Let R be a ring and $p$ a fixed prime number. Then $I_p = \{r \in R $ : additive order of $r$ is a power of $p$ $\}$ is an ideal of $R$.
Approach: Pick $r_1,r_2 \in I_p$ and $r \in R$. Then, $a^pr_1 = 0 = b^pr_2$. We want to show that $r_1 - r_2$ has additive order power of $p$ and $rr_1$ has the additive order power of $p$. Well, since $a^p(rr_1) = 0$, then $rr_1 \in I_P$. But, Im not seeing how to show $r_1 - r_2 $ is in $I_p$.