Let $ I $ be a closed ideal of a commutative unital Banach algebra such that $\mathcal{A}/I\ncong \mathbb{C}$.
Let $[a]\in \mathcal{A}/I$ such that $[a]\notin \mathbb{C}[1]$ and let $\lambda\in \sigma ([a])$. Then, $a-\lambda1\notin I$ and $I+\mathcal{A}(a-\lambda1)$ is an ideal in $\mathcal{A}$ such that $I+\mathcal{A}(a-\lambda1)\not = \mathcal{A}.$
Therefore, $ I $ is not an maximal ideal.
I have already proven that $ a- \lambda1 \notin I $ and that $ I + \mathcal{A} (a- \lambda1) $ is an ideal. But I have not been able to prove that $ I + \mathcal{A} (a- \lambda1) \not = \mathcal{A} $, an idea is to show that $ 1 \notin I + \mathcal {A} (a- \lambda1) $. How can I do this? How can I conclude that $ I $ is not a maximal ideal? Thanks so much for reading.
Assume that $I+\mathcal A(a-\lambda 1) = \mathcal A$. Then, there are $x\in I$ and $y\in A$ such that: $$1 = x+(a-\lambda 1)y$$ Reducing this modulo $I$ yields: $$[1] = ([a]-\lambda[1])[y]$$ Therefore, $[a]-\lambda[1]\in\mathcal A/I$ is invertible, i. e. $\lambda\in\sigma([a])$. That is a contradiction to the choice of $\lambda$.
Thus, $I+\mathcal A(a-\lambda 1) \neq \mathcal A$. Since $[a]\notin\mathbb C[1]$, $[a]-\lambda[1]\neq [0]$, i. e. $a-\lambda\notin I$.
This implies that $I$ is not a maximal ideal because of the following proper inclusions of ideals: $$I\subsetneq I+\mathcal A(a-\lambda 1)\subsetneq\mathcal A$$