NOTES ON CUBIC EQUATIONS

MARKUS ROST

For a cubic element (or a triangle) we define its “basic line”. It leads to the normalization (4) of cubic equations. The method works in all characteristics for generic cubic elements. We describe for this 1-parameter family the discriminant extension and the corresponding variant of Cardano’s formula.

1. Basic invariants. Let F be a field and let K be a cubic extension of F. We consider the functions

T, Q, N, A, B, D, M, f, g: K → F, δ, φ: K → K

defined as follows: For x ∈ K the polynomial

P_{x}(r) = r^{3} − T (x)r^{2} + Q(x)r − N(x)

is the characteristic polynomial of x. In other words, T (x) is the trace of x, N(x) is the norm of x, and, for invertible x,

Q(x) = T (x^{−1})N(x)

Moreover

δ(x) = \frac{dP_{x}(r)}{dr}|_{r=x} = 3x^{2} − 2T (x)x + Q(x)
φ(x) = 3x − T (x)
Δ(x) = N(φ(x)^{2} − 4δ(x))
A(x) = N(φ(x))
B = \frac{Δ − A^{2}}{4}
D(x) = T (δ(x))
M = QT − 9N
f = \frac{T D − M}{D}
g = −\frac{A}{D}

One finds

D = T^{2} − 3Q
A = −(3M − 2T D)
3f + g = T
2f + g = \frac{M}{D}

Date: August 24 2003, April 27 2004, Dec 17 2005, Mar 20 2008.

The polynomial Δ(x) is the discriminant of x.

Remark 1. Suppose K = F × F × F. Then for x = (x_{0}, x_{1}, x_{2}) one has

D(x) = x_{0}^{2} + x_{1}^{2} + x_{2}^{2} − x_{0}x_{1} − x_{1}x_{2} − x_{2}x_{0}
= (x_{0} + ζx_{1} + ζ^{2}x_{2})(x_{0} + ζ^{2}x_{1} + ζx_{2})

where ζ is subject to

1 + ζ + ζ^{2} = 0

Remark 2. Note that 1 + ζ + ζ^{2} = 0 means that ζ is a cube root of unity (and is primitive as long as char F ≠ 3). Thus, if F = C (complex numbers) in Remark 1, then D(x) = 0 if and only if the Euclidean triangle x_{0}, x_{1}, x_{2} is equilateral.

Remark 3. Suppose K = F × F × F. Then for x = (x_{0}, x_{1}, x_{2}) one has

A(x) = (2x_{0} − x_{1} − x_{2})(2x_{1} − x_{2} − x_{0})(2x_{2} − x_{0} − x_{1})

Lemma. For x ∈ K with D(x) ≠ 0 and a, b ∈ F one has

(1)  D(ax + b) = a^{2}D(x)
(2)  g(ax + b) = ag(x)
(3)  f(ax + b) = af(x) + b

Proof. Claim (1) is easy to check, for instance using Remark 1, or by using a similar property for δ(x).

As for (2), note that T (x) − 3x is invariant under translations x ↦ x + b. The same is true for A(x) and also for D by (1). The claim is now clear from deg g = 1.

Claim (3) follows easily from (2) and 3f + g = T. □

2. The basic line.

Definition. For x ∈ K with D(x) ≠ 0 the function

ℓ_{x}(s) = f(x) + sg(x), s ∈ F

is called the basic line of x.

Clearly one has

ℓ_{ax+b}(s) = aℓ_{x}(s) + b

Remark 4. Let x_{0}, x_{1}, x_{2} be an Euclidean triangle which is not equilateral. Then ℓ_{(x_{0},x_{1},x_{2})}(s) (understanding K = C × C × C) determines a natural line associated to the triangle.

Apart from the degenerate cases D(x) = 0 and A(x) = 0, the basic line is defined and non-degenerate. We thus can normalize it by means of an affine transformation x ↦ ax + b, so that f(x) = 0 and g(x) = 1. This yields the following normalization for the basic parameters:

T = 1, 9N − 4Q + 1 = 0

This gives

(T, Q, N) = (1, 9t − 2, 4t − 1)

with t ∈ F and we get the following normalized form of a cubic equation

(4)  x^{3} − x^{2} + (9t − 2)x − (4t − 1) = 0

Remark 5. One finds

t = \frac{B}{A^{2}}

The invariant t is the basic modulus for triangles up to affine transformations. I think it can serve as a toy model for the j-invariant for elliptic curves.

Remark 6. For t = 0 equation (4) becomes x^{3} − x^{2} − 2x + 1 = 0 which has the complex roots x = 2 cos \frac{π}{7} , 2 cos \frac{3π}{7} , 2 cos \frac{5π}{7} .

Remark 7. After the change of variables

y = 3x − 1

one gets the family

y^{3} − 3Dy + D = 0

with D = 7 − 27t.

3. The discriminant. The discriminant of the cubic equation (4) is

(1 − 4t)(27t − 7)^{2}

The discriminant algebra is after a normalization simply given by the quadratic equation

u^{2} − u + t = 0

In fact, one may check that the linear fractional transformation

Φ(x) = \frac{ux + 3t − 1}{x + u − 1}

defines an F [u]-automorphism of F [u][x] of order 3. In particular, Φ(x) and Φ^{2}(x) are the conjugates of x.

4. Cardano’s formula. Explicit solutions of (4) in terms of radicals are given by:

D = 7 − 27t
w^{2} + w + D = 0
w + \overline{w} = −1
w\overline{w} = D
α^{3} = w^{2}\overline{w}
\overline{α}^{3} = \overline{w}^{2}w
α\overline{α} = D
x = \frac{1 + α + \overline{α}}{3}

This gives

x = \frac{1}{6} (2 − \sqrt[3]{(1 − E)(1 + \sqrt{E})} − \sqrt[3]{(1 − E)(1 − \sqrt{E})})

with

E = 1 − 4D = 27(4t − 1)

Here are further related formulas:

X := 3x − 1
D (\frac{1}{X + w} ) = D (\frac{1}{X + \overline{w}} ) = 0
N(X + w) = 27(4t − 1)w
Y := \frac{X + w}{X + \overline{w}}
Y^{3} = \frac{w}{\overline{w}}

5. Appendix. (Added April 2004)

Here are more remarks: Let

ϕ_{t}: K → K
ϕ_{t}(x) = (1 − 3t)x + tT (x)

Then

D(x) = Q(x) − Q(ϕ_{1}(x))
D(x) = \frac{−1}{2} \frac{d}{dt}Q(ϕ_{t}(x))|_{t=0}
f(x) = \frac{\frac{d}{dt}N(ϕ_{t}(x))|_{t=0}}{\frac{d}{dt}Q(ϕ_{t}(x))|_{t=0}}

f(x) = \frac{N(x) − N(ϕ_{1}(x))}{Q(x) − Q(ϕ_{1}(x))}

This is perhaps helpful to give a more geometric definition of f.

Moreover one has

f(x_{1}, x_{2}, x_{3}) = x_{1} ⇒ x_{2} = x_{3}

This is perhaps helpful to characterize f.

Can one draw any analogies between x ↦ f(x) and the orthocenter of a Euclidean triangle?

Department of Mathematics, The Ohio State University, 231 W 18th Avenue, Columbus, OH 43210, USA

E-mail address: rost@math.ohio-state.edu
URL: http://www.math.ohio-state.edu/~rost