Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to show that the maximal ideal space of the Wiener algebra $W$ is $ \{ M_z : z \in \mathbb{T} \}$ where $M_z = \{ g \in W : g(z)=0 \}$

Could you please help me?

share|cite|improve this question
Also, the symbol $\mathbb{T}$ for the unit circle is not very common; it wouldn't hurt to introduce it. – joriki Apr 6 '11 at 10:34
Done =).. I still need a help =\ – saba Apr 6 '11 at 10:52
Yes -- I was helping you by making it easier for others who might be able to help you to understand your question :-) – joriki Apr 6 '11 at 10:58
You need to show three things: a) Each $M_z$ is an ideal. b) Each of these ideals is maximal. c) There are no other maximal ideals. Which of these are you having difficulties with, and what have you tried? – joriki Apr 6 '11 at 11:16
You may have misunderstood my comment about $\mathbb{T}$ -- I didn't mean to say you should make the T blackboard bold (though it's good that you did); I meant that people might not know that this refers to the unit circle and you should introduce it by defining it. The point of all these comments is that there are a lot of people here (like myself) who don't specifically know much about the Wiener algebra and the notation used in its context, but know enough about maximal ideals to be able to help you nevertheless. – joriki Apr 6 '11 at 11:37


The Wiener algebra is a commutative Banach algebra.

To see that the $M_z$ is a maximal ideal, write it as the kernel of a character.

To see that every maximal ideal $M$ is of the form $M_z$ for some $z$, consider the image of the identity function under the quotient map $\phi\colon W\to W/M\cong\mathbb C$.

share|cite|improve this answer
One could add that the fact that $\mathbb C$ is the unique commutative Banach field is implicitly being invoked in your third step. – Matt E Apr 6 '11 at 19:22
Rasmus-Could you please explain more? – saba Apr 8 '11 at 12:46
@saba: Yes, what confuses you? Are you familiar with the character theory of commutative Banach algebras? – Rasmus Apr 8 '11 at 12:54
No, I am not familiar with the character theory of commutative Banach algebras. – saba Apr 8 '11 at 13:15
Your problem is a good opportunity to become so. – Rasmus Apr 8 '11 at 13:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.