Integrally closed domains, where is the reduced hypothesis used?
$begingroup$
The question is to show, given a local domain $A$, with maximal ideal $(pi)$, $B$ a domain containing $A$, $S=A-(pi)$. If $S^{-1}B$ is integrally closed, and $B/pi B$ reduced, then $B$ is integrally closed.
My proof is to take $xinmathrm{Quot}(B)$ integral over $B$, this will be integral over $S^{-1}B$, so will be contained in $S^{-1}B$, thus there exists $sin A-(pi)$ with $sxin B$.
But in $A$, $s$ is a unit, so is a unit in $B$, so $xin B$, and we are done.
This argument doesn't use that the maximal ideal is principle, or the reduced hypothesis, so I suspect there's an error somewhere, but I cant see where.
abstract-algebra proof-verification commutative-algebra localization
edited Dec 16 '18 at 16:18
user26857
39.3k124183
asked Dec 16 '18 at 13:26
user277182user277182
456212
$endgroup$
add a comment |
$begingroup$
The question is to show, given a local domain $A$, with maximal ideal $(pi)$, $B$ a domain containing $A$, $S=A-(pi)$. If $S^{-1}B$ is integrally closed, and $B/pi B$ reduced, then $B$ is integrally closed.
My proof is to take $xinmathrm{Quot}(B)$ integral over $B$, this will be integral over $S^{-1}B$, so will be contained in $S^{-1}B$, thus there exists $sin A-(pi)$ with $sxin B$.
But in $A$, $s$ is a unit, so is a unit in $B$, so $xin B$, and we are done.
This argument doesn't use that the maximal ideal is principle, or the reduced hypothesis, so I suspect there's an error somewhere, but I cant see where.
abstract-algebra proof-verification commutative-algebra localization
edited Dec 16 '18 at 16:18
user26857
39.3k124183
asked Dec 16 '18 at 13:26
user277182user277182
456212
$endgroup$
2
$begingroup$
I have a guess: $S={1,pi,pi^2,dots}$.
$endgroup$
– user26857
Dec 16 '18 at 15:34
3
$begingroup$
From what you said $S$ consists of units of $B$, so $S^{-1}B = B$. Like user26857, I suspect that you have $S$ backwards
$endgroup$
– Badam Baplan
Dec 16 '18 at 15:37
1
$begingroup$
Thanks, I'm guessing there is a typo somewhere in the question, since it is stated as $A-(pi)$, but this doesn't make much sense. $A-(0)$ makes more sense, given that tensor products aren't used in the book, possibly explaining the weird notation.
$endgroup$
– user277182
Dec 17 '18 at 0:59
$begingroup$
I don't think it's supposed to be $S = A - (0)$ because in that case, taking $B$ to be any ring between $A$ and its field of fractions $K$, you would have $S^{-1}B = K$ and the assumption that $S^{-1}B$ is integrally closed would become trivial. In particular, you'd have to exclude the case $A=B$ for any nonintegrally closed $A$. Taking $S$ as in @user26857 does work out and is a fun exercise.
$endgroup$
– Badam Baplan
Dec 17 '18 at 3:33
$begingroup$
As an example take a valuation domain of rank greater than $1$, where the two rings of fraction are different.
$endgroup$
– R.C.Cowsik
Dec 17 '18 at 7:09
add a comment |
$begingroup$
The question is to show, given a local domain $A$, with maximal ideal $(pi)$, $B$ a domain containing $A$, $S=A-(pi)$. If $S^{-1}B$ is integrally closed, and $B/pi B$ reduced, then $B$ is integrally closed.
My proof is to take $xinmathrm{Quot}(B)$ integral over $B$, this will be integral over $S^{-1}B$, so will be contained in $S^{-1}B$, thus there exists $sin A-(pi)$ with $sxin B$.
But in $A$, $s$ is a unit, so is a unit in $B$, so $xin B$, and we are done.
This argument doesn't use that the maximal ideal is principle, or the reduced hypothesis, so I suspect there's an error somewhere, but I cant see where.
abstract-algebra proof-verification commutative-algebra localization
edited Dec 16 '18 at 16:18
user26857
39.3k124183
asked Dec 16 '18 at 13:26
user277182user277182
456212
$endgroup$
The question is to show, given a local domain $A$, with maximal ideal $(pi)$, $B$ a domain containing $A$, $S=A-(pi)$. If $S^{-1}B$ is integrally closed, and $B/pi B$ reduced, then $B$ is integrally closed.
