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Maximal ideals of C(X) and zero set ultrafilters

Statement

Let $X$ be a Tychonoff space and let $C (X) = C (X, R)$ be the ring of all continuous real functions with pointwise operations (Zero set filter and zero set ultrafilter). Then the two assignments

$M ⟼ Z [M] : = {Z (f) : f \in M}, U ⟼ M_{U} : = {f \in C (X) : Z (f) \in U}$

are mutually inverse bijections between the set of maximal ideals of $C (X)$ (Prime ideals and maximal ideals in a commutative ring) and the set of z-ultrafilters on $X$ ; that is, $Z [M_{U}] = U$ for every z-ultrafilter and $M_{Z [M]} = M$ for every maximal ideal.

No choice principle is used: the argument is a theorem of ZF; functions may be unbounded and no norm on $C (X)$ is involved.

Facts & Assumptions

Given: A Tychonoff space $X$ , the ring $C (X)$ of all continuous real functions with pointwise operations, and the family $Z (X)$ of zero sets.

[L1]

$Z (X)$ is closed under finite intersections with $Z (f) \cap Z (g) = Z (f^{2} + g^{2})$ ; $X = Z (0)$ and $\emptyset = Z (1)$ ; a z-filter is a family of zero sets containing $X$ , omitting $\emptyset$ , closed under finite intersections and upward closed in $Z (X)$ ; a z-ultrafilter is a maximal z-filter (Zero set filter and zero set ultrafilter).

[L2]

An ideal of the commutative ring $C (X)$ is a subgroup closed under multiplication by arbitrary elements, and it is maximal when it is maximal among proper ideals; the ring has unit the constant function $1$ , so an ideal is proper exactly when it omits $1$ (Prime ideals and maximal ideals in a commutative ring).

[L3]

If $f \in C (X)$ has $Z (f) = \emptyset$ then $f (x) \neq 0$ for all $x$ , so $1 / f$ is continuous and $f$ is invertible in $C (X)$ (Zero set filter and zero set ultrafilter).

Proof

technique · direct

1.1

Let $M$ be a maximal ideal of $C (X)$ . Then $Z [M]$ is a z-filter: it contains $X = Z (0)$ because $0 \in M$ , it omits $\emptyset$ because $Z (f) = \emptyset$ would make $f$ invertible by [L3] and force $1 \in M$ by [L2], and it is closed under finite intersections because $Z (f) \cap Z (g) = Z (f^{2} + g^{2})$ with $f^{2} + g^{2} \in M$ [L1].

L1L2L3algebra

1.2

For a z-ultrafilter $U$ the family $M_{U}$ is a proper ideal: it contains $0$ since $Z (0) = X \in U$ ; it is closed under addition because $Z (f) \cap Z (g) = Z (f^{2} + g^{2}) \subseteq Z (f + g)$ and $U$ is upward closed; it is closed under multiplication by $h \in C (X)$ because $Z (f) \subseteq Z (h f)$ ; and it is proper because $1 \in M_{U}$ would give $\emptyset = Z (1) \in U$ .

L1L2algebra

1.3

A separation property of z-ultrafilters. If $U$ is a z-ultrafilter and $Z \in Z (X)$ with $Z \notin U$ , then there is $Z^{'} \in U$ with $Z \cap Z^{'} = \emptyset$ : otherwise $Z$ meets every member of $U$ , and then $U^{'} : = {Z^{''} \in Z (X) : Z^{''} \supseteq Z \cap Z^{'} for some Z^{'} \in U}$ is a z-filter: it contains $Z$ (for any $Z^{'} \in U$ one has $Z \supseteq Z \cap Z^{'}$ ), so it is nonempty; it omits $\emptyset$ , because $\emptyset = Z^{''} \supseteq Z \cap Z^{'}$ would force $Z \cap Z^{'} = \emptyset$ , contrary to the standing assumption that $Z$ meets every member of $U$ ; it is upward closed by definition; and it is closed under finite intersections because $Z_{1}^{''} \supseteq Z \cap Z_{1}^{'}$ and $Z_{2}^{''} \supseteq Z \cap Z_{2}^{'}$ give $Z_{1}^{''} \cap Z_{2}^{''} \supseteq Z \cap (Z_{1}^{'} \cap Z_{2}^{'})$ with $Z_{1}^{'} \cap Z_{2}^{'} \in U$ ; since $U^{'} \supseteq U$ and $Z \in U^{'} ∖ U$ , this contradicts the maximality of $U$ .

