If $I,J,K$ are ideals of a ring $R$ and $I+J=J+K=K+I=R$ then $IJ+JK+KI=R$
0
$begingroup$
I have made some progress using the First Isomorphism theorem but would like your opinion on how this should be done
abstract-algebra ring-theory ideals
asked Dec 23 '18 at 17:15
Andjela TodorovicAndjela Todorovic
42
$endgroup$
add a comment |
0
$begingroup$
I have made some progress using the First Isomorphism theorem but would like your opinion on how this should be done
abstract-algebra ring-theory ideals
asked Dec 23 '18 at 17:15
Andjela TodorovicAndjela Todorovic
42
$endgroup$
4
$begingroup$
$$R subseteq I+J subseteq I+J+K subseteq R$$ so all those are equal.
$endgroup$
– Crostul
Dec 23 '18 at 17:17
$begingroup$
You changed the question making all answers invalid. Don't, please.
$endgroup$
– egreg
Dec 24 '18 at 11:36
$begingroup$
are you sure this holds with no further requirements? i can prove it when $I, J, K$ commute (that is, $IJ = JI$ etc.).
$endgroup$
– darij grinberg
Dec 24 '18 at 13:30
$begingroup$
@darijgrinberg Yes, I totally missed that part, R is a commutative ring with neutral.
$endgroup$
– Andjela Todorovic
Dec 24 '18 at 14:32
$begingroup$
@egreg I have accidentally posted the result of another question, which made the problem trivial.
$endgroup$
– Andjela Todorovic
Dec 24 '18 at 14:34
add a comment |
0
0
0
$begingroup$
I have made some progress using the First Isomorphism theorem but would like your opinion on how this should be done
abstract-algebra ring-theory ideals
asked Dec 23 '18 at 17:15
Andjela TodorovicAndjela Todorovic
42
$endgroup$
I have made some progress using the First Isomorphism theorem but would like your opinion on how this should be done
abstract-algebra ring-theory ideals
abstract-algebra ring-theory ideals
asked Dec 23 '18 at 17:15
Andjela TodorovicAndjela Todorovic
42
asked Dec 23 '18 at 17:15
Andjela TodorovicAndjela Todorovic
42
edited Dec 24 '18 at 11:30
Andjela Todorovic
asked Dec 23 '18 at 17:15
Andjela TodorovicAndjela Todorovic
42
asked Dec 23 '18 at 17:15
Andjela TodorovicAndjela Todorovic
42
asked Dec 23 '18 at 17:15
Andjela TodorovicAndjela Todorovic
42
42
4
$begingroup$
$$R subseteq I+J subseteq I+J+K subseteq R$$ so all those are equal.
$endgroup$
– Crostul
Dec 23 '18 at 17:17
$begingroup$
You changed the question making all answers invalid. Don't, please.
$endgroup$
– egreg
Dec 24 '18 at 11:36
$begingroup$
are you sure this holds with no further requirements? i can prove it when $I, J, K$ commute (that is, $IJ = JI$ etc.).
$endgroup$
– darij grinberg
Dec 24 '18 at 13:30
$begingroup$
@darijgrinberg Yes, I totally missed that part, R is a commutative ring with neutral.
$endgroup$
– Andjela Todorovic
Dec 24 '18 at 14:32
$begingroup$
@egreg I have accidentally posted the result of another question, which made the problem trivial.
$endgroup$
– Andjela Todorovic
Dec 24 '18 at 14:34
add a comment |
4
$begingroup$
$$R subseteq I+J subseteq I+J+K subseteq R$$ so all those are equal.
$endgroup$
– Crostul
Dec 23 '18 at 17:17
$begingroup$
You changed the question making all answers invalid. Don't, please.
$endgroup$
– egreg
Dec 24 '18 at 11:36
$begingroup$
are you sure this holds with no further requirements? i can prove it when $I, J, K$ commute (that is, $IJ = JI$ etc.).
$endgroup$
– darij grinberg
Dec 24 '18 at 13:30
$begingroup$
@darijgrinberg Yes, I totally missed that part, R is a commutative ring with neutral.
$endgroup$
– Andjela Todorovic
Dec 24 '18 at 14:32
$begingroup$
@egreg I have accidentally posted the result of another question, which made the problem trivial.
$endgroup$
– Andjela Todorovic
Dec 24 '18 at 14:34
4
4
$begingroup$
$$R subseteq I+J subseteq I+J+K subseteq R$$ so all those are equal.
$endgroup$
– Crostul
Dec 23 '18 at 17:17
$begingroup$
$$R subseteq I+J subseteq I+J+K subseteq R$$ so all those are equal.
$endgroup$
– Crostul
Dec 23 '18 at 17:17
$begingroup$
You changed the question making all answers invalid. Don't, please.
$endgroup$
– egreg
Dec 24 '18 at 11:36
$begingroup$
You changed the question making all answers invalid. Don't, please.
$endgroup$
– egreg
Dec 24 '18 at 11:36
$begingroup$
are you sure this holds with no further requirements? i can prove it when $I, J, K$ commute (that is, $IJ = JI$ etc.).
