If $Q inmathrmSpec(B)$ is the unique prime ideal lying over $P inmathrmSpec(A)$, then $B_P=B_Q$. The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)intersections of powers of primes lying over a prime in a Galois extensionComputing the “lying over”, “going up”, “going down” ideals.If $mathopmathrmSpecA$ is not connected then there is a nontrivial idempotentExistence of prime ideal lying over $p implies pS cap R=p $?intersection of localization with the base fieldQuestion about correspondence of ideals in rings of fractions (if $S^-1q'$ is a maximal ideal containing $S^-1b,$ then must $q'$ contain $b$?)A question of the uniqueness of the prime ideal lying over in an integral extension (Corollary 5.9 in Atiyah's Commutative Algebra)$B subseteq A$ be an integral extension of integral domains where $B$ is normal; if $P in Spec A$ and $Q=Pcap B$, then $dim A_P=dim B_Q$?If $B$ is integral over $A$ and there is only one $P$ over $mathfrak p$, then $B_P=B_mathfrak p$?Checking whether certain lying-over prime is isolated
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If $Q inmathrmSpec(B)$ is the unique prime ideal lying over $P inmathrmSpec(A)$, then $B_P=B_Q$.
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)intersections of powers of primes lying over a prime in a Galois extensionComputing the “lying over”, “going up”, “going down” ideals.If $mathopmathrmSpecA$ is not connected then there is a nontrivial idempotentExistence of prime ideal lying over $p implies pS cap R=p $?intersection of localization with the base fieldQuestion about correspondence of ideals in rings of fractions (if $S^-1q'$ is a maximal ideal containing $S^-1b,$ then must $q'$ contain $b$?)A question of the uniqueness of the prime ideal lying over in an integral extension (Corollary 5.9 in Atiyah's Commutative Algebra)$B subseteq A$ be an integral extension of integral domains where $B$ is normal; if $P in Spec A$ and $Q=Pcap B$, then $dim A_P=dim B_Q$?If $B$ is integral over $A$ and there is only one $P$ over $mathfrak p$, then $B_P=B_mathfrak p$?Checking whether certain lying-over prime is isolated
$begingroup$
Let $A subseteq B$ be an integral extension of integral domains. Suppose $P in mathrmSpec(A)$ and $Q in mathrmSpec(B)$ is the unique prime ideal lying above $P$. Prove that $B_Q=B_P$.
My attempt: since $P=Qcap A subseteq Q$, we have $B_Q subseteq B_P$. For the converse, suppose $b/s in B_P$. Then $s notin P$. Since $P=Q cap A$, we have $s notin Q cap A$. That is, $s notin Q$ or $s notin A$. If it's the former, then that's what we want. If it's the latter, how should I arrive at a contradiction?
I don't know how to use the condition that $B$ is integral over $A$. If $s$ satisfies the equation $s^n+a_n-1s^n-1+cdots+a_1s+a_0=0$ for some $a_n-1,ldots,a_0 in A$. What can that do for me?
Can anyone give me a hint? Thank you!
commutative-algebra localization integral-extensions
edited Apr 8 at 8:50
user26857
39.5k124284
asked Apr 8 at 8:13
bbwbbw
53739
$endgroup$
add a comment |
$begingroup$
Let $A subseteq B$ be an integral extension of integral domains. Suppose $P in mathrmSpec(A)$ and $Q in mathrmSpec(B)$ is the unique prime ideal lying above $P$. Prove that $B_Q=B_P$.
My attempt: since $P=Qcap A subseteq Q$, we have $B_Q subseteq B_P$. For the converse, suppose $b/s in B_P$. Then $s notin P$. Since $P=Q cap A$, we have $s notin Q cap A$. That is, $s notin Q$ or $s notin A$. If it's the former, then that's what we want. If it's the latter, how should I arrive at a contradiction?
I don't know how to use the condition that $B$ is integral over $A$. If $s$ satisfies the equation $s^n+a_n-1s^n-1+cdots+a_1s+a_0=0$ for some $a_n-1,ldots,a_0 in A$. What can that do for me?
Can anyone give me a hint? Thank you!
commutative-algebra localization integral-extensions
edited Apr 8 at 8:50
user26857
39.5k124284
asked Apr 8 at 8:13
bbwbbw
53739
$endgroup$
$begingroup$
Actually it is obvious that $B_Psubseteq B_Q$. The other containement has to be proved.
