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?
 A: 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 $4$.
A: Hint, by difference of squares:
$$X^{12}-1=(X^6-1)(X^6+1)=(X^3-1)(X^3+1)(X^6+1)$$
Now, use the formula for sum and difference of cubes, and you get two linear terms, three quadratics and a fourth degree.
Then simply check if you can factor those or not.
A: Since $\text{char}(\mathbb{F}_4) = 2$, we have $x^6 + 1 = x^6 - 1$ and $x^3 + 1 = x^3 - 1$. This means:
$$x^{12} - 1 = (x^6 + 1 )(x^6 - 1) = (x^6 - 1)^2 = \left((x^3+1)(x^3-1)\right)^2 = (x^3-1)^4$$
Another way to look at this is for any finite field $F$ with $\text{char}(F) = p$, the map
$$f: F \ni x \mapsto x^p \in F$$ 
is known to be a field automorphism (Frobenius automorphism) . This again implies:
$$x^{12} - 1 = f(f(x^3)) - 1 = f(f(x^3 -1)) = (x^3-1)^4$$
Back to the factor $x^3 - 1$, for any finite group $F$, it is known that $F_\times$, the multiplicative submonoid of its non-zero elements, forms a cyclic group. Let $\omega$ be
any generator for $F_\times$, then $F_\times = \{ 1, \omega, \ldots, \omega^{|F|-2} \}$ and
$$x^{|F|-1} - 1 = \prod_{i=0}^{|F|-2} (x - \omega^{i})$$
Apply this to $\mathbb{F}_4$ and take any $\omega \in \mathbb{F}_4 \setminus \{ 0, 1 \}$, we have:
 $$(x^{12}-1) = (x^3-1)^4 = (x - 1)^4(x-\omega)^4(x-\omega^2)^4$$
A: The roots of $x^{12} -1 $ are the 12-th roots of unity.
$\mathbb{F}_4$ has characteristic $2$, so the factor of $4$ is redundant; each of the cube roots of unity appears as a factor four times. The same fact could also be discovered by observing $x^{12} - 1 = (x^3)^4 - 1^4 = (x^3 - 1)^4$.
The cube roots of unity in characteristic lie in the smallest field $\mathbb{F}_{2^n}$ such that $3 \mid 2^n - 1$. That is $n=2$, so they lie in $\mathbb{F}_4$.
Therefore,
$$x^{12} - 1 = (x-1)^4 (x-\omega)^4 (x-\omega^2)^4 $$
where $\omega$ is a primitive cube root of unity. Both elements of $\mathbb{F}_4 \setminus \mathbb{F}_2$ happen to be such things.
