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Let $\mathcal{A}$ be a non (necessarily) unital commutative Banach algebra, and let $$ M_{\mathcal{A}} = \{ \phi:\mathcal{A} \to \mathbb{C} : \phi \mbox{ is multiplicative and not trivial}\} $$ and $$ \mathrm{Max}(\mathcal{A})=\{ I \lhd \mathcal{A} : I \mbox{ maximal} \}.$$ If $\mathcal{A}$ is unital, it is well known that there is a bijection between $M_{\mathcal{A}}$ and $\mathrm{Max}(\mathcal{A})$ sending each functional to its kernel (the inverse is given by the quotient and the Gelfand-Mazur theorem).

My question is,

is this still a bijection in the non-unital case?

I'm aware that if $\mathcal{A}$ is a commutative C*-algebra it is still a bijection. Also that the restriction gives a bijection from $M_{\tilde{\mathcal{A}}} \setminus \{ \pi:\tilde{\mathcal{A}} \to \mathbb{C} \}$ to $M_{\mathcal{A}}$; but this fact don't seem enough to conclude the result. I haven't been able to find a source for this.

Thanks in advance.

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    $\begingroup$ The correspondence fails when $\mathcal A$ is non-commutative, even if it is a C$^*$-algebra (canonical example: $B(H)$ has no characters but it has one unique---so maximal---ideal). $\endgroup$
    – Martin Argerami
    Commented Jul 13, 2016 at 1:24
  • $\begingroup$ I know about the non-commutative case, I'm asking about the non-unital but commutative case. I realized that I've failed to describe that in my original question, so I've edited it to reflect this. $\endgroup$
    – sjvega
    Commented Jul 13, 2016 at 15:14

2 Answers 2

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The kernels of nonzero homomorphisms to $\mathbb C$ are modular ideals, terminology that might help you find more references.

Without any further restriction on the algebras, using the zero product is a way to provide trivial counterexamples. E.g., take $\mathbb C$ with the $0$ product, which has maximal ideal $\{0\}$ and no nonzero homomorphisms to $\mathbb C$.

Googling led me to the following maybe more interesting example, Example 1.3 in this Feinstein and Somerset article: Take $C[0,1]$ with its usual Banach space structure but with multiplication $(f\diamond g)(t) = f(t)g(t)t$. Then the ideal of functions vanishing at $0$ is maximal but not the kernel of a nonzero homomorphism to $\mathbb C$.

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Linear-multiplicative functionals (aka characters) on complex Banach algebras are automatically continuous, so their kernels are closed. (You will find a slick proof of this fact on p. 181 of Allan's and Dales' Introduction to Banach Spaces and Algebras.) However, in the non-unital case it may well happen that a maximal ideal is dense.

The right notion to look at is the notion of a maximal modular ideal. Such ideals are bijectively associated to (kernels of) characters.

You may also be interested in this thread.

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