# ‎‎If $A$ contains ‎an ‎idempotent $e‎$ (‎‎$‎e‎\neq ‎‎0,1‎‎$‎) , then $‎\Omega(A)‎$ ‎is ‎disconnected

If $A$‎ ‎be a‎ ‎unital ‎abelian ‎Banach ‎algebra ‎and ‎contains ‎an ‎idempotent $e$‎ ‎(that ‎is ‎‎$‎e=‎e‎^{‎2‎}‎‎$‎) ‎other ‎than $0$‎ ‎and $1$‎ ,‎ ‎then help me to show that ‎‎$‎\Omega(A)‎$ ‎is ‎disconnected.‎

$‎\Omega(A)‎$ is the set of characters on $A$, that is, the set of non-zero homomorphisms from $A$ to $\Bbb C$.

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‎‎$‎\Omega(A)=\{‎\tau: ‎A‎\longrightarrow‎‎ \mathbb{‎C}‎~‎|‎~‎‎\tau~‎‎is~‎‎a‎~‎non-‎zero~‎‎homomorphism ‎\}‎$ –  Ali Qurbani Jan 16 '13 at 13:37

Let $0\neq e\neq 1$ be an idempotent in $A$. Then prove that $\hat{e}:\Omega(A)\rightarrow \{0,1\}$ defined by $\phi\mapsto \phi(e)$ is continuous and onto.
Let $x_0\in A\setminus\{\mathbf 1,0\}$ such that $x_0^2=x_0$.
Let $S_1:=\{h\in \Omega(A),h(x_0)=0\}$ and $S_2:=\{h\in \Omega(A),h(x_0)=1\}$. This form a partition of $\Omega(A)$. As these sets are closed, what we have to show is that $S_1$ is not empty.
If it was, then by this thread, the spectrum of $x_0$ would be reduced to $0$. The formula of spectral radius gives a contradiction.