Character on Banach Algebra is Surjective/Proof 2

Theorem

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

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

Then $\phi$ is surjective.

Proof

As $\phi$ is non-zero, there exists an $x_0 \in A$ such that:

$\map \phi {x_0} \in \C \setminus \set 0$

Thus, for each $a \in \C$:

\(\ds \frac a {\map \phi {x_0} } x_0\)

\(\in\)

\(\ds A\)

as $A$ is a $\C$-algebra

and:

\(\ds \map \phi {\frac a {\map \phi {x_0} } x_0}\)

\(=\)

\(\ds \frac a {\map \phi {x_0} } \map \phi {x_0}\)

as $\phi$ is a $\C$-algebra homomorphism

\(\ds \)

\(=\)

\(\ds a\)

$\blacksquare$