1 Introduction
The reducibility of polynomials over finite fields is important for many applications, including coding theory and cryptography. The well-known Berlekamp Algorithm [Reference Berlekamp2] provides a method for factorising such polynomials. A more challenging problem is to decide reducibility for parametrised families of polynomials. For example, Dickson [Reference Dickson5] determined for which nonzero parameters
$\alpha ,\beta $
, in the finite field
$\mathbb {F}_{p^n}$
with
$p>3$
a prime, the polynomial
$x^3+\alpha x+\beta $
is reducible over
$\mathbb {F}_{p^n}$
. The analogous characterisation was obtained by Williams [Reference Williams9] for finite fields of characteristic
$2$
and
$3$
. Polynomials of this form are trinomials as they have exactly three nontrivial terms. Von zur Gathen [Reference von zur Gathen8] has considered the irreducibility of trinomials over finite fields and formulated some conjectures on their distribution. This line of research was continued by Ahmadi in [Reference Ahmadi1], who solved some of these conjectures and also proved irreducibility results for trinomials of the form
$x^d+\alpha x^k+\beta $
, where
$\alpha , \beta \in \mathbb {F}_q^\times $
with even degree d.
Results of this flavour have applications in group theory. For example, in the course of classifying the conjugacy classes of the simple group of Lie type
$G_2(q)$
, Chang [Reference Chang4] required knowledge of the parameters
$\zeta \in \mathbb {F}_q$
for which
$x^3-3x-\zeta $
is irreducible over
$\mathbb {F}_q$
. Similarly, the determination of the conjugacy classes in
$F_4(2^n)$
required Shinoda [Reference Shinoda7] to classify those parameters
$\zeta \in \mathbb {F}_{2^n}$
for which
$x^3+x+\zeta $
is irreducible and, at the same time,
$x^2+\zeta x+\zeta ^2 +1$
is reducible.
This note is also motivated by a problem in group theory (see [Reference Pan6]): the suborbit classification of the (primitive) actions of
$G_2(q)$
requires a detailed analysis of the reducibility of
$x^2+\zeta x-\zeta $
and
$x^3-3x^2-\zeta $
with parameter
$\zeta \in \mathbb {F}_q$
. We consider a slightly more general case and state our main result as follows.
Theorem 1.1. Let q be a prime power,
$\gamma \in \mathbb {F}_q, \zeta \in \mathbb {F}_q^\times $
, and
$$ \begin{align*}P = x^2 + \zeta x - \zeta \quad \text{and} \quad Q = x^3 - \gamma x^2 - \zeta.\end{align*} $$
For
$q \equiv 0,1 \bmod 3$
, if
$\gamma = 3$
or
$\gamma ^2+3\gamma +9=0$
, then at least one of P and Q is reducible. If
$q \equiv 2 \bmod 3$
and
$\gamma = 3$
, then there are
$\tfrac {1}{3}(q+1)$
elements
$\zeta \in \mathbb {F}_q^\times $
such that both P and Q are irreducible.
2 Proof of Theorem 1.1
Throughout, let
$q=p^n$
be a prime power. If
$p=3$
, then, by assumption, we have
$\gamma = 0$
and Q always has a root since
$c\mapsto c^3$
is a Frobenius automorphism of
$\mathbb {F}_{3^n}$
. For
$p \ne 3$
, by construction, P and Q are trinomials over
$\mathbb {F}_q$
.
