Alphabeta Math
PropositionStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24
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The coefficient formula and discriminant of the quartic resolvent

Statement

For

f(x)=x4+ax3+bx2+cx+d,

the resolvent of The resolvent cubic of a monic quartic is

Rf(y)=y3by2+(ac4d)y(a2d+c24bd).

A monic quartic and its resolvent cubic have the same discriminant.

Facts & Assumptions

Given: Four roots α1,,α4 in a splitting field, their elementary symmetric functions e1=a, e2=b, e3=c, e4=d, and the discriminant convention of The discriminant of a monic polynomial as the coefficient expression of Δn2.

[L1]

Every symmetric polynomial has a unique expression Q(e1,,en) (Fundamental theorem of symmetric polynomials: unique expression as a polynomial in e1,,en).

Proof

technique · direct
1.1

For the three pairing roots β1,β2,β3, direct expansion gives β1+β2+β3=e2=b, β1β2+β1β3+β2β3=e1e34e4=ac4d, and β1β2β3=e12e4+e324e2e4=a2d+c24bd. These are symmetric identities licensed by [L1], and substitution in i(yβi) gives the displayed formula.

L1algebra
1.2

The differences factor as β1β2=(α2α3)(α1α4), β1β3=(α2α4)(α1α3), and β2β3=(α3α4)(α1α2).

algebra
2.1

Multiplying the squares of the three identities in step 1.2 uses each of the six differences αiαj exactly once. The root-product formulas for the two discriminants therefore give Disc(Rf)=Disc(f). The identity remains valid when coefficients or root differences vanish.

step 1.2algebra

Depends on

Used by

Cited to discharge well-definedness by The resolvent cubic of a monic quartic.

Dependency tree · two levels

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Sources