$endgroup$
– darij grinberg
Dec 24 '18 at 13:30
$begingroup$
are you sure this holds with no further requirements? i can prove it when $I, J, K$ commute (that is, $IJ = JI$ etc.).
$endgroup$
– darij grinberg
Dec 24 '18 at 13:30
$begingroup$
@darijgrinberg Yes, I totally missed that part, R is a commutative ring with neutral.
$endgroup$
– Andjela Todorovic
Dec 24 '18 at 14:32
$begingroup$
@darijgrinberg Yes, I totally missed that part, R is a commutative ring with neutral.
$endgroup$
– Andjela Todorovic
Dec 24 '18 at 14:32
$begingroup$
@egreg I have accidentally posted the result of another question, which made the problem trivial.
$endgroup$
– Andjela Todorovic
Dec 24 '18 at 14:34
$begingroup$
@egreg I have accidentally posted the result of another question, which made the problem trivial.
$endgroup$
– Andjela Todorovic
Dec 24 '18 at 14:34
add a comment |
2 Answers
2
active
oldest
votes
1
$begingroup$
If $I + J = R$, then $ K subset I + J$. Hence, $I + J + K = I + J = R$.
answered Dec 23 '18 at 17:18
GregGreg
183112
$endgroup$
add a comment |
1
$begingroup$
This seems pretty trivial. Of course $I+J+Ksubset R$. But $R= I+Jsubset I+J+K$.
answered Dec 23 '18 at 17:21
Chris CusterChris Custer
14.2k3827
$endgroup$
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
});
}
});
draft saved
draft discarded
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%2f3050513%2fif-i-j-k-are-ideals-of-a-ring-r-and-ij-jk-ki-r-then-ijjkki-r%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
If $I + J = R$, then $ K subset I + J$. Hence, $I + J + K = I + J = R$.
answered Dec 23 '18 at 17:18
GregGreg
183112
$endgroup$
add a comment |
1
$begingroup$
If $I + J = R$, then $ K subset I + J$. Hence, $I + J + K = I + J = R$.
answered Dec 23 '18 at 17:18
GregGreg
183112
$endgroup$
add a comment |
1
1
1
$begingroup$
If $I + J = R$, then $ K subset I + J$. Hence, $I + J + K = I + J = R$.
answered Dec 23 '18 at 17:18
GregGreg
183112
$endgroup$
If $I + J = R$, then $ K subset I + J$. Hence, $I + J + K = I + J = R$.
answered Dec 23 '18 at 17:18
GregGreg
183112
answered Dec 23 '18 at 17:18
GregGreg
183112
answered Dec 23 '18 at 17:18
GregGreg
183112
answered Dec 23 '18 at 17:18
GregGreg
183112
183112
add a comment |
add a comment |
1
$begingroup$
This seems pretty trivial. Of course $I+J+Ksubset R$. But $R= I+Jsubset I+J+K$.
answered Dec 23 '18 at 17:21
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
1
$begingroup$
This seems pretty trivial. Of course $I+J+Ksubset R$. But $R= I+Jsubset I+J+K$.
answered Dec 23 '18 at 17:21
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
1
1
1
$begingroup$
This seems pretty trivial. Of course $I+J+Ksubset R$. But $R= I+Jsubset I+J+K$.
answered Dec 23 '18 at 17:21
Chris CusterChris Custer
14.2k3827
$endgroup$
This seems pretty trivial. Of course $I+J+Ksubset R$. But $R= I+Jsubset I+J+K$.
answered Dec 23 '18 at 17:21
Chris CusterChris Custer
14.2k3827
answered Dec 23 '18 at 17:21
Chris CusterChris Custer
14.2k3827
answered Dec 23 '18 at 17:21
Chris CusterChris Custer
14.2k3827
answered Dec 23 '18 at 17:21
Chris CusterChris Custer
14.2k3827
14.2k3827
add a comment |
add a comment |
draft saved
draft discarded
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.
draft saved
draft discarded
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%2f3050513%2fif-i-j-k-are-ideals-of-a-ring-r-and-ij-jk-ki-r-then-ijjkki-r%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
4
$begingroup$
$$R subseteq I+J subseteq I+J+K subseteq R$$ so all those are equal.
$endgroup$
– Crostul
Dec 23 '18 at 17:17
$begingroup$
You changed the question making all answers invalid. Don't, please.
$endgroup$
– egreg
Dec 24 '18 at 11:36
$begingroup$
are you sure this holds with no further requirements? i can prove it when $I, J, K$ commute (that is, $IJ = JI$ etc.).
$endgroup$
– darij grinberg
Dec 24 '18 at 13:30
$begingroup$
@darijgrinberg Yes, I totally missed that part, R is a commutative ring with neutral.
$endgroup$
– Andjela Todorovic
Dec 24 '18 at 14:32
$begingroup$
@egreg I have accidentally posted the result of another question, which made the problem trivial.
$endgroup$
– Andjela Todorovic
Dec 24 '18 at 14:34