Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 10 / 5 / c

A character of an algebra is a nonzero multiplicative complex linear functional $\varphi :A\to \mathbb{C}$. Nonzeroness and multiplicativity force $\varphi (1)=1$, so it is onto $\mathbb{C}$. Its kernel is an ideal, and the quotient is the field $\mathbb{C}$, making the kernel maximal.

Continuity is automatic. The element $a-\varphi (a)1$ cannot be invertible, since applying $\varphi$ to an inverse equation would give $0=1$. On the other hand $a-\lambda 1$ is invertible whenever $|\lambda |>\Vert a\Vert$, by the Neumann series. Hence $|\varphi (a)|\le \Vert a\Vert$, the automatic continuity of characters bound.

Conversely let $M$ be a maximal ideal. Its closure is an ideal, because multiplication is continuous. That closure is proper: otherwise there would be $m\in M$ with $\Vert 1-m\Vert <1$, making $m$ invertible by the Neumann series and forcing $M=A$. Maximality gives $\overline{M}=M$. The quotient $A/M$ is consequently a Banach algebra. It is a division algebra: for any $a\notin M$, the larger ideal $M+Aa$ equals $A$, so the class of $a$ has an inverse. By the Gelfand-Mazur theorem, $A/M\cong \mathbb{C}$, and its quotient map gives a continuous character of an algebra with kernel $M$.

If two algebra characters have the same kernel, then $a-\varphi (a)1$ lies in the second kernel, giving $\psi (a)=\varphi (a)$ for every $a$. Thusis a bijection. This proves that maximal ideals of a commutative complex unital Banach algebra are character kernels.