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.
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!)