If
$p> 3$
, then rewriting
$x = X+3^{-1} \gamma $
yields
$Q = X^3 -3^{-1} \gamma ^2 X - (\zeta +2{\cdot }27^{-1}\gamma ^3)$
. Thus, in this case, [Reference Dickson5, Theorems 1 and 3] applies, and P and Q are both irreducible only if we have
$\zeta ^2+4\zeta $
is a nonsquare and
$-\zeta (4\gamma ^3+27\zeta )$
is a square in
$\mathbb {F}_q$
, and
${2^{-1}(\mu \sqrt {-3}+\zeta )+27^{-1}\gamma ^3}$
is a noncube in the field extension
$\mathbb {F}_q(\sqrt {-3})$
, where
$\mu \in \mathbb {F}_q$
with
$81\mu ^2=-\zeta (4\gamma ^2+27\zeta )$
. This answers the irreducibility question, but it remains to count. Unfortunately, a similar result cannot be easily deduced from Williams’ work [Reference Williams9] for
$p=2$
. In the following, we present a uniform proof for all p (excluding
$p=3$
as mentioned above). Since P and Q are reducible if and only if they have a root, for a fixed
$\gamma \in \mathbb {F}_q^\times $
, we consider
$$ \begin{align*}\mathcal{S}_1 = \{\zeta \in \mathbb{F}_q^\times \mid P \text{ has roots in } \mathbb{F}_q^\times\} \quad\text{and}\quad \mathcal{S}_2 = \{\zeta \in \mathbb{F}_q^\times \mid Q \text{ has roots in } \mathbb{F}_q^\times\}.\end{align*} $$
We will determine the size of these sets separately; then determine
$|\mathcal{S}_1 \cup \mathcal{S}_2|$
from
$|\mathcal {S}_1 \cup \mathcal {S}_2| = |\mathcal {S}_1| + |\mathcal {S}_2| - |\mathcal {S}_1 \cap \mathcal {S}_2|$
.
The polynomial P is reducible if and only if there are
$u,v\in \mathbb {F}_q^\times $
such that
$P = (x - u)(x - v)$
, which is equivalent to
$uv = -\zeta $
and
$u + v = -\zeta $
. Thus, P is reducible if and only if
$\zeta = {v^2}{(1 - v)^{-1}}$
for some
$v\in \mathbb {F}_q\backslash \{0,1\}$
. If we define
$f\colon \mathbb {F}_q\backslash \{0, 1\} \to \mathbb {F}_q^\times $
by
$f(c)={c^2}{(1 - c)^{-1}}$
, then
$|\mathcal {S}_1|=|\mathrm {Im~} f|$
. Note that
$f(s)=f(t)$
with
$s,t\notin \{0,1\}$
if and only if
$s = t$
or
$t = {s}{(s-1)^{-1}}$
. Since
$s = {s}{(s-1)^{-1}}$
if and only if q is odd and
$s=2$
,
$$ \begin{align*} |\mathcal{S}_1| = |\mathrm{Im~} f|= \begin{cases} \tfrac12(q - 2) &\text{if } q \equiv 0 \bmod 2,\\ \tfrac12(q - 1) &\text{otherwise}. \end{cases}\end{align*} $$
Now, we consider
$\mathcal {S}_2$
. For a fixed
$\gamma \in \mathbb {F}_q^\times $
, we know that Q has a root
$u\in \mathbb {F}_q$
if and only if
$\zeta =u^3-\gamma u^2$
. Since
$\zeta \ne 0$
, it follows that
$|\mathcal {S}_2| = |\mathrm {Im~} g|$
, where
$g \colon \mathbb {F}_q\setminus \{0,\gamma \} \to \mathbb {F}_q^\times $
is the map
$g(c)=c^3 - \gamma c^2$
. Note that
$g(s) = g(t)$
for
$s,t\notin \{0,\gamma \}$
if and only if
$s = t$
or
$t = ks$
for some
$k \in \mathbb {F}_q\backslash \{0,1\}$
