Topologically Noetherian Banach algebras

Abstract

In this work, we introduce and study the properties of topologically Noetherian Banach algebras. In particular, we prove, if every prime closed ideal of a commutative Banach algebra A is maximal, then A is finite dimensional. Finally, we show that if every maximal ideal of a Banach algebra is generated by idempotent element then the Banach algebra is finite dimensional.

Keyword:

• Noetherian and topological Banach algebras

Public Interest Statement

In this paper, we modify the ascending chain introducing the so-called topologically Noetherian Banach algebra which inhered several of the main properties of Noetherian Banach algebra. In particular, we prove that if every prime closed ideal of a commutative Banach algebra A is maximal, then A is finite dimensional.

1. Introduction

Grauert and Remmert (1971) proved if every closed ideal in a commutative Banach algebra A is finitely generated, then A is finite dimensional. Sinclair and Tullo (1974) obtained a non-commutative version of this result. Ferreira and Tomassini (1978) improved Grauert and Remmert’s result by showing that the statement is also true if one replaces “closed ideals” by “maximal ideals” in the Shilov boundary of A. Dales and Żelazko (2012) shorten their proof, with some extensions. For notations and terminologies we follow Bowers and Kalton (2014).

Let A be a Banach algebra with identity over a field of complex number. Let I be a closed ideal of A. Then I is said to be irreducible if it is not a finite intersection of closed ideals of A properly containing I, otherwise, I is termed reducible.

A closed ideal I of a commutative Banach algebra A is called primary if the conditions ab ∊ I and a ∉ I together imply bⁿ ∊ I, for some positive integer n.

An element e in A is said to be idempotent, if e² = e.

A left ideal I of A is said to be topologically finitely generated (t.f.g.) if there exist x₁, x₂, …, x_n ∊ I such that $\overline{I}=\overline{<x_1,x_2,\dots ,x_n>}$. We say that A is a topologically Noetherian (T.N) Banach algebra if for every ascending chain I₁ ⊆ I₂ ⊆ … ⊆ I_n ⊆ … of closed left ideals of A, there exists n ∊ Z⁺ such that I_m = I_n for all m ≥ n.

It is clear that any left Noetherian and any simple Banach algebra is topologically Noetherian.

2. Basic properties

It is easy to prove the following:

Proposition 2.1

Let A be a Banach algebra. Then the following conditions are equivalent:

(a)	A is topologically Noetherian.
(b)	Every closed left ideal of A is topologically finitely generated.
(c)	Every non-empty family of closed left ideals of A, has a maximal element

Proposition 2.2

If A is a topologically Noetherian Banach algebra and I is a closed of A, then A/I is topologically Noetherian.

Next we prove a partial converse of Proposition 1.2.

Proposition 2.3

If A is a commutative Banach algebra and A/I is a topologically Noetherian for all closed 0 ≠ I ◁ A, then A is topologically Noetherian.

Proof

Let I₁ ⊆ I₂ ⊆ I₃ ⊆ … be any ascending chain of closed ideals of A. Then $I_2/I_1\subseteq I_3/I_1\subseteq \dots$ is an ascending chain of closed ideals of A/I₁. But A/I₁ is topologically Noetherian, hence there exists n ∊ Z⁺ such that $I_n/I_1=I_m/I_1$ for all m ≥ n. Hence I_m = I_n for all m ≥ n and A is topologically Noetherian.  □

Theorem 2.4

Let I be a closed ideal of Banach algebra A. If I and A/I are topologically Noetherian, then A is topologically Noetherian.

Proof

Let I₁ ⊆ I₂ ⊆ … be any ascending chain of closed left ideals of A. Then I₁ ∩ I ⊆ I₂ ∩ I ⊆ … is an ascending chain of closed left ideals of I and $\frac{I_1+I}{I}\subseteq \frac{I_2+I}{I}\subseteq \dots$ is an ascending chain of closed left ideals of A/I. But I and A/I are topologically Noetherian, hence there exists $r,s\in {\mathit{Z}}^{+}$ such that I_r ∩ I = I_n ∩ I for all r ≥ n and $\frac{I_s+I}{I}=\frac{I_n+I}{I}$ for all s ≥ n which implies that I_s + I = I_n + I for all s ≥ n. Let m = max{r, s}. Then I_m ∩ I = I_n ∩ I and I_m + I = I_n + I for all m ≥ n. But I_n ⊆ I_m, hence I_m = I_m ∩ (I_m + I) = I_m ∩ (I_n + I) = I_n + (I_m ∩ I) = I_n + (I_n ∩ I) = I_n for all m ≥ n by the modular law. Therefore, A is topologically Noetherian.  □

Corollary 2.5

A finite direct sum of topologically Noetherian Banach algebras is topologically Noetherian.

3. The ideal structures and some classification

In this section, we study the ideal structures of ideals in topologically Noetherian Banach algebra.

First we prove the following.

Proposition 3.1

Let A be a unital topologically Noetherian Banach algebra and I is a closed ideal of A. Then $\sqrt{I}$ = {x ∊ A:∃n ≥ 1, xⁿ ∊ I} is closed.

Proof

It is enough to show that $\sqrt{I}$ is the intersection of closed prime ideals of A.

