Surjectivity of the Gelfand Transform map This is a very basic question on understanding the Gelfand Transform...
Let $A$ be a unital commutative Banach Algebra.
Let $0\neq e\neq 1$ be an idempotent in $A$.
Let $\Omega(A)$ be the set of non-zero homomorphisms from $A$ to $\Bbb C$.
Show the map $\hat{e}:\Omega(A)\rightarrow \{0,1\}$ defined by $\phi\mapsto \phi(e)$ is surjective.
I see that we have $\phi(e ) = \phi(e^2) = \phi(e)^2= \cdots = \phi(e^n)= \phi(e)^n$.
Is this enough to show surjectivity? I'm a bit confused by the definitions.
Thank you for your help.
 A: The characters of $A$ (the nonzero homomorphisms $A\to\mathbb{C}$) are in a bijective correspondence with the maximal ideals of $A$. More precisely, the maximal ideals are exactly the kernels of the characters of $A$.
Now, since $e$ is an idempotent we have $e^2=e$, and so $e(e-1)=0$. If $e$ was invertible then we would get $e=1$, a contradiction. Thus $e$ is not invertible, and so it belongs to some maximal ideal $M\subseteq A$. (follows from Zorn's lemma) But this means $e$ belongs to the kernel of some character $\phi$, i.e there is $\phi\in\Omega(A)$ such that $\phi(e)=0$.
Also, since $e\ne 0$, the equality $e(e-1)=0$ implies that $e-1$ also cannot be invertible. So $e-1$ also belongs some maximal ideal, i.e to the kernel of some character $\varphi$. But then we get:
$0=\varphi(e-1)=\varphi(e)-\varphi(1)=\varphi(e)-1$
And so $\varphi(e)=1$ for this $\varphi$. So indeed there is a character $\varphi\in\Omega(A)$ such that $\varphi(e)=1$.
Alright, so your map $\Omega(A)\to\{0,1\}$ is indeed surjective.