such that
$$ \begin{align*} g(ks)-g(s)=s^2(k-1) (k^2 s+k s-\gamma k+s-\gamma) = 0. \end{align*} $$
Since
$s^2(k-1)\ne 0$
, the latter is equivalent to
$\ell _s(k)=0$
, where
$$ \begin{align*} \ell_s(k)=k^2+r_sk+r_s\quad\text{with}\quad r_s=1-\gamma s^{-1} \ne 0.\end{align*} $$
Note that any such k satisfies
$ks\notin \{0,\gamma \}$
(for otherwise,
$r_s=0$
or
$\ell _s=1$
, which is a contradiction). Thus, we are interested in
$$ \begin{align*}{\kappa(s)=|\{k\in\mathbb{F}_q\setminus\{0,1\}\mid \ell_s(k)=0\}|},\end{align*} $$
which informs us of
$\mathrm {Im~} g$
. Suppose
$\ell _s(k)$
has roots
$u,v$
; note that
$u,v\notin \{0,-1\}$
. Then,
$u+v=-r_s$
and
$uv=r_s$
, so
$r_s=-v^2(1+v)^{-1}$
. Moreover,
$k=1$
is a root of
$\ell _s(k)$
if and only if q is odd and
$r_s = -2^{-1}$
. We have
$u=v$
if and only if
$p=3$
and
$r_s=1$
, or
$q \equiv \pm 1\bmod 6$
and
$s\in \{-3^{-1} \gamma ,\gamma \}$
. With such notation, since
$s \ne \gamma $
by assumption,
$$ \begin{align*}\kappa(s) = \begin{cases} 0 & \text{if } r_s \notin \{-{v^2}{(1+v)^{-1}}\mid v \in \mathbb{F}_q\backslash\{0,-1\}\};\\ 1 &\text{if } q \equiv \pm 1\bmod 6 \text{ and } r_s\in\{4, -2^{-1}\};\\ 2 & \text{if } r_s \in \{-{v^2}{(1+v)^{-1}}\mid v \in \mathbb{F}_q\backslash\{0,-1\}\}\setminus\{1,4,-2^{-1}\}; \end{cases}\end{align*} $$
recall that
$p=2$
is allowed, in which case
$-2^{-1}$
does not occur. Since the map g restricted to the subset
$\mathcal {K}_1=\{s\in \mathbb {F}_q\backslash \{0,\gamma \}\mid \kappa (s)=0\}$
is injective, restricted on
$\mathcal {K}_2=\{s\in \mathbb {F}_q\backslash \{0,\gamma \}\mid \kappa (s)=1\}$
is
$2$
-to-
$1$
, restricted to
$\mathcal {K}_3=\{s\in \mathbb {F}_q\backslash \{0,\gamma \}\mid \kappa (s)=2\}$
is
$3$
-to-
$1$
, and
$\mathcal {K}_1\sqcup \mathcal {K}_2\sqcup \mathcal {K}_3 = \mathbb {F}_q\backslash \{0,\gamma \}$
,
$$ \begin{align*}|\mathrm{Im~} g| =|\mathcal{K}_1|+\tfrac12|\mathcal{K}_2|+\tfrac13|\mathcal{K}_3|.\end{align*} $$
More specifically, observe that for any
${c \ne 1}$
, there exists
$s = \gamma (1-c)^{-1}$
such that
$r_s=c$
. Also, by construction,
$r_s\ne 1$
and
$1 \in \{-{v^2}{(1+v)^{-1}}\mid v \in \mathbb {F}_q\backslash \{0,-1\}\}$
if and only if
$q \equiv 0,1 \bmod 3$
. Moreover, by symmetry,
$$ \begin{align*}|\mathcal{K}_1| = |\{-{v^2}{(1+v)^{-1}}\mid v \in \mathbb{F}_q\backslash\{0,-1\}\}|=|\mathrm{Im~} f|\end{align*} $$
and
$$ \begin{align*} |\mathcal{K}_3|&=|\{-{v^2}{(1+v)^{-1}}\mid v \in \mathbb{F}_q\backslash\{0,-1\}\}\setminus\{1,4,-2^{-1}\}|\\ &= \begin{cases} \tfrac12(q - 2) &\text{if } q \equiv 2 \bmod 6,\\ \tfrac12(q - 2)-1 &\text{if } q \equiv 4 \bmod 6,\\ \tfrac12(q - 1)-3 &\text{if } q \equiv 1 \bmod 6,\\ \tfrac12(q - 1)-2 &\text{if } q \equiv 5 \bmod 6.\\ \end{cases} \end{align*} $$
