∈𝑆(𝐿).

As an easy observation of Lemma 2.7, we can give the following two results.

Proposition 2.8 (see cf. [5, Proposition 2]). Let 𝐿 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]). Let 𝐿 be a compactly generated lattice. Then the following are equivalent.(i)Rad(𝐿)/0 has finite uniform dimension.(ii)There exists a positive integer 𝑘 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∨⋯∨𝑠_𝑘+1∈𝑆(𝐿), and 𝑢((𝑠₁∨𝑠₂∨⋯∨𝑠_𝑘+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 𝑘_𝑖 of Rad(𝐿)/0 such that 𝑘_𝑖≤𝑥_𝑖. So by Lemma 2.7, 𝑘_𝑖∈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. Let 𝐿 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∨⋯∨𝑎_𝑛 such that the join is independent and each 𝑎_𝑖 is atom in 𝐿. Since each 𝑎_𝑖/0 is Artinian, (𝑎₁∨𝑎₂∨⋯∨𝑎_𝑛)/0 is Artinian, and hence 𝐿 is Artinian.

Lemma 3.2. Let 𝐿 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. Let 𝑓 be a finitely generated element of Rad(𝐿)/0. Then 𝑓≤Rad(𝐿)=⋁_𝐼{𝑠_𝑖∣𝑠_𝑖∈𝑆(𝐿)} implies that 𝑓≤⋁_𝐹{𝑠_𝑖∣𝑠_𝑖∈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. Let 𝐿 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. Let 𝑘<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. Let 𝑎 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. Since 𝐿 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)}𝑎_𝑖 for some elements 𝑎_𝑖(1≤𝑖≤𝑛) of 𝑈. Therefore 𝑎 is a finitely generated element of 𝐿.

Lemma 3.5. Let 𝐿 be a compactly generated lattice which satisfies 𝐷𝐶𝐶 on small elements. Suppose that the set Ω=𝑎_𝑖∣0≤𝑎_𝑖(𝐿)(𝐿)≤RadandRad𝑎_𝑖isnotfinitelycogenerated (3.1) is nonempty. Then:(1)the set Ω has a minimal member 𝑝 which is a small element of 𝐿;(2)if Soc(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) Let Γ be any chain in Ω. Let 𝑐=⋀_{𝑐𝑖∈Γ}𝑐_𝑖. If 𝑐∉Ω, then Rad(𝐿)/𝑐 is finitely cogenerated. Therefore 𝑐=𝑐_𝑖 for some 𝑐_𝑖∈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 that 𝑠 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, 1=𝑝∨𝜐𝑠∨𝜐=𝑝∨𝜐𝑠∨𝑝∨𝜐≅𝑠𝑝∨𝜐=𝑠𝑠∧𝑝∨𝜐𝑝∨(𝑠∧𝜐). (3.2) 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]). Let 𝐿 be a compactly generated lattice. Then the following are equivalent.(i)Rad(𝐿)/0 is Artinian.(ii)For every small element 𝑎 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 Ω=𝑎_𝑖∣0≤𝑎_𝑖(𝐿)(𝐿)≤RadandRad𝑎_𝑖isnotfinitelycogenerated (3.3) 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. Let 𝐿 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 𝑥₁≥𝑥₂≥⋯ (3.4) of small elements of 𝐿. Put 𝑥=⋀_𝑖≥1𝑥_𝑖. Thus 𝑥 is small in 𝐿. By assumption, 1/𝑥 is finitely cogenerated. So there exists an integer 𝑛 such that 𝑥=⋀^𝑛_𝑖=1𝑥_𝑖=𝑥_𝑛. Hence 𝐿 has DCC on small elements. By Theorem 3.7, Rad(𝐿)/0 is Artinian.

Let 𝑎 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. Let 𝐿 be a compactly generated supplemented lattice with 𝐷𝐶𝐶 on supplement elements. Then 1/Rad(𝐿) is a finitely generated semiatomic lattice.

By using Theorem 3.7 and Lemma 3.9, we get the following theorem.

Theorem 3.10. Let 𝐿 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 that 𝐿 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.

Acknowledgments

The second author has been supported by the TÜBİTAK (The Scientific and Technological Research Council of Turkey) within the program numbered 2218. She would like to thank TÜBİTAK for their support. Also the authors would like to thank Professor Patrick F. Smith (Glasgow University) for his helpful comments on the paper.