CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passaudited 2026-09-04

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.

Z_p is not the integral closure of Z in Q_p

Statement refuted

The ring $Z_{p}$ is the integral closure of $Z$ inside $Q_{p}$ .

Facts & Assumptions

Given: The ring $Z_{p} \subseteq Q_{p}$ .

[L1]

$Z_{p}$ sits inside $Q_{p}$ and every element of $Z_{p}$ has a unique digit expansion (The p-adic completion agrees with the fraction field of Z_p, Z_p is the valuation ring of Q_p, Every p-adic number has a unique digit expansion).

[L2]

Being integral over $Z$ means satisfying a monic polynomial with integer coefficients (Integral elements over a commutative ring and algebraic integers).

Counterexample

technique · direct

1.1

The map $(ε_{n})_{n \geq 0} ⟼ \sum_{n = 0}^{\infty} ε_{n} p^{n}$ from ${0, 1}^{N}$ to $Z_{p}$ is injective by uniqueness of digit expansions in [L1]. Therefore $Z_{p}$ is uncountable.

L1givenalgebra

1.2

The subset of $Q_{p}$ consisting of elements integral over $Z$ is countable: there are only countably many monic polynomials with integer coefficients, and each has only finitely many roots in the field $Q_{p}$ .

L2algebra

2.1

Hence some element of $Z_{p}$ is not integral over $Z$ . That element lies in $Z_{p} \subseteq Q_{p}$ by [L1], so $Z_{p}$ cannot equal the integral closure of $Z$ in $Q_{p}$ .

step 1.1step 1.2L1L2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

13 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

Andrew V. Sutherland, 18.782 Lecture 8, Remark 8.2 (standard reference, not scraped)