We therefore deduce that
$$ \begin{align*}|\mathcal{S}_2| = |\mathrm{Im~} g| =\begin{cases} q - 2 - \tfrac12{(q - 2)} + \tfrac16{(q-2)} = \tfrac23{(q-2)} &\text{if } q \equiv 2 \bmod 6,\\ q - 2 - \tfrac12{(q - 2)} + 1 + \tfrac16{(q-4)} = \tfrac23(q-1) &\text{if } q \equiv 4 \bmod 6,\\ q - 2 - \tfrac12{(q - 1)} + 1 + \tfrac22 + \tfrac16{(q-7)} = \tfrac23(q-1) &\text{if } q \equiv 1 \bmod 6,\\ q - 2 - \tfrac12{(q - 1)} + \tfrac22 + \tfrac16{(q-5)} = \tfrac23(q-2) &\text{if } q \equiv 5 \bmod 6. \end{cases} \end{align*} $$
In summary,
$$ \begin{align*}|\mathcal{S}_2| = |\mathrm{Im~} g| = \begin{cases} \tfrac23(q-1) &\text{ if } q \equiv 1 \bmod 3,\\\tfrac23(q-2) &\mbox{ if } q \equiv 2 \bmod 3. \end{cases} \end{align*} $$
It remains to examine
$\mathcal {S}_1 \cap \mathcal {S}_2$
. With the same setup as above,
$\zeta \in \mathcal {S}_1 \cap \mathcal {S}_2$
if and only if there exist
$s \in \mathbb {F}_q \backslash \{0, \gamma \}$
and
$t \in \mathbb {F}_q\backslash \{0, 1\}$
such that
$\zeta = f(t) = g(s)$
, where
$f,g$
are as defined above. The latter equality is equivalent to
$h(t)=0$
, where
$$ \begin{align*}h(t)=t^2 + (s^3 - \gamma s^2)t - (s^3 - \gamma s^2).\end{align*} $$
When
$p \ne 2$
, the quadratic equation
$h(t)=0$
in t holds for some
$t \in \mathbb {F}_q\backslash \{0, 1\}$
if and only if
$$ \begin{align*}c(s)=s^2 (s-\gamma) (s^3-\gamma s^2+4)\end{align*} $$
is a square in
$\mathbb {F}_q$
. In particular, given
$s \notin \{0, \gamma \}$
, we see
$c(s) = 0$
if and only if
$-4=s^3-\gamma s^2$
. If
$\gamma ^2+3\gamma +9=0$
, then
$q\equiv 1 \bmod 3$
and
$\gamma = (-3\pm 3w)2^{-1}$
for
$w \in \mathbb {F}_q^\times $
such that
$-3=w^2$
, and
$$ \begin{align*}c(s)=s^2 (s-\gamma) (s+3^{-1}\gamma)(s-2{\cdot}3^{-1}\gamma)^2.\end{align*} $$
Note that when
$\gamma =3$
, the same factorisation exists; that is,
$$ \begin{align*}c(s)=s^2 (s-\gamma) (s+1)(s-2)^2.\end{align*} $$
For
$p \ne 2$
, if
$q \equiv 1 \bmod 3$
, then let
$\gamma \in \{3, \ (-3+3w)2^{-1}, \ (-3-3w)2^{-1}\}$
; if
$q \equiv 2 \bmod 3$
, then let
$\gamma = 3$
. Recall that if
$p \ne 2$
, then
$\kappa (s)=2$
if and only if the discriminant
$s^{-2}(\gamma -s)(\gamma +3s)$
of
$\ell _s(k)$
is a square in
$\mathbb {F}_q^\times $
, and
$\kappa (s)=0$
if and only if
$(\gamma -s)(\gamma +3s)$
is a nonsquare in
$\mathbb {F}_q^\times $
. Since
$-3$
is a square in
$\mathbb {F}_q^\times $
if
$q \equiv 2 \bmod 3$
, and is a nonsquare if
$q \equiv 1 \bmod 3$
and
$-3(s-\gamma )(s+3^{-1}\gamma )=(\gamma -s)(3s+\gamma )$
, it follows that
