LemmaStatement: Literature-sourcedProof: Literature-sourcedPipeline-generatedaudited 2026-09-06

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Banaschewski's prime obstruction

Statement

Let $X$ be finite and let $A \subseteq Z^{X}$ be a subring. If $1_{X} \notin A$ , then there are $x \in X$ and a prime $p$ such that $a (x) \in p Z$ for every $a \in A$ . In particular, this applies to the integer-valued hyperelementary permutation subring evaluated on conjugacy classes.

Facts & Assumptions

[F1]

The cited prerequisite is The character of a permutation representation counts fixed points.

Proof

Given: $A$ is closed under pointwise multiplication and addition.

1.1

For $x \in X$ , the value set $A_{x} = {a (x) : a \in A}$ is an ideal $n_{x} Z$ . If every $A_{x}$ contained $1$ , choose $a_{x} \in A$ with $a_{x} (x) = 1$ and form $\prod_{x \in X} (1_{X} - a_{x}) = 0$ .

F1givenassume-hypcontrapositive-reduce

2.1

Expanding the finite product would put $1_{X}$ in $A$ , a contradiction. Hence some $A_{x} = n_{x} Z$ has $n_{x} > 1$ ; any prime divisor $p$ of $n_{x}$ has the stated property. Permutation characters are integer-valued fixed-point counts, so their span is a subring of this form. ∎

step 1.1discharge-contrapositive

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 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

Wen-Wei Li, Yanqi Lake Lectures on Algebra I, Lemma 14.3.4 (standard reference, not scraped)