# Roots of a polynomial and finite additive subgroup

Suppose $F$ is a field with characteristic $p$ and $f(x)\in F[x]$ Then$f(x)=x^{p^m}+a_1x^{p^{m-1}}+\cdots +a_mx \iff$ its roots form a finite subgroup of the additive group of the splitting field.

For $\Longrightarrow$ direction, I am having a hard time to prove $-b$ is a root if $b$ is root. I know the p-power will distribute but then I will have the polynomial $f(x)=(-1)^{p^m}b^{p^m}+(-1)^{p^{m-1}}a_1b^{p^{m-1}}+\cdots +(-1)a_mb$ which doesnt say that $-b$ is a root.

Next, I am having a hard time proving the other direction for which I would like some hints to proceed. Suppose I assume that roots of a polynomial $g(x)$ form a finite subgroup and $g(x)=\sum_{i=0}^{n}{c_ix^i}$. I am not sure how I can relate the assumption to prove that the $i's$ in $g(x)$ are prime powers without a constant term.

• If $p=2$, then $-b=b$. If $p$ is odd, then $(-1)^{p^k}=(-1)$ for all $k$. – Berci Mar 27 '15 at 23:23

Hints I think that you missed a part of the equivalence: all roots have the same multiplicity $p^{e}$.
$(\Rightarrow)$ you need to prove that each root has multiplicity some $p^e$
$(\Leftarrow)$ Let us prove that $f$ is a $p$–polynomial. We know that $K$ has characteristic $p$, it contains $F_p$ as a subﬁeld. the roots of $f$ are closed under addition, they form a vector space $V$ over $F_p$. Prove the statement by induction on $\dim_{F_p}(V )$.
• What if $V=\{0\}$
• if $V$ is of dimension $n$ then there exist $V'$ a vector space of dimension $n-1$ and $v_0\in V$ such that $V=v_0 F_p+V'$, and now apply induction hypothesis on $g(x)=\prod_{v\in V'}(x-v)^{p^e}$ and observe that $$f(x)=\prod_{j\in F_p}\prod_{v\in V'}(x-(v_0j+v))^{p^e}=\prod_{j\in F_p}(g(x)-jg(v_0))=g(x)^{p}-g(x)g(v_0)$$