$$ \begin{align*} \mathcal{S}_1 \cap \mathcal{S}_2 & = \{-4\} \cup \{s^3 - \gamma s^2 \mid s \in \mathbb{F}_q\backslash\{0,\gamma\} \text { and }c(s) \text{ is a square in } \mathbb{F}_q^\times\}\\ & = \{-4\} \cup \begin{cases} \{s^3 - \gamma s^2 \mid s \in \mathbb{F}_q\backslash\{0,\gamma\} \text{ and } \kappa(s)=2\} \quad \text{ if }q \equiv 1 \bmod 6,\\ \{s^3 - \gamma s^2 \mid s \in \mathbb{F}_q\backslash\{0,\gamma\} \text{ and } \kappa(s)=0\} \quad \text{ if }q \equiv 5 \bmod 6.\\ \end{cases} \end{align*} $$
Now, consider the case where
$p = 2$
and
$q = 2^n$
. The trace map over
$\mathbb {F}_{2^n}$
is defined by
$$ \begin{align*} \operatorname{\mathrm{tr}} \colon \mathbb{F}_{2^n} \to \{0, 1\}, \quad s \mapsto \bigg(\sum_{i = 0}^{n - 1} s^{(2^i)}\bigg) \bmod 2. \end{align*} $$
It follows that
$\operatorname {\mathrm {tr}}(x)=\operatorname {\mathrm {tr}}(x^2)$
and
$\operatorname {\mathrm {tr}}(x)+\operatorname {\mathrm {tr}}(y)=\operatorname {\mathrm {tr}}(x+y)$
for all
$x,y\in \mathbb {F}_{2^n}$
. We now consider
$\gamma = 1$
or
$\gamma ^2+\gamma +1=0$
. Note that there exists
$\gamma \in \mathbb {F}_q^\times $
such that
$\gamma ^2+\gamma +1=0$
if and only if
$q \equiv 1 \bmod 3$
and
$\gamma ^3 =1$
. Moreover, it is well known that
$x^2+\alpha x + \beta $
for
$\alpha , \beta \in \mathbb {F}_q^\times $
is reducible over
$\mathbb {F}_q$
if and only if
$\operatorname {\mathrm {tr}}(a^{-2}b)=0$
(see [Reference Berlekamp3, Theorem 6.69]). Thus,
$h(t)=0$
has:
-
• two solutions if and only if
$\operatorname {\mathrm {tr}}({s^{-2}(s + \gamma )^{-1}}) = 0$
and -
• no solution if and only if
$\operatorname {\mathrm {tr}}({s^{-2}(s + \gamma )^{-1}}) = 1$
.
However, when
$p = 2$
, the quadratic
$\ell _s(k)=0$
has two solutions if and only if
$\operatorname {\mathrm {tr}}({s}{(s + \gamma )^{-1}}) = 0$
and no solution if and only if
$\operatorname {\mathrm {tr}}({s}{(s + \gamma )^{-1}}) = 1$
. Observe that if
$\gamma ^3=1$
, then
$1+s^3=\gamma ^3+s^3=(\gamma +s)(\gamma ^2+s^2+ \gamma s)$
and so
$$ \begin{align*}\operatorname{\mathrm{tr}}({s^{-2}(s + \gamma)^{-1}}) + \operatorname{\mathrm{tr}}({s}{(s + \gamma)^{-1}}) = \operatorname{\mathrm{tr}}(\gamma^2s^{-2}) + \operatorname{\mathrm{tr}}(\gamma s^{-1}) + \operatorname{\mathrm{tr}}(1) = \operatorname{\mathrm{tr}}(1).\end{align*} $$
Since
$\operatorname {\mathrm {tr}}(1)=0$
if and only if
$q \equiv 1 \bmod 3$
, and
$\operatorname {\mathrm {tr}}(1)=1$
if and only if
$q\equiv 2 \bmod 3$
, it follows that
$$ \begin{align*} \mathcal{S}_1 \cap \mathcal{S}_2 & = \{s^3 + \gamma s^2 \mid s \in \mathbb{F}_q\backslash\{0,\gamma\} \text { and } \operatorname{\mathrm{tr}}({s^{-2}(s + 1)^{-1}}) = 0\}\\ & = \begin{cases} \{s^3 + \gamma s^2 \mid s \in \mathbb{F}_q\backslash\{0,\gamma\} \text{ and } \kappa(s)=2\} & \text{if } q \equiv 4 \bmod 6,\\ \{s^3 + \gamma s^2 \mid s \in \mathbb{F}_q\backslash\{0,\gamma\} \text{ and } \kappa(s)=0\} & \text{if } q \equiv 2 \bmod 6. \end{cases} \end{align*} $$
