Spectra of operators

Question

Please help me proof a theorem: If $\mathfrak{U}$ is a complex, commutative Banach algebra with identity and $x\in\mathfrak{U}$, then

$$ \sigma(x)=\{\phi(x):\phi \text{ is a homomorphism of } \mathfrak {U} \text{ onto } \mathbb{C}\} $$

Answer 1

First, some elementary facts about onto homomorphisms $h$. We have $h(0)=0$ and $h(e)=1$. If $x$ is invertible then $\phi(x)\neq 0$.

If $\lambda=\phi(x)$ for such a homomorphism, then $\phi(x-\phi(x)e)=0$ which implies that $x-\phi(x)e$ is not invertible hence $\lambda\in \sigma(x)$.

If $\lambda\neq\phi(x)$ for all $\phi$, we have to show that $x-\lambda e$ is invertible. We have that $x-\lambda e\notin\ker\phi$ for all $\phi$ an onto homomorphism. In Rudin's book Functional analysis, it's shown that each maximal ideal is the kernel of an onto homomorphism and each strict ideal is contained in a maximal ideal. Can you conclude from that?

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The result actually says that $\widehat x$, the Gelfand transform of $x$, defined on the set of onto homomorphisms by $\widehat x(\phi):=\phi(x)$, has the spectrum of $x$ a codomain.