 Solution

For every character of an algebra $\varphi$, the element $a-\varphi (a)1$ cannot be invertible, so $\varphi (a)\in {\sigma }_A(a)$. Conversely, if $\lambda \in {\sigma }_A(a)$, the principal ideal $A(a-\lambda 1)$ is proper: otherwise $b(a-\lambda 1)=1$ for some $b$, which in this commutative algebra gives an inverse. Part (b) puts that ideal inside a maximal ideal $M$, and the corresponding algebra character from part (c) has $\varphi (a)=\lambda$. ThereforeHere $\hat{a}(\varphi )=\varphi (a)$ is the Gelfand transform. The spectrum equals character values in a commutative Banach algebra result does not require the Gelfand transform to be injective: the algebra need not be semisimple.