Does every ID with subring which has a unity have an unity?
2
$begingroup$
For an arbitrary ID (integral domain) $R$ with subring $S$, assume that $S$ has an unity. Then does $R$ have a unity too? If not, please provide a counter-example.
ring-theory integral-domain
edited Aug 6 '15 at 12:43
man_in_green_shirt
8381028
asked Aug 6 '15 at 12:09
jawlangjawlang
30715
$endgroup$
add a comment |
2
$begingroup$
For an arbitrary ID (integral domain) $R$ with subring $S$, assume that $S$ has an unity. Then does $R$ have a unity too? If not, please provide a counter-example.
ring-theory integral-domain
edited Aug 6 '15 at 12:43
man_in_green_shirt
8381028
asked Aug 6 '15 at 12:09
jawlangjawlang
30715
$endgroup$
add a comment |
2
2
2
0
$begingroup$
For an arbitrary ID (integral domain) $R$ with subring $S$, assume that $S$ has an unity. Then does $R$ have a unity too? If not, please provide a counter-example.
ring-theory integral-domain
edited Aug 6 '15 at 12:43
man_in_green_shirt
8381028
asked Aug 6 '15 at 12:09
jawlangjawlang
30715
$endgroup$
For an arbitrary ID (integral domain) $R$ with subring $S$, assume that $S$ has an unity. Then does $R$ have a unity too? If not, please provide a counter-example.
ring-theory integral-domain
ring-theory integral-domain
edited Aug 6 '15 at 12:43
man_in_green_shirt
8381028
asked Aug 6 '15 at 12:09
jawlangjawlang
30715
edited Aug 6 '15 at 12:43
man_in_green_shirt
8381028
asked Aug 6 '15 at 12:09
jawlangjawlang
30715
edited Aug 6 '15 at 12:43
man_in_green_shirt
8381028
edited Aug 6 '15 at 12:43
man_in_green_shirt
8381028
edited Aug 6 '15 at 12:43
man_in_green_shirt
8381028
8381028
asked Aug 6 '15 at 12:09
jawlangjawlang
30715
asked Aug 6 '15 at 12:09
jawlangjawlang
30715
asked Aug 6 '15 at 12:09
jawlangjawlang
30715
30715
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
1
$begingroup$
If $S$ is nontrivial (that is, has a nonzero identity $e$) then yes.
A nonzero idempotent in a domain must act as the identity for the domain.
To see this, just examine what $e(ex-x)$ and $(xe-x)e$ must be, where $x$ is an arbitrary element of the domain.
This came up under slightly different circumstances here.
edited Apr 13 '17 at 12:20
Community♦
1
answered Aug 6 '15 at 12:47
rschwiebrschwieb
106k12102249
$endgroup$
$begingroup$
If $e$ is the identity, then won't $ex=x$ and therefore $ex-x=0$?
$endgroup$
– man_in_green_shirt
Aug 6 '15 at 12:55
1
$begingroup$
@man_in_green_shirt: As a general rule, it's unproductive to assume that which you are attempting to prove.
$endgroup$
– WillO
Aug 6 '15 at 12:57
$begingroup$
Ok, but I don't see the significance of $e(ex-x)$. Ok, if you multiply $e$ by $e(ex-x)$, then you get $e(ex-x)$ because $e$ is idempotent. So if any element can be written as $ex-x$, then $e$ is an identity. However, how can that be possible when $ex-x$ is always $0$?
$endgroup$
– man_in_green_shirt
Aug 6 '15 at 13:07
1
$begingroup$
@man_in_green_shirt: What's being claimed here is not that any element can be written as $ex-x$, but that every element can be written as $x$.
$endgroup$
– WillO
Aug 6 '15 at 13:29
$begingroup$
@man_in_green_shirt The conclusion is that $ex-x=xe-x=0$, and therefore $e$ is the identity. It is not the hypothesis...
$endgroup$
– rschwieb
Aug 6 '15 at 14:01
|
show 1 more 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%2f1386528%2fdoes-every-id-with-subring-which-has-a-unity-have-an-unity%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
1
$begingroup$
If $S$ is nontrivial (that is, has a nonzero identity $e$) then yes.
A nonzero idempotent in a domain must act as the identity for the domain.
To see this, just examine what $e(ex-x)$ and $(xe-x)e$ must be, where $x$ is an arbitrary element of the domain.
This came up under slightly different circumstances here.
edited Apr 13 '17 at 12:20
Community♦
1
answered Aug 6 '15 at 12:47
rschwiebrschwieb
106k12102249
$endgroup$
$begingroup$
If $e$ is the identity, then won't $ex=x$ and therefore $ex-x=0$?
