Let $q=p^s$ for some prime $p$ and $s\in \mathbb{Z}$. My problem is as follows.
The polynomial $f(x)-c$ over $\mathbb{F}_q$, where $f(x)=\prod_{v\in V}(x-v)$ for some proper subgroup $V$ of $\mathbb{F}_q^+$ and $c\in \mathbb{F}_q.$ Can we determine the degree of its irreducible factors over $\mathbb{F}_q$?
Let $|V|=p^e$. For any $c\in \{f(y):y\in \mathbb{F}_q\}$, $f(x)-c$ can be factor into $p^e$ linear factors, since $V$ is a subgroup of $\mathbb{F}_q^+$, we have $f(x+v_0)=f(x)$ for any $v_0\in V$.
However, when $c\notin \{f(y):y\in \mathbb{F}_q\}$, what is the degree of its irreducible factors over $\mathbb{F}_q$?
My process:
(1) I have known that $f(x)-c$ 's irreducible factors has the same degree by the following argument:
Assume that $x_0$ is one of its root in its spliting field $K$ over $\mathbb{F}_q$. Then we have $\mathbb{F}_q(x_0)=K$ since $\{x_0+v:v\in V\}$ are its all roots and $v\in \mathbb{F}_q$. By $[K:\mathbb{F}_q]=[\mathbb{F}_q(x_0):\mathbb{F}_q]$, the minimal polynomial of $x_0$ over $\mathbb{F}_q$ is of degree $[K:\mathbb{F}_q]$. Note that $x_0$ is chose arbitrarily. The proof is over.
(2) By the algebraic caculator MAGMA. I have some examples that $f(x)-c$ factor into irreducible factors of degree $p$ when $c\notin \{f(y):y\in \mathbb{F}_q\}$. I tried to prove that it factor into irreducible polynomials of degree $p$. But no result is got until now.
Thanks for any advice!
Write $F$ for the Frobenius map $x \mapsto x^q$. If $\alpha$ is a root of this polynomial then $F(\alpha) = \alpha + v$ for some $v \in V$. Repeatedly applying the Frobenius map then gives $F^k(\alpha) = \alpha + kv$, from which it follows that if $\alpha \not \in \mathbb{F}_q$ (equivalently, if $v \neq 0$) then $\alpha$ has exactly $p$ conjugates, namely the translations $\alpha + kv, 0 \le k \le p-1$ by the elements of the cyclic subgroup of $V$ generated by $v$.