Irreducible 3rd degree polynomials over $\mathbb{Z}_3$ field? I want to find all irreducible polynomials over $\mathbb{Z}_3$ field which have the form
$x^3 + a_2 x^2 +a_1 x + a_0$.
My thought process:


*

*Third degree polynomial is irreducible if it has no roots within
$\mathbb{Z}_3$.

*$a_0 \neq 0$, otherwise $0 \in \mathbb{Z}_3$ is a root of the
polynomial in question.


However this leaves me with 18 polynomials and 54 substitutions for $x$ to check - though definitely possible not exactly what I'm looking for on a little sheet of paper. How can I further simplify this work? 

Apparently the answer is
$$
\begin{align*}
& x^3+ 2*x^2+  1 \\
& x^3+ 2*x^2+ x+ 1 \\
& x^3+  2*x+ 1 \\
& x^3+ x^2+ 2*x+ 1 \\
& x^3+ x^2+  2 \\
& x^3+ x^2+ x+ 2 \\
& x^3+  2*x+ 2 \\
& x^3+ 2*x^2+ 2*x+ 2
\end{align*}
$$
but I still fail to see any inherent regularities.
 A: A cubic polynomial in $\mathbb{Z}_3[x]$ is irreducible if and only if it has no roots in $\mathbb{Z}_3 = \{0,1,2\}$. (Since if $f$ is reducible then it must have a factor of degree $1$, which immediately gives a root in $\mathbb{Z}_3$.) So T. Bongers' comment is a necessary and sufficient condition on the coefficients for $x^3+a_2x^2+a_1x+a_0$ to be irreducible.
More abstractly, we could start with the finite field $\mathbb{F}_{3^3}$. This can be constructed as the splitting field of $X^{27}-X$ over $\mathbb{F}_3$. (This is your $\mathbb{Z}_3$, but I much prefer to use $\mathbb{F}_3$ here.) 
Any irreducible cubic polynomial $f(x) \in \mathbb{F}_3[x]$ splits into linear factors in $\mathbb{F}_{3^3}$, essentially because the field obtained by adjoining one root of $f$ to $\mathbb{F}_3$ is a finite field of order $27$, so isomorphic to $\mathbb{F}_{3^3}$. Hence the irreducible cubic polynomials over $\mathbb{F}_3$ are exactly the irreducible cubic factors of $X^{27}-X$. Of course one must still do quite a bit of calculation to find these factors.
