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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 $I_p$.

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You're misinterpreting the statement $r$ has additive order a power of $p$: it means $p^k r = 0$ for some $k \in \mathbb{N}$. – Michael Joyce Nov 16 '12 at 5:33
up vote 2 down vote accepted

Let $x, y \in I_p$. There exist positive integers $k, l$ such that $p^k x = 0, p^l y = 0$. Let $m = max(k, l)$. Then $p^m(x + y) = p^m x + p^m y = 0$. Hence $x + y \in I_p$.

Let $a \in R$. $p^k(ax) = a(p^k x) = 0$. Hence $ax \in I_p$. $p^k(xa) = (p^k x)a = 0$. Hence $xa \in I_p$.

Therefore $I_p$ is an ideal of $R$.

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it's probably worth noting explicitly where we're using primeness: namely, once you know that $p^k x = 0$, that tells you $x$ has order dividing $p^k$, hence of the form $p^i$ for some $i \leqslant k$ – uncookedfalcon Nov 16 '12 at 6:42

Hint $\ $ Let $\rm\:M\subset \Bbb N\:$ be any set such that $\rm\:jk\in M\iff j,k\in M,\:$ i.e. $\rm\:M\:$ is a saturated submonoid of $\rm\:\Bbb N,\:$ and let $\rm\:I_M = \{r\in R:\,$ additive order of $\rm\,r\,$ is $\rm\in M \}.\:$ If $\rm\:r,s\in I_M\:$ then $\rm\:jr = 0 = ks,\,\ j,k\in M,\:$ so $\rm\ jk(r-s) = k(jr)-j(ks)=0\:$ hence $\rm\: ord(r\!-\!s)\mid jk\in M\:$ $\Rightarrow$ $\rm\:ord(r\!-\!s)\in M\:$ $\Rightarrow$ $\rm\:r\!-\!s\in I_M.$

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