$endgroup$
– user26857
Apr 8 at 13:50
add a comment |
$begingroup$
Let $A subseteq B$ be an integral extension of integral domains. Suppose $P in mathrmSpec(A)$ and $Q in mathrmSpec(B)$ is the unique prime ideal lying above $P$. Prove that $B_Q=B_P$.
My attempt: since $P=Qcap A subseteq Q$, we have $B_Q subseteq B_P$. For the converse, suppose $b/s in B_P$. Then $s notin P$. Since $P=Q cap A$, we have $s notin Q cap A$. That is, $s notin Q$ or $s notin A$. If it's the former, then that's what we want. If it's the latter, how should I arrive at a contradiction?
I don't know how to use the condition that $B$ is integral over $A$. If $s$ satisfies the equation $s^n+a_n-1s^n-1+cdots+a_1s+a_0=0$ for some $a_n-1,ldots,a_0 in A$. What can that do for me?
Can anyone give me a hint? Thank you!
commutative-algebra localization integral-extensions
edited Apr 8 at 8:50
user26857
39.5k124284
asked Apr 8 at 8:13
bbwbbw
53739
$endgroup$
Let $A subseteq B$ be an integral extension of integral domains. Suppose $P in mathrmSpec(A)$ and $Q in mathrmSpec(B)$ is the unique prime ideal lying above $P$. Prove that $B_Q=B_P$.
My attempt: since $P=Qcap A subseteq Q$, we have $B_Q subseteq B_P$. For the converse, suppose $b/s in B_P$. Then $s notin P$. Since $P=Q cap A$, we have $s notin Q cap A$. That is, $s notin Q$ or $s notin A$. If it's the former, then that's what we want. If it's the latter, how should I arrive at a contradiction?
I don't know how to use the condition that $B$ is integral over $A$. If $s$ satisfies the equation $s^n+a_n-1s^n-1+cdots+a_1s+a_0=0$ for some $a_n-1,ldots,a_0 in A$. What can that do for me?
Can anyone give me a hint? Thank you!
commutative-algebra localization integral-extensions
commutative-algebra localization integral-extensions
edited Apr 8 at 8:50
user26857
39.5k124284
asked Apr 8 at 8:13
bbwbbw
53739
edited Apr 8 at 8:50
user26857
39.5k124284
asked Apr 8 at 8:13
bbwbbw
53739
edited Apr 8 at 8:50
user26857
39.5k124284
edited Apr 8 at 8:50
user26857
39.5k124284
edited Apr 8 at 8:50
user26857
39.5k124284
39.5k124284
asked Apr 8 at 8:13
bbwbbw
53739
asked Apr 8 at 8:13
bbwbbw
53739
asked Apr 8 at 8:13
bbwbbw
53739
53739
$begingroup$
Actually it is obvious that $B_Psubseteq B_Q$. The other containement has to be proved.
$endgroup$
– user26857
Apr 8 at 13:50
add a comment |
$begingroup$
Actually it is obvious that $B_Psubseteq B_Q$. The other containement has to be proved.
$endgroup$
– user26857
Apr 8 at 13:50
$begingroup$
Actually it is obvious that $B_Psubseteq B_Q$. The other containement has to be proved.
$endgroup$
– user26857
Apr 8 at 13:50
$begingroup$
Actually it is obvious that $B_Psubseteq B_Q$. The other containement has to be proved.
$endgroup$
– user26857
Apr 8 at 13:50
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Look at $B_P|_A_P$ which is an integral extension. The fact that only one prime lies over $PA_P$ implies $B_P$ is local. Also observe that $QB_P$ has to be the unique maximal ideal of $B_P$. Thus anything outside $Q$ is a unit in $B_P$ and hence $B_Q$ =$B_P$
answered Apr 8 at 8:45
Soumik GhoshSoumik Ghosh
1,298112
$endgroup$
$begingroup$
Why does $B_P$ is local implies $B_Q=B_P$? I couldn't follow this step.
$endgroup$
– bbw
Apr 8 at 9:09
$begingroup$
@bbw $B_Psubset B_Q$ since you are inverting more elements. Now take any element $s$ with $s in B-Q$. So $s$ is outside $QB_P$ and hence a unit in $B_P$ Thus all elements of $B-Q$ are already inverted.
