ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-22

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Free maximal ideals of C(N) and beta N

Example

Assume the Axiom of Choice (The Axiom of Choice). Let $N$ be discrete, so that $C (N, R) = R^{N}$ is the ring of all real sequences. Then there is a maximal ideal of $R^{N}$ which is not of the form ${f : f (n) = 0}$ for any $n \in N$ : the cofinite filter on $N$ extends to a free ultrafilter $U$ , and

$M_{U} = {f : Z (f) \in U}$

is a maximal ideal that is free. Consequently the maximal ideals of the ring of all continuous real functions on $N$ are in set-theoretic bijection with the points of the Stone–Čech compactification $β N$ , while the fixed ideals correspond exactly to $N$ (Gelfand-Kolmogorov for rings of continuous functions), and $β N ∖ N \neq \emptyset$ . No topology on the maximal-ideal set is asserted or reconstructed here.

Facts & Assumptions

Given: The Axiom of Choice, the discrete space $N$ , the ring $R^{N} = C (N, R)$ of all real sequences, and the cofinite filter on $N$ .

[L1]

For a Tychonoff space $X$ the maximal ideals of $C (X, R)$ are exactly the ideals $M_{p} = {f : p \in {\overline{Z (f)}}^{β X}}$ with $p \in β X$ unique, and $M_{p}$ is fixed if and only if $p \in X$ ; for $p \in X$ one has $M_{p} = {f : f (p) = 0}$ (Gelfand-Kolmogorov for rings of continuous functions, The Axiom of Choice).

[L2]

For discrete $N$ every subset is a zero set, because the characteristic function of any subset is continuous, and the z-filters are exactly the ordinary filters on $N$ ; the maximal ideals correspond to the filters that are maximal, and the fixed maximal ideals are the $M_{n} = {f : f (n) = 0}$ (Maximal ideals of C(X) and zero set ultrafilters).

[L3]

Under the Axiom of Choice every proper filter on a set extends to an ultrafilter (The ultrafilter lemma, from the Axiom of Choice: every filter extends to an ultrafilter, The Axiom of Choice).

[L4]

A subset of a compact Hausdorff space that is compact is closed, and $N$ with the discrete topology is not compact since the cover by singletons has no finite subcover; $N$ is dense in $β N$ . [algebra]

Verification

technique · direct

1.1

The cofinite filter $F = {A \subseteq N : N ∖ A finite}$ is a proper filter: it contains $N$ , omits $\emptyset$ (whose complement is infinite), and is closed under finite intersections and upward inclusion.

algebra

1.2

By [L3] extend $F$ to an ultrafilter $U$ ; since $F \subseteq U$ , no singleton belongs to $U$ (a singleton has infinite complement, so it is not in the cofinite filter, and its complement is in $U$ , so the singleton is not), that is, $U$ is free.

1.1L3algebra

2.1

$M_{U} = {f : Z (f) \in U}$ is a maximal ideal of $R^{N}$ by [L2], and it is not fixed: if $M_{U} = M_{n} = {f : f (n) = 0}$ for some $n$ , then for the characteristic function $f : = 1_{{n}}$ one has $Z (f) = N ∖ {n} \in U$ by freeness, so $f \in M_{U}$ while $f (n) = 1 \neq 0$ , so $f \notin M_{n}$ , a contradiction.

step 1.1step 1.2L2algebra

3.1

By [L1] the maximal ideals of $C (N, R)$ are the $M_{p}$ with $p \in β N$ uniquely determined, and by [step 2.1] there is a maximal ideal that is not fixed. By [L1] its point $p$ lies outside $N$ , so $β N ∖ N \neq \emptyset$ . This proves the point-set parametrisation claimed in the example; [L1] supplies no topology on the maximal-ideal set, and none is inferred.

step 2.1L1

4.1

Equivalently, $β N \neq N$ directly: $N$ is dense in the compact space $β N$ by [L4], so if $β N = N$ then $N$ would be compact, contradicting [L4].

L4algebra∎

Remarks

The unbounded function $n \mapsto n$ is never evaluated at infinity. Both the ideal $M_{U}$ and the identification $M_{p} = {f : p \in \overline{Z (f)}}$ use only zero sets; the witness $1_{{n}}$ is bounded, and no value of an unbounded sequence at a point of $β N ∖ N$ is asserted.
Free ultrafilters on $N$ exist under AC, and the resulting free maximal ideals are the algebraic shadow of the points at infinity of $β N$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

18 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, after Theorem 1, pp. 448–449. This endpoint was inaccessible in the current run; the exact local proof and failed recovery record are in the Batch 4 coverage ledger. (standard reference, not scraped)