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I have a short question regarding operator algebras. Given an abelian Banach algebra $\mathcal{A}$. Assume that $\phi \in \big\{ \phi : \phi \text{ is a non-zero linear multiplicative functional} \big\} := \Sigma_{\mathcal{A}}$, let $\mathcal{M} = \text{ker}\phi$.

Why does it follows that $\mathbb{C} \cong \phi(\mathcal{A}) \cong \mathcal{A}/ \text{ker}~\phi$?

Thanks for any assistance.

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$\phi(A)\neq\lbrace 0 \rbrace$ is a (complex) subspace of $\mathbb C$ and therefore equal to $\mathbb C$. The second part is a general fact about quotients: Whenever $\phi: X\to Y$ is a surjective linear map between vector spaces then $\tilde T: X/\mathrm{ker} T \to Y$ is a linear bijection.

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    $\begingroup$ Also note that $\tilde{T}$ is multiplicative because $\phi$ is, and $\tilde{T}$ is continuous since $X/\ker(\phi)$ is finite dimensional (since $\mathbb{C}$ is). Hence one gets an isomorphism of Banach algebras (and not just of vector spaces). $\endgroup$
    – Prahlad Vaidyanathan
    Commented Sep 30, 2015 at 9:37

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