Irreducible factors for $x^q-x-a$ in $\mathbb{F}_p$. If $\mathbb{F}_q$ is a finite field with $q=p^m$ elements where $p$ is prime, if $a\in\mathbb{F}_p^\times$, ¿Which are the irreducible factors of $f_a=x^q-x-a$ ?
Attempt: If $\alpha$ is a root for $f_a$ then $\alpha+b$ is a root for $f_a$ for each $b\in\mathbb{F}_q$. If $m=1$ I show that $f_a$ is irreducible.
 A: Let $\alpha\in\overline{\Bbb{F}}_p$ be a root of $f_a$. We want to find the minimal polynomial of $\alpha$. To that end we need to figure out how many conjugates $\alpha$ has in $\overline{\Bbb{F}}_p$. As $\alpha$ is a zero of $f_a$
$$
\alpha^q=\alpha+a,\quad \alpha^{q^2}=\alpha^q+a^q=\alpha+2a,\quad \alpha^{q^3}=\alpha+3a
$$
et cetera. Therefore $\alpha^{q^p}=\alpha+pa=\alpha$. This already tells us that $\alpha$ has exactly $p$ conjugates over $\Bbb{F}_q$, and that its minimal polynomial over $\Bbb{F}_q$ is
$$
m_{\alpha, q}=\prod_{j=0}^{p-1}(x-\alpha-ja).
$$
It is known that $P(x):=x^p-x=\prod_{j=0}^{p-1}(x-j)$, so 
$$
x^p-a^{p-1}x=a^pP(\frac xa)=\prod_{j=0}^{p-1}(x-ja).
$$
Therefore
$$
m_{\alpha, q}(x)=(x-\alpha)^p-a^{p-1}(x-\alpha)=x^p-a^{p-1}x-(\alpha^p-a^{p-1}\alpha)=x^p-x-P(\alpha),
$$
because $a^{p-1}=1$ by Little Fermat. 
So over $\Bbb{F}_q$ the polynomial $f_a(x)$ splits into a product of $p^{m-1}$ factors of degree $p$. As the derivative $f'_a(x)=1$ is coprime to $f_a$, the factors are distinct, so the constant term $P(\alpha)$ takes $p^{m-1}$ distinct values in $\Bbb{F}_q$. Next we identify the set of those constant terms. To that end we recall the definition of the trace function from $\Bbb{F}_q$ to $\Bbb{F}_p$:
$$
tr(z)=z+z^p+z^{p^2}+\cdots+z^{p^{m-1}}=\sum_{j=0}^{m-1}z^{p^j}.
$$
Trace a surjective function and linear over $\Bbb{F}_p$. Therefore it takes all the values in the prime field exactly $p^{m-1}$ times. But
$$
tr(P(\alpha))=\sum_{j=0}^{m-1}P(\alpha)^{p^j}=\sum_{j=0}^{m-1}(\alpha^{p^{j+1}}-\alpha^{p^j})=\alpha^{p^m}-\alpha=a,
$$
because the sum telescopes. Therefore we get the factorization
$$
f_a(x)=\prod_{z\in\Bbb{F}_q, tr(z)=a}(x^p-x-z)
$$
in the ring $\Bbb{F}_q[x]$.
Over $\Bbb{F}_p$ the picture is more complicated. The question is how those factors combine to form irreducible polynomials in $\Bbb{F}_p[x]$. 

Quick Edit explaining what I think happens over the smaller field. I'm fairly sure that this is correct, but I don't have the time to fill in everything right now.
Above we saw that over $\Bbb{F}_q$ the factors are of the form $x^p-x-z$ such that $z\in\Bbb{F}_q$ satisfies the trace condition $tr(z)=a$. To get the irreducible factors in $\Bbb{F}_p[x]$ we simply need to multiply together conjugates factors of this type. So if the conjugates of $z$ are $z$, $z^p$,
$z^{p^2}$, $\ldots$, $z^{p^{k-1}}$ (,$z^{p^k}=z$), then $k\mid m$, and the minimal polynomial of $z$ is
$$
m_z(x)=(x-z)(x-z^p)\cdots(x-z^{p^{k-1}})\in\Bbb{F}_p[x].
$$
Now the corresponding factor of $f_a(x)$ is a similar product of the factors
$(x^p-x-z^{p^i})$, $i=0,1,\ldots,k-1.$ This product clearly equals $m_z(x^p-x)$.
Thus the factorization of $f_a(x)$ in the ring $\Bbb{F}_p[x]$ is
$$
f_a(x)=\prod_{z\in D}m_z(x^p-x),
$$
where $D$ is a set of representatives of Galois conjugacy classes of elements of $\Bbb{F}_q$ with the property $tr(z)=a$.
Example: Let $p=2$, $m=3$, $a=1$, so $f_a(x)=f_1(x)=x^8-x-1$.
In the field $\Bbb{F}_8$ there are four elements with trace $=1$. Namely $z_1=1$ and the three roots ($z_2,z_3,z_4$)of the irreducible polynomial $x^3+x^2+1\in\Bbb{F}_2[x]$. The respective minimal polynomials are thus $m_1(x)=x+1$ and
$m_2(x)=x^3+x^2+1$. Therefore the above theory gives the factorization
$$
f_1(x)=x^8+x+1=m_1(x^2+x)m_2(x^2+x)=(x^2+x+1)(x^6+x^5+x^3+x^2+1),
$$
which checks out.