My proof is to take $xinmathrm{Quot}(B)$ integral over $B$, this will be integral over $S^{-1}B$, so will be contained in $S^{-1}B$, thus there exists $sin A-(pi)$ with $sxin B$.
But in $A$, $s$ is a unit, so is a unit in $B$, so $xin B$, and we are done.
This argument doesn't use that the maximal ideal is principle, or the reduced hypothesis, so I suspect there's an error somewhere, but I cant see where.
abstract-algebra proof-verification commutative-algebra localization
abstract-algebra proof-verification commutative-algebra localization
edited Dec 16 '18 at 16:18
user26857
39.3k124183
asked Dec 16 '18 at 13:26
user277182user277182
456212
edited Dec 16 '18 at 16:18
user26857
39.3k124183
asked Dec 16 '18 at 13:26
user277182user277182
456212
edited Dec 16 '18 at 16:18
user26857
39.3k124183
edited Dec 16 '18 at 16:18
user26857
39.3k124183
edited Dec 16 '18 at 16:18
user26857
39.3k124183
39.3k124183
asked Dec 16 '18 at 13:26
user277182user277182
456212
asked Dec 16 '18 at 13:26
user277182user277182
456212
asked Dec 16 '18 at 13:26
user277182user277182
456212
456212
2
$begingroup$
I have a guess: $S={1,pi,pi^2,dots}$.
$endgroup$
– user26857
Dec 16 '18 at 15:34
3
$begingroup$
From what you said $S$ consists of units of $B$, so $S^{-1}B = B$. Like user26857, I suspect that you have $S$ backwards
$endgroup$
– Badam Baplan
Dec 16 '18 at 15:37
1
$begingroup$
Thanks, I'm guessing there is a typo somewhere in the question, since it is stated as $A-(pi)$, but this doesn't make much sense. $A-(0)$ makes more sense, given that tensor products aren't used in the book, possibly explaining the weird notation.
$endgroup$
– user277182
Dec 17 '18 at 0:59
$begingroup$
I don't think it's supposed to be $S = A - (0)$ because in that case, taking $B$ to be any ring between $A$ and its field of fractions $K$, you would have $S^{-1}B = K$ and the assumption that $S^{-1}B$ is integrally closed would become trivial. In particular, you'd have to exclude the case $A=B$ for any nonintegrally closed $A$. Taking $S$ as in @user26857 does work out and is a fun exercise.
$endgroup$
– Badam Baplan
Dec 17 '18 at 3:33
$begingroup$
As an example take a valuation domain of rank greater than $1$, where the two rings of fraction are different.
$endgroup$
– R.C.Cowsik
Dec 17 '18 at 7:09
add a comment |
2
$begingroup$
I have a guess: $S={1,pi,pi^2,dots}$.
$endgroup$
– user26857
Dec 16 '18 at 15:34
3
$begingroup$
From what you said $S$ consists of units of $B$, so $S^{-1}B = B$. Like user26857, I suspect that you have $S$ backwards
$endgroup$
– Badam Baplan
Dec 16 '18 at 15:37
1
$begingroup$
Thanks, I'm guessing there is a typo somewhere in the question, since it is stated as $A-(pi)$, but this doesn't make much sense. $A-(0)$ makes more sense, given that tensor products aren't used in the book, possibly explaining the weird notation.
$endgroup$
– user277182
Dec 17 '18 at 0:59
$begingroup$
I don't think it's supposed to be $S = A - (0)$ because in that case, taking $B$ to be any ring between $A$ and its field of fractions $K$, you would have $S^{-1}B = K$ and the assumption that $S^{-1}B$ is integrally closed would become trivial. In particular, you'd have to exclude the case $A=B$ for any nonintegrally closed $A$. Taking $S$ as in @user26857 does work out and is a fun exercise.
$endgroup$
– Badam Baplan
Dec 17 '18 at 3:33
$begingroup$
As an example take a valuation domain of rank greater than $1$, where the two rings of fraction are different.