L1algebra

2.1

With $M$ maximal as in [step 1.1], $Z [M]$ is upward closed in $Z (X)$ : let $f \in M$ and let $h \in C (X)$ satisfy $Z (f) \subseteq Z (h)$ ; if $h \notin M$ , then maximality gives $1 = m + a h$ for some $m \in M$ and $a \in C (X)$ , and the function $w : = f^{2} + m^{2} \in M$ satisfies $w > 0$ everywhere, because at a point with $f (x) = 0$ one has $h (x) = 0$ and then $m (x) = 1 - a (x) h (x) = 1$ , while at a point with $f (x) \neq 0$ one has $w (x) \geq f (x)^{2} > 0$ ; hence $1 / w$ is continuous and $1 = w \cdot (1 / w) \in M$ , contradicting the properness of $M$ . So $h \in M$ , and $Z [M]$ is a z-filter by [step 1.1].

step 1.1L1L2algebra

2.2

For a z-ultrafilter $U$ the ideal $M_{U}$ is maximal: let $N \supseteq M_{U}$ be a proper ideal and let $h \in N$ ; if $Z (h) \notin U$ then by [step 1.3] there is $Z (g) \in U$ with $Z (g) \cap Z (h) = \emptyset$ , so $g \in M_{U} \subseteq N$ and $g^{2} + h^{2} \in N$ ; but $Z (g^{2} + h^{2}) = \emptyset$ , so $g^{2} + h^{2}$ is invertible by [L3] and $1 \in N$ , contradicting properness. Hence $Z (h) \in U$ and $h \in M_{U}$ , so $N = M_{U}$ .

step 1.2step 1.3L1L2L3

3.1

For a maximal ideal $M$ the z-filter $Z [M]$ is maximal: if $W \supseteq Z [M]$ is a z-filter and $h \in C (X)$ has $Z (h) \in W$ , then either $h \in M$ and hence $Z (h) \in Z [M]$ , or $h \notin M$ and maximality gives $1 = m + a h$ with $m \in M$ , so that $Z (m) \cap Z (h) = \emptyset$ (a common zero would give $1 = 0$ ); now $Z (m) \in Z [M] \subseteq W$ and $Z (h) \in W$ , so $\emptyset \in W$ by closure under intersections, contradicting that $W$ is a z-filter. Hence every member of $W$ is a member of $Z [M]$ , and $Z [M] = W$ .

step 1.1step 2.1L1L2algebra

3.2

The assignments are inverse: for a maximal ideal $M$ , $f \in M_{Z [M]}$ means $Z (f) = Z (m)$ for some $m \in M$ , hence $Z (m) \subseteq Z (f)$ and $f \in M$ by the upward-closure argument of [step 2.1]; conversely $f \in M$ gives $Z (f) \in Z [M]$ ; so $M_{Z [M]} = M$ . For a z-ultrafilter $U$ , $Z (f) \in Z [M_{U}]$ means $f \in M_{U}$ , that is, $Z (f) \in U$ ; so $Z [M_{U}] = U$ .

step 1.2step 2.1

4.1

By [step 3.1] the assignment $M \mapsto Z [M]$ sends maximal ideals to z-ultrafilters, by [step 2.2] the assignment $U \mapsto M_{U}$ sends z-ultrafilters to maximal ideals, and by [step 3.2] the two are inverse; hence they are mutually inverse bijections.

step 2.2step 3.1step 3.2∎

Remarks

The two ingredients of maximality. The forward direction uses that a maximal ideal is prime-like through the identity $1 = m + a h$ ; the reverse direction uses the separation property [step 2.2] of z-ultrafilters, which is a repackaging of maximality for z-filters.
No normality or compactness. The argument uses only the ring structure of $C (X)$ and the lattice identity for zero sets; Tychonoffness is used only to have the class of spaces for which the later $β X$ statements are formulated.

Depends on

Used by

Free maximal ideals of C(N) and beta N Example
Gelfand-Kolmogorov for rings of continuous functions Theorem

Dependency tree · two levels

16 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

L. Gillman, M. Henriksen and M. Jerison, On a Theorem of Gelfand and Kolmogoroff Concerning Maximal Ideals in Rings of Continuous Functions (1954) — §2, Theorem 1 proof, pp. 448–449. This endpoint was inaccessible in the current run; the complete local alternative and failed recovery record are in the Batch 4 coverage ledger. (standard reference, not scraped)