Proposition 14

Let A be a commutative Banach algebra without radical. In order that A admit a unit element, it is necessary and sufficient that $\mathsf{X}(A)$ be compact.

The condition is necessary (I, p. 29, corollaire). Suppose $\mathsf{X}(A)$ compact. Let $\widetilde{A}$ be the Banach algebra deduced from A by adjunction of a unit element, and identify $\mathsf{X}'(A)$ with $\mathsf{X}(\widetilde{A})$. The complement of $\mathsf{X}(A)$ in $\mathsf{X}(\widetilde{A})$ reduces to the character $\chi_0$ of $\widetilde{A}$ whose kernel is A. The subsets $\mathsf{X}(A)$ and $\{\chi_0\}$ are open in $\mathsf{X}(\widetilde{A})$. By Prop. 13, there exists an element $j\in A$ such that $\chi (j) = 1$ for $\chi \in \mathsf{X}(A)$, and $\chi_0(j) = 0$. Hence $j\in A$.

Let then $x$ be in A. We have $\chi (jx) =\chi (x)$ for every $\chi \in \mathsf{X}(A)$, hence $jx=x$ since A is without radical. Thus, $j$ is a unit element of A.