$endgroup$
– R.C.Cowsik
Dec 17 '18 at 7:09
2
2
$begingroup$
I have a guess: $S={1,pi,pi^2,dots}$.
$endgroup$
– user26857
Dec 16 '18 at 15:34
$begingroup$
I have a guess: $S={1,pi,pi^2,dots}$.
$endgroup$
– user26857
Dec 16 '18 at 15:34
3
3
$begingroup$
From what you said $S$ consists of units of $B$, so $S^{-1}B = B$. Like user26857, I suspect that you have $S$ backwards
$endgroup$
– Badam Baplan
Dec 16 '18 at 15:37
$begingroup$
From what you said $S$ consists of units of $B$, so $S^{-1}B = B$. Like user26857, I suspect that you have $S$ backwards
$endgroup$
– Badam Baplan
Dec 16 '18 at 15:37
1
1
$begingroup$
Thanks, I'm guessing there is a typo somewhere in the question, since it is stated as $A-(pi)$, but this doesn't make much sense. $A-(0)$ makes more sense, given that tensor products aren't used in the book, possibly explaining the weird notation.
$endgroup$
– user277182
Dec 17 '18 at 0:59
$begingroup$
Thanks, I'm guessing there is a typo somewhere in the question, since it is stated as $A-(pi)$, but this doesn't make much sense. $A-(0)$ makes more sense, given that tensor products aren't used in the book, possibly explaining the weird notation.
$endgroup$
– user277182
Dec 17 '18 at 0:59
$begingroup$
I don't think it's supposed to be $S = A - (0)$ because in that case, taking $B$ to be any ring between $A$ and its field of fractions $K$, you would have $S^{-1}B = K$ and the assumption that $S^{-1}B$ is integrally closed would become trivial. In particular, you'd have to exclude the case $A=B$ for any nonintegrally closed $A$. Taking $S$ as in @user26857 does work out and is a fun exercise.
$endgroup$
– Badam Baplan
Dec 17 '18 at 3:33
$begingroup$
I don't think it's supposed to be $S = A - (0)$ because in that case, taking $B$ to be any ring between $A$ and its field of fractions $K$, you would have $S^{-1}B = K$ and the assumption that $S^{-1}B$ is integrally closed would become trivial. In particular, you'd have to exclude the case $A=B$ for any nonintegrally closed $A$. Taking $S$ as in @user26857 does work out and is a fun exercise.
$endgroup$
– Badam Baplan
Dec 17 '18 at 3:33
$begingroup$
As an example take a valuation domain of rank greater than $1$, where the two rings of fraction are different.
$endgroup$
– R.C.Cowsik
Dec 17 '18 at 7:09
$begingroup$
As an example take a valuation domain of rank greater than $1$, where the two rings of fraction are different.
$endgroup$
– R.C.Cowsik
Dec 17 '18 at 7:09
add a comment |
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$begingroup$
I have a guess: $S={1,pi,pi^2,dots}$.
$endgroup$
– user26857
Dec 16 '18 at 15:34
3
$begingroup$
From what you said $S$ consists of units of $B$, so $S^{-1}B = B$. Like user26857, I suspect that you have $S$ backwards
$endgroup$
– Badam Baplan
Dec 16 '18 at 15:37
1
$begingroup$
Thanks, I'm guessing there is a typo somewhere in the question, since it is stated as $A-(pi)$, but this doesn't make much sense. $A-(0)$ makes more sense, given that tensor products aren't used in the book, possibly explaining the weird notation.
$endgroup$
– user277182
Dec 17 '18 at 0:59
$begingroup$
I don't think it's supposed to be $S = A - (0)$ because in that case, taking $B$ to be any ring between $A$ and its field of fractions $K$, you would have $S^{-1}B = K$ and the assumption that $S^{-1}B$ is integrally closed would become trivial. In particular, you'd have to exclude the case $A=B$ for any nonintegrally closed $A$. Taking $S$ as in @user26857 does work out and is a fun exercise.
$endgroup$
– Badam Baplan
Dec 17 '18 at 3:33
$begingroup$
As an example take a valuation domain of rank greater than $1$, where the two rings of fraction are different.
$endgroup$
– R.C.Cowsik
Dec 17 '18 at 7:09