TITLE: Correspondence between maximal ideals and multiplicative functionals of a non unital, commutative Banach algebra.
QUESTION [11 upvotes]: 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.

REPLY [5 votes]: 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.