Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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
The formulation is rather confusing -- I think what you mean is what you get when you replace the equals sign after the "$W$" by "is"? – joriki Apr 6 '11 at 10:31
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
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
@Jesse Madnick: What do you mean? say nothing about boundary functions, one may of course consider the analytic Wiener algebra = space of absolutely convergent taylor series = $\ell^1(\mathbb{N})$. – AD. Apr 29 '11 at 19:18

1 Answer 1


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.