TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-16

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For monic $f$ , $Res (f, g) = \prod_{i} g (α_{i})$ and it vanishes exactly when $f$ and $g$ have a common root

Statement

Let $F$ be a field, let $f (t) = t^{n} + a_{1} t^{n - 1} + \dots + a_{n} \in F [t]$ be monic, and let $g \in F [t]$ . If $f$ splits in an extension with roots $α_{1}, \dots, α_{n}$ , then

$Res (f, g) = \prod_{i = 1}^{n} g (α_{i}) .$

If $n > 0$ , this value is zero if and only if $f$ and $g$ have a common root in some extension field of $F$ . For $n = 0$ , $f = 1$ and the resultant is $1$ .

Facts & Assumptions

Given: A field $F$ , a monic polynomial $f$ of degree $n$ , and a polynomial $g$ .

[L1]

The monic resultant is obtained by expressing the symmetric formal product $\prod_{i} g (x_{i})$ in the elementary symmetric polynomials and substituting the signed coefficients of $f$ (The monic resultant $Res (f, g)$ from the symmetric coefficient expression of $\prod_{i} g (x_{i})$ ).

[L2]

Vieta's formulas identify those elementary symmetric values with the signed coefficients of a split monic polynomial (Vieta's formulas identify the coefficients of a split monic polynomial with elementary symmetric functions of its roots).

[L3]

An element is a root of $g$ exactly when its evaluation $g (a)$ is zero (Evaluation and roots of a polynomial in a commutative target ring).

[L4]

Every nonzero polynomial over a field has a splitting field (Every nonzero polynomial over a field has a splitting field).

Proof

technique · direct

1.1givenL1L2

By [L1], write the formal symmetric product as $Q_{g} (e_{1}, \dots, e_{n})$ . Evaluating at the roots of $f$ and using [L2] gives $Res (f, g) = \prod_{i} g (α_{i})$ .

2.1step 1.1L4algebra

Assume $n > 0$ . In a splitting field of the nonzero polynomial $f$ and, when $g \neq 0$ , of $f g$ , the product in step 1.1 is zero exactly when $g (α_{i}) = 0$ for some $i$ , since the extension is a field.

3.1step 2.1L3

By [L3], the condition in step 2.1 says exactly that some root $α_{i}$ of $f$ is also a root of $g$ . If $g = 0$ , every root of the positive-degree polynomial $f$ is common and every factor in step 1.1 is zero.

4.1L1∎

If $n = 0$ , then $f = 1$ and [L1] defines the resultant as the empty product $1$ .

Depends on

Used by

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

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

For monic ff, Res⁡(f,g)=∏ig(αi)Res(f,g)=∏i​g(αi​) and it vanishes exactly when ff and gg have a common root