Exercise 15, § 4

Let $X$ be a completely regular space and let $\beta X$ be its Stone-Čech compactification. Endow the ring $A = \mathscr{C}(X; \mathbf{R})$ with the topology $\mathscr{T}$ defined in Exercise 14. For each $f \in A$, let $V(f)$ be the closure in $\beta X$ of the subset $\overline{f}(0)$ of $X$. For each subset $M$ of $A$, let $V(M)$ denote $\bigcap_{f \in M} V(f)$. For each subset $Y$ of $\beta X$, let $\mathfrak{J}(Y)$ denote the set of all $f \in A$ such that $Y \subset V(f)$, and write $\mathfrak{J}(x)$ in place of $\mathfrak{J}(\{x\})$. We have $V(f) = \varnothing$ if and only if $f$ is a unit in $A$.

a) If $f, g$ are two elements of $A$ such that $\overline{f}(0) \cap \overline{g}(0) = \varnothing$, show that $V(f) \cap V(g) = \varnothing$. [Consider the function $|f|/(|f| + |g|)$. Deduce that, for each subset $M$ of $A$, we have $V(M) = V(a)$, where $a$ is the ideal generated by $M$ in $A'$.

b) Show that, for each subset $Y$ of $\beta X$, $\mathfrak{J}(Y)$ is an ideal of $A$ [use a)] and that $V(\mathfrak{J}(Y))$ is the closure $\overline{Y}$ of $Y$ in $\beta X$.

c) Let $u \in A$ be such that $u(x) > 0$ for all $x \in X$, and let $g \in A$. Show that there exists $f \in A$ such that $|f - g| \leq u$ and such that $V(f)$ is a neighbourhood of $V(g)$. [Define $f$ to be equal to $g + u$ if $g(x) + u(x) < 0$, to $g - u$ if $g(x) - u(x) > 0$, and to 0 otherwise; consider the function $f' = \inf((u + g)^+, (u - g)^-)$ and use a).]

d) Let $a$ be an ideal of $A$. Show that if $f \in A$ is such that $V(f)$ is a neighbourhood of $V(a)$, then $f \in a$. [Note first that there exists $g \in a$ such that $V(g)$ is contained in the interior of $V(f)$, and then consider the function $h$ defined on $X$ which is equal to $f(x)/g(x)$ if $f(x) \neq 0$, and equal to $0$ if $f(x) = 0$.]

e) Deduce from c) and d) that $\mathfrak{J}(V(a)) = \overline{a}$ for every ideal $a$ of $A$. [Show first that $V(M) = V(\overline{M})$ for all subsets $M$ of $A$, arguing by contradiction.] Deduce that, for each subset $Y$ of $\beta X$, $\mathfrak{J}(Y)$ is a closed ideal of $A$. [Note that $\mathfrak{J}(Y) = \mathfrak{J}(V(\mathfrak{J}(Y)))$.]

f) Use e) to show that $x \to \mathfrak{J}(x)$ is a bijection of $\beta X$ onto the set of maximal ideals (necessarily closed) of $A$. An ideal of $A$ is closed if and only if it is an intersection of maximal ideals. The ring $A$ has no radical.