$endgroup$
– Soumik Ghosh
Apr 8 at 9:20
add a comment |
$begingroup$
Clearly $B_Psubseteq B_Q$. In order to prove that $B_P=B_Q$ notice that $B_Q$ is the localization of $B_P$ at $QB_P$. Now let's prove that $B_P$ is a local ring and its maximal ideal is $QB_P$. Any prime ideal of $B_P$ is of the form $Q'B_P$ with $Q'cap Asubseteq P$. Since $Q$ is the only prime ideal lying over $P$ we have $Q'cap Asubsetneq P$ provided $Q'ne Q$. By Going Up there is a prime ideal $Q''$ of $B$ (strictly) containing $Q'$ and lying over $P$. By hypothesis $Q''=Q$, and thus $Q'subsetneq Q$ proving that $QB_P$ is the only maximal ideal.
answered Apr 8 at 14:06
user26857user26857
39.5k124284
$endgroup$
$begingroup$
Why is $B_Q$ the localization of $B_P$ at $QB_P$? $B_Q$ is $(B setminus Q)^-1 B$ and the other is $(B_P setminus QB_P)^-1B_P$. And I still don't understand why $B_P$ is local $Rightarrow$ $B_P=B_Q$. Can you explain a little bit more?
$endgroup$
– bbw
Apr 8 at 17:36
$begingroup$
If $S=Asetminus P$, then $Qcap S=emptyset$ and a well known property of localization says that $(S^-1B)_S^-1Q$ is isomorphic to $B_Q$. But $S^-1B=B_P$, and $S^-1Q=QB_P$.
$endgroup$
– user26857
Apr 8 at 20:07
$begingroup$
If $R$ is a local ring with maximal ideal $mathfrak m$ then $R_mathfrak m=R$ since every element in $Rsetminusmathfrak m$ is invertible in $R$.
$endgroup$
– user26857
Apr 8 at 20:09
$begingroup$
Thank you for you explanation!
$endgroup$
– bbw
Apr 8 at 23:38
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Look at $B_P|_A_P$ which is an integral extension. The fact that only one prime lies over $PA_P$ implies $B_P$ is local. Also observe that $QB_P$ has to be the unique maximal ideal of $B_P$. Thus anything outside $Q$ is a unit in $B_P$ and hence $B_Q$ =$B_P$
answered Apr 8 at 8:45
Soumik GhoshSoumik Ghosh
1,298112
$endgroup$
$begingroup$
Why does $B_P$ is local implies $B_Q=B_P$? I couldn't follow this step.
$endgroup$
– bbw
Apr 8 at 9:09
$begingroup$
@bbw $B_Psubset B_Q$ since you are inverting more elements. Now take any element $s$ with $s in B-Q$. So $s$ is outside $QB_P$ and hence a unit in $B_P$ Thus all elements of $B-Q$ are already inverted.
$endgroup$
– Soumik Ghosh
Apr 8 at 9:20
add a comment |
$begingroup$
Look at $B_P|_A_P$ which is an integral extension. The fact that only one prime lies over $PA_P$ implies $B_P$ is local. Also observe that $QB_P$ has to be the unique maximal ideal of $B_P$. Thus anything outside $Q$ is a unit in $B_P$ and hence $B_Q$ =$B_P$
answered Apr 8 at 8:45
Soumik GhoshSoumik Ghosh
1,298112
$endgroup$
$begingroup$
Why does $B_P$ is local implies $B_Q=B_P$? I couldn't follow this step.
$endgroup$
– bbw
Apr 8 at 9:09
$begingroup$
@bbw $B_Psubset B_Q$ since you are inverting more elements. Now take any element $s$ with $s in B-Q$. So $s$ is outside $QB_P$ and hence a unit in $B_P$ Thus all elements of $B-Q$ are already inverted.
