Kernel of Character on Unital Commutative Banach Algebra is Maximal Ideal/Proof 2

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Theorem

Let $\struct {A, \norm {\, \cdot \,} }$ be a unital commutative Banach algebra over $\C$.

Let $\phi : A \to \C$ be a character on $A$.


Then $\ker \phi$ is a maximal ideal of $A$.


Proof

Let $I$ be an ideal of $A$ such that:

$\ker \phi \subsetneq I$

We need to show $I = A$.

That is, we need to show:

${\mathbf 1}_A \in I$


Let:

$x \in I \setminus \ker \phi$

Then:

$\map \phi x \ne 0$

Thus we can define:

$\ds \tilde x := {\map \phi x}^{-1} x$

Then:

\(\ds \map \phi { {\mathbf 1}_A - \tilde x}\) \(=\) \(\ds \map \phi { {\mathbf 1}_A } - \map \phi {\tilde x}\)
\(\ds \) \(=\) \(\ds 1 - \map \phi {\tilde x}\) Definition of Unital Algebra Homomorphism
\(\ds \) \(=\) \(\ds 1 - \map \phi { {\map \phi x}^{-1} x}\)
\(\ds \) \(=\) \(\ds 1 - {\map \phi x}^{-1} \map \phi x\)
\(\ds \) \(=\) \(\ds 0\)

Thus:

${\mathbf 1}_A = \underbrace{ {\mathbf 1}_A - \tilde x}_{\in \ker \phi} + \underbrace{\paren { {\map \phi x}^{-1} {\mathbf 1}_A} }_{\in A} \underbrace{x}_{\in I} \in I$

$\blacksquare$