Would using the Subring test be good here?











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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.






abstract-algebra ring-theory







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















up vote
1
down vote

favorite












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.






abstract-algebra ring-theory







share|cite|improve this question






















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













up vote
1
down vote

favorite









up vote
1
down vote

favorite











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.






abstract-algebra ring-theory







share|cite|improve this question













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.






abstract-algebra ring-theory




abstract-algebra ring-theory






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share|cite|improve this question











share|cite|improve this question




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asked 15 hours ago









Raul Quintanilla Jr.

442




442












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


















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
















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




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










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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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    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!)






    share|cite|improve this answer

























      up vote
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      accepted










      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!)






      share|cite|improve this answer























        up vote
        0
        down vote



        accepted







        up vote
        0
        down vote



        accepted






        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!)






        share|cite|improve this answer












        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!)







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 13 hours ago









        Alex Sanger

        474




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