Together with our discussion for odd q above, we have shown that
$$ \begin{align*} |\mathcal{S}_1 \cap \mathcal{S}_2| = \begin{cases} \tfrac{1}{6}(q - 1) \quad\text{if } q \equiv 1 \bmod 6,\\ \tfrac{1}{2}(q - 1) \quad\text{if } q \equiv 5 \bmod 6,\\ \tfrac{1}{6}(q - 4) \quad\text{if } q \equiv 4 \bmod 6,\\ \tfrac{1}{2}(q - 2) \quad\text{if } q \equiv 2 \bmod 6. \end{cases} \end{align*} $$
In summary, applying the Principle of Inclusion-Exclusion, we see that
$$ \begin{align*}|\mathcal{S}_1 \cup \mathcal{S}_2| = \begin{cases} q - 1 &\text{if } q \equiv 1 \bmod 3 \text{ and } \gamma =3 \text{ or } \gamma^2+3\gamma+9=0,\\ \tfrac{2}{3}(q - 2) &\text{if } q \equiv 2 \bmod 3 \text{ and } \gamma = 3. \end{cases} \end{align*} $$
This completes the proof.
3 Further discussion
Although our proof for the main result excludes the case
$p=3$
, our discussion for
$\mathcal {S}_2$
remains valid for any
$\gamma \in \mathbb {F}_q^\times $
in all characteristics. More specifically, from the discussion above, we also obtain the following result.
Corollary 3.1. Let q be a prime power,
$\zeta \in \mathbb {F}_q^\times $
and
$$ \begin{align*}Q = x^3 - \gamma x^2 - \zeta.\end{align*} $$
Then, for each
$\gamma \in \mathbb {F}_q^\times $
,
$$ \begin{align*}|\{\zeta \in \mathbb{F}_q^\times \mid Q \text{ is reducible}\}|= \begin{cases} \tfrac13(2q-3) &\text{if } q \equiv 0 \bmod 3,\\ \tfrac23(q-1) &\text{if } q \equiv 1 \bmod 3,\\ \tfrac23(q-2) &\text{if } q \equiv 2 \bmod 3. \end{cases} \end{align*} $$
The next result is a corollary of [Reference Williams9, Theorem 1] that turns out to be useful in finding suborbit representatives in primitive
$G_2(q)$
-actions, where
$G_2(q)$
is the finite simple group of exceptional Lie type
$G_2$
over
$\mathbb {F}_q$
. In the case where
$p=2$
, setting
$x = y+1$
yields the depressed form of Q, namely,
$D = y^3+\gamma y+\zeta $
, which has the same reducibility as Q. In particular, by assumption, we have
$\gamma ^3=1$
if
$q \equiv 1\bmod 3$
and
$\gamma = 1$
if
$q \equiv 2\bmod 3$
. For the sake of completeness, we include a restatement of [Reference Williams9, Theorem 1].
Theorem 3.2 [Reference Williams9, Theorem 1].
Let
$q = 2^{n}$
for some positive integer n. Let
$D=x^3+\gamma x + \zeta $
, where
$\gamma \in \mathbb {F}_q, \zeta \in \mathbb {F}_q^\times $
. Let
$t_1,t_2$
denote the roots of
$t^2+\zeta t +\gamma ^3 = 0$
. Then,
$t_1, t_2$
lie in
$\mathbb {F}_q$
if
$\operatorname {\mathrm {tr}}(\gamma ^3 \zeta ^{-2})=0$
and in
$\mathbb {F}_{q^2}$
otherwise.