$endgroup$
– man_in_green_shirt
Aug 6 '15 at 12:55
1
$begingroup$
@man_in_green_shirt: As a general rule, it's unproductive to assume that which you are attempting to prove.
$endgroup$
– WillO
Aug 6 '15 at 12:57
$begingroup$
Ok, but I don't see the significance of $e(ex-x)$. Ok, if you multiply $e$ by $e(ex-x)$, then you get $e(ex-x)$ because $e$ is idempotent. So if any element can be written as $ex-x$, then $e$ is an identity. However, how can that be possible when $ex-x$ is always $0$?
$endgroup$
– man_in_green_shirt
Aug 6 '15 at 13:07
1
$begingroup$
@man_in_green_shirt: What's being claimed here is not that any element can be written as $ex-x$, but that every element can be written as $x$.
$endgroup$
– WillO
Aug 6 '15 at 13:29
$begingroup$
@man_in_green_shirt The conclusion is that $ex-x=xe-x=0$, and therefore $e$ is the identity. It is not the hypothesis...
$endgroup$
– rschwieb
Aug 6 '15 at 14:01
|
show 1 more comment
1
$begingroup$
If $S$ is nontrivial (that is, has a nonzero identity $e$) then yes.
A nonzero idempotent in a domain must act as the identity for the domain.
To see this, just examine what $e(ex-x)$ and $(xe-x)e$ must be, where $x$ is an arbitrary element of the domain.
This came up under slightly different circumstances here.
edited Apr 13 '17 at 12:20
Community♦
1
answered Aug 6 '15 at 12:47
rschwiebrschwieb
106k12102249
$endgroup$
$begingroup$
If $e$ is the identity, then won't $ex=x$ and therefore $ex-x=0$?
$endgroup$
– man_in_green_shirt
Aug 6 '15 at 12:55
1
$begingroup$
@man_in_green_shirt: As a general rule, it's unproductive to assume that which you are attempting to prove.
$endgroup$
– WillO
Aug 6 '15 at 12:57
$begingroup$
Ok, but I don't see the significance of $e(ex-x)$. Ok, if you multiply $e$ by $e(ex-x)$, then you get $e(ex-x)$ because $e$ is idempotent. So if any element can be written as $ex-x$, then $e$ is an identity. However, how can that be possible when $ex-x$ is always $0$?
$endgroup$
– man_in_green_shirt
Aug 6 '15 at 13:07
1
$begingroup$
@man_in_green_shirt: What's being claimed here is not that any element can be written as $ex-x$, but that every element can be written as $x$.
$endgroup$
– WillO
Aug 6 '15 at 13:29
$begingroup$
@man_in_green_shirt The conclusion is that $ex-x=xe-x=0$, and therefore $e$ is the identity. It is not the hypothesis...
$endgroup$
– rschwieb
Aug 6 '15 at 14:01
|
show 1 more comment
1
1
1
$begingroup$
If $S$ is nontrivial (that is, has a nonzero identity $e$) then yes.
A nonzero idempotent in a domain must act as the identity for the domain.
To see this, just examine what $e(ex-x)$ and $(xe-x)e$ must be, where $x$ is an arbitrary element of the domain.
This came up under slightly different circumstances here.
edited Apr 13 '17 at 12:20
Community♦
1
answered Aug 6 '15 at 12:47
rschwiebrschwieb
106k12102249
$endgroup$
If $S$ is nontrivial (that is, has a nonzero identity $e$) then yes.
A nonzero idempotent in a domain must act as the identity for the domain.
To see this, just examine what $e(ex-x)$ and $(xe-x)e$ must be, where $x$ is an arbitrary element of the domain.
This came up under slightly different circumstances here.
edited Apr 13 '17 at 12:20
Community♦
1
answered Aug 6 '15 at 12:47
rschwiebrschwieb
106k12102249
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
1
answered Aug 6 '15 at 12:47
rschwiebrschwieb
106k12102249
answered Aug 6 '15 at 12:47
rschwiebrschwieb
106k12102249
answered Aug 6 '15 at 12:47
rschwiebrschwieb
106k12102249
106k12102249
$begingroup$
If $e$ is the identity, then won't $ex=x$ and therefore $ex-x=0$?
$endgroup$
– man_in_green_shirt
Aug 6 '15 at 12:55
1
$begingroup$
@man_in_green_shirt: As a general rule, it's unproductive to assume that which you are attempting to prove.
