PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck 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

$f (x) = x^{4} + a x^{3} + b x^{2} + c x + d,$

the resolvent of The resolvent cubic of a monic quartic is

$R_{f} (y) = y^{3} - b y^{2} + (a c - 4 d) y - (a^{2} d + c^{2} - 4 b d) .$

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

Facts & Assumptions

Given: Four roots $α_{1}, \dots, α_{4}$ in a splitting field, their elementary symmetric functions $e_{1} = - a$ , $e_{2} = b$ , $e_{3} = - c$ , $e_{4} = d$ , and the discriminant convention of The discriminant of a monic polynomial as the coefficient expression of $Δ_{n}^{2}$ .

[L1]

Every symmetric polynomial has a unique expression $Q (e_{1}, \dots, e_{n})$ (Fundamental theorem of symmetric polynomials: unique expression as a polynomial in $e_{1}, \dots, e_{n}$ ).

Proof

technique · direct

1.1L1algebra

For the three pairing roots $β_{1}, β_{2}, β_{3}$ , direct expansion gives $β_{1} + β_{2} + β_{3} = e_{2} = b$ , $β_{1} β_{2} + β_{1} β_{3} + β_{2} β_{3} = e_{1} e_{3} - 4 e_{4} = a c - 4 d$ , and $β_{1} β_{2} β_{3} = e_{1}^{2} e_{4} + e_{3}^{2} - 4 e_{2} e_{4} = a^{2} d + c^{2} - 4 b d$ . These are symmetric identities licensed by [L1], and substitution in $\prod_{i} (y - β_{i})$ gives the displayed formula.

1.2algebra

The differences factor as $β_{1} - β_{2} = (α_{2} - α_{3}) (α_{1} - α_{4})$ , $β_{1} - β_{3} = (α_{2} - α_{4}) (α_{1} - α_{3})$ , and $β_{2} - β_{3} = (α_{3} - α_{4}) (α_{1} - α_{2})$ .

2.1step 1.2algebra∎

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

Depends on

Used by

A quartic solved through its resolvent cubic Example
x⁴-10x²+1 has Galois group V₄ over ℚ Example
x⁴-x-1 has Galois group S₄ over ℚ Example
x⁴+8x+12 has Galois group A₄ over ℚ Example
x⁴+x³+x²+x+1 has Galois group C₄ over ℚ Example
The five-case resolvent classification of an irreducible quartic Galois group Theorem

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

Dependency tree · two levels

9 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, v5.10, Lemma 4.9 (standard reference, not scraped)
K. Conrad, Galois Groups of Cubics and Quartics, Theorem 3.4 (standard reference, not scraped)