Kernel of Character on Unital Commutative Banach Algebra is Maximal Ideal

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 1

From Kernel of Ring Homomorphism is Ideal, $\ker \phi$ is an ideal of $A$.

From the First Ring Isomorphism Theorem, we have:

$\phi \sqbrk A$ and $\dfrac A {\ker \phi}$ are isomorphic as rings.

From Character on Banach Algebra is Surjective, we have that $\phi \sqbrk A = \C$.

Hence:

$\dfrac A {\ker \phi} \cong \C$

That is:

$\dfrac A {\ker \phi}$ is a field.

From Maximal Ideal iff Quotient Ring is Field, we conclude that $\ker \phi$ is a maximal ideal of $A$.

$\blacksquare$

Proof 2

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:

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

$\blacksquare$