# Functional Analysis question

1. Apr 12, 2006

### 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 Im 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 Im 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.

2. Apr 12, 2006

### 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 Im 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?

3. Apr 12, 2006

### 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?

4. Apr 12, 2006

### 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. Im 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...

Last edited: Apr 12, 2006
5. Apr 19, 2006

### Oxymoron

Ive been given a hint on how Im 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. :(