Let $R$ be a ring and $m$ be a fixed integer.

Let $S$ = {$r \in R| mr = 0_R$}.

Prove that $S$ is a subring of $R$.

I'm fairly sure that I can show this using the Subring Test which says that I need to only show that the subset $S$ is closed under subtraction and multiplication, but I I'm not sure how to do that here.

Any help would be greatly appreciated.

  • $\begingroup$ For closure under addition, just notice that $mr + ms = m(r+s)$. Can you do something similar for multiplication? $\endgroup$ – Nick Nov 19 '18 at 0:49

You are indeed correct that the subring test applies here. For closure under subtraction, let $r,s \in S$; we need to show $m(r-s)=0_R$. Now, $m(r-s)=mr-ms=0_R-0_R=0_R$ by the distributive property and the fact that $mr=0_R$ and $ms=0_R$ (since $r$ and $s$ are in $S$.) For closure under multiplication, we need to show $m(rs)=0_R$. For this note that $m(rs)=(mr)s=0_Rs=0_R$ (if you can't see why rearranging the brackets in the last step is justified, remember that $m$ is an integer, so in effect we are adding $rs$ to itself $m$ times, or $-m$ times, if $m<0$. So if we factor out an $s$ from the sum... you should be able to fill in the details!)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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