$R$ is a noetherian domain and $K=operatorname{Frac}(R)$. If $S$ subring s.t. $Rsubset Ssubset K$, is $S$...
2
$textbf{Q:}$ Suppose $R$ is noetherian domain and $K=operatorname{Frac}(R)$. If $S$ is a subring of $K$ s.t. $Rsubset Ssubset K$, is $S$ noetherian? I suspect this will be the case in general. One can consider $(R:_R I)$ with $S$-ideal $Isubset S$. There is no good reason that $(R:_R I)neq 0$. Furthermore, there is no good reason that $S$ being localization of $R$ in general unless $R$ is Bézout.
abstract-algebra commutative-algebra
edited Nov 26 at 21:35
Bernard
118k639112
asked Nov 26 at 21:32
user45765
2,5192721
add a comment |
2
$textbf{Q:}$ Suppose $R$ is noetherian domain and $K=operatorname{Frac}(R)$. If $S$ is a subring of $K$ s.t. $Rsubset Ssubset K$, is $S$ noetherian? I suspect this will be the case in general. One can consider $(R:_R I)$ with $S$-ideal $Isubset S$. There is no good reason that $(R:_R I)neq 0$. Furthermore, there is no good reason that $S$ being localization of $R$ in general unless $R$ is Bézout.
abstract-algebra commutative-algebra
edited Nov 26 at 21:35
Bernard
118k639112
asked Nov 26 at 21:32
user45765
2,5192721
Is it not true that every intermediate ring is obtained by a localization? Say, by localizing $R$ at the set of units of $S$? Maybe it is not true. But if it were true, then $S$ would be Noetherian (as a localization of $R$) . Maybe $S$ can be weird so that the localization at its units is properly contained in $S$.
– rschwieb
Nov 26 at 21:43
1
@rschwieb I think I asked a question on this before and someone gave a concrete example that an intermediate ring is not localization. However, not being localization does not say anything about noetherian property of a subring that is not localization. I think there is a loop hole in my logic somewhere. I was expecting the quotient ideal non-trivial but I cannot see any good reason that it is not trivial.
– user45765
Nov 26 at 21:47
5
The answer to both claims (every subring of field of fractions is a localization and every intermediate ring is noetherian) is no: math.stackexchange.com/questions/287245/…
– freakish
Nov 26 at 21:51
@freakish glad I asked: thanks for the link
– rschwieb
Nov 27 at 1:04
add a comment |
2
2
2
$textbf{Q:}$ Suppose $R$ is noetherian domain and $K=operatorname{Frac}(R)$. If $S$ is a subring of $K$ s.t. $Rsubset Ssubset K$, is $S$ noetherian? I suspect this will be the case in general. One can consider $(R:_R I)$ with $S$-ideal $Isubset S$. There is no good reason that $(R:_R I)neq 0$. Furthermore, there is no good reason that $S$ being localization of $R$ in general unless $R$ is Bézout.
abstract-algebra commutative-algebra
edited Nov 26 at 21:35
Bernard
118k639112
asked Nov 26 at 21:32
user45765
2,5192721
$textbf{Q:}$ Suppose $R$ is noetherian domain and $K=operatorname{Frac}(R)$. If $S$ is a subring of $K$ s.t. $Rsubset Ssubset K$, is $S$ noetherian? I suspect this will be the case in general. One can consider $(R:_R I)$ with $S$-ideal $Isubset S$. There is no good reason that $(R:_R I)neq 0$. Furthermore, there is no good reason that $S$ being localization of $R$ in general unless $R$ is Bézout.
abstract-algebra commutative-algebra
abstract-algebra commutative-algebra
edited Nov 26 at 21:35
Bernard
118k639112
asked Nov 26 at 21:32
user45765
2,5192721
edited Nov 26 at 21:35
Bernard
118k639112
asked Nov 26 at 21:32
user45765
2,5192721
edited Nov 26 at 21:35
Bernard
118k639112
edited Nov 26 at 21:35
Bernard
118k639112
edited Nov 26 at 21:35
Bernard
118k639112
118k639112
asked Nov 26 at 21:32
user45765
2,5192721
asked Nov 26 at 21:32
user45765
2,5192721
asked Nov 26 at 21:32
user45765
2,5192721
2,5192721
Is it not true that every intermediate ring is obtained by a localization? Say, by localizing $R$ at the set of units of $S$? Maybe it is not true. But if it were true, then $S$ would be Noetherian (as a localization of $R$) . Maybe $S$ can be weird so that the localization at its units is properly contained in $S$.
