# Would using the Subring test be good here?

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.

• For closure under addition, just notice that $mr + ms = m(r+s)$. Can you do something similar for multiplication? – 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!)