Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $I_p$.

share|improve this question
2  
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
add comment

2 Answers

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 $R$.

share|improve this answer
1  
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
add comment

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 $\rm\: ord(r\!-\!s)\mid jk\in M\:$ $\Rightarrow$ $\rm\:ord(r\!-\!s)\in M\:$ $\Rightarrow$ $\rm\:r\!-\!s\in I_M.$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.