Maximal Ideal of non-unital Banach algebra I've studied Gelfand's theory of commutative Banach algebra in textbooks on Functional Analysis like Rudin's or Peter Lax's. It seems to me that when concerning maximal ideal of Banach algebra, these textbooks only deal with unital Banach algebra.
On the other hand, $l_1(\mathbb{C})$ is an example of non-unital Banach algebra (multiplication defined by $((xy)_n)=(x_n\cdot y_n)$) and I found that maximal ideal of $l_1(\mathbb{C})$ is in one-to-one correspondence with integer $\mathbb{Z}$. Apparently, $A_i=\{(x_n)|x_i=0\}$ is a maximal ideal of $l_1(\mathbb{C})$.
I wonder whether all maximal ideals can be written in the form of $A_i$.
Update: As Ryszard points out in his answer, any maximal ideal other than $A_i\ (i\in\mathbb{Z}^+)$, must contain the ideal $A_f$ which consists of the elements with finitely many nonzero coordinates, and must not be closed.
 A: Let $M$ be a closed maximal ideal of $\ell^1.$
We will show that $M\subset M_n=\{x\in \ell^1\,:\,x_n=0\}$ for some $n.$
Assume that for every $n$ there is $y^{(n)}\in M$ such that $y^{(n)}_n\neq 0.$ Then
$$y^{(n)}\delta_n =y^{(n)}_n\delta_n\in M$$
Hence $\delta_n\in M$ for every $n.$ Thus
$$\sum_{n=-N}^Na_n\delta_n\in M$$ for any $N$ and any coefficients $a_n.$ As $M$ is closed then $\ell^1=M,$ a contradiction.
The closedness assumption seems essential, but there is no complete proof. Below there are some arguments in this direction. Let $M_f=\mathcal{F}(\mathbb{Z})$ denote the elements with finitely many nonzero coordinates. Then $M_f$ is an ideal, not contained in any ideal $M_n.$ Let $y=\{2^{-|n|}\}_{n\in \mathbb{Z}}.$ Let $$\mathcal{M}_y=\{M\,:\, M\ {\rm ideal},\ M_f\subset M,\ y\notin M\}$$   The family $ \mathcal{M}_y$ is nonempty as $M_f\in \mathcal{M}_y,$ and partially ordered with respect to the inclusion. Moreover every chain in $\mathcal{M}_y$ is bounded by the union of ideals in this chain. By Zorn's lemma the family contains a maximal element $M.$ We have $M\subsetneq \ell^1$ as $y\notin M$ and $M\neq M_n$ for any $n$ as $M_f\subset M.$ Assume $M\subsetneq N,$ where $N$ is an ideal in $\ell^1.$ Then $y\in N.$ In order to show that $M$ is a maximal ideal in $\ell^1$ we have to prove that $N=\ell^1.$ In case of unital algebra the role of $y$ is played by the unit $e.$ Therefore the conclusion is obvious.
