Bourbaki

Exercise 15, § 2

Commutative Algebra, II, 126

Bourbaki marks this one as harder than the rest.

Let XX be a completely regular topological space and βX\beta X its Stone-Čech compactification (General Topology, Chapter IX, § 1, Exercise 7). For all x∈Xx \in X, let g(x)\mathfrak{g}(x) denote the maximal ideal of the ring C(X;R)\mathscr{C}(X; \mathbf{R}) consisting of those f∈C(X;R)f \in \mathscr{C}(X; \mathbf{R}) such that xx belongs to the closure of the sub set f‾(0)\overline{f}(0) of XX (General Topology, Chapter X, § 4, Exercise 15); let u(x)\mathfrak{u}(x) denote the ideal consisting of those f∈C(X;R)f \in \mathscr{C}(X; \mathbf{R}) such that f‾(0)\overline{f}(0) is the trace on XX of a neighbourhood of xx in XX; u(x)\mathfrak{u}(x) is equal to its radical.

An ideal aa of C(X;R)\mathscr{C}(X; \mathbf{R}) is called isolated (resp. absolutely isolated) if, for all f≥0f \geq 0 on aa, every gg such that 0≤g≤f0 \leq g \leq f (resp. ∣g∣≤f|g| \leq f) belongs to aa (cf. Algebra, Chapter IV, § 1, Exercise 4); then C(X;R)/a\mathscr{C}(X; \mathbf{R})/a possesses an order structure (resp. a lattice-order structure) compatible with its ring structure and for which the elements ≥0\geq 0 are the classes mod. aa of the elements ≥0\geq 0 to C(X;R)\mathscr{C}(X; \mathbf{R}) (loc. cit.).

(a) Show that an ideal aa of C(X;R)\mathscr{C}(X; \mathbf{R}) which is equal to its radical is absolutely isolated (if f∈af \in a and ∣g∣≤∣f∣|g| \leq |f|, consider the function equal to 0 when f(x)=0f(x) = 0 and to (g(x))2/f(x)(g(x))^2/f(x) when f(x)≠0f(x) \neq 0). In this case for C(X;R)/a\mathscr{C}(X; \mathbf{R})/a to be totally ordered, it is necessary and sufficient that aa be prime; the set of prime ideals of C(X;R)/a\mathscr{C}(X; \mathbf{R})/a is then totally ordered by inclusion.

(b) For an isolated ideal aa of C(X;R)\mathscr{C}(X; \mathbf{R}) to be such that C(X;R)/a\mathscr{C}(X; \mathbf{R})/a is totally ordered, it is necessary that aa be contained in a single maximal ideal g(x)\mathfrak{g}(x) and then aa necessarily contains u(x)\mathfrak{u}(x); it is sufficient that aa contain a prime ideal and then aa is absolutely isolated (note that the relation fg=0fg = 0 then implies f∈af \in a or g∈ag \in a).

(c) Take X=RX = \mathbf{R}, x=0x = 0; show that the principal ideal generated by the function t↦∣t∣t \mapsto |t| is contained in g(0)\mathfrak{g}(0) and contains u(0)\mathfrak{u}(0) but is not isolated; the principal ideal generated by the identity mapping t↦tt \mapsto t is contained in g(0)\mathfrak{g}(0), contains u(0)\mathfrak{u}(0) and is isolated but not absolutely isolated.

(d) Show that, if pp and qq are two prime ideals in C(X;R)\mathscr{C}(X; \mathbf{R}), then pq=p∩qpq = p \cap q (and in particular p2=pp^2 = p) and p+qp + q is prime or equal to C(X;R)\mathscr{C}(X; \mathbf{R}). (If p+q≠C(X;R)p + q \neq \mathscr{C}(X; \mathbf{R}), observe that p+qp + q is absolutely isolated and equal to its radical and use (a).)

(e) For u(x)\mathfrak{u}(x) to be prime, it is necessary and sufficient that, for every function f∈C(X;R)f \in \mathscr{C}(X; \mathbf{R}), the closure in βX\beta X of one of the two sets f1‾([0,+∞[)\overline{f^1}([0, +\infty[), f1‾([−∞,0])\overline{f^1}([-\infty, 0]) is a neighbourhood of xx. Deduce an example where u(x)\mathfrak{u}(x) is prime and not maximal (consider the topological space associated with a non-trivial ultrafilter; cf. General Topology, Chapter I, § 6, no. 5).

(f) Suppose that x∈Xx \in X. For U(x)=F(x)\mathscr{U}(x) = \mathfrak{F}(x), it is necessary and sufficient that every countable intersection of neighbourhoods of xx in XX be a neighbourhood of xx in XX. (Cf. General Topology, Chapter I, § 7, Exercise 7.)

(g) Show that C(X;R)\mathscr{C}(X; \mathbf{R}) is canonically identified with the total ring of fractions of C∞(X;R)\mathscr{C}^\infty(X; \mathbf{R}).

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.