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If $B$ is integral over $A$ and there is only one $P$ over $mathfrak p$, then $B_P=B_mathfrak p$?
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I am stuck in the Exercise 5.3 of Matsumura's Commutative Algebra:
$newcommandpmathfrak p$
$newcommandspoperatornameSpec$
Let $B$ be a ring, $A$ be a subring and $pinsp(A)$. Suppose that $B$ is integral over $A$ and that there is only one prime ideal $P$ of $B$ over $p$. Then $B_P=B_p$, where $B_p=Botimes_A A_p$.
In addition, this book gives a hint: show that $B_p$ is a local ring with maximal ideal $PB_p$. I am really confused about this problem, especially in the following points:
What is the exact meaning of $B_P=B_p$? Can I say $B_Psubset B_p$ since $psubset P$?
Even if I have showed that $B_p$ has a unique maximal ideal $PB_p$, how does this fact imply $B_P=B_p$?
I know that the going-up theorem holds for $Ahookrightarrow B$ since $B$ is integral over $A$, but how can the condition that there is only one prime ideal $P$ such that $Pcap A=p$ can be used to show that $(B_p, PB_p)$ is a local ring?
Thus I would like to ask for some explanation and further hints. Thanks in advance...
commutative-algebra
asked Jul 17 at 16:22
josephz
1,3832419
add a comment |Â
up vote
2
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I am stuck in the Exercise 5.3 of Matsumura's Commutative Algebra:
$newcommandpmathfrak p$
$newcommandspoperatornameSpec$
Let $B$ be a ring, $A$ be a subring and $pinsp(A)$. Suppose that $B$ is integral over $A$ and that there is only one prime ideal $P$ of $B$ over $p$. Then $B_P=B_p$, where $B_p=Botimes_A A_p$.
In addition, this book gives a hint: show that $B_p$ is a local ring with maximal ideal $PB_p$. I am really confused about this problem, especially in the following points:
What is the exact meaning of $B_P=B_p$? Can I say $B_Psubset B_p$ since $psubset P$?
Even if I have showed that $B_p$ has a unique maximal ideal $PB_p$, how does this fact imply $B_P=B_p$?
I know that the going-up theorem holds for $Ahookrightarrow B$ since $B$ is integral over $A$, but how can the condition that there is only one prime ideal $P$ such that $Pcap A=p$ can be used to show that $(B_p, PB_p)$ is a local ring?
Thus I would like to ask for some explanation and further hints. Thanks in advance...
commutative-algebra
asked Jul 17 at 16:22
josephz
1,3832419
1
I'm guessing the equal sign in $B_P = B_mathfrak p$ is a ``natural" isomorphism, which I suspect arises from the universal properties of localization and tensor product. If you follow the hint, then I think u.p. of localization gives map $B_mathfrak pto B_P$
– cat
Jul 25 at 19:19
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am stuck in the Exercise 5.3 of Matsumura's Commutative Algebra:
$newcommandpmathfrak p$
$newcommandspoperatornameSpec$
Let $B$ be a ring, $A$ be a subring and $pinsp(A)$. Suppose that $B$ is integral over $A$ and that there is only one prime ideal $P$ of $B$ over $p$. Then $B_P=B_p$, where $B_p=Botimes_A A_p$.
In addition, this book gives a hint: show that $B_p$ is a local ring with maximal ideal $PB_p$. I am really confused about this problem, especially in the following points:
What is the exact meaning of $B_P=B_p$? Can I say $B_Psubset B_p$ since $psubset P$?
Even if I have showed that $B_p$ has a unique maximal ideal $PB_p$, how does this fact imply $B_P=B_p$?
I know that the going-up theorem holds for $Ahookrightarrow B$ since $B$ is integral over $A$, but how can the condition that there is only one prime ideal $P$ such that $Pcap A=p$ can be used to show that $(B_p, PB_p)$ is a local ring?
Thus I would like to ask for some explanation and further hints. Thanks in advance...
commutative-algebra
asked Jul 17 at 16:22
josephz
1,3832419
I am stuck in the Exercise 5.3 of Matsumura's Commutative Algebra:
$newcommandpmathfrak p$
$newcommandspoperatornameSpec$
Let $B$ be a ring, $A$ be a subring and $pinsp(A)$. Suppose that $B$ is integral over $A$ and that there is only one prime ideal $P$ of $B$ over $p$. Then $B_P=B_p$, where $B_p=Botimes_A A_p$.
