Exercise 15, § 4

Let $X$ be a completely regular space, $\beta X$ its Stone-Čech compactification; equip the ring $A = \mathscr{C}(X; \mathbf{R})$ with the topology $\mathscr{T}$ defined in exerc. 14. For every $f \in A$, let $V(f)$ be the closure in $\beta X$ of the subset $f^{-1}(0)$ of $X$. For every subset $M$ of $A$, denote by $V(M)$ the intersection of the $V(f)$ for $f \in M$. For every subset $Y$ of $\beta X$, denote by $\mathscr{F}(Y)$ the set of the $f \in A$ such that $Y \subset V(f)$; we shall write $\mathscr{F}(x)$ instead of $\mathscr{F}(\{x\})$. In order that $V(f) = \varnothing$, it is necessary and sufficient that $f$ be invertible in $A$. a) If $f, g$ are two functions of $A$ such that $f^{-1}(0) \cap g^{-1}(0) = \varnothing$, now prove that one has also $V(f) \cap V(g) = \varnothing$ (consider the bounded continuous function $|f|/(|f| + |g|)$). Deduce that, for every subset $M$ of $A$, one has $V(M) = V(a)$, where $a$ is the ideal generated by $M$ in $A$. b) Now prove that for every subset $Y$ of $\beta X$, $\mathscr{F}(Y)$ is an ideal of $A$ (use $a$), and that one has $V(\mathscr{F}(Y)) = \overline{Y}$ (closure of $Y$ in $\beta X$). c) Let $u$ be a function of $A$ such that $u(x) > 0$ for every $x \in X$, $g$ a function of $A$. Now prove that there exists $f \in A$ such that $|f - g| \leq u$ and that $V(f)$ is a neighbourhood of $V(g)$. (Now prove that the function $f$ equal to $g + u$ if $g(x) + u(x) < 0$, to $g - u$ if $g(x) - u(x) > 0$, to 0 elsewhere, answers the question; one will consider the function $f' = \inf((u + g)^+, (u - g)^-)$ and use $a$). d) Let $a$ be an ideal of $A$; now prove that if $f \in A$ is such that $V(f)$ is a neighbourhood of $V(a)$, one has $f \in a$. (First remark that there exists $g \in a$ such that $V(g)$ is contained in the interior of $V(f)$, then consider the function $h$ defined in $X$, equal to $f(x)/g(x)$ if $f(x) \neq 0$, to 0 if $f(x) = 0$. e) Deduce from c) and d) that for every ideal $a$ of $A$, $\mathscr{F}(V(a)) = \overline{a}$. (One will first prove that for every subset $M$ of $A$, $V(M) = V(\overline{M})$, by reasoning by contradiction). Deduce that for every subset $Y$ of $\beta X$, $\mathscr{F}(Y)$ is a closed ideal of $A$ (remark that $\mathscr{F}(Y) = \mathscr{F}(V(\mathscr{F}(Y)))$). f) Conclude from e) that $x \mapsto \mathscr{F}(x)$ is a bijection of $\beta X$ onto the set of maximal ideals (necessarily closed) of $A$. For an ideal of $A$ to be closed, it is necessary and sufficient that it be an intersection of maximal ideals. The ring $A$ is without radical.