1
$\begingroup$

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?

  • 0
    Also, the symbol $\mathbb{T}$ for the unit circle is not very common; it wouldn't hurt to introduce it.2011-04-06
  • 0
    Done =).. I still need a help =\2011-04-06
  • 0
    Yes -- I was helping you by making it easier for others who might be able to help you to understand your question :-)2011-04-06
  • 2
    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?2011-04-06
  • 3
    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

1 Answers 1

3

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

  • 0
    One could add that the fact that $\mathbb C$ is the unique commutative Banach field is implicitly being invoked in your third step.2011-04-06
  • 0
    Rasmus-Could you please explain more?2011-04-08
  • 0
    @saba: Yes, what confuses you? Are you familiar with the character theory of commutative Banach algebras?2011-04-08
  • 0
    No, I am not familiar with the character theory of commutative Banach algebras.2011-04-08
  • 2
    Your problem is a good opportunity to become so. http://en.wikipedia.org/wiki/Banach_algebra#Ideals_and_characters2011-04-08
  • 0
    I almost get it :\.. but how to show that every maximal ideal $M$ is of the form $M_z$ for some $z$? Please... I am afraid that I will have to give up ? :S2011-04-08
  • 0
    Well, *if* $M$ is of the form $M_z$ then the character $W\to W/M$ takes the identity function on the circle to $z$, right? So, if you start with a maximal ideal $M$ then the image of the identity function in $W/M$ is your considate for $z$.2011-04-09
  • 0
    @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