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.
abstract-algebra ring-theory
asked Nov 19 at 0:47
Raul Quintanilla Jr.
702
add a comment |
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 Nov 19 at 0:47
Raul Quintanilla Jr.
702
For closure under addition, just notice that $mr + ms = m(r+s)$. Can you do something similar for multiplication?
– Nick
Nov 19 at 0:49
add a comment |
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 Nov 19 at 0:47
Raul Quintanilla Jr.
702
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 Nov 19 at 0:47
Raul Quintanilla Jr.
702
asked Nov 19 at 0:47
Raul Quintanilla Jr.
702
asked Nov 19 at 0:47
Raul Quintanilla Jr.
702
asked Nov 19 at 0:47
Raul Quintanilla Jr.
702
asked Nov 19 at 0:47
Raul Quintanilla Jr.
702
702
For closure under addition, just notice that $mr + ms = m(r+s)$. Can you do something similar for multiplication?
– Nick
Nov 19 at 0:49
add a comment |
For closure under addition, just notice that $mr + ms = m(r+s)$. Can you do something similar for multiplication?
– Nick
Nov 19 at 0:49
For closure under addition, just notice that $mr + ms = m(r+s)$. Can you do something similar for multiplication?
– Nick
Nov 19 at 0:49
For closure under addition, just notice that $mr + ms = m(r+s)$. Can you do something similar for multiplication?
– Nick
Nov 19 at 0:49
add a comment |
1 Answer
1
active
oldest
votes
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 Nov 19 at 3:30
Alex Sanger
865
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004354%2fwould-using-the-subring-test-be-good-here%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 Nov 19 at 3:30
Alex Sanger
865
add a comment |
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 Nov 19 at 3:30
Alex Sanger
865
add a comment |
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 Nov 19 at 3:30
Alex Sanger
865
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 Nov 19 at 3:30
Alex Sanger
865
answered Nov 19 at 3:30
Alex Sanger
865
answered Nov 19 at 3:30
Alex Sanger
865
answered Nov 19 at 3:30
Alex Sanger
865
865
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004354%2fwould-using-the-subring-test-be-good-here%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
For closure under addition, just notice that $mr + ms = m(r+s)$. Can you do something similar for multiplication?
– Nick
Nov 19 at 0:49