-
(a) D has three distinct roots in
$\mathbb {F}_q$
if and only if
$\operatorname {\mathrm {tr}}(\gamma ^3 \zeta ^{-2}) = \operatorname {\mathrm {tr}}(1)$
and
$t^2 + \zeta t+\gamma ^3$
has roots
$t_1, t_2$
that are cubes in
$\mathbb {F}_q$
(n even) or in
$\mathbb {F}_{q^2}$
(n odd). -
(b) D has precisely one root in
$\mathbb {F}_q$
if and only if
$\operatorname {\mathrm {tr}}(\gamma ^3 \zeta ^{-2}) \ne \operatorname {\mathrm {tr}}(1)$
. -
(c) D has no root in
$\mathbb {F}_q$
if and only if
$\operatorname {\mathrm {tr}}(\gamma ^3 \zeta ^{-2}) = \operatorname {\mathrm {tr}}(1)$
and
$t^2 + \zeta t+\gamma ^3$
has roots
$t_1, t_2$
that are noncubes in
$\mathbb {F}_q$
(n even) or in
$\mathbb {F}_{q^2}$
(n odd).
Corollary 3.3. Let
$q = 2^{n}$
for some positive integer n and let
$\zeta \in \mathbb {F}_q^\times $
. Let
$\xi $
be a primitive element of
$\mathbb {F}_{q^2}^\times $
and
$$ \begin{align*}\epsilon=\begin{cases} 1 & \text{if } q \equiv 1 \bmod 3,\\ -1 & \text{if } q \equiv 2 \bmod 3. \end{cases}\end{align*} $$
Let
$D=x^3 + \gamma x + \zeta $
, where
$\zeta \in \mathbb {F}_q^\times $
and
$\gamma = 1$
if n is odd, or
$\gamma \in \mathbb {F}_q$
such that
$\gamma ^3=1$
if n is even. Then, the following hold.
-
(a) D is irreducible over
$\mathbb {F}_q$
if and only if
$\zeta = \eta ^{3j+1} + \eta ^{-3j-1}$
for some j with
$0 \le j < \tfrac 13(q-\epsilon )$
, where
$\eta = \xi ^{q+\epsilon }$
. -
(b) D has precisely one root in
$\mathbb {F}_q$
if and only if
$\zeta = \nu ^k + \nu ^{-k}$
for some k with
$0 < k \le \lfloor \tfrac {1}{2}(q+\epsilon )\rfloor $
, where
$\nu = \xi ^{q-\epsilon }$
. -
(c) If D is irreducible over
$\mathbb {F}_q$
, then
$x^2+\zeta (\zeta +1)^{-1} x + \zeta (\zeta + 1)^{-1}$
is reducible over
$\mathbb {F}_q$
. -
(d) If
$r \in \mathbb {F}_q^\times $
is a root of D, then
$x^2+\zeta x+\zeta $
and
$x^2 + rx + r$
have the same reducibility over
$\mathbb {F}_q$
.
Proof. (a) It follows from Theorem 3.2(c) that D is irreducible if and only if
$\operatorname {\mathrm {tr}}(\zeta ^{-2}) = \operatorname {\mathrm {tr}}(1)$
and the roots of
$x^2 + \zeta x + 1$
(in
$\mathbb {F}_q$
if n is even or in
$\mathbb {F}_{q^2}$
if n is odd) are noncubes. Moreover,
$t\ne 0$
is a root of
$x^2 + \zeta x + 1$
if and only if
$t^{-1}$
is also a root and
$t + t^{-1} = \zeta $
. Since
$\zeta \in \mathbb {F}_q^\times $
, it follows that
$t^q+t^{-q}=t+t^{-1}$
, and so either
$t\in \mathbb {F}_q^\times $
or
$t \in \mathbb {F}_{q^2}\backslash \mathbb {F}_q$
and
$t^{q+1}=1$
. Such t is a noncube if and only if
$t = \eta ^{3j\pm 1}$
for some
$j \in \mathbb {Z}$
, where
$\eta = \xi ^{q+\epsilon }$
. However, if
$t = \eta ^{3j-1}$
, then
$t^{-1}=\eta ^{-3j+1}=\eta ^{3j'+1}$
for some
$j'=-j\in \mathbb {Z}$
. Thus, the irreducibility criteria in Theorem 3.2(c) is equivalent to
$\zeta = \eta ^{3j+1} + \eta ^{-3j-1}$
for some
$j \in \mathbb {Z}$
.
