Functional Analysis question

[Oxymoron]

I have a commutative Banach algebra A with identity 1. If A contains an element e such that e^2 = e and e is neither 0 nor 1 (I think this also means to say that it contains a non-trivial idempotent), then the maximal ideal space of A is disconnected.

Currently I am trying to show this but I am not getting very far. Here is a summary of what I think I may need to show this:

Because the question involves the maximal ideal space I am assuming I have to use the Gelfand transform somewhere. In particular it might be interesting to see what the Gelfand transform of the idempotent element e is.

 

[Oxymoron]

Question: Does the idempotent e allow us to decompose the Banach algebra A into a direct sum of ideals:

A = I_1\oplus I_2

If so, I could write the first ideal as being the set generated by e, ie.

I_1 = eA = \{x\in A\,:\,x = ex\}

and

I_2 = \{x\in A\,:\,ex = 0\}

although I am not too sure about the relevancy of these two ideals, or if they are at all constructible.

Although, what I am sure about is that if I can create such ideals then it is clear that they are disjoint (closed) ideals and that for any element of the Banach algebra, x,

x = ex+(1-ex)

which is an element of each ideal, and so

A = I_1\oplus I_2

The problem remains: Does the idempotent allow me to construct these two disjoint ideals? What is the motivation behind such a decomposition? Does such a decomposition even help me show that the maximal ideals space \Delta_A is disconnected?

 

[matt grime]

Forget functional analysis, it's not important.

Just let A be a ring, m is a maximal ideal iff A/m is a field or at least an integral domain, right? So what do e and (1-e) quotient out to if m is maximal and e is an idempotent (forget non-trivial too; that's an unnecessry proof by contradiction) and what do integral domains (fields) not have?

 

[Oxymoron]

Matt, thankyou for this new angle for approaching my problem. I must say I understand a little bit more about algebra than I do about functional analysis.


An integral domain is a commutative ring with unity 1\neq 0 and does not have any divisors of 0. All fields are integral domains.

If M is my maximal ideal in A, then since A is commutative with unity, A/M is also a nonzero commutative ring with unity IF M = A.

Let (a+M)\in A/M, with a\neq M so that a+M is not the additive identity element of A/M. Suppose that a has no multiplicative inverse in A/M. Then the set

(A/M)(a+M) = \{(b+M)(a+M)\,|\,(b+M)\in A/M\}

does not contain 1+M. So (A/M)(a+M) is an ideal of A/M.

It is non-trivial because a\notin M, and it is proper because it does not contain 1+M. But this contradicts our assumption that M is a maximal ideal, therefore a+M must have a multiplicative inverse in A/M.


This is how one of my proofs from algebra last year went. I am not sure how I will get to say that the quotient'ing' of the two elements e and e-1 will give me disjoint subsets other than one will not contain elements of the other. I believe I am still lacking the crucial information to make that call. I will keep thinking...

 

[Oxymoron]

Ive been given a hint on how I am meant to do this problem:

Consider the case A = C(X) where X is compact. Then apply this case to \hat{e} (the Gelfand transform of e), and you will want to prove that \hat{e} is neither 0 or 1. The spectral radius formula and the observation that (1-e)^2 = 1-e should help.


Im still struggling to show that \Delta_A is not connected. :(