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.



Under what conditions is the polynomial of degree $6$ irreducible?
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 $Q(s\_1, s\_2)$.)


