CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16

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For monic $f$ , if $g - g_{1} = q f$ , then $Res (f, g) = Res (f, g_{1})$

Statement

Let $F$ be a field, let $f \in F [t]$ be monic, and let $g, g_{1}, q \in F [t]$ satisfy

$g - g_{1} = q f .$

Then

$Res (f, g) = Res (f, g_{1}) .$

Facts & Assumptions

Given: Polynomials $f, g, g_{1}, q$ satisfying the identity in the Statement, with $f$ monic.

[L1]

If $f$ has roots $α_{i}$ in a splitting field, then $Res (f, h) = \prod_{i} h (α_{i})$ for every polynomial $h$ (For monic $f$ , $Res (f, g) = \prod_{i} g (α_{i})$ and it vanishes exactly when $f$ and $g$ have a common root).

[L2]

Polynomial evaluation is a ring homomorphism and an element $a$ is a root of $f$ exactly when $f (a) = 0$ (Evaluation and roots of a polynomial in a commutative target ring).

Proof

technique · direct

1.1

For every root $α_{i}$ of $f$ , evaluate $g - g_{1} = q f$ to obtain $g (α_{i}) - g_{1} (α_{i}) = q (α_{i}) f (α_{i}) = 0$ , hence $g (α_{i}) = g_{1} (α_{i})$ .

givenL2algebra

2.1

Apply [L1] to $g$ and $g_{1}$ and multiply the equal values from step 1.1 to obtain equality of the resultants.

step 1.1L1

3.1

If $\deg f = 0$ , then $f = 1$ and both resultants are the empty product $1$ ; the argument remains valid.

L1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 20 results over 10 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

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

For monic ff, if g−g1=qfg−g1​=qf, then Res⁡(f,g)=Res⁡(f,g1)Res(f,g)=Res(f,g1​)