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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.

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  • $\begingroup$ For closure under addition, just notice that $mr + ms = m(r+s)$. Can you do something similar for multiplication? $\endgroup$
    – Nick
    Nov 19, 2018 at 0:49

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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!)

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