I am studying the spectral theory of operators on Banach and Hilbert spaces, reading through Conway's A Course in Functional Analysis. In section VII.4, Conway states Proposition 4.11 as a consequence of the spectral mapping theorem.
Proposition 4.11 (existence of nontrivial idempotents)
Let $\mathcal{A}$ be a Banach algebra with identity. Suppose that $a \in \mathcal{A}$ with $\sigma(a) = F_1 \cup F_2$, where $F_1,F_2$ are disjoint, non-empty closed subsets of $\mathbb{C}$. Then there exists a nontrivial idempotent $e \in \mathcal{A}$ such that
(i) If $b \in \mathcal{A}$ commutes with $a$, then $b$ commutes with $e$.
(ii) If $a_1 = ae$ and $a_2 = a(1-e)$, then $a = a_1 + a_2$ and $a_1a_2 = a_2a_1 = 0$.
(iii) $\sigma(a_1) = F_1 \cup \{0\}$, $\sigma(a_2) = F_2 \cup \{0\}$.
Conway comments that a "nicer" conclusion for this proposition would be that $\sigma(a_i) = F_i$ for $i = 1,2$, and that this conclusion can be obtained in a certain sense:
I would like to show that the set $\mathcal{A}_1 \equiv \mathcal{A}e$ is also a Banach algebra, and that if $a_1$ is considered as an element of of $\mathcal{A}_1$, then it's spectrum as an element of $\mathcal{A}_1$ is equal to $F_1$.
My first hurdle is figuring out how to properly define multiplication in $\mathcal{A_1}$ since it's not immediately clear that, for $a, b \in \mathcal{A}$, we have $aebe \in \mathcal{A_1}$. The next step would be to properly define the norm on $\mathcal{A_1}$. Finally one needs to demonstrate that $\sigma_{\mathcal{A}_1}(a_1) = F_1$. I imagine that $\sigma_{\mathcal{A}_1} \subseteq \sigma_{\mathcal{A}}(a_1)$ can be done with a fairly straightforward argument, but then a little more work will be required to show that if $0 \notin F_1$, then $a_1$ is invertible in $\mathcal{A}_1$.
Any hints our solutions are greatly appreciated.