Integral Closure Flat
$begingroup$
Let $R$ be an integral domain and $A$ it's integral closure. Futhermore $A$ is a finitely generated $R$-module.
I want to show that if $A$ is flat over $R$ then already $R = A$ holds.
My attempts: I tried (without success) following: Assume $R neq A$. Let $a in A backslash R$. Since $A$ integral closure there exist a minimal $n$ with
(*) $$a^n +r_{n-1}a^{n-1} +... +r_0 =0$$
Since $R$ integral then $R subset Frac(R)$ and the multiplication map $m: R to A, r mapsto ar$ is injective Since $A$ flat it should be stay injective after tensoring with $-otimes A$.
Now I can multipy $a$ also with all elements from $A$. My hope was to deduce the contradiction using (*) by multiplying with a appropiate polynomial in $a$. The problem is the "constant" term r_0.
I think that I'm very close with this proof to the solution but I don't see the final step. Can anybody help. BTW: Up to now I don't used the finitely generated property.
Source: Liu's "Algebraic Geometry". Futhermore: Could the statement be true without the finiteness assumtion. Liu's says yes but how to show?
ring-theory commutative-algebra flatness
asked Dec 6 '18 at 0:40
KarlPeterKarlPeter
6081315
$endgroup$
|
show 5 more comments
$begingroup$
Let $R$ be an integral domain and $A$ it's integral closure. Futhermore $A$ is a finitely generated $R$-module.
I want to show that if $A$ is flat over $R$ then already $R = A$ holds.
My attempts: I tried (without success) following: Assume $R neq A$. Let $a in A backslash R$. Since $A$ integral closure there exist a minimal $n$ with
(*) $$a^n +r_{n-1}a^{n-1} +... +r_0 =0$$
Since $R$ integral then $R subset Frac(R)$ and the multiplication map $m: R to A, r mapsto ar$ is injective Since $A$ flat it should be stay injective after tensoring with $-otimes A$.
Now I can multipy $a$ also with all elements from $A$. My hope was to deduce the contradiction using (*) by multiplying with a appropiate polynomial in $a$. The problem is the "constant" term r_0.
I think that I'm very close with this proof to the solution but I don't see the final step. Can anybody help. BTW: Up to now I don't used the finitely generated property.
Source: Liu's "Algebraic Geometry". Futhermore: Could the statement be true without the finiteness assumtion. Liu's says yes but how to show?
ring-theory commutative-algebra flatness
asked Dec 6 '18 at 0:40
KarlPeterKarlPeter
6081315
$endgroup$
$begingroup$
See math.stackexchange.com/q/482908/660
$endgroup$
– Pierre-Yves Gaillard
Dec 6 '18 at 14:09
$begingroup$
and math.stackexchange.com/questions/385364/…
$endgroup$
– Badam Baplan
Dec 6 '18 at 23:44
$begingroup$
@BadamBaplan: Hi. Thank you for the reference. You refering to Hagen Knaf's answer, don't you? The only not satisfying point there is that for localizations wrt a prime ideal $mathfrak p$ he assumes that there exist always a $x_{mathfrak p}in B_{mathfrak p}$ with $B_{mathfrak p}=x_{mathfrak p}A_{mathfrak p}$. Could you explain why this is true?
$endgroup$
– KarlPeter
Dec 7 '18 at 0:06
$begingroup$
Actually I much prefer the other answer there, which does not use the assumption that the integral closure is an f.g. $R$-module.
$endgroup$
– Badam Baplan
Dec 7 '18 at 1:11
$begingroup$
The answer you mention is using the fact that f.g. flat modules are locally cyclic. My feeling is that that is not the right answer to this question, since it has nothing to do with the integral closure. That approach shows that a domain has no proper f.g. flat overrings, but the second approach shows that a domain has no proper integral flat overrings. The second result implies the first.
$endgroup$
– Badam Baplan
Dec 7 '18 at 1:21
|
show 5 more comments
$begingroup$
Let $R$ be an integral domain and $A$ it's integral closure. Futhermore $A$ is a finitely generated $R$-module.
I want to show that if $A$ is flat over $R$ then already $R = A$ holds.
My attempts: I tried (without success) following: Assume $R neq A$. Let $a in A backslash R$. Since $A$ integral closure there exist a minimal $n$ with
(*) $$a^n +r_{n-1}a^{n-1} +... +r_0 =0$$
Since $R$ integral then $R subset Frac(R)$ and the multiplication map $m: R to A, r mapsto ar$ is injective Since $A$ flat it should be stay injective after tensoring with $-otimes A$.
Now I can multipy $a$ also with all elements from $A$. My hope was to deduce the contradiction using (*) by multiplying with a appropiate polynomial in $a$. The problem is the "constant" term r_0.
I think that I'm very close with this proof to the solution but I don't see the final step. Can anybody help. BTW: Up to now I don't used the finitely generated property.
