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