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Show that $S^-1A = B$ for integral domains and $S = xin Asetminus0: x^-1in B$ [duplicate]
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Subrings of fraction fields [closed]
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Let $A subset B$ be commutative integral domains with $operatornameQuot(A) = operatornameQuot(B).$
Now consider the multiplicatively closed subset $S = xin Asetminus0: x^-1in B$.
I want to show that $S^-1A = B$. I would appreciate any help.
ring-theory commutative-algebra localization
edited Jul 31 at 17:09
Bernard
110k635102
asked Jul 31 at 16:57
mathcourse
355110
marked as duplicate by Claude Leibovici, John Ma, max_zorn, Taroccoesbrocco, BLAZE Aug 4 at 5:09
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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up vote
1
down vote
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This question already has an answer here:
Subrings of fraction fields [closed]
4 answers
Let $A subset B$ be commutative integral domains with $operatornameQuot(A) = operatornameQuot(B).$
Now consider the multiplicatively closed subset $S = xin Asetminus0: x^-1in B$.
I want to show that $S^-1A = B$. I would appreciate any help.
ring-theory commutative-algebra localization
edited Jul 31 at 17:09
Bernard
110k635102
asked Jul 31 at 16:57
mathcourse
355110
marked as duplicate by Claude Leibovici, John Ma, max_zorn, Taroccoesbrocco, BLAZE Aug 4 at 5:09
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
Isn't $A = K[x,y]$, $B=K[x,y, y/x]$ a counter-example? As far as I can tell, it seems that $S = A^*$ in this case, so that $S^-1 A = A neq B$ in this case.
– Marc Paul
Jul 31 at 20:35
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This question already has an answer here:
Subrings of fraction fields [closed]
4 answers
Let $A subset B$ be commutative integral domains with $operatornameQuot(A) = operatornameQuot(B).$
Now consider the multiplicatively closed subset $S = xin Asetminus0: x^-1in B$.
I want to show that $S^-1A = B$. I would appreciate any help.
ring-theory commutative-algebra localization
edited Jul 31 at 17:09
Bernard
110k635102
asked Jul 31 at 16:57
mathcourse
355110
This question already has an answer here:
Subrings of fraction fields [closed]
4 answers
Let $A subset B$ be commutative integral domains with $operatornameQuot(A) = operatornameQuot(B).$
Now consider the multiplicatively closed subset $S = xin Asetminus0: x^-1in B$.
I want to show that $S^-1A = B$. I would appreciate any help.
This question already has an answer here:
Subrings of fraction fields [closed]
4 answers
ring-theory commutative-algebra localization
edited Jul 31 at 17:09
Bernard
110k635102
asked Jul 31 at 16:57
mathcourse
355110
edited Jul 31 at 17:09
Bernard
110k635102
edited Jul 31 at 17:09
Bernard
110k635102
edited Jul 31 at 17:09
Bernard
110k635102
110k635102
asked Jul 31 at 16:57
mathcourse
355110
asked Jul 31 at 16:57
mathcourse
355110
asked Jul 31 at 16:57
mathcourse
355110
355110
marked as duplicate by Claude Leibovici, John Ma, max_zorn, Taroccoesbrocco, BLAZE Aug 4 at 5:09
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Claude Leibovici, John Ma, max_zorn, Taroccoesbrocco, BLAZE Aug 4 at 5:09
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
Isn't $A = K[x,y]$, $B=K[x,y, y/x]$ a counter-example? As far as I can tell, it seems that $S = A^*$ in this case, so that $S^-1 A = A neq B$ in this case.
– Marc Paul
Jul 31 at 20:35
add a comment |Â
1
Isn't $A = K[x,y]$, $B=K[x,y, y/x]$ a counter-example? As far as I can tell, it seems that $S = A^*$ in this case, so that $S^-1 A = A neq B$ in this case.
– Marc Paul
Jul 31 at 20:35
1
1
Isn't $A = K[x,y]$, $B=K[x,y, y/x]$ a counter-example? As far as I can tell, it seems that $S = A^*$ in this case, so that $S^-1 A = A neq B$ in this case.
– Marc Paul
Jul 31 at 20:35
Isn't $A = K[x,y]$, $B=K[x,y, y/x]$ a counter-example? As far as I can tell, it seems that $S = A^*$ in this case, so that $S^-1 A = A neq B$ in this case.
– Marc Paul
Jul 31 at 20:35
add a comment |Â
1 Answer
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It suffices to prove $Bsubseteq S^-1A$, since we already have $Asubseteq Bsubseteq mathrmFrac(A)$.
For any $xin B$, $x=fracab$ for some $a,bin A$. By definition, correct.
answered Aug 1 at 20:25
Li Guanyu
144
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
-1
down vote
It suffices to prove $Bsubseteq S^-1A$, since we already have $Asubseteq Bsubseteq mathrmFrac(A)$.
For any $xin B$, $x=fracab$ for some $a,bin A$. By definition, correct.
answered Aug 1 at 20:25
Li Guanyu
144
add a comment |Â
up vote
-1
down vote
It suffices to prove $Bsubseteq S^-1A$, since we already have $Asubseteq Bsubseteq mathrmFrac(A)$.
For any $xin B$, $x=fracab$ for some $a,bin A$. By definition, correct.
answered Aug 1 at 20:25
Li Guanyu
144
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
It suffices to prove $Bsubseteq S^-1A$, since we already have $Asubseteq Bsubseteq mathrmFrac(A)$.
For any $xin B$, $x=fracab$ for some $a,bin A$. By definition, correct.
answered Aug 1 at 20:25
Li Guanyu
144
It suffices to prove $Bsubseteq S^-1A$, since we already have $Asubseteq Bsubseteq mathrmFrac(A)$.
For any $xin B$, $x=fracab$ for some $a,bin A$. By definition, correct.
answered Aug 1 at 20:25
Li Guanyu
144
answered Aug 1 at 20:25
Li Guanyu
144
answered Aug 1 at 20:25
Li Guanyu
144
answered Aug 1 at 20:25
Li Guanyu
144
144
add a comment |Â
add a comment |Â
1
Isn't $A = K[x,y]$, $B=K[x,y, y/x]$ a counter-example? As far as I can tell, it seems that $S = A^*$ in this case, so that $S^-1 A = A neq B$ in this case.
– Marc Paul
Jul 31 at 20:35