THE IRREDUCIBILITY OF A CUBIC OVER 𝔽_{q}

DAVID G. A. JACKSON

1. Introduction

It has been over 450 years since Cardano first gave his method for systematically determining the roots of a cubic equation. Finite fields have been around for at least 150 years. One would have thought that the question of deciding whether a cubic polynomial over a finite field is reducible would have been studied to death. Apparently not so.

The attention of the author was drawn to this subject when studying a paper of Chang [2] on the classification of the conjugacy classes of a Chevalley group of type G_{2} over a finite field. On pages 197–198 of this paper Chang implicitly gives necessary and sufficient conditions for the polynomial x^{3} − 3x − ζ to be irreducible over the field 𝔽_{q} for q ≡ −1 mod 3. In the author’s PhD thesis [4], a generalization of this result was needed which dealt with a general cubic, and this was subsequently extended further to include the case q ≡ 1 mod 3. At this point Tonny Springer pitched in and gave a reformulation of the main result which was aesthetically quite pleasing. It is amusing that these results do not appear in the standard literature as they are quite elementary and could easily have been written down 100 years ago.

2. A refresher on Cardano’s method

It is worthwhile reminding the reader on how Cardano solved his cubics. Cardano would have dealt with polynomials whose coefficients were real numbers. The coefficients of our cubics are however to be chosen from the finite field k := 𝔽_{q} (where q is some power of a prime p). It will be necessary to restrict the characteristic p of this field to be different from 2 or 3. Let our cubic be

g(y) := y^{3} + a_{2}y^{2} + a_{1}y + a_{0}

where a_{0}, a_{1}, a_{2} ∈ k. By the substitution y = x − a_{2}/3, we are reduced to finding the roots of the cubic

f(x) := x^{3} + ξx + η

with ξ, η ∈ k. It is worthwhile noting at this point that if η = 0 then f(x) is clearly reducible and it is well-known (see for example [3, p. 251]) that f has multiple roots (in k) if and only if 4ξ^{3} + 27η^{2} = 0. Cardano’s formulae (see for example [3, p. 258]) state that the three roots of f are

x_{0} = v + w, x_{1} = ωv + ω^{−1}w, x_{−1} = ω^{−1}v + ωw,

where ω ∈ F_{q^{2}} is a primitive cube root of unity and v, w are such that vw = −\frac{1}{3}ξ and t = v^{3}, u = w^{3} are the roots of the auxiliary quadratic polynomial

X^{2} + ηX − \frac{1}{27}ξ^{3}.

It should be noted that ω is given by the formula

ω = \frac{−1 + \sqrt{−3}}{2}

in which \sqrt{−3} stands for any root in 𝔽_{q^{2}} of the polynomial x^{2} + 3. If q ≡ 1 mod 3 then ω in fact lies in 𝔽_{q}. A neat way of seeing this is to note that ω^{q} = ω^{q mod 3} and hence that ω^{q} = ω if and only if q ≡ 1 mod 3. But ω^{q} = ω if and only if ω ∈ k since the elements of k are precisely the q roots of the polynomial x^{q} − x.

Also note that t ∈ k if and only if u ∈ k. It makes sense therefore to make the following two definitions.

ϵ_{f} := \begin{cases}1 & if t, u ∈ k, \\ −1 & otherwise; \end{cases}

k_{ϵ} := \begin{cases}k & if ϵ = 1, \\ 𝔽_{q^{2}} & if ϵ = −1. \end{cases}

For simplicity we will write k_{f} instead of k_{ϵ_{f}}.

THE IRREDUCIBILITY OF A CUBIC OVER 𝔽_{q} 3

3. The Main Result: Version 1

Theorem 1. Let f(x) = x^{3} + ξx + η with ξ, η ∈ k. Let t be a root of the quadratic polynomial

X^{2} + ηX − \frac{1}{27}ξ^{3}

(with the understanding that we choose the root −η in case ξ = 0). Then f is irreducible over k if and only if the following conditions are satisfied:

(a) η(4ξ^{3} + 27η^{2}) ≠ 0,

(b) q ≡ ϵ_{f} mod 3,

(c) t ∉ (k_{f}^{×})^{3}.

Proof. Case 1. ξ = 0.

First observe that in this case t = −η ∈ k and so ϵ_{f} = 1. Now f is irreducible if and only if t = −η is a noncube in k. Since noncubes exist in k if and only if q ≡ 1 mod 3 we are done in this case.

Case 2. ξ ≠ 0, ϵ_{f} = 1.

Suppose first that f is irreducible. We have seen above that this implies that (a) holds. A consequence of (a) is that t ≠ u and both t, u are nonzero.