$endgroup$
– Soumik Ghosh
Apr 8 at 9:20
add a comment |
$begingroup$
Look at $B_P|_A_P$ which is an integral extension. The fact that only one prime lies over $PA_P$ implies $B_P$ is local. Also observe that $QB_P$ has to be the unique maximal ideal of $B_P$. Thus anything outside $Q$ is a unit in $B_P$ and hence $B_Q$ =$B_P$
answered Apr 8 at 8:45
Soumik GhoshSoumik Ghosh
1,298112
$endgroup$
Look at $B_P|_A_P$ which is an integral extension. The fact that only one prime lies over $PA_P$ implies $B_P$ is local. Also observe that $QB_P$ has to be the unique maximal ideal of $B_P$. Thus anything outside $Q$ is a unit in $B_P$ and hence $B_Q$ =$B_P$
answered Apr 8 at 8:45
Soumik GhoshSoumik Ghosh
1,298112
answered Apr 8 at 8:45
Soumik GhoshSoumik Ghosh
1,298112
answered Apr 8 at 8:45
Soumik GhoshSoumik Ghosh
1,298112
answered Apr 8 at 8:45
Soumik GhoshSoumik Ghosh
1,298112
1,298112
$begingroup$
Why does $B_P$ is local implies $B_Q=B_P$? I couldn't follow this step.
$endgroup$
– bbw
Apr 8 at 9:09
$begingroup$
@bbw $B_Psubset B_Q$ since you are inverting more elements. Now take any element $s$ with $s in B-Q$. So $s$ is outside $QB_P$ and hence a unit in $B_P$ Thus all elements of $B-Q$ are already inverted.
$endgroup$
– Soumik Ghosh
Apr 8 at 9:20
add a comment |
$begingroup$
Why does $B_P$ is local implies $B_Q=B_P$? I couldn't follow this step.
$endgroup$
– bbw
Apr 8 at 9:09
$begingroup$
@bbw $B_Psubset B_Q$ since you are inverting more elements. Now take any element $s$ with $s in B-Q$. So $s$ is outside $QB_P$ and hence a unit in $B_P$ Thus all elements of $B-Q$ are already inverted.
$endgroup$
– Soumik Ghosh
Apr 8 at 9:20
$begingroup$
Why does $B_P$ is local implies $B_Q=B_P$? I couldn't follow this step.
$endgroup$
– bbw
Apr 8 at 9:09
$begingroup$
Why does $B_P$ is local implies $B_Q=B_P$? I couldn't follow this step.
$endgroup$
– bbw
Apr 8 at 9:09
$begingroup$
@bbw $B_Psubset B_Q$ since you are inverting more elements. Now take any element $s$ with $s in B-Q$. So $s$ is outside $QB_P$ and hence a unit in $B_P$ Thus all elements of $B-Q$ are already inverted.
$endgroup$
– Soumik Ghosh
Apr 8 at 9:20
$begingroup$
@bbw $B_Psubset B_Q$ since you are inverting more elements. Now take any element $s$ with $s in B-Q$. So $s$ is outside $QB_P$ and hence a unit in $B_P$ Thus all elements of $B-Q$ are already inverted.
$endgroup$
– Soumik Ghosh
Apr 8 at 9:20
add a comment |
$begingroup$
Clearly $B_Psubseteq B_Q$. In order to prove that $B_P=B_Q$ notice that $B_Q$ is the localization of $B_P$ at $QB_P$. Now let's prove that $B_P$ is a local ring and its maximal ideal is $QB_P$. Any prime ideal of $B_P$ is of the form $Q'B_P$ with $Q'cap Asubseteq P$. Since $Q$ is the only prime ideal lying over $P$ we have $Q'cap Asubsetneq P$ provided $Q'ne Q$. By Going Up there is a prime ideal $Q''$ of $B$ (strictly) containing $Q'$ and lying over $P$. By hypothesis $Q''=Q$, and thus $Q'subsetneq Q$ proving that $QB_P$ is the only maximal ideal.
answered Apr 8 at 14:06
user26857user26857
39.5k124284
$endgroup$
$begingroup$
Why is $B_Q$ the localization of $B_P$ at $QB_P$? $B_Q$ is $(B setminus Q)^-1 B$ and the other is $(B_P setminus QB_P)^-1B_P$. And I still don't understand why $B_P$ is local $Rightarrow$ $B_P=B_Q$. Can you explain a little bit more?
$endgroup$
– bbw
Apr 8 at 17:36
$begingroup$
If $S=Asetminus P$, then $Qcap S=emptyset$ and a well known property of localization says that $(S^-1B)_S^-1Q$ is isomorphic to $B_Q$. But $S^-1B=B_P$, and $S^-1Q=QB_P$.
