Assume $A$ is a commutative unital Banach algebra and $\tau : A \to \mathbb C$ is a character. I can prove that $I = \mathrm{ker}(\tau)$ is a maximal ideal using some basic abstract aglebra. The problem is, there is an alternative argument that I do not understand. Could someone please help me understand it? Here goes:
For every $a \in A$: $a-\tau (a) \in I$. Therefore that $I + \mathbb C 1 = A$. Therefore $I$ is maximal.
Why does $a-\tau (a) \in I$ for all $a $ imply $I + \mathbb C 1 = A$?
I assume $\tau:A \to \Bbb C$ is non-trivial, that is, there is some $a \in A$ with $\tau(a) \ne 0$. Then $\tau(a) = \tau(1_A a) = \tau(1_A)\tau(a)$, which leads directly to $\tau(1_A) = 1 \in \Bbb C$.
We also observe that
$\tau \left (\dfrac{a}{\tau(a)} \right ) = \dfrac{\tau(a)}{\tau(a)} = 1, \tag 1$
another way to see the existence of elements $c \in A$ with $\tau(c) = 1 \in \Bbb C$.
Note that the expression $a - \tau(a)$ means, in this context, $a - \tau(a)1_A$, where $1_A \in A$ is the unit element of $A$. Clearly $I + \Bbb C 1_A \subset A$. For any $a \in A$, $a - \tau(a)1_A \in I$ since
$\tau(a - \tau(a)1_A) = \tau(a) - \tau(a)\tau(1_A) = \tau(a) - \tau(a)1 = 0; \tag 2$
that is, for some $b \in I$, $a - \tau(a)1_A = b$. Then $a = b + \tau(a)1_A \in I + \Bbb C 1_A$ since $\tau(a) \in \Bbb C$, so $A \subset I + \Bbb C 1_A$. Thus in fact $A = I + \Bbb C 1_A$. That's why.
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!
