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We know that a unital commutative Banach algebra $\mathcal{A}$ has a compact maximal ideal space $M_{\mathcal{A}}$. But I am curious about the converse. If the maximal ideal space $M_{\mathcal{A}}$ is compact, can we say that the commutative Banach algebra $\mathcal{A}$ is unital?

Let $\mathcal{A}$ be a non-unital commutative Banach algebra. We know that $\mathcal{A}$ can be viewed as an maximal ideal in $\tilde{\mathcal{A}}$ (its unitalization) and $M_{\mathcal{A}}$ can be identified with $M_{\tilde{\mathcal{A}}}\setminus \{\phi_0\}$, where $\phi_0$ is the multiplication functional which associate to the ideal $\mathcal{A}$. Because of this, we can say that $M_{\mathcal{A}}$ is a locally compact space. But several notes in the internet farther says that it is non-compact. How can we confirm that $\phi_0$ is not an isolated point in $M_{\tilde{\mathcal{A}}}$?

My attempt:

To say that $\phi_0$ is not an isolated point in $M_{\tilde{\mathcal{A}}}$. It is equivalent to say that there's a net $\{\phi_{\alpha}\}$ in $M_{\mathcal{A}}$ converge to $0$ weakly. But I have no idea to do this.

But If $\mathcal{A}$ is a commutative C*-algebra with compact maximal ideal space $M_{\mathcal{A}}$, it is easy to derive the unitalness of $\mathcal{A}$ by the fact that $\mathcal{A}$ is isometrically *-isomorphic to $C_0(M_{\mathcal{A}})=C(M_{\mathcal{A}})$, since $M_{\mathcal{A}}$ is compact.

Can any one enlighten me for advance?

If I'm wrong, please give me some references or counterexamples if possible.

Thanks a lot!

yoyo
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  • Locally compact corresponds to (not necessarily unital) commutative $C^*$-algebra. Compact corresponds to unital. – user399601 Jun 23 '17 at 03:18
  • Sorry~I have edited it. Let $\mathcal{A}$ be non-unital in second paragraph. – yoyo Jun 23 '17 at 06:05
  • So I am asking why compact maximal ideal space implies unital? – yoyo Jun 23 '17 at 06:08
  • @yoyo The corresponding algebra is $C_0(X) = C(X)$ which contains $1$. But you mentioned that in the question already. – user399601 Jun 23 '17 at 06:28
  • @user399601 My question is "If the maximal ideal space $M_{\mathcal{A}}$ is compact, can we say that the commutative Banach algebra $\mathcal{A}$ is unital?". Your argument $C(X)$ do not relevant to $\mathcal{A}$. You can see the answer below by Eric Wofsey. The answer is false. – yoyo Jun 23 '17 at 10:32
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    Sorry, I was thinking that the whole question was about $C^*$-algebras – user399601 Jun 23 '17 at 15:30

1 Answers1

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This is false. For instance, let $\mathcal{A}$ be $1$-dimensional with every product in $\mathcal{A}$ equal to zero. Then $\mathcal{A}$ is not unital, but it has only one maximal ideal, namely $\{0\}$, so its maximal ideal space is compact.

(Here I am assuming that by "maximal ideal" you mean "maximal proper linear subspace closed under multiplication by elements of $\mathcal{A}$". Note that with this definition, it is not true that $M_{\mathcal{A}}$ can be identified with $M_{\tilde{\mathcal{A}}}\setminus \{\phi_0\}$: in the example above, the maximal ideal $\{0\}$ of $\mathcal{A}$ does not correspond to any maximal ideal of $\tilde{\mathcal{A}}$. If you define "maximal ideal" instead as "kernel of a nonzero algebra homomorphism $\mathcal{A}\to\mathbb{C}$" then the example still works, since $\mathcal{A}$ has no maximal ideals in this sense. If you define "maximal ideal" in the usual sense of ideals in a ring (which in particular does not require an ideal to be a linear subspace), then the example works once again, since $\mathcal{A}$ has no maximal ideals in this sense either.)

Eric Wofsey
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  • Thank you for your perfact example! And you point out an important thing! Yes, I need to re-confirm the definition of the "maximal ideal". – yoyo Jun 23 '17 at 09:06