Character on Unital Banach Algebra is Uniquely Identified by Kernel

Theorem

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

Let $\phi, \psi : A \to \C$ be characters on $A$ such that:

$\ker \phi = \ker \psi$

Then $\phi = \psi$.

From Character on Unital Banach Algebra is Unital Algebra Homomorphism, we have $\map \phi { {\mathbf 1}_A} = 1$ and $\map \psi { {\mathbf 1}_A} = 1$.

Let $x \in A$.

We have:

$\ds \map \phi {x - \map \phi x {\mathbf 1}_A}$

$=$

$\ds \map \phi x - \map \phi {\map \phi x {\mathbf 1}_A}$

$\ds $

$=$

$\ds \map \phi x - \map \phi x \map \phi { {\mathbf 1}_A}$

$\ds $

$=$

$\ds \map \phi x - \map \phi x$

$\ds $

$=$

$\ds 0$

So we have $x - \map \phi x {\mathbf 1}_A \in \ker \phi$.

Since $\ker \phi = \ker \psi$, we have $x - \map \phi x {\mathbf 1}_A \in \ker \psi$.

So, we have:

$\map \psi {x - \map \phi x {\mathbf 1}_A} = 0$

We therefore obtain that:

$\ds 0$

$=$

$\ds \map \psi {x - \map \phi x {\mathbf 1}_A}$

$\ds $

$=$

$\ds \map \psi x - \map \psi {\map \phi x {\mathbf 1}_A}$

$\ds $

$=$

$\ds \map \psi x - \map \phi x \map \psi { {\mathbf 1}_A}$

$\ds $

$=$

$\ds \map \psi x - \map \phi x$

So we have $\map \phi x = \map \psi x$.

Since $x \in A$ was arbitrary, we obtain $\phi = \psi$.

$\blacksquare$