Finite fields: factorization of the trace function over the base field Let $q$ be a prime power, and $m$ a positive integer. The trace function from $GF(q^m)$ to $GF(q)$ is defined to be the mapping
$$Tr : GF(q^m) \rightarrow GF(q) $$
$$Tr(x) = x+x^q+x^{q^2}+\cdots+x^{q^{m-1}}.$$
With that in mind, I'm interested in the polynomial factorization of the polynomial $x+x^q+x^{q^2}+\cdots+x^{q^{m-1}}$ over $GF(q)$. What can we say about this factorization? Can we give it explicitly?
 A: Notice, your map $b\mapsto b^q-b$ has a nice property: $a^q-a=b^q-b$ if and only if $(a-b)^q=a-b$, that is, if and only if $a-b$ is an element of $\mathbb F_q$. 
This means this map send $\mathbb F_{q^m}$ to $q^{m-1}$ distinct values, and therefore, these are all the roots of $Tr(x)$.
This in turn means that $Tr(x)$ is a factor of $x^{q^m}-x$, which is easy to see since $$x+Tr(x)^q = Tr(x)+x^{q^m}$$
This means more, that $$x^{q^m}-x=\prod_{a\in\mathbb F_q} (Tr(x)-a)$$
That's sort of obvious in retrospect, given that every element of $\mathbb F_{q^m}$ must be a root of the right side, and the degree of the right side is the same as the degree of the left side.
My strongly intuitive conjecture is:

Prime $\pi(x)\in\mathbb F_q[x]$ divides $Tr(x)$ if and only if the the degree of $\pi$ divides $m$ and the second coefficient of $\pi(x)$ is zero. 

I don't have a proof yet, but I suspect this is true.
This is because the "trace function" is actually a linear algebra trace of a matrix. This might only work if $m$ is relatively prime to $q$, though.
A: Supplementing Thomas Andrews' answer as follows.
$Tr(x)$ has $q^{m-1}$ zeros. All in the field $GF(q^m)$, and all simple (as Thomas commented). Therefore its irreducible factors are the minimal polynomials (over $GF(q)$) of its zeros.
Consider an element $\alpha\in GF(q^m)$. Its minimal polynomial is
$$
m_\alpha(x)=\prod(x-\alpha^{q^i})=x^d+a_1x^{d-1}+\cdots+a_d,
$$
where $d=[GF(q)(\alpha):GF(q)]$ is some factor of $m$. As Thomas observed
$$
Tr(\alpha)=\frac md\cdot a_1,
$$
so $Tr(\alpha)=0$ if and only if either $a_1=0$ or $p\mid\dfrac md$.
All the irreducible polynomials of degree $d$, $d\mid m$ are minimal polynomials of some elements $\alpha\in GF(q^m)$. Therefore

An irreducible polynomial $m(x)=x^d+a_1x^{d-1}+\cdots+a_d\in GF(q)[x]$, $d\mid m$, is a factor of $Tr(x)$ if and only if either $p\mid(m/d)$ or $a_1=0$. This accounts for all the irreducible factors of $Tr(x)$.

