Existence of "simple" irreducible polynomial of degree 12 in a finite field

Question

Assume that we have a finite field $\mathbb{F}_p$, where $p$ is prime, $p \equiv 1\ (\textrm{mod}\ 4)$ and $p \equiv 1\ (\textrm{mod}\ 3)$. I was looking for irreducible polynomial in a form $X^{12} + a$, where $a \in \mathbb{F}_p$ and it turned out that it's very simple to find one: in all cases I tested it's sufficient to take $a \in \{2,5,6\}$. The problem is that I can't seem to figure out why this is the case. I'd be happy if someone could explain how to show the existence of irreducible polynomial in such form (I guess the reason why small $a$ is sufficient will follow from the reasoning).

Answer 1

$x^{12}-a$ is reducible mod $p=12k+1$ if and only if $a=0$ or $a^k$'s order is a proper divisor of 12.

To see this, recall that a monic polynomial of degree $d$ is irreducible mod $p$ if and only if $\gcd(f(x), g_i(x)) = 1$ for all $g_i(x) = x^{p^i}-x$, $1\leq i < d$.

Fixing $i$, and working in $\mathbb{Z}_p[x^{12}-a]$,

\begin{align*} g_i(x) &\equiv x^{(12k+1)^i}-x\\ &\equiv x\cdot x^{12ki} \cdot x^{\sum_{j=2}^i (12k)^j \binom{i}{j}}-x\\ &\equiv xa^{ki}-x\\ &\equiv x\left[a^{ki}-1\right]. \end{align*}

The third line follows from $$\sum_{j=2}^i (12k)^j\binom{i}{j} = 12\cdot 12k\cdot k\sum_{j=2}^i (12k)^{j-2}\binom{i}{j},$$ so $$x^{\sum_{j=2}^i (12k)^j \binom{i}{j}} = a^{(p-1)\cdot k\sum_{j=2}^i (12k)^{j-2}\binom{i}{j}} = 1^{k\sum_{j=2}^i (12k)^{j-2}\binom{i}{j}} = 1.$$

So if $a^k$'s order properly divides $12$, $x^{12}-a$ divides $g_i(x)$ for some $i<12$ and $f(x)$ reducible. Conversely, if $a^k$'s order does not properly divide 12, then $\gcd(x^{12}-a,g_i(x)) = 1$ for all $1\leq i < 12$ and $x^{12}-a$ is irreducible.

EDIT: I think the above is correct now. Notice that $a$ being a primitive root is sufficient but not necessary for irreducibility; one example is $a=5, p=1453.$