I'm trying to solve the following exercise. I did the first two points( hope they're right), but have no idea on how to solve the last one.
Let $f(x)= x^8 -x$ in $\mathbb{F}_3[x]$.
Find:
its splitting field over $\mathbb{F}_3$
the number of irreducible factors of $f$ in $\mathbb{F}_3[x]$
decompose $f$ in irreducible factors over $\mathbb{F}_{81}$.
My attempt:
I write $f=x(x^7-1)$. Now, I know that $x^7-1$ has splitting field $\mathbb{F}_{3^t}$, where $t$ is the smallest number (integer) such that $7|3^t-1$. I find $t=6$, therefore the splitting field of $f$ over $\mathbb{F}_3$ is $\mathbb{F}_{3^6}$.
To find the number of irreducible factors, I compute the 3-cyclotomic cosets modulo 7, which are
$$\mathcal{C_0} = \{0 \}, \mathcal{C_1}=\{ 1,3,2,4,6,5 \}$$ Therefore, I have two irreducible factors for $x^7-1$, therefore I have $3$ irreducible factors for $f$, they are: $$x,x-1,(x-\alpha^1) \cdots (x-\alpha^5)$$ where $\alpha$ is a $7-$th root of unity.
- For the last question, I don't think I should explicitely construct $\mathbb{F}_{81}$. Is there any trick to solve it? I really can't figure it out.
