Proposition 4

Let A be a commutative regular Banach algebra without radical. Let F be a closed subset of $\mathsf{X}(A)$. The set of ideals I of A such that V(I) = F, ordered by inclusion, has a greatest element, namely Υ(F), and a smallest element, namely the set J of $x\in A$ such that $\mathscr{G}(x)$ has compact support disjoint from F.

By construction, Υ(F) is an ideal of A such that V(Υ(F)) contains F, and it is the greatest ideal of A having this property. Since F is closed, there exists an ideal I of A such that V(I) = F; hence $I\subset \Upsilon (F)$, whence $V(\Upsilon (F))\subset V(I) = F$, so that V(Υ(F)) = F. This proves the first assertion.

The set J is an ideal of A and V(J) contains F. Let us show that V(J) = F. Let $\chi \in \mathsf{X}(A)$ not belonging to F. Let U be a compact neighbourhood of $\chi$ not meeting F (TG, I, p. 65, cor. of prop. 9). By assertion (ii) of prop. 1 of I, p. 88, there exists $x\in A$ such that $\mathscr{G}(x)$ is equal to 1 at $\chi$ and to 0 outside U. We then have $x\in J$ and therefore $\chi \notin V(J)$. This shows that $V(J)\subset F$ and therefore V(J) = F.

Finally, let I be an ideal of A such that V(I) = F. Let us show that $J\subset I$. Let $x\in J$ and let C be the support of $\mathscr{G}(x)$; the subset C is a compact subset of $\mathsf{X}(A)$ disjoint from F. By Prop. 3, there exists an element $u\in I$ such that $\mathscr{G}(u) = 1$ on C. We then have $\mathscr{G}(x) =\mathscr{G}(ux)$, and therefore $x=ux$ since A is without radical (Prop. 8 of I, p. 38). Consequently, we have $x\in I$, which shows that $J\subset I$.