Exercise 16, § 4

We retain the hypotheses and notation of Exercise 15.

a) Let $a$ be a closed ideal in the ring $A$. Show that in the quotient ring $A/a$ the set $P$ of canonical images of functions $f \geq 0$ of $A$ is the set of elements $\geq 0$ with respect to an ordering which makes $A/a$ a lattice-ordered ring [observe, using Exercise 15 e), that if $f$ and $g$ belong to $a$, then so do $|f| + |g|$ and every function $h \in A$ such that $|h| \leq |f|$]. The canonical image in $A/a$ of the set of constant functions is isomorphic to $\mathbf{R}$ as an ordered field.

b) In particular, for each $x \in \beta X$, the field $A/\mathfrak{J}(x)$ is canonically endowed with an ordered field structure. The maximal ideal $\mathfrak{J}(x)$ is said to be real if $A/\mathfrak{J}(x)$ is isomorphic to $\mathbf{R}$, hyper-real otherwise. For $\mathfrak{J}(x)$ to be hyper-real it is necessary and sufficient that $x \in \beta X - X$ and that there is a unit $f \in A$ such that $\lim_{y \to x, y \in X} f(y) = 0$. Deduce that all the maximal ideals of $A$ are real if and only if $X$ is pseudo-compact (Chapter IX, § 1, Exercise 21).

c) For each $x \in \beta X$, let $\varphi_x$ be the canonical homomorphism $A \to A/\mathfrak{J}(x)$. For $\varphi_x(f)$ to be $\geq 0$ in $A/\mathfrak{J}(x)$, it is necessary and sufficient that there should exist $g \in \mathfrak{J}(x)$ such that $f(y) \geq 0$ in $\overline{g}^{-1}(0)$ [observe that the relation $\varphi_x(f) \geq 0$ is equivalent to $f \equiv |f| \pmod{\mathfrak{J}(x)}$]. For $\varphi_x(f)$ to be $> 0$, it is necessary and sufficient that there should exist $g \in \mathfrak{J}(x)$ such that $f(y) > 0$ in $\overline{g}^{-1}(0)$. [Note that, if $f \notin \mathfrak{J}(x)$, there exists $g \in \mathfrak{J}(x)$ such that $\overline{f}^{-1}(0) \cap \overline{g}^{-1}(0) = \emptyset$, by using Exercise 15.]

d) Show that if $\mathfrak{J}(x)$ is hyper-real, the transcendence degree of the field $A/\mathfrak{J}(x)$ over $\mathbf{R}$ is at least equal to $c = \mathrm{Card}(\mathbf{R})$. [Let $f > 0$ be a unit of $A$ such that $\varphi_x(f) = u$ is infinitely large with respect to $\mathbf{R}$. Show that as $r$ runs through the elements of a base of $\mathbf{R}$ over $\mathbf{Q}$ (consisting of numbers $> 0$), the elements $\varphi_x(f^r)$ of $A/\mathfrak{J}(x)$ are algebraically independent.]

\* e) For each point $a = (a_1, \ldots, a_n) \in \mathbf{R}^n$, let $f_a(X)$ denote the polynomial $X^n + a_1 X^{n-1} + \cdots + a_n$. For each real number $\xi$, let $v(\xi)$ be the sum of the multiplicities of the zeros $z$ of $f_a$ in $\mathbf{C}$ such that $R(z) = \xi$; and for each integer $k$ such that $1 \leq k \leq n$, let $\rho_k(a)$ be the smallest real number $\xi$ such that $\sum_{\eta \leq \xi} v(\eta) \geq k$. Show that each of the functions $\rho_k$ is continuous on $\mathbf{R}^n$ (use Rouché’s theorem).

f) Show that, for each $x \in \beta X$, the ordered field $A/\mathfrak{J}(x)$ is real-closed. [Let $F(X) = X^n + f_1 X^{n-1} + \cdots + f_n$ be a polynomial of odd degree with coefficients in $A$; the functions $g_k(y) = \rho_k(f_1(y), \ldots, f_n(y))$, where the $\rho_k$ are the functions defined in e), are continuous on $X$, and for each $y \in X$ there is an index $k$ such that $F(g_k(y)) = 0$; deduce that the product $F(g_1) \cdots F(g_n)$ belongs to $\mathfrak{J}(x)$.]

g) Let $X$ be an infinite discrete space and let $y \to F_y$ be a bijection of $X$ onto the set of finite subsets of $X$; for each $z \in X$ let $M_z$ be the set of all $y \in X$ such that $z \in F_y$; these sets form a base of a filter $\mathfrak{F}$ on $X$. Let $x$ be a point of $\beta X$ such that the corresponding ultrafilter on $X$ (Chapter I, § 9, Exercise 27) is finer than $\mathfrak{F}$. Let $B$ be a subset of $A$ such that $\mathrm{Card}(B) \leq \mathrm{Card}(X)$, and let $y \to g_y$ be a surjective mapping $X \to B$. For each $z \in X$ put $f(z) = 1 + \sup_{y \in F_z} g_y(z)$. Show that $f(z) > g_y(z)$ for all $z \in M_y$, and deduce that $\varphi_x(f) > \varphi_x(g_y)$ for all $y \in Y$ [use c)]. Conclude that $\mathrm{Card}(A/\mathfrak{J}(x)) > \mathrm{Card} X$.