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

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.

Computing the monic resultant of two quadratics from roots and coefficients

Example

For

$f (t) = t^{2} + a t + b, g (t) = t^{2} + c t + d$

over a field,

$Res (f, g) = (d - b)^{2} - a (c - a) (d - b) + b (c - a)^{2} .$

Facts & Assumptions

Given: Monic quadratics $f, g$ and roots $α, β$ of $f$ in a splitting field.

[L1]

The monic resultant is $Res (f, g) = g (α) g (β)$ and vanishes exactly when the two polynomials have a common root (For monic $f$ , $Res (f, g) = \prod_{i} g (α_{i})$ and it vanishes exactly when $f$ and $g$ have a common root).

[L2]

Vieta's formulas give $α + β = - a$ and $α β = b$ (Vieta's formulas identify the coefficients of a split monic polynomial with elementary symmetric functions of its roots).

Verification

technique · direct

1.1givenalgebra

Since $α^{2} = - a α - b$ and $β^{2} = - a β - b$ , put $u = c - a$ and $v = d - b$ to obtain $g (α) = u α + v$ and $g (β) = u β + v$ .

2.1step 1.1L2algebra

Multiply and use [L2]: $g (α) g (β) = u^{2} α β + u v (α + β) + v^{2} = b u^{2} - a u v + v^{2}$ .

3.1step 2.1L1algebra

Substitution of $u = c - a$ and $v = d - b$ gives the displayed formula, and [L1] identifies it as the resultant.

4.1step 3.1L1algebra∎

For $f = (t - 1) (t - 2)$ and $g = (t - 2) (t - 4)$ the formula gives $0$ , as the shared root predicts. For the same $f$ and $g = t^{2} + 1$ it gives $10$ , so over $Q$ the polynomials have no common root.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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

J. S. Milne, Fields and Galois Theory, Proposition 4.35 (standard reference, not scraped)