Norm of the projection onto a maximal ideal

Question

Let $A$ be a complex Banach algebra and $\chi \ne 0$ be a complex character. Consider the quotient space $\hat A = \dfrac A {\ker \chi} \simeq \Bbb C$. If $\hat x \in \hat A$, how can one quickly prove that $|\hat x| \le \| x \|$, i.e. that the norm of the natural projection is $\le 1$ (is it exactly $1$?)?

Answer 1

Every complex character is continuous, so $\ker(\chi)$ is a closed ideal.

Now in general, when $A$ is a normed linear space and $F<A$ is a closed subspace, then the quotient map $$ \pi : A\to A/F $$ has norm $\leq 1$ because of the definition of the quotient norm as $$ \|\pi(x)\| = \inf\{\|x-y\| : y\in F\} \leq \|x\| $$ and $\|\pi\|=1$ if $F \neq A$ (which it is in this case since $\chi \neq 0$).