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Cardano's formula for $x^{3} - 3 x - 1$ from the Lagrange resolvent

Example

Let

$u^{3} = \frac{1 + \sqrt{- 3}}{2}, v^{3} = \frac{1 - \sqrt{- 3}}{2}, u v = 1.$

Then

$x : = u + v$

is a root of $x^{3} - 3 x - 1$ . The two nontrivial Lagrange resolvents are $3 u$ and $3 v$ ; equivalently, $u$ and $v$ are the normalized resolvents obtained after division by $3$ .

Facts & Assumptions

Given: The depressed cubic $f (x) = x^{3} - 3 x - 1$ and the displayed radicals.

[L1]

The Lagrange resolvent is the weighted sum attached to a cyclic action and a chosen root of unity (The Lagrange resolvent attached to a cyclic action and a root of unity).

[L2]

In the cyclic cubic situation over a field containing the cube roots of unity, the resolvent eigenvectors lie in a radical extension (If $μ_{n} \subseteq F$ and $char F ∤ n$ , then a degree- $n$ extension is cyclic exactly when it is $F (α)$ with $α^{n} \in F$ and $x^{n} - α^{n}$ irreducible).

Verification

technique · direct

1.1givenalgebra

The displayed cube roots satisfy $u^{3} + v^{3} = 1, u^{3} v^{3} = 1,$ so our choice $u v = 1$ is compatible. Now $(u + v)^{3} = u^{3} + v^{3} + 3 u v (u + v) = 1 + 3 (u + v) .$ Therefore $x^{3} - 3 x - 1 = (u + v)^{3} - 3 (u + v) - 1 = 0.$

2.1L1L2step 1.1algebra∎

Put $x_{0} = u + v$ , $x_{1} = ω u + ω^{2} v$ , and $x_{2} = ω^{2} u + ω v$ , and let $σ$ cycle $x_{0}, x_{1}, x_{2}$ . The definition [L1] gives $R_{σ, ω} (x_{0}) = x_{0} + ω^{2} x_{1} + ω x_{2} = 3 u,$ and $R_{σ, ω^{2}} (x_{0}) = x_{0} + ω x_{1} + ω^{2} x_{2} = 3 v .$ Thus $u$ and $v$ are the two nontrivial resolvents divided by $3$ , while step 1.1 is the load-bearing check that their symmetric combination is a genuine root of the cubic.

Depends on

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Sources

J. Ash, Basic Abstract Algebra, cubic formulas in Galois theory (standard reference, not scraped)
J. S. Milne, Fields and Galois Theory, v5.10, cyclic cubic examples (standard reference, not scraped)

Cardano's formula for x3−3x−1x3−3x−1 from the Lagrange resolvent