# Suppose that A is a unital commutative Banach algebra. It is a nice application of the Shilov

Suppose that $A$ is a unital commutative Banach algebra. It is a nice application of the Shilov idempotent theorem that if the spectrum of $A$ is totally disconnected, then $A$ is regular.
Can we show that idempotents in $A$ are linearly dense if the spectum of $A$ is totally disconnected?
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rideonthebussp
A totally disconnected spectrum doesn't imply a dense span of the idempotents.
Let's look first at the silly example ${A}_{0}={\mathbb{C}}^{2}$, endowed with the norm $‖\left(x,y\right){‖}_{{A}_{0}}=|x|+|y|$ and the multiplication $\left(u,v\right)\cdot \left(x,y\right)=\left(ux,uy+vx\right)$, so ${A}_{0}\cong \mathbb{C}\left[X\right]/\left({X}^{2}\right)$, endowed with the ${\ell }^{1}$-norm. One verifies that this is a unital commutative Banach algebra, and its only idempotent elements are ${e}_{0}=\left(1,0\right)$ and $0$, so the linear span of the idempotents is not dense. The spectrum of ${A}_{0}$ is a singleton (the only unital algebra homomorphism to $\mathbb{C}$ is the first coordinate projection), so it's totally disconnected (and connected).
We use that to construct a less silly example, in which the spectrum is totally disconnected and not connected. Let ${A}_{1}$ be a unital commutative Banach algebra with (nonempty) totally disconnected spectrum, for example ${A}_{1}=C\left(K,\mathbb{C}\right)$, where $K$ is the Cantor set (${A}_{0}$ would also work). Let $A={A}_{0}×{A}_{1}$, with componentwise multiplication, and endow it with the norm $‖\left(x,y\right){‖}_{A}=max\left\{‖x{‖}_{{A}_{0}},‖y{‖}_{{A}_{1}}\right\}$. That makes $A$ a unital commutative Banach algebra, and the idempotents in $A$ are the pairs $\left(x,y\right)$ where $x$ is an idempotent in ${A}_{0}$ and $y$ an idempotent in ${A}_{1}$. So the span of idempotents is not dense, $\left(\left(0,1\right),0\right)$ does not belong to the closure of that span. Since $\left\{0\right\}×{A}_{1}$ are ideals in $A$, it follows that every unital homomorphism $A\to \mathbb{C}$ is either ${\psi }_{0}:\left(\left(u,v\right),w\right)↦u$ or $\left(x,y\right)↦\lambda \left(y\right)$ with $\lambda \in \mathrm{\Delta }\left({A}_{1}\right)$. Thus $\mathrm{\Delta }\left(A\right)\cong \mathrm{\Delta }\left({A}_{1}\right)\cup \left\{{\psi }_{0}\right\}$, where ${\psi }_{0}$ is an isolated point, so totally disconnected.
It would be interesting to investigate whether the presence of nilpotent elements is the only thing that can make the span of idempotents non-dense if the spectrum is totally disconnected.