In addition, this book gives a hint: show that $B_p$ is a local ring with maximal ideal $PB_p$. I am really confused about this problem, especially in the following points:
What is the exact meaning of $B_P=B_p$? Can I say $B_Psubset B_p$ since $psubset P$?
Even if I have showed that $B_p$ has a unique maximal ideal $PB_p$, how does this fact imply $B_P=B_p$?
I know that the going-up theorem holds for $Ahookrightarrow B$ since $B$ is integral over $A$, but how can the condition that there is only one prime ideal $P$ such that $Pcap A=p$ can be used to show that $(B_p, PB_p)$ is a local ring?
Thus I would like to ask for some explanation and further hints. Thanks in advance...
commutative-algebra
asked Jul 17 at 16:22
josephz
1,3832419
asked Jul 17 at 16:22
josephz
1,3832419
asked Jul 17 at 16:22
josephz
1,3832419
asked Jul 17 at 16:22
josephz
1,3832419
1,3832419
1
I'm guessing the equal sign in $B_P = B_mathfrak p$ is a ``natural" isomorphism, which I suspect arises from the universal properties of localization and tensor product. If you follow the hint, then I think u.p. of localization gives map $B_mathfrak pto B_P$
– cat
Jul 25 at 19:19
add a comment |Â
1
I'm guessing the equal sign in $B_P = B_mathfrak p$ is a ``natural" isomorphism, which I suspect arises from the universal properties of localization and tensor product. If you follow the hint, then I think u.p. of localization gives map $B_mathfrak pto B_P$
– cat
Jul 25 at 19:19
1
1
I'm guessing the equal sign in $B_P = B_mathfrak p$ is a ``natural" isomorphism, which I suspect arises from the universal properties of localization and tensor product. If you follow the hint, then I think u.p. of localization gives map $B_mathfrak pto B_P$
– cat
Jul 25 at 19:19
I'm guessing the equal sign in $B_P = B_mathfrak p$ is a ``natural" isomorphism, which I suspect arises from the universal properties of localization and tensor product. If you follow the hint, then I think u.p. of localization gives map $B_mathfrak pto B_P$
– cat
Jul 25 at 19:19
add a comment |Â
1 Answer
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I hope my thinking is right. Here is what I have come up with.
- $(B_mathfrakp,PB_mathfrakp)$ is local:
If we rewrite $B_mathfrakp = S^-1B$ and $A_mathfrakp = S^-1A$ with $S = Asetminus mathfrakp$, we see that since $A subseteq B$ is integral, $A_mathfrakp = S^-1A subseteq S^-1B = B_mathfrakp$ is therefore also integral (and it is still an inclusion!). Let $Q in operatornameSpec B_mathfrak p$ be a prime ideal. Then, since $(A_mathfrak p , mathfrak p A_mathfrak p)$ is local, we have $Q cap A_mathfrak p subseteq mathfrak p A_mathfrak p$, and by going-up, there exists a $Q^prime in operatornameSpec B_mathfrak p$ such that $Q subseteq Q^prime$ and $Q^prime cap A_mathfrak p = mathfrak p A_mathfrak p$.
By ideal correspondence, $Q^prime$ must be of the form $Q^prime = P^prime B_mathfrak p $ with $P^prime in operatornameSpec B$ and $P^prime cap A = mathfrak p$.
By our assumption, we must have $P^prime = P$.
This concludes that $(B_mathfrak p , PB_mathfrak p)$ is local. - Since $P cap A = mathfrak p$, we have an inclusion $Asetminus mathfrak p subseteq B setminus P$
which should imply $requireencloseenclosehorizontalstrikeB_mathfrak p = (Asetminus mathfrak p)^-1B subseteq (B setminus P)^-1B = B_P$EDIT: which gives us a homomorphism of rings $B_mathfrak p = (Asetminus mathfrak p)^-1B rightarrow (B setminus P)^-1B = B_P$ (a priori it is not injective, since we could have zero-divisors in our multiplicative subsets).
