Under what conditions is the polynomial of degree $6$ irreducible? Let $k$ be a perfect field of characteristic $p \neq 2,3$ such that $\omega := \sqrt[3]{1} \in k$, where $\omega \neq 1$. Consider an absolutely irreducible (not necessarily homogenous) quadratic polynomial $Q \in k[s_1, s_2]$ in two variables $s_1, s_2$. Under what conditions is the polynomial $R(t_1,t_2) := Q(t_1^3, t_2^3)$ (of degree $6$) absolutely irreducible (or at least irreducible over $k$) ?
Thank you in advance.
 A: The following argument might help reduce the problem to one in elimination of variables, which can be solved using a computer algebra system.
Write $S = k[s_1,s_2]$, $T = k[t_1, t_2]$ and $\phi : S \rightarrow T$,
$s_i \mapsto t_i^3$.
The ramification locus of $\phi$ is defined by $t_1t_2$.
Since $Q(s_1, s_2)$ is irreducible and of degree $2$, it does not
ramify in $T$; hence $R(t_1, t_2) = Q(t_1^3, t_2^3)$ is a product of
distinct irreducible polynomials in $T$.
Since the extension $k(s_1,s_2) \subset k(t_1,t_2)$ of fraction fields
is Galois with Galois group $G := \mathbb{Z}/3 \oplus \mathbb{Z}/3$ and $S$
and $T$ are integrally closed, the going-down theorem for integral
extensions applies.
Hence $G$ acts transitively on the irreducible factors of $R(t_1, t_2)$.
By degree considerations, the stabilizer of an irreducible factor of $R(t_1, t_2)$ cannot be
$\langle 1 \rangle$.
If the stabilizer is $G$, then $R(t_1, t_2)$ is irreducible.
Therefore assume that the stabilizer is isomorphic to $\mathbb{Z}/3$,
and that $\sigma \in G$ generates the quotient group. Again, by degree
considerations, the irreducible factors of $R(t_1, t_2)$ are of degree two.
Write
$Q(s_1,s_2) = as_1^2 + bs_1s_2 + cs_2^2 + ds_1 + es_2 + f$.
Let $f(t_1, t_2) = a_1t_1^2 + b_1t_1t_2 + c_1t_2^2 + d_1t_1 + e_1t_2 + f_1$
be an irreducible factor of $R(t_1, t_2) = Q(t_1^3, t_2^3)$.
As an example, suppose that $\sigma (t_1) = \omega t_1$ and $\sigma(t_2)
= t_2$. Since
$$
R(t_1, t_2) =                                                                   f(t_1, t_2)\cdot \sigma(f(t_1, t_2)) \cdot \sigma(\sigma(f(t_1, t_2)) =
f(t_1,t_2) \cdot f(\omega t_1,t_2) \cdot f(\omega^2 t_1,t_2)
$$
we will get six polynomials of the form
$a-g_1(a_1, \ldots, f_1), b-g_2(a_1, \ldots, f_1), \ldots$,
equating the coefficients of the monomials in $t_1,t_2$.
Now a computer algebra system can be used to eliminate the variables
$a_1, \ldots, f_1$, and get the relations between $a, \ldots, f$.
Let $Z$ be the  algebraic set defined by these relations.
Now impose the condition that there exists $\tau \in G$ such that $\tau$ fixes
$f(t_1, t_2)\cdot \sigma(f(t_1, t_2)) \cdot \sigma(\sigma(f(t_1, t_2))$. Remove the corresponding points from $Z$. This
will give
a set of $Q(s_1, s_2)$ for which $R(t_1, t_2)$ is irreducible.
(One might have to repeat this for other possible choices of the stabilizer of $R(t_1, t_2)$, to get the set of all such $Q(s_1, s_2)$.)