$endgroup$
– user26857
Apr 8 at 20:07
$begingroup$
If $R$ is a local ring with maximal ideal $mathfrak m$ then $R_mathfrak m=R$ since every element in $Rsetminusmathfrak m$ is invertible in $R$.
$endgroup$
– user26857
Apr 8 at 20:09
$begingroup$
Thank you for you explanation!
$endgroup$
– bbw
Apr 8 at 23:38
add a comment |
$begingroup$
Clearly $B_Psubseteq B_Q$. In order to prove that $B_P=B_Q$ notice that $B_Q$ is the localization of $B_P$ at $QB_P$. Now let's prove that $B_P$ is a local ring and its maximal ideal is $QB_P$. Any prime ideal of $B_P$ is of the form $Q'B_P$ with $Q'cap Asubseteq P$. Since $Q$ is the only prime ideal lying over $P$ we have $Q'cap Asubsetneq P$ provided $Q'ne Q$. By Going Up there is a prime ideal $Q''$ of $B$ (strictly) containing $Q'$ and lying over $P$. By hypothesis $Q''=Q$, and thus $Q'subsetneq Q$ proving that $QB_P$ is the only maximal ideal.
answered Apr 8 at 14:06
user26857user26857
39.5k124284
$endgroup$
$begingroup$
Why is $B_Q$ the localization of $B_P$ at $QB_P$? $B_Q$ is $(B setminus Q)^-1 B$ and the other is $(B_P setminus QB_P)^-1B_P$. And I still don't understand why $B_P$ is local $Rightarrow$ $B_P=B_Q$. Can you explain a little bit more?
$endgroup$
– bbw
Apr 8 at 17:36
$begingroup$
If $S=Asetminus P$, then $Qcap S=emptyset$ and a well known property of localization says that $(S^-1B)_S^-1Q$ is isomorphic to $B_Q$. But $S^-1B=B_P$, and $S^-1Q=QB_P$.
$endgroup$
– user26857
Apr 8 at 20:07
$begingroup$
If $R$ is a local ring with maximal ideal $mathfrak m$ then $R_mathfrak m=R$ since every element in $Rsetminusmathfrak m$ is invertible in $R$.
$endgroup$
– user26857
Apr 8 at 20:09
$begingroup$
Thank you for you explanation!
$endgroup$
– bbw
Apr 8 at 23:38
add a comment |
$begingroup$
Clearly $B_Psubseteq B_Q$. In order to prove that $B_P=B_Q$ notice that $B_Q$ is the localization of $B_P$ at $QB_P$. Now let's prove that $B_P$ is a local ring and its maximal ideal is $QB_P$. Any prime ideal of $B_P$ is of the form $Q'B_P$ with $Q'cap Asubseteq P$. Since $Q$ is the only prime ideal lying over $P$ we have $Q'cap Asubsetneq P$ provided $Q'ne Q$. By Going Up there is a prime ideal $Q''$ of $B$ (strictly) containing $Q'$ and lying over $P$. By hypothesis $Q''=Q$, and thus $Q'subsetneq Q$ proving that $QB_P$ is the only maximal ideal.
answered Apr 8 at 14:06
user26857user26857
39.5k124284
$endgroup$
Clearly $B_Psubseteq B_Q$. In order to prove that $B_P=B_Q$ notice that $B_Q$ is the localization of $B_P$ at $QB_P$. Now let's prove that $B_P$ is a local ring and its maximal ideal is $QB_P$. Any prime ideal of $B_P$ is of the form $Q'B_P$ with $Q'cap Asubseteq P$. Since $Q$ is the only prime ideal lying over $P$ we have $Q'cap Asubsetneq P$ provided $Q'ne Q$. By Going Up there is a prime ideal $Q''$ of $B$ (strictly) containing $Q'$ and lying over $P$. By hypothesis $Q''=Q$, and thus $Q'subsetneq Q$ proving that $QB_P$ is the only maximal ideal.
answered Apr 8 at 14:06
user26857user26857
39.5k124284
edited Apr 8 at 14:12
answered Apr 8 at 14:06
user26857user26857
39.5k124284
answered Apr 8 at 14:06
user26857user26857
39.5k124284
answered Apr 8 at 14:06
user26857user26857
39.5k124284
39.5k124284
$begingroup$
Why is $B_Q$ the localization of $B_P$ at $QB_P$? $B_Q$ is $(B setminus Q)^-1 B$ and the other is $(B_P setminus QB_P)^-1B_P$. And I still don't understand why $B_P$ is local $Rightarrow$ $B_P=B_Q$. Can you explain a little bit more?
