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The discriminant of $x^{3} + p x + q$ is $- 4 p^{3} - 27 q^{2}$

Example

For the monic polynomial $f (t) = t^{3} + p t + q$ over any field,

$Disc (f) = - 4 p^{3} - 27 q^{2} .$

The formula is an integer polynomial identity, so its specializations in characteristics two and three are included.

Facts & Assumptions

Given: A field $F$ , the polynomial $f (t) = t^{3} + p t + q$ , and roots $α, β, γ$ in a splitting field.

[L1]

For a monic cubic, $Res (f, f^{'}) = - Disc (f)$ (For monic $f$ of degree $n$ , $Res (f, f^{'}) = (- 1)^{n (n - 1) / 2} Disc (f)$ ).

[L2]

Vieta's formulas give $α + β + γ = 0$ , $α β + α γ + β γ = p$ , and $α β γ = - q$ (Vieta's formulas identify the coefficients of a split monic polynomial with elementary symmetric functions of its roots).

Verification

technique · direct

1.1

Since $f^{'} (t) = 3 t^{2} + p$ , the root-product formula inside [L1] gives $Res (f, f^{'}) = \prod_{r \in {α, β, γ}} (3 r^{2} + p)$ .

givenL1algebra

2.1

Expanding this product gives $27 (α β γ)^{2} + 9 p (α^{2} β^{2} + α^{2} γ^{2} + β^{2} γ^{2}) + 3 p^{2} (α^{2} + β^{2} + γ^{2}) + p^{3}$ .

step 1.1algebra

3.1

By [L2], $α^{2} + β^{2} + γ^{2} = - 2 p$ and $α^{2} β^{2} + α^{2} γ^{2} + β^{2} γ^{2} = p^{2}$ , while $(α β γ)^{2} = q^{2}$ . Substitution in step 2.1 gives $Res (f, f^{'}) = 4 p^{3} + 27 q^{2}$ .

step 2.1L2algebra

4.1

Apply [L1] to obtain $Disc (f) = - 4 p^{3} - 27 q^{2}$ . Every calculation used integer coefficients, so reduction to any field characteristic is valid.

step 3.1L1∎

Depends on

Used by

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Sources

J. S. Milne, Fields and Galois Theory, Example 4.37 (standard reference, not scraped)

The discriminant of x3+px+qx3+px+q is −4p3−27q2−4p3−27q2