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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?

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    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.2011-04-06

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Hints:

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 $\phi\colon W\to W/M\cong\mathbb C$.

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    @Matt E: Even though it doesn't make a difference in this case, I thought I'd mention that the result is true without assuming commutativity (i.e., $\mathbb{C}$ is the unique Banach division algebra). I admit that this is not an apt point here, because the commutativity is needed to conclude that a simple algebra is a division algebra.2011-04-10