$endgroup$
– bbw
Apr 8 at 17:36
$begingroup$
If $S=Asetminus P$, then $Qcap S=emptyset$ and a well known property of localization says that $(S^-1B)_S^-1Q$ is isomorphic to $B_Q$. But $S^-1B=B_P$, and $S^-1Q=QB_P$.
$endgroup$
– user26857
Apr 8 at 20:07
$begingroup$
If $R$ is a local ring with maximal ideal $mathfrak m$ then $R_mathfrak m=R$ since every element in $Rsetminusmathfrak m$ is invertible in $R$.
$endgroup$
– user26857
Apr 8 at 20:09
$begingroup$
Thank you for you explanation!
$endgroup$
– bbw
Apr 8 at 23:38
add a comment |
$begingroup$
Why is $B_Q$ the localization of $B_P$ at $QB_P$? $B_Q$ is $(B setminus Q)^-1 B$ and the other is $(B_P setminus QB_P)^-1B_P$. And I still don't understand why $B_P$ is local $Rightarrow$ $B_P=B_Q$. Can you explain a little bit more?
$endgroup$
– bbw
Apr 8 at 17:36
$begingroup$
If $S=Asetminus P$, then $Qcap S=emptyset$ and a well known property of localization says that $(S^-1B)_S^-1Q$ is isomorphic to $B_Q$. But $S^-1B=B_P$, and $S^-1Q=QB_P$.
$endgroup$
– user26857
Apr 8 at 20:07
$begingroup$
If $R$ is a local ring with maximal ideal $mathfrak m$ then $R_mathfrak m=R$ since every element in $Rsetminusmathfrak m$ is invertible in $R$.
$endgroup$
– user26857
Apr 8 at 20:09
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Thank you for you explanation!
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– bbw
Apr 8 at 23:38
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Why is $B_Q$ the localization of $B_P$ at $QB_P$? $B_Q$ is $(B setminus Q)^-1 B$ and the other is $(B_P setminus QB_P)^-1B_P$. And I still don't understand why $B_P$ is local $Rightarrow$ $B_P=B_Q$. Can you explain a little bit more?
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– bbw
Apr 8 at 17:36
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Why is $B_Q$ the localization of $B_P$ at $QB_P$? $B_Q$ is $(B setminus Q)^-1 B$ and the other is $(B_P setminus QB_P)^-1B_P$. And I still don't understand why $B_P$ is local $Rightarrow$ $B_P=B_Q$. Can you explain a little bit more?
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– bbw
Apr 8 at 17:36
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If $S=Asetminus P$, then $Qcap S=emptyset$ and a well known property of localization says that $(S^-1B)_S^-1Q$ is isomorphic to $B_Q$. But $S^-1B=B_P$, and $S^-1Q=QB_P$.
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– user26857
Apr 8 at 20:07
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If $S=Asetminus P$, then $Qcap S=emptyset$ and a well known property of localization says that $(S^-1B)_S^-1Q$ is isomorphic to $B_Q$. But $S^-1B=B_P$, and $S^-1Q=QB_P$.
$endgroup$
– user26857
Apr 8 at 20:07
$begingroup$
If $R$ is a local ring with maximal ideal $mathfrak m$ then $R_mathfrak m=R$ since every element in $Rsetminusmathfrak m$ is invertible in $R$.
$endgroup$
– user26857
Apr 8 at 20:09
$begingroup$
If $R$ is a local ring with maximal ideal $mathfrak m$ then $R_mathfrak m=R$ since every element in $Rsetminusmathfrak m$ is invertible in $R$.
$endgroup$
– user26857
Apr 8 at 20:09
$begingroup$
Thank you for you explanation!
$endgroup$
– bbw
Apr 8 at 23:38
$begingroup$
Thank you for you explanation!
$endgroup$
– bbw
Apr 8 at 23:38
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Actually it is obvious that $B_Psubseteq B_Q$. The other containement has to be proved.
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– user26857
Apr 8 at 13:50