$endgroup$
– WillO
Aug 6 '15 at 12:57
$begingroup$
Ok, but I don't see the significance of $e(ex-x)$. Ok, if you multiply $e$ by $e(ex-x)$, then you get $e(ex-x)$ because $e$ is idempotent. So if any element can be written as $ex-x$, then $e$ is an identity. However, how can that be possible when $ex-x$ is always $0$?
$endgroup$
– man_in_green_shirt
Aug 6 '15 at 13:07
1
$begingroup$
@man_in_green_shirt: What's being claimed here is not that any element can be written as $ex-x$, but that every element can be written as $x$.
$endgroup$
– WillO
Aug 6 '15 at 13:29
$begingroup$
@man_in_green_shirt The conclusion is that $ex-x=xe-x=0$, and therefore $e$ is the identity. It is not the hypothesis...
$endgroup$
– rschwieb
Aug 6 '15 at 14:01
|
show 1 more comment
$begingroup$
If $e$ is the identity, then won't $ex=x$ and therefore $ex-x=0$?
$endgroup$
– man_in_green_shirt
Aug 6 '15 at 12:55
1
$begingroup$
@man_in_green_shirt: As a general rule, it's unproductive to assume that which you are attempting to prove.
$endgroup$
– WillO
Aug 6 '15 at 12:57
$begingroup$
Ok, but I don't see the significance of $e(ex-x)$. Ok, if you multiply $e$ by $e(ex-x)$, then you get $e(ex-x)$ because $e$ is idempotent. So if any element can be written as $ex-x$, then $e$ is an identity. However, how can that be possible when $ex-x$ is always $0$?
$endgroup$
– man_in_green_shirt
Aug 6 '15 at 13:07
1
$begingroup$
@man_in_green_shirt: What's being claimed here is not that any element can be written as $ex-x$, but that every element can be written as $x$.
$endgroup$
– WillO
Aug 6 '15 at 13:29
$begingroup$
@man_in_green_shirt The conclusion is that $ex-x=xe-x=0$, and therefore $e$ is the identity. It is not the hypothesis...
$endgroup$
– rschwieb
Aug 6 '15 at 14:01
$begingroup$
If $e$ is the identity, then won't $ex=x$ and therefore $ex-x=0$?
$endgroup$
– man_in_green_shirt
Aug 6 '15 at 12:55
$begingroup$
If $e$ is the identity, then won't $ex=x$ and therefore $ex-x=0$?
$endgroup$
– man_in_green_shirt
Aug 6 '15 at 12:55
1
1
$begingroup$
@man_in_green_shirt: As a general rule, it's unproductive to assume that which you are attempting to prove.
$endgroup$
– WillO
Aug 6 '15 at 12:57
$begingroup$
@man_in_green_shirt: As a general rule, it's unproductive to assume that which you are attempting to prove.
$endgroup$
– WillO
Aug 6 '15 at 12:57
$begingroup$
Ok, but I don't see the significance of $e(ex-x)$. Ok, if you multiply $e$ by $e(ex-x)$, then you get $e(ex-x)$ because $e$ is idempotent. So if any element can be written as $ex-x$, then $e$ is an identity. However, how can that be possible when $ex-x$ is always $0$?
$endgroup$
– man_in_green_shirt
Aug 6 '15 at 13:07
$begingroup$
Ok, but I don't see the significance of $e(ex-x)$. Ok, if you multiply $e$ by $e(ex-x)$, then you get $e(ex-x)$ because $e$ is idempotent. So if any element can be written as $ex-x$, then $e$ is an identity. However, how can that be possible when $ex-x$ is always $0$?
$endgroup$
– man_in_green_shirt
Aug 6 '15 at 13:07
1
1
$begingroup$
@man_in_green_shirt: What's being claimed here is not that any element can be written as $ex-x$, but that every element can be written as $x$.
$endgroup$
– WillO
Aug 6 '15 at 13:29
$begingroup$
@man_in_green_shirt: What's being claimed here is not that any element can be written as $ex-x$, but that every element can be written as $x$.
$endgroup$
– WillO
Aug 6 '15 at 13:29
$begingroup$
@man_in_green_shirt The conclusion is that $ex-x=xe-x=0$, and therefore $e$ is the identity. It is not the hypothesis...
$endgroup$
– rschwieb
Aug 6 '15 at 14:01
$begingroup$
@man_in_green_shirt The conclusion is that $ex-x=xe-x=0$, and therefore $e$ is the identity. It is not the hypothesis...
$endgroup$
– rschwieb
Aug 6 '15 at 14:01
|
show 1 more 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%2f1386528%2fdoes-every-id-with-subring-which-has-a-unity-have-an-unity%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
