CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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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.

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Power sums need not generate the symmetric ring in characteristic two

Statement refuted

For every field and every number of variables, the first $n$ power sums generate the symmetric-polynomial ring.

Facts & Assumptions

Given: Two variables over $F_{2}$ .

[L1]

Newton's identities include $p_{1} = e_{1}$ and $p_{2} = e_{1} p_{1} - 2 e_{2}$ (Newton's identities: $k e_{k} = \sum_{i = 1}^{k} (- 1)^{i - 1} e_{k - i} p_{i}$ ).

[L2]

The elementary symmetric polynomials $e_{1}, e_{2}$ are algebraically independent over the coefficient field (The elementary symmetric polynomials are algebraically independent over the coefficient ring).

[L3]

For every prime $p$ , the ring $Z / p$ is a field (For every prime $p$ , the two operations on $Z / p$ make it a field).

Counterexample

technique · direct

1.1givenL1L3algebra

Over $F_{2}$ , [L1] gives $p_{1} = e_{1}$ and $p_{2} = e_{1} p_{1} = e_{1}^{2}$ , because $2 = 0$ .

2.1step 1.1

Hence every polynomial in $p_{1}, p_{2}$ belongs to the proper subring $F_{2} [e_{1}]$ .

3.1step 2.1L2

The element $e_{2}$ is not in $F_{2} [e_{1}]$ , since an equality $e_{2} = Q (e_{1})$ would be a nonzero polynomial relation between $e_{1}$ and $e_{2}$ , contrary to [L2].

4.1step 3.1∎

Thus $p_{1}, p_{2}$ do not generate the two-variable symmetric ring over $F_{2}$ , refuting the universal statement and showing why the factorial-unit hypothesis is necessary.

Depends on

Used by

Nothing in the library uses this result yet.

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

K. Conrad, Symmetric Polynomials, Section 3 (standard reference, not scraped)