ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-6-sol)audited 2026-09-27

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.

Resultant of two binary linear forms

Example

Let $k$ be a field and let $F = a X + b Y$ and $G = c X + d Y$ with $a, b, c, d \in k$ be binary forms of nominated degree one. Then ${Res}_{1, 1} (F, G) = a d - b c,$ and this element of $k$ vanishes exactly when $F$ and $G$ have a common point of $P^{1} (K)$ for an algebraic closure $K$ of $k$ . The two zero-form cases $F = 0$ and $G = 0$ are included: then $a d - b c = 0$ and the two forms do have a common projective zero.

Facts & Assumptions

Given: A field $k$ , coefficients $a, b, c, d \in k$ , the linear forms $F = a X + b Y$ , $G = c X + d Y$ of nominated degree $1$ , and an algebraic closure $K$ of $k$ .

[L1]

For $d = e = 1$ the Sylvester map is $(A, B) \mapsto A F + B G$ from $k [X, Y]_{0} \oplus k [X, Y]_{0}$ to $k [X, Y]_{1}$ , and ${Res}_{1, 1} (F, G)$ is the determinant of its matrix in the ordered bases $1$ of each copy of $k [X, Y]_{0}$ and $X, Y$ of the target (Sylvester resultant of two positive-degree binary forms, For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

[L2]

For a field $k$ , an algebraic closure $K$ , and homogeneous $F, G$ of nominated positive degrees, $Res (F, G) = 0$ if and only if $F$ and $G$ vanish together at some point of $P^{1} (K)$ ; the zero forms are allowed (The binary Sylvester resultant detects a common geometric projective root, An algebraic closure of a field, projective space points).

Verification

technique · direct

1.1

The domain $k [X, Y]_{0} \oplus k [X, Y]_{0}$ has basis $(1, 0), (0, 1)$ , and $Φ (1, 0) = F = a X + b Y$ , $Φ (0, 1) = G = c X + d Y$ are the two columns of the matrix in the target basis $X, Y$ ; hence the Sylvester matrix is $(\begin{array}{cc} a & c \\ b & d \end{array})$ and its determinant is ${Res}_{1, 1} (F, G) = a d - b c$ .

L1algebra

1.2

If $a d - b c = 0$ and $(c, d) \neq (0, 0)$ , then the nonzero vector $(d, - c) \in K^{2}$ satisfies $a \cdot d + b \cdot (- c) = a d - b c = 0$ and $c \cdot d + d \cdot (- c) = 0$ , so $[d : - c] \in P^{1} (K)$ is a common zero of $F$ and $G$ . If $(c, d) = (0, 0)$ then $G = 0$ vanishes at every point. When also $(a, b) = (0, 0)$ , $F = 0$ too and any point, such as $[1 : 0]$ , is common; otherwise $(b, - a) \in K^{2}$ is nonzero and $F (b, - a) = a b - b a = 0$ , so $[b : - a] \in P^{1} (K)$ is common.

algebragiven

1.3

Conversely, if $F$ and $G$ vanish together at a point $[x : y]$ of $P^{1} (K)$ , then $(x, y) \neq (0, 0)$ is a nonzero vector orthogonal to both coefficient vectors $(a, b)$ and $(c, d)$ , so those two vectors are linearly dependent and $a d - b c = 0$ .

algebragiven

2.1

Steps 1.1-1.3 show that $a d - b c = 0$ if and only if $F$ and $G$ have a common zero in $P^{1} (K)$ ; this agrees with the general criterion [L2], and the zero-form cases are covered by step 1.2.

L2step 1.1step 1.2step 1.3∎

Depends on

Used by

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Dependency tree · two levels

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Sources

J. S. Milne, Algebraic Geometry v6.10, Sylvester determinant and Proposition 7.28, pp. 166-167 (standard reference, not scraped)