the Banach algebra, invertible element Let $A$ be an abelian Banach algebra with identity, $a, b \in A$. How can one show:


*

*$a \in A$ is invertible if only if  for all characters $\varphi‎ ‎\in‎ ‎\Omega(A)$‎, one has $\varphi(a)‎\neq ‎0‎$‎.

*‎if ‎$ ‎\lambda‎ ‎\neq ‎0‎ $‎‎, ‎then ‎‎‎‎‎‎‎$ \{a, b \in A : ab -ba = ‎\lambda1\} =:  \phi‎ $ ‎‎‎is empty.
 A: *

*If $\varphi$ is a character, that is a homomorphism into $\mathbb C$ that is not the zero map, you have $\varphi(ab)=\varphi(a)\varphi(b)$ for all $a,b\in A$. Since you have $\varphi(a)=\varphi(1a)=\varphi(1)\varphi(a)$ and $\varphi\neq0$ you must have $\varphi(1)=1$.
Suppose now $a$ is invertible, given the previous information, what can you see if you look at $\varphi(a)\varphi(a^{-1})$? Is there any way to see $\varphi(a)\neq0$?
If for any non-invertible $a$ there exists a character $\psi$ so that $\psi(a)=0$ then you get:

$\varphi(a)\neq0$ for all characters $\implies$ $a$ is invertible

so you need to find a character for a non-invertible element that sends the element to zero. First you have to know that there is a one-to-one correspondence between maximal ideals in $A$ and characters of $A$. This correspondence works in that the character associated to $I$ is the projection map $A\to A/I$ where $A/I$ is canonically isomorphic to $\mathbb C$.
Now if $a$ is not invertible, the ideal generated by $a$ is a proper ideal. Every proper ideal is contained in a maximal ideal, so you have a maximal ideal $I$ that contains $a$. Can you see why the projection $A\to A/I$ will in this case send $a$ to $0$?

*Your algebra is abelian. This means $ab-ba$ is always equal to zero for any $a,b\in A$. How can this fact be used to solve part 2?