Note that if
$j - k = \tfrac 13(q-\epsilon )$
, then
$\eta ^{3j+1} + \eta ^{-3j-1} = \eta ^{3k+1} + \eta ^{-3k-1}$
; thus, for
$\zeta \in \mathbb {F}_q^\times $
, the trinomial D is irreducible if and only if
$\zeta \in \{\eta ^{3j+1} + \eta ^{-3j-1}\mid 0 \le j < \tfrac 13(q-\epsilon )\}$
; there are precisely
$\tfrac 13(q-\epsilon )$
such irreducible polynomials.
(b) From Theorem 3.2(b), we see that D has precisely one root in
$\mathbb {F}_q^\times $
if and only if
${\operatorname {\mathrm {tr}}(\zeta ^{-2})\ne \operatorname {\mathrm {tr}}(1)}$
. Since
$\operatorname {\mathrm {tr}}(1)\equiv n-1 \bmod 2$
, it follows that
$\operatorname {\mathrm {tr}}(\zeta ^{-2})\ne \operatorname {\mathrm {tr}}(1)$
if and only if the quadratic
$x^2+\zeta x+1$
has two roots
$t, t^{-1} \in \mathbb {F}_{q^2}\backslash \mathbb {F}_q$
if
$q \equiv 1 \bmod 3$
, or two roots
$t, t^{-1} \in \mathbb {F}_q^\times $
if
$q \equiv 2\bmod 3$
; in both cases,
$t + t^{-1} = \zeta $
. As seen above, such roots t satisfy
$t^{q+\epsilon }=1$
. That is, D has precisely one root in
$\mathbb {F}_q$
if and only if
$\zeta = \nu ^k + \nu ^{-k}$
, where
$\nu = \xi ^{q-\epsilon }$
, for some
$0 < k < q + \epsilon $
. Moreover, if
$i + j = q + \epsilon $
, then
$\nu ^i + \nu ^{-i} = \nu ^{j} + \nu ^{-j}$
. From this, we can conclude that
$x^2+\zeta x+1$
is irreducible over
$\mathbb {F}_q$
if and only if
$\zeta \in \{\nu ^k + \nu ^{-k} \mid 0 < k \le \lfloor {(q+\epsilon )}/{2}\rfloor \}$
.
(c) The quadratic
$x^2+\zeta (\zeta +1)^{-1} x + \zeta (\zeta + 1)^{-1}$
is reducible if and only if
$\operatorname {\mathrm {tr}}(1+\zeta ^{-1})=0$
. Since
$\operatorname {\mathrm {tr}}(\zeta ^{-2})=\operatorname {\mathrm {tr}}(1)$
by Theorem 3.2(c) and
$\operatorname {\mathrm {tr}}(\zeta ^{-1})=\operatorname {\mathrm {tr}}(\zeta ^{-2})$
, the claim follows.
(d) By assumption, we have
$r(r + 1)^2 = \zeta $
. Since
$$ \begin{align*}\operatorname{\mathrm{tr}}({r^{-1}(r + 1)^{-2}}) + \operatorname{\mathrm{tr}}({r}^{-1}) = \operatorname{\mathrm{tr}}({(r+1)^{-1}}) + \operatorname{\mathrm{tr}}({(r+1)^{-2}}) = 0,\end{align*} $$
it follows that
$\operatorname {\mathrm {tr}}(\zeta ^{-1})=\operatorname {\mathrm {tr}}(r^{-1})$
, namely,
$x^2 + x + {r}^{-1}$
is reducible over
$\mathbb {F}_q$
if and only if
$x^2 + x + \zeta ^{-1}$
is reducible over
$\mathbb {F}_q$
, which proves the claim.
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