I think $B_P = B_mathfrak p$ shortly follows. I hope I didn't overlook anything. Good luck!
answered Jul 25 at 18:28
Berber
657
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I hope my thinking is right. Here is what I have come up with.
- $(B_mathfrakp,PB_mathfrakp)$ is local:
If we rewrite $B_mathfrakp = S^-1B$ and $A_mathfrakp = S^-1A$ with $S = Asetminus mathfrakp$, we see that since $A subseteq B$ is integral, $A_mathfrakp = S^-1A subseteq S^-1B = B_mathfrakp$ is therefore also integral (and it is still an inclusion!). Let $Q in operatornameSpec B_mathfrak p$ be a prime ideal. Then, since $(A_mathfrak p , mathfrak p A_mathfrak p)$ is local, we have $Q cap A_mathfrak p subseteq mathfrak p A_mathfrak p$, and by going-up, there exists a $Q^prime in operatornameSpec B_mathfrak p$ such that $Q subseteq Q^prime$ and $Q^prime cap A_mathfrak p = mathfrak p A_mathfrak p$.
By ideal correspondence, $Q^prime$ must be of the form $Q^prime = P^prime B_mathfrak p $ with $P^prime in operatornameSpec B$ and $P^prime cap A = mathfrak p$.
By our assumption, we must have $P^prime = P$.
This concludes that $(B_mathfrak p , PB_mathfrak p)$ is local. - Since $P cap A = mathfrak p$, we have an inclusion $Asetminus mathfrak p subseteq B setminus P$
which should imply $requireencloseenclosehorizontalstrikeB_mathfrak p = (Asetminus mathfrak p)^-1B subseteq (B setminus P)^-1B = B_P$EDIT: which gives us a homomorphism of rings $B_mathfrak p = (Asetminus mathfrak p)^-1B rightarrow (B setminus P)^-1B = B_P$ (a priori it is not injective, since we could have zero-divisors in our multiplicative subsets).
I think $B_P = B_mathfrak p$ shortly follows. I hope I didn't overlook anything. Good luck!
answered Jul 25 at 18:28
Berber
657
add a comment |Â
up vote
1
down vote
accepted
I hope my thinking is right. Here is what I have come up with.
- $(B_mathfrakp,PB_mathfrakp)$ is local:
If we rewrite $B_mathfrakp = S^-1B$ and $A_mathfrakp = S^-1A$ with $S = Asetminus mathfrakp$, we see that since $A subseteq B$ is integral, $A_mathfrakp = S^-1A subseteq S^-1B = B_mathfrakp$ is therefore also integral (and it is still an inclusion!). Let $Q in operatornameSpec B_mathfrak p$ be a prime ideal. Then, since $(A_mathfrak p , mathfrak p A_mathfrak p)$ is local, we have $Q cap A_mathfrak p subseteq mathfrak p A_mathfrak p$, and by going-up, there exists a $Q^prime in operatornameSpec B_mathfrak p$ such that $Q subseteq Q^prime$ and $Q^prime cap A_mathfrak p = mathfrak p A_mathfrak p$.
By ideal correspondence, $Q^prime$ must be of the form $Q^prime = P^prime B_mathfrak p $ with $P^prime in operatornameSpec B$ and $P^prime cap A = mathfrak p$.
By our assumption, we must have $P^prime = P$.
This concludes that $(B_mathfrak p , PB_mathfrak p)$ is local. - Since $P cap A = mathfrak p$, we have an inclusion $Asetminus mathfrak p subseteq B setminus P$
which should imply $requireencloseenclosehorizontalstrikeB_mathfrak p = (Asetminus mathfrak p)^-1B subseteq (B setminus P)^-1B = B_P$EDIT: which gives us a homomorphism of rings $B_mathfrak p = (Asetminus mathfrak p)^-1B rightarrow (B setminus P)^-1B = B_P$ (a priori it is not injective, since we could have zero-divisors in our multiplicative subsets).
I think $B_P = B_mathfrak p$ shortly follows. I hope I didn't overlook anything. Good luck!
answered Jul 25 at 18:28
Berber
657
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I hope my thinking is right. Here is what I have come up with.