Source: Liu's "Algebraic Geometry". Futhermore: Could the statement be true without the finiteness assumtion. Liu's says yes but how to show?
ring-theory commutative-algebra flatness
asked Dec 6 '18 at 0:40
KarlPeterKarlPeter
6081315
$endgroup$
Let $R$ be an integral domain and $A$ it's integral closure. Futhermore $A$ is a finitely generated $R$-module.
I want to show that if $A$ is flat over $R$ then already $R = A$ holds.
My attempts: I tried (without success) following: Assume $R neq A$. Let $a in A backslash R$. Since $A$ integral closure there exist a minimal $n$ with
(*) $$a^n +r_{n-1}a^{n-1} +... +r_0 =0$$
Since $R$ integral then $R subset Frac(R)$ and the multiplication map $m: R to A, r mapsto ar$ is injective Since $A$ flat it should be stay injective after tensoring with $-otimes A$.
Now I can multipy $a$ also with all elements from $A$. My hope was to deduce the contradiction using (*) by multiplying with a appropiate polynomial in $a$. The problem is the "constant" term r_0.
I think that I'm very close with this proof to the solution but I don't see the final step. Can anybody help. BTW: Up to now I don't used the finitely generated property.
Source: Liu's "Algebraic Geometry". Futhermore: Could the statement be true without the finiteness assumtion. Liu's says yes but how to show?
ring-theory commutative-algebra flatness
ring-theory commutative-algebra flatness
asked Dec 6 '18 at 0:40
KarlPeterKarlPeter
6081315
asked Dec 6 '18 at 0:40
KarlPeterKarlPeter
6081315
asked Dec 6 '18 at 0:40
KarlPeterKarlPeter
6081315
asked Dec 6 '18 at 0:40
KarlPeterKarlPeter
6081315
asked Dec 6 '18 at 0:40
KarlPeterKarlPeter
6081315
6081315
$begingroup$
See math.stackexchange.com/q/482908/660
$endgroup$
– Pierre-Yves Gaillard
Dec 6 '18 at 14:09
$begingroup$
and math.stackexchange.com/questions/385364/…
$endgroup$
– Badam Baplan
Dec 6 '18 at 23:44
$begingroup$
@BadamBaplan: Hi. Thank you for the reference. You refering to Hagen Knaf's answer, don't you? The only not satisfying point there is that for localizations wrt a prime ideal $mathfrak p$ he assumes that there exist always a $x_{mathfrak p}in B_{mathfrak p}$ with $B_{mathfrak p}=x_{mathfrak p}A_{mathfrak p}$. Could you explain why this is true?
$endgroup$
– KarlPeter
Dec 7 '18 at 0:06
$begingroup$
Actually I much prefer the other answer there, which does not use the assumption that the integral closure is an f.g. $R$-module.
$endgroup$
– Badam Baplan
Dec 7 '18 at 1:11
$begingroup$
The answer you mention is using the fact that f.g. flat modules are locally cyclic. My feeling is that that is not the right answer to this question, since it has nothing to do with the integral closure. That approach shows that a domain has no proper f.g. flat overrings, but the second approach shows that a domain has no proper integral flat overrings. The second result implies the first.
$endgroup$
– Badam Baplan
Dec 7 '18 at 1:21
|
show 5 more comments
$begingroup$
See math.stackexchange.com/q/482908/660
$endgroup$
– Pierre-Yves Gaillard
Dec 6 '18 at 14:09
$begingroup$
and math.stackexchange.com/questions/385364/…
$endgroup$
– Badam Baplan
Dec 6 '18 at 23:44
$begingroup$
@BadamBaplan: Hi. Thank you for the reference. You refering to Hagen Knaf's answer, don't you? The only not satisfying point there is that for localizations wrt a prime ideal $mathfrak p$ he assumes that there exist always a $x_{mathfrak p}in B_{mathfrak p}$ with $B_{mathfrak p}=x_{mathfrak p}A_{mathfrak p}$. Could you explain why this is true?
$endgroup$
– KarlPeter
Dec 7 '18 at 0:06
$begingroup$
Actually I much prefer the other answer there, which does not use the assumption that the integral closure is an f.g. $R$-module.
$endgroup$
– Badam Baplan
Dec 7 '18 at 1:11
$begingroup$
The answer you mention is using the fact that f.g. flat modules are locally cyclic. My feeling is that that is not the right answer to this question, since it has nothing to do with the integral closure. That approach shows that a domain has no proper f.g. flat overrings, but the second approach shows that a domain has no proper integral flat overrings. The second result implies the first.
$endgroup$
– Badam Baplan
Dec 7 '18 at 1:21
$begingroup$
See math.stackexchange.com/q/482908/660
$endgroup$
– Pierre-Yves Gaillard
Dec 6 '18 at 14:09
$begingroup$
See math.stackexchange.com/q/482908/660
$endgroup$
– Pierre-Yves Gaillard
Dec 6 '18 at 14:09
$begingroup$
and math.stackexchange.com/questions/385364/…
$endgroup$
– Badam Baplan
Dec 6 '18 at 23:44
$begingroup$
and math.stackexchange.com/questions/385364/…
$endgroup$
– Badam Baplan
Dec 6