So suppose that $x\notin \sqrt{I}$ and consider the family ℑ of all closed ideals J of A such that I ⊂ J and $x\notin \sqrt{I}$. Let M be a maximal element of ℑ. Suppose that there exist a, b ∊ A such that a, b ∉ M and ab ∊ M. Let N = {c ∊ A:cb ∊ M},then since N is a closed ideal of A and M ⊂ N it follows that N ∉ ℑ, so there is an n ∊ Z⁺ such that xⁿ ∊ N. Hence, xⁿc ∊ M. Arguing the same way with K = {d ∊ A:dxⁿ ∊ M},we get m ∊ Z⁺ such that x^m+n ∊ M. Hence x ∊ M which is a contradiction. Therefore M is prime.  □

An immediate consequence of Proposition 3.1, we have the following:

Corollary 3.2

Let A be a unital topologically Noetherian Banach algebra. Then nil(A)={ x ∊ A:∃n ≥ 1, xⁿ = 0}.

It is well known that any ideal in a commutative Noetherian ring contains a finite product of primes, see Cohn, (1979) page 404. A similar result holds for a commutative T.N Banach algebras as we prove in the following.

Proposition 3.3

Let A be a commutative T.N. Banach algebra and I is a closed ideal of A. Then

(a)	There are prime closed ideals P1, P2,…, Pm of A such that $\prod _{i=1}^m{P}_i\subseteq I.$
(b)	There exists m ∊ Z+ such that ${\left(\sqrt{I}\right)}^m\subseteq I$.

Proof

(a)

Let ℑ be the family of ideals in A which does not contain a product of closed prime ideals and suppose that ℑ ≠ φ. Then ℑ has a maximal element say J. Hence, J is not a prime ideal in A, so there are a, b∊ A such that ab ∊ J. With a ∉ I and b ∉ I. Let K = J + < a > , L = J + < b > . Then $\overline{K}$ and $\overline{L}$ are closed ideals of A and $J\subset \overline{K,}J\subset \overline{L}$. Hence, each one of them contains a finite product of closed prime ideals. But $\overline{K}\overline{L}\subset J$, hence J contains a finite product of closed prime ideals which is a contradiction. Therefore ℑ = φ.

(b)

Since $\prod _{i=1}^m{P}_i\subset I$ and $\sqrt{I}\subseteq P_i$ for all I, it follows that ${\left(\sqrt{I}\right)}^m\subseteq I$.  □

Next, we study primary ideals in T.N. Banach algebra.

Theorem 3.4

Let A be a commutative topologically Noetherian Banach algebra, then every closed ideal in A is a finite intersection of primary closed ideals.

Proof

It is easy to show that every closed ideal of A is a finite intersection of irreducible closed ideals, so it is enough to prove that every irreducible closed ideal is primary. Let I be an irreducible closed ideal of A. We need to show that I is primary. Since the homomorphic image of a T.N. Banach algebra is a T.N, it is enough to show that 0 is a closed primary ideal. It sufficient to show that if xy = 0∊ R, then either x = 0 or yⁿ = 0 for some n. First we claim that 〈x〉 ∩ 〈yⁿ〉 = 0.

To prove this, we consider the ascending chain

$\text{ann}⟨y⟩\subseteq \text{ann}⟨y^2⟩\subseteq \dots \subseteq \text{ann}⟨y^n⟩\subseteq \dots$ of closed ideals of A, since A is a T.N, then there exists n∊ Z⁺ such that $\text{ann}⟨y^n⟩=\text{ann}⟨y^{n+1}⟩$. Now if a ∊ 〈x〉 ∩ 〈yⁿ〉, then a = bx = cyⁿ for some b, c so ay = 0 since (bx)y = b(xy) = 0. Hence cyyⁿ = cyⁿ⁺¹ = 0. So $c\in \text{ann}⟨y^{n+1}⟩=\text{ann}⟨y^n⟩$. Therefore a = cyⁿ = 0 and 〈x〉 ∩ 〈yⁿ〉 = 0. Now by irreducibility of the zero ideal, it follows that 〈x〉 = 0 or 〈yⁿ〉 = 0. Hence x = 0 or yⁿ = 0.  □

Now we raise the following question:

Is any topologically Noetherian Banach algebra finite dimensional?

Note that, if A = C([0,1]), then every maximal ideal is t.f.g., but A is neither T.N. nor finite dimensional, however we have the following:

Theorem 3.5

Let A be a T.N. commutative Banach algebra such that every closed prime ideal of A is maximal, then A is finite dimensional.

Proof

By Proposition 3.3a, there are closed prime ideals P₁, P₂, …, P_m of A such that $\prod _{i=1}^m{P}_i=0.$ Note that P_i, P_j are comaximal, therefore by Bourbaki, (1974, prop.7, p. 108), $\prod _{i=1}^m{P}_i=0\Rightarrow \bigcap _{i=1}^m{P}_i=0$. But $A\cong A\bigcap _{i=1}^m{P}_i$ is imbedded in $\text{Dr}\prod _{i=1}^{m}A/P_i$, and A/P_i is finite dimensional, hence A is finite dimensional.   □

Corollary 3.6

Let A be a T.N. commutative Banach algebra. If every maximal ideal of A generated by idempotent element, then A is finite dimensional.

Proof

Let I be a closed primary ideal in A but not maximal ideal of A. Then I ⊂ M, where M is a maximal ideal of A. But M = < e >, where e² = e, and 0 ≠ e ≠ 1, hence e(1 − e) = 0 ∊ I and e ∉ I. Therefore (1 − e)ⁿ ∊ I ⊂ M. But M is a prime ideal of A, hence (1 − e) ∊ M. Therefore 1 ∊ M and M = A, which is a contradiction. Hence I is a maximal ideal of A. Therefore every closed prime ideal of A is maximal, and by Theorem 3.5, A is finite dimensional.  □

Additional information

Funding

Funding. The authors received no direct funding for this research.

Notes on contributors

Falih A.M. Aldosray

Falah Aldosary and Rachid El Harti are professors of mathematics in Umm-Al Qura University. They published several papers in national and international journals of mathematics.