- $(B_mathfrakp,PB_mathfrakp)$ is local:
If we rewrite $B_mathfrakp = S^-1B$ and $A_mathfrakp = S^-1A$ with $S = Asetminus mathfrakp$, we see that since $A subseteq B$ is integral, $A_mathfrakp = S^-1A subseteq S^-1B = B_mathfrakp$ is therefore also integral (and it is still an inclusion!). Let $Q in operatornameSpec B_mathfrak p$ be a prime ideal. Then, since $(A_mathfrak p , mathfrak p A_mathfrak p)$ is local, we have $Q cap A_mathfrak p subseteq mathfrak p A_mathfrak p$, and by going-up, there exists a $Q^prime in operatornameSpec B_mathfrak p$ such that $Q subseteq Q^prime$ and $Q^prime cap A_mathfrak p = mathfrak p A_mathfrak p$.
By ideal correspondence, $Q^prime$ must be of the form $Q^prime = P^prime B_mathfrak p $ with $P^prime in operatornameSpec B$ and $P^prime cap A = mathfrak p$.
By our assumption, we must have $P^prime = P$.
This concludes that $(B_mathfrak p , PB_mathfrak p)$ is local. - Since $P cap A = mathfrak p$, we have an inclusion $Asetminus mathfrak p subseteq B setminus P$
which should imply $requireencloseenclosehorizontalstrikeB_mathfrak p = (Asetminus mathfrak p)^-1B subseteq (B setminus P)^-1B = B_P$EDIT: which gives us a homomorphism of rings $B_mathfrak p = (Asetminus mathfrak p)^-1B rightarrow (B setminus P)^-1B = B_P$ (a priori it is not injective, since we could have zero-divisors in our multiplicative subsets).
I think $B_P = B_mathfrak p$ shortly follows. I hope I didn't overlook anything. Good luck!
answered Jul 25 at 18:28
Berber
657
I hope my thinking is right. Here is what I have come up with.
- $(B_mathfrakp,PB_mathfrakp)$ is local:
If we rewrite $B_mathfrakp = S^-1B$ and $A_mathfrakp = S^-1A$ with $S = Asetminus mathfrakp$, we see that since $A subseteq B$ is integral, $A_mathfrakp = S^-1A subseteq S^-1B = B_mathfrakp$ is therefore also integral (and it is still an inclusion!). Let $Q in operatornameSpec B_mathfrak p$ be a prime ideal. Then, since $(A_mathfrak p , mathfrak p A_mathfrak p)$ is local, we have $Q cap A_mathfrak p subseteq mathfrak p A_mathfrak p$, and by going-up, there exists a $Q^prime in operatornameSpec B_mathfrak p$ such that $Q subseteq Q^prime$ and $Q^prime cap A_mathfrak p = mathfrak p A_mathfrak p$.
By ideal correspondence, $Q^prime$ must be of the form $Q^prime = P^prime B_mathfrak p $ with $P^prime in operatornameSpec B$ and $P^prime cap A = mathfrak p$.
By our assumption, we must have $P^prime = P$.
This concludes that $(B_mathfrak p , PB_mathfrak p)$ is local. - Since $P cap A = mathfrak p$, we have an inclusion $Asetminus mathfrak p subseteq B setminus P$
which should imply $requireencloseenclosehorizontalstrikeB_mathfrak p = (Asetminus mathfrak p)^-1B subseteq (B setminus P)^-1B = B_P$EDIT: which gives us a homomorphism of rings $B_mathfrak p = (Asetminus mathfrak p)^-1B rightarrow (B setminus P)^-1B = B_P$ (a priori it is not injective, since we could have zero-divisors in our multiplicative subsets).
I think $B_P = B_mathfrak p$ shortly follows. I hope I didn't overlook anything. Good luck!
answered Jul 25 at 18:28
Berber
657
edited Jul 31 at 14:24
answered Jul 25 at 18:28
Berber
657
answered Jul 25 at 18:28
Berber
657
answered Jul 25 at 18:28
Berber
657
657
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1
I'm guessing the equal sign in $B_P = B_mathfrak p$ is a ``natural" isomorphism, which I suspect arises from the universal properties of localization and tensor product. If you follow the hint, then I think u.p. of localization gives map $B_mathfrak pto B_P$
– cat
Jul 25 at 19:19