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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.

Wembley Inter
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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.

Ryszard Szwarc
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  • I wonder how to prove the first line? Is maximal ideal non-unital Banach algebra still always closed? – Wembley Inter Sep 07 '22 at 22:55
  • It is the kernel of a continuous multiplicative linear functional.By continuity you get that the kernel is closed – Ryszard Szwarc Sep 07 '22 at 22:56
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    As in this question, such correspondence might not exist on specific non-unital Banach algebra. – Wembley Inter Sep 07 '22 at 23:14
  • You are right. I will double check the details and re-edit tomorrow, whether in your case the correspondence holds. – Ryszard Szwarc Sep 07 '22 at 23:29
  • I wonder if my answer is OK now. If I remember well from old times by Zorn's lemma every ideal is contained in a maximal ideal. – Ryszard Szwarc Sep 08 '22 at 05:28
  • Thanks! The problem you raised in the last sentence is exactly what I'm wondering about before posting this on stackexchange. – Wembley Inter Sep 08 '22 at 06:22
  • Thanks for the update! I'm wondering that while $M_y$ is the maximal element in $\mathcal{M}_y$, is it that possible that $M_y$ is not a maximal ideal of $l_1$ or even for several $y$, $M_y$ might be contained in the same maximal ideal of $l_1$. – Wembley Inter Sep 08 '22 at 13:31
  • To me, I even doubt that whether there exists a maximal ideal containing the $M_f$ you defined. – Wembley Inter Sep 08 '22 at 13:46
  • I understand such process, but after you prove the existence of maximal element in this family, what is its relation to the maximal ideal of $l_1$? – Wembley Inter Sep 08 '22 at 14:17
  • You are right. For every ideal $N$ such that $M\subsetneq N$ we have $y\in N.$ What's missing is a proof (if possible) that $N=\ell^1.$ – Ryszard Szwarc Sep 08 '22 at 14:25