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
asked yesterday
Raul Quintanilla Jr.
442
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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
asked yesterday
Raul Quintanilla Jr.
442
For closure under addition, just notice that $mr + ms = m(r+s)$. Can you do something similar for multiplication?
– Nick
yesterday
add a comment |
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
asked yesterday
Raul Quintanilla Jr.
442
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
asked yesterday
Raul Quintanilla Jr.
442
asked yesterday
Raul Quintanilla Jr.
442
asked yesterday
Raul Quintanilla Jr.
442
asked yesterday
Raul Quintanilla Jr.
442
asked yesterday
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
yesterday
add a comment |
For closure under addition, just notice that $mr + ms = m(r+s)$. Can you do something similar for multiplication?
– Nick
yesterday
For closure under addition, just notice that $mr + ms = m(r+s)$. Can you do something similar for multiplication?
– Nick
yesterday
For closure under addition, just notice that $mr + ms = m(r+s)$. Can you do something similar for multiplication?
– Nick
yesterday
add a comment |
1 Answer
1
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oldest
votes
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!)
answered 23 hours ago
Alex Sanger
474
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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!)
answered 23 hours ago
Alex Sanger
474
add a comment |
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!)
answered 23 hours ago
Alex Sanger
474
add a comment |
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!)
answered 23 hours ago
Alex Sanger
474
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!)
answered 23 hours ago
Alex Sanger
474
answered 23 hours ago
Alex Sanger
474
answered 23 hours ago
Alex Sanger
474
answered 23 hours ago
Alex Sanger
474
474
add a comment |
add a comment |
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For closure under addition, just notice that $mr + ms = m(r+s)$. Can you do something similar for multiplication?
– Nick
yesterday