Since ϵ_{f} = 1, we have t ∈ k and (v^{q})^{3} = t^{q} = t. We cannot have v^{q} = v (that is t ∈ (k^{×})^{3}), otherwise the root x_{0} would lie in k. We thus find that (c) holds and v^{−1}v^{q} is a primitive cube root which we may assume to be ω. Since vw ∈ k we must have v^{q}w^{q} = vw and hence w^{q} = ω^{−1}w. If q ≡ −1 mod 3 we would have

x_{−1}^{q} = ω^{−q+1}v + ω^{q−1}w = x_{−1}

and f would be reducible. This shows that (a), (b) and (c) hold if f is irreducible.

Now suppose that (a), (b) and (c) hold. We see from (c) that we cannot have v^{q} = v. We may then again assume that v^{q} = ωv and w^{q} = ω^{−1}w. It is now straightforward to check that, from (b), x_{i}^{q} ≠ x_{i} for i = 0, 1, −1. For example, if x_{0}^{q} = x_{0} then

ωv + ω^{−1}w = v + w

and hence w/v = ω (remember that from (a), v ≠ 0). Raising to the qth power then gives w/v = ω^{q+2}. Hence ω^{q+2} = ω which contradicts (b). The other two cases are

4 DAVID G. A. JACKSON

dealt with in a similar fashion. Hence f is irreducible and we have settled the case ϵ_{f} = 1. Case 3. ξ ≠ 0, ϵ_{f} = −1.

We now have k_{f} ≠ k. The operation F of raising to the qth power takes f to the polynomial (x^{q})^{3} + ξx^{q} + η and hence F induces a permutation σ on the roots of f. The operation F^{2} (that is, raising to the q^{2}th power) induces the permutation σ^{2} on the roots. It is easy to see that σ has a fixed point if and only if σ^{2} has a fixed point. Hence f is irreducible over k if and only it is irreducible over k_{f}.

Working over k_{f} we see, by replacing q by q^{2} and ϵ_{f} by 1 in the argument for ϵ_{1}, that f is irreducible if (a), (b) and (c) hold (notice that k_{f} has q^{2} elements and that q^{2} ≡ 1 mod 3).

Now suppose f is irreducible over k and hence over k_{f}. As for the case ϵ_{f} = 1 we find that (a) and (c) hold (for (c) we show that v^{q^{2}} ≠ v otherwise x_{0} ∈ k_{f}). The proof that (b) holds is, however, slightly different. Since t, u ∈ k_{f}\k are roots of a quadratic polynomial over k, we have t^{q} = u and u^{q} = t. Hence (v^{q})^{3} = t^{q} = u = w^{3}. If v^{q} = w then w^{q} = v (since vw ∈ k) and x_{0} ∈ k. We may then assume that v^{q} = ωw and w^{q} = ω^{−1}v. We want to show that q ≡ −1 mod 3. Suppose to the contrary that q ≡ 1 mod 3. Then

x_{1}^{q} = ω^{−q−1}v + ω^{q+1}w = x_{1}

and hence f is reducible, a contradiction. This finishes the proof.

It should be noted that an argument of the same sort shows that f has three distinct roots in k if and only if 4ξ^{3} + 27η^{2} ≠ 0, (b) holds and (c) doesn’t hold.

4. The Main Result: Version 2

In the case ξ = −3 it is possible to reformulate the statement of the Main Theorem in a manner that readily generalizes to the case of an arbitrary cubic. To state the other formulation requires the following lemma.

Lemma 2. Let ϵ = ±1 and suppose that q ≡ ϵ mod 3. Then t is a noncube in k_{ϵ} such that t^{q−ϵ} = 1 if and only if t = ν^{3j±1} where ν = ζ^{q+ϵ} and ζ generates the group 𝔽_{q^{2}}^{×}.

THE IRREDUCIBILITY OF A CUBIC OVER 𝔽_{q} 5

Proof. If ϵ = 1 then ν generates 𝔽_{q}^{×} and the condition t^{q−ϵ} = 1 is equivalent to t ∈ 𝔽_{q}. Since q ≡ 1 mod 3, t is a noncube in 𝔽_{q} if and only if t = ν^{3j±1} for some j ∈ ℤ.

Now suppose ϵ = −1. Since q ≡ −1 mod 3, q^{2} ≡ 1 mod 3 and hence ζ^{i} is a cube in 𝔽_{q^{2}} if and only if 3 | i. Hence ν^{3j±1} is a noncube in 𝔽_{q^{2}}. We have

ν^{(3j±1)(q+1)} = ζ^{(q−1)(q+1)(3j±1)} = ζ^{(q^{2}−1)(3j±1)} = 1.

Hence ν^{3j±1} satisfies the conditions. Conversely, let t = ξ^{i} be such that t^{q+1} = 1. Then t^{q+1} = ζ^{i(q+1)} = 1 implies that q^{2} − 1 | i(q + 1) which in turn implies that q − 1 | i and hence t = ζ^{(q−1)k} = ν^{k} for some k. Also, t is a noncube if and only if 3 ∤ (q − 1)k ⟺ 3 ∤ k and hence t = ν^{3j±1} for some j ∈ ℤ.

