= Solution
For every <character of an algebra> $\phi$, the element $a-\phi(a)1$ cannot be invertible, so $\phi(a)\in\sigma_A(a)$. Conversely, if $\lambda\in\sigma_A(a)$, the principal <ideal> $A(a-\lambda1)$ is proper: otherwise $b(a-\lambda1)=1$ 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 $\phi(a)=\lambda$. Therefore
$$
\boxed{\sigma_A(a)=\{\phi(a):\phi\in\Phi_A\}=\widehat a(\Phi_A).}
$$
Here $\widehat a(\phi)=\phi(a)$ 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.
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