Theorem 2

Let A be a commutative Banach algebra. The mapping $\chi \mapsto$ Ker($\chi$ ) is a bijection of $\mathsf{X}(A)$ onto the set J(A) of regular maximal ideals of A.

Let I be a regular maximal ideal of A. It is closed (I, p. 22, cor. 2), hence $A/I$ is a Banach algebra. Since it is a field (A, VIII, p. 426, prop. 2), the Gelfand-Mazur theorem (I, p. 26, cor. 2) implies that $A/I$ is of dimension 1 over $\mathbf{C}$. Hence I is of codimension 1 in A. The theorem then follows from Lemma 3 of I, p. 10.

It may happen that a nonzero commutative Banach algebra A, without a unit element, has no maximal ideal; then $\mathsf{X}'(A)$ is reduced to $\{0\}($cf. I, p. 186, exercise 31).

This theorem shows that the sets J(A) and $\widehat{A}$ of I, p. 11, which are identified since A is commutative, may also be identified with the set $\mathsf{X}(A)$. There is therefore reason to consider on $\mathsf{X}(A)$ the weak topology, and the Jacobson topology which is less fine (I, p. 15, prop. 5). The closed sets V(M) of the Jacobson topology, for $M\subset A$, are the sets

of $\mathsf{X}(A)$, which will sometimes still be denoted by V(M). Similarly, for every subset M of $\mathsf{X}(A)$, we shall denote by Υ(M) the ideal intersection of the kernels of the $\chi \in M$; it is equal to the set $\Upsilon (M')$ defined in I, p. 13 for the subset $M'\subset J(A)$ corresponding to M when J(A) and $\mathsf{X}(A)$ are identified.

When a topological notion is used in $\mathsf{X}(A)$ without specifying which topology is meant, it will always be the weak topology.