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TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16
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For monic f, Res(f,g)=ig(αi) and it vanishes exactly when f and g have a common root

Statement

Let F be a field, let f(t)=tn+a1tn1++anF[t] be monic, and let gF[t]. If f splits in an extension with roots α1,,αn, then

Res(f,g)=i=1ng(αi).

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 ig(xi) in the elementary symmetric polynomials and substituting the signed coefficients of f (The monic resultant Res(f,g) from the symmetric coefficient expression of ig(xi)).

[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
1.1

By [L1], write the formal symmetric product as Qg(e1,,en). Evaluating at the roots of f and using [L2] gives Res(f,g)=ig(αi).

givenL1L2
2.1

Assume n>0. In a splitting field of the nonzero polynomial f and, when g0, of fg, the product in step 1.1 is zero exactly when g(αi)=0 for some i, since the extension is a field.

step 1.1L4algebra
3.1

By [L3], the condition in step 2.1 says exactly that some root αi 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.

step 2.1L3
4.1

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

L1

Depends on

Used by

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