$R$ is a noetherian domain and $K=operatorname{Frac}(R)$. If $S$ subring s.t. $Rsubset Ssubset K$, is $S$...
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$textbf{Q:}$ 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
117k637109
asked Nov 26 at 21:32
user45765
2,4522721
add a comment |
up vote
2
down vote
favorite
$textbf{Q:}$ 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
117k637109
asked Nov 26 at 21:32
user45765
2,4522721
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 |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
$textbf{Q:}$ 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
117k637109
asked Nov 26 at 21:32
user45765
2,4522721
$textbf{Q:}$ 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
117k637109
asked Nov 26 at 21:32
user45765
2,4522721
edited Nov 26 at 21:35
Bernard
117k637109
asked Nov 26 at 21:32
user45765
2,4522721
edited Nov 26 at 21:35
Bernard
117k637109
edited Nov 26 at 21:35
Bernard
117k637109
edited Nov 26 at 21:35
Bernard
117k637109
117k637109
asked Nov 26 at 21:32
user45765
2,4522721
asked Nov 26 at 21:32
user45765
2,4522721
asked Nov 26 at 21:32
user45765
2,4522721
2,4522721
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 |
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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