– rschwieb
Nov 26 at 21:43
1
@rschwieb I think I asked a question on this before and someone gave a concrete example that an intermediate ring is not localization. However, not being localization does not say anything about noetherian property of a subring that is not localization. I think there is a loop hole in my logic somewhere. I was expecting the quotient ideal non-trivial but I cannot see any good reason that it is not trivial.
– user45765
Nov 26 at 21:47
5
The answer to both claims (every subring of field of fractions is a localization and every intermediate ring is noetherian) is no: math.stackexchange.com/questions/287245/…
– freakish
Nov 26 at 21:51
@freakish glad I asked: thanks for the link
– rschwieb
Nov 27 at 1:04
add a comment |
Is it not true that every intermediate ring is obtained by a localization? Say, by localizing $R$ at the set of units of $S$? Maybe it is not true. But if it were true, then $S$ would be Noetherian (as a localization of $R$) . Maybe $S$ can be weird so that the localization at its units is properly contained in $S$.
– rschwieb
Nov 26 at 21:43
1
@rschwieb I think I asked a question on this before and someone gave a concrete example that an intermediate ring is not localization. However, not being localization does not say anything about noetherian property of a subring that is not localization. I think there is a loop hole in my logic somewhere. I was expecting the quotient ideal non-trivial but I cannot see any good reason that it is not trivial.
– user45765
Nov 26 at 21:47
5
The answer to both claims (every subring of field of fractions is a localization and every intermediate ring is noetherian) is no: math.stackexchange.com/questions/287245/…
– freakish
Nov 26 at 21:51
@freakish glad I asked: thanks for the link
– rschwieb
Nov 27 at 1:04
Is it not true that every intermediate ring is obtained by a localization? Say, by localizing $R$ at the set of units of $S$? Maybe it is not true. But if it were true, then $S$ would be Noetherian (as a localization of $R$) . Maybe $S$ can be weird so that the localization at its units is properly contained in $S$.
– rschwieb
Nov 26 at 21:43
Is it not true that every intermediate ring is obtained by a localization? Say, by localizing $R$ at the set of units of $S$? Maybe it is not true. But if it were true, then $S$ would be Noetherian (as a localization of $R$) . Maybe $S$ can be weird so that the localization at its units is properly contained in $S$.
– rschwieb
Nov 26 at 21:43
1
1
@rschwieb I think I asked a question on this before and someone gave a concrete example that an intermediate ring is not localization. However, not being localization does not say anything about noetherian property of a subring that is not localization. I think there is a loop hole in my logic somewhere. I was expecting the quotient ideal non-trivial but I cannot see any good reason that it is not trivial.
– user45765
Nov 26 at 21:47
@rschwieb I think I asked a question on this before and someone gave a concrete example that an intermediate ring is not localization. However, not being localization does not say anything about noetherian property of a subring that is not localization. I think there is a loop hole in my logic somewhere. I was expecting the quotient ideal non-trivial but I cannot see any good reason that it is not trivial.
– user45765
Nov 26 at 21:47
5
5
The answer to both claims (every subring of field of fractions is a localization and every intermediate ring is noetherian) is no: math.stackexchange.com/questions/287245/…
– freakish
Nov 26 at 21:51
The answer to both claims (every subring of field of fractions is a localization and every intermediate ring is noetherian) is no: math.stackexchange.com/questions/287245/…
– freakish
Nov 26 at 21:51
@freakish glad I asked: thanks for the link
– rschwieb
Nov 27 at 1:04
@freakish glad I asked: thanks for the link
– rschwieb
Nov 27 at 1:04
add a comment |
active
oldest
votes
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%2f3014956%2fr-is-a-noetherian-domain-and-k-operatornamefracr-if-s-subring-s-t%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
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.
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%2f3014956%2fr-is-a-noetherian-domain-and-k-operatornamefracr-if-s-subring-s-t%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
Is it not true that every intermediate ring is obtained by a localization? Say, by localizing $R$ at the set of units of $S$? Maybe it is not true. But if it were true, then $S$ would be Noetherian (as a localization of $R$) . Maybe $S$ can be weird so that the localization at its units is properly contained in $S$.
– rschwieb
Nov 26 at 21:43
1
@rschwieb I think I asked a question on this before and someone gave a concrete example that an intermediate ring is not localization. However, not being localization does not say anything about noetherian property of a subring that is not localization. I think there is a loop hole in my logic somewhere. I was expecting the quotient ideal non-trivial but I cannot see any good reason that it is not trivial.
– user45765
Nov 26 at 21:47
5
The answer to both claims (every subring of field of fractions is a localization and every intermediate ring is noetherian) is no: math.stackexchange.com/questions/287245/…
– freakish
Nov 26 at 21:51
@freakish glad I asked: thanks for the link
– rschwieb
Nov 27 at 1:04