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!
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.)