Exercise 16, § 4

Les hypotheses and notations are those of exerc. 15. a) Let $a$ be a closed ideal in the ring $A$. Now prove that in the quotient ring $A/a$, the set $P$ of the canonical images of the functions $f \geq 0$ of $A$ is the set of the elements $\geq 0$ for an ordered structure making $A/a$ a lattice-ordered ring (observe, using exerc. 15 e) that if $f, g$ belong to $a$, the same is true of $|f| + |g|$ and of 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 $\mathbf{R}$ for its ordered field structure.

b) In particular, for every $x \in \beta X$, the field $A/\mathscr{F}(x)$ is endowed canonically with an ordered field structure. One says that the maximal ideal $\mathscr{F}(x)$ is real if $A/\mathscr{F}(x)$ is isomorphic to $\mathbf{R}$, hyper-real in the contrary case. For $\mathscr{F}(x)$ to be hyper-real, it is necessary and sufficient that $x \in \beta X - X$ and that there exists an invertible function $f \in A$ such that $\lim_{y \to x, y \in X} f(y) = 0$. Deduce that for all the maximal ideals of $A$ to be real, it is necessary and sufficient that $X$ be weierstrassian (IX, p. 88, exerc. 20).

c) For every $x \in \beta X$, let $\varphi_x$ be the canonical homomorphism $A \to A/\mathscr{F}(x)$. For $\varphi_x(f) \geq 0$ in $A/\mathscr{F}(x)$, it is necessary and sufficient that there exists $g \in \mathscr{F}(x)$ such that $f(y) \geq 0$ in $g(0)$ (remark that the relation $\varphi_x(f) \geq 0$ is equivalent to $f \equiv |f|$ (mod. $\mathscr{F}(x)$)). For $\varphi_x(f) > 0$, it is necessary and sufficient that there exists $g \in \mathscr{F}(x)$ such that $f(y) > 0$ in $g(0)$ (remark that if $f \notin \mathscr{F}(x)$, there exists $g \in \mathscr{F}(x)$ such that $f(0) \cap g(0) = \varnothing$, using exerc. 15).

d) Now prove that if $\mathscr{F}(x)$ is hyper-real, the degree of transcendence (A, V, § 14) of the field $A/\mathscr{F}(x)$ over $\mathbf{R}$ is at least equal to $c = \mathrm{Card}(\mathbf{R})$. (Let $f > 0$ be an invertible element of $A$ such that $\varphi_x(f) = u$ is infinitely large with respect to $\mathbf{R}$ (A, VI, § 2, exerc. 1). Prove that when $r$ runs through the elements of a basis of $\mathbf{R}$ over $\mathbf{Q}$ (formed of numbers $> 0$), the elements $\varphi_x(f^r)$ of $A/\mathscr{F}(x)$ are algebraically independent).

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

f) Prove that for every $x \in \beta X$, the ordered field $A/\mathscr{F}(x)$ is maximal (A, VI, § 2, No. 5). (Let $F(X) = X^n + f_1 X^{n-1} + \cdots + f_n$ be a polynomial of odd degree whose coefficients belong to $A$; 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 in $X$ and for every $y \in X$, there is an index $k$ such that $F(g_k(y)) = 0$; deduce that the product $F(g_1) \ldots F(g_n)$ belongs to $\mathscr{F}(x)$.*

g) Let $X$ be an infinite discrete space, $y \mapsto F_y$ a bijection of $X$ onto the set of finite subsets of $X$; for every $z \in X$, let $M_z$ be the set of the $y \in X$ such that $z \in F_y$; these sets form a basis of a filter $\mathscr{F}$ on $X$; let $x$ be a point of $\beta X$ such that the corresponding ultrafilter on $X$ (I, p. 110, exerc. 27) is finer than $\mathscr{F}$. Let $B$ be a subset of $A$ such that $\mathrm{Card}(B) \leq \mathrm{Card}(X)$, and let $y \mapsto g_y$ be a surjective mapping $X \to B$. For every $z \in X$, put $f(z) = 1 + \sup_{y \in F_z} g_y(z)$. Now prove that $f(z) > g_y(z)$ for every $z \in M_y$, and deduce that $\varphi_x(f) > \varphi_x(g_y)$ for every $y \in Y$ (use c)). Conclude that $\mathrm{Card}(A/\mathscr{F}(x)) > \mathrm{Card}(X)$.