<s>[INST] Polynomial factorization over finite fields How can i factorize the polynomial $x^{12}-1$ as product of irreducibles polynomials over $\mathbb{F}_4$? Anyone can help me?
 [/INST] Let $\alpha$ and $\beta = \alpha^2 = \alpha+1$ denote the two elements of $\mathbb F_4-\mathbb F_2$. Then, $\alpha$ and $\beta$ are roots of $x^2+x+1$ and
we have that
$$x^3-1 = (x-1)(x^2+x+1) = (x-1)(x-\alpha)(x-\beta).$$
In a field of characteristic $p$, $(a-b)^{p^i} = a^{p^i} - b^{p^i}$ and so we have
that
$$x^{12} - 1 = (x^3-1)^4 = (x-1)^4(x-\alpha)^4(x-\beta)^4$$ 
factors into $12$ linear factors that are three distinct polynomials
each occurring with multiplicity $4$.

Modified approach using cyclotomic polynomials
In a field of characteristic $p$, $(a-b)^{p^i} = a^{p^i} - b^{p^i}$ and so we have
that
$x^{12} - 1 = (x^3-1)^4$. But, $$x^3-1 = Q_1(x)Q_3(x) = (x-1)(x^2+x+1)$$
where $Q_1(x) = x-1$ and $Q_3(x) = x^2+x+1$ are cyclotomic polynomials.
Also, $\mathbb F_4$ is the splitting field of $x^3-1$ since the
multiplicative group $\mathbb F_4^{\times}$ is a cyclic group of order
$3$, and so $x^3-1 = (x-1)(x-\alpha)(x-\beta)$ where 
$\alpha, \beta \in \mathbb F_4 - \mathbb F_2$. So,
$x^{12}-1 = (x^3-1)^4$ factors into $12$ linear factors over $\mathbb F_4$,
and these factors are three distinct polynomials
each occurring with multiplicity $4$.
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