The radical of $kG$-modules $\DeclareMathOperator\Rad{Rad}$Let $k$ be a finite field of $p$ elements. Let $G$ be an elementary abelian p-group and $V$ a $kG$-module corresponding to the representation $\alpha:G\rightarrow \mathrm{GL}_{n}(k)$. Denote the radical of $V$ by $\Rad(V)$ which is defined to be the intersection of all maximal submodules of $V$. Note that $\Rad^{i}(V)=\Rad(  \Rad^{i-1}(V) )$. Now, let $s<p$ and suppose that $(\alpha(g)-1)^{s}=0$ for all   $ g \in G $ . In fact, I'd conjecture that $\Rad^{s}(V)=0$. I think we should start  from the case  $ s=p-1$ but I don't know how to do this.
I would appreciate any hints and comments. Thank you in advance!
 A: The answer is yes: $\text{Rad}(kG)^s$ is generated as an ideal by $(g-1)^s$ for $G$ an elementary abelian $p$-group and $s \leq p-1$.
Lemma: Let $V$ be a $k$-vector space and let $s \leq p-1$. Then $\text{Sym}^s(V)$ is spanned by the elements $v^s$ for $v \in V$.
Proof: $\text{Sym}^s(V)$ is clearly spanned by products of the form $v_1 v_2 \cdots v_s$. We have
$$v_1 v_2 \cdots v_s = \frac{1}{s!} \sum_{c_1, c_2, \ldots, c_s \in \{ 0,1 \}} (-1)^{s-\sum c_i} (c_1 v_1 + c_2 v_2 + \cdots + c_s v_s)^s.$$
We used that $s \leq p-1$ in order to be allowed to divide by $s!$. $\square$
We now answer the question. Let $(g_i)_{i \in I}$ be a set of generators for $G$ and put $t_i = g_i-1$. Then $kG \cong k[t_i : i \in I]/\langle t_i^p : i \in I  \rangle$. The radical $R$ is the ideal $\langle t_i \rangle$. We want to show that $R^s$ is generated by the elements $(1-\prod_i g_i^{a_i})^s$, where $a_i$ is a finitely supported function $I \to \mathbb{Z}/p \mathbb{Z}$. By Nakayama's lemma, it is enough to show that $R^s/R^{s+1}$ is generated by these elements. We have
$$(1-\prod_i g_i^{a_i})^s = (1-\prod_i (1+t_i)^{a_i})^s \equiv \left( \sum a_i t_i \right)^s \bmod R^{s+1}.$$
So our generators are $s$-th powers of linear forms, and $R^s/R^{s+1}$ is the degree $s$ polynomials; this is exactly what our lemma addresses. $\square$
A: Under these assumptions, note that $\mathrm{Rad}(kG)$ coincides with the augmentation ideal $\omega(kG)$ of $kG$. Now, as remarked by Jeremy Rickard, the question is equivalent to whether the elements $(g-1)^s$ with $g\in G$ generate $\mathrm{Rad}^s(kG)$ as an ideal of $kG$. Theorem 3.7 in Section 5.3 of the book [D.S. Passman: The algebraic structure of group rings] provides an explicite description of $Rad(kG)^s$ for all $s>0$. As proved in the beautiful David's answer, $Rad(kG)^s$ is in fact generated by the elements of the form $(g-1)^s$ with $g\in G$ in the given situation.