Using this lemma, we arrive at the following reformulation in the case ξ = −3.

Corollary 3. Suppose q ≡ ϵ mod 3. If η ∈ 𝔽_{q} then the polynomial x^{3} − 3x − η is irreducible over 𝔽_{q} if and only if η = ν^{3j+1} + ν^{−3j−1} for some j ∈ ℤ where ν = ζ^{q+ϵ} and ζ generates the group 𝔽_{q^{2}}^{×}.

Proof. Suppose f(x) = x^{3} − 3x − η is irreducible. Then η ≠ 0, ±2 and hence ξη(4ξ^{3} + 27η^{2}) ≠ 0. The auxiliary quadratic becomes X^{2} − ηX + 1 and hence u = t^{−1} and η = t + t^{−1}. Since η ∈ k we must either have t^{q−1} = 1 or t^{q+1} = 1. Since η ≠ 0, ±2 we see that t ≠ ±1 and hence, if t^{q+1} = 1 then t ∉ k. We thus have t^{q−ϵ_{f}} = 1. Theorem 1 implies that q ≡ ϵ_{f} mod 3 and hence ϵ_{f} = ϵ. Lemma 2 then implies that t = ν^{3j±1} for some j ∈ ℤ and hence η = ν^{3j+1} + ν^{−3j−1}.

Conversely, suppose η = ν^{3j+1} + ν^{−3j−1} for some j ∈ ℤ. Then η ≠ 0 and if η = t + t^{−1} then t = ν^{±(3j+1)} = ν^{3k±1} for a suitable k ∈ ℤ. Lemma 2 then implies that (i) t ≠ ±1; (ii) if q ≡ 1 mod 3 then t ∈ k and t is a noncube in k and (iii) if q ≡ −1 mod 3 then t ∉ k and t is a noncube in k_{ϵ}. From (i) and the fact that η ≠ 0 it follows that condition (a) of Theorem 1 is satisfied. From (ii) and (iii) it follows that conditions (b) and (c) are satisfied. Hence f is irreducible and we are done.

We can now generalize this result to deal with an arbitrary cubic over 𝔽_{q}.

Theorem 4. Let ζ generate 𝔽_{q^{2}}^{×} and let ν = ζ^{q±1} according as q ≡ ±1 mod 3. Let f(x) = a_{0}x^{3} + 3a_{1}x^{2} + 3a_{2}x + a_{3} with a_{0} ≠ 0 be a cubic polynomial over 𝔽_{q}. If

6 DAVID G. A. JACKSON

α = a_{0}a_{2} − a_{1}^{2} and β = a_{0}^{2}a_{3} − 3a_{0}a_{1}a_{2} + 2a_{1}^{3} then f(x) is irreducible over 𝔽_{q} if and only if one of the following two conditions hold.

(a) α = 0 and β is a noncube in 𝔽_{q} (impossible if q ≡ −1 mod 3);

(b) α ≠ 0, β ≠ 0 and −\frac{β^{2}}{α^{3}} − 2 = ν^{3j+1} + ν^{−3j−1} for some j ∈ ℤ.

Proof. By a standard argument (see, for example, [1, 12.1]) we form the polynomial

f_{1}(x) := a_{0}^{2}f(\frac{x − a_{1}}{a_{0}} ) = x^{3} + 3αx + β

and f(x) is irreducible if and only if f_{1}(x) is. If β = 0, it is clear that f_{1}(x) is reducible, so assume from now on that β ≠ 0. Then f_{1}(x) is irreducible if and only if f_{2}(x) := x^{3}f_{1}(x^{−1}) = βx^{3} + 3αx^{2} + 1 is irreducible, that is, if and only if

f_{3}(x) := β^{2}f_{2}(\frac{x − α}{β}) = x^{3} − 3α^{2}x + (2α^{3} + β^{2})

is irreducible. If α = 0 then f_{3}(x) and hence f(x) is reducible if and only if β^{2} and hence β is a cube in 𝔽_{q}. So assume that α ≠ 0. Then f_{3}(x) is irreducible if and only if f_{4}(x) := α^{−3}f_{3}(αx) = x^{3} − 3x + (\frac{β^{2}}{α^{3}} + 2) is irreducible. By Corollary 3, this is irreducible if and only if −\frac{β^{2}}{α^{2}} − 2 is of the form ν^{3j+1} + ν^{−3j−1} for some j ∈ ℤ.

References

[1] J. W. Archbold. Algebra. Pitman, Bath, 1964.

[2] Bomshik Chang. The conjugate classes of Chevalley groups of type (G_{2}). J. Algebra, 9:190–211, 1968.

[3] N. Jacobson. Basic Algebra I. W. H. Freeman and Company, San Francisco, 1974.

[4] D. G. A. Jackson. On nilpotent orbits of type G_{2}. PhD Thesis, Sydney University, 1997.

School of Mathematics and Statistics

University of Sydney Australia 2006