# Let A be a commutative Banach algebra. Let &#x03C7;<!-- χ --> 1 </msub> an

Let $A$ be a commutative Banach algebra. Let ${\chi }_{1}$ and ${\chi }_{2}$ be characters of $A$.
I am having some difficulty seeing why the following statement is true:
If $\mathrm{ker}{\chi }_{1}=\mathrm{ker}{\chi }_{2}$, then since ${\chi }_{1}\left(\mathbf{\text{1}}\right)={\chi }_{2}\left(\mathbf{\text{1}}\right)=\mathbf{\text{1}}$, we have that ${\chi }_{1}={\chi }_{2}$.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Miles Mueller
Suppose a is an element of the algebra, and ${\chi }_{1}\left(a\right)=\lambda$. Then $a-\lambda \mathbf{1}\in \mathrm{ker}{\chi }_{1}=\mathrm{ker}{\chi }_{2}$ so
$0={\chi }_{2}\left(a-\lambda \mathbf{1}\right)={\chi }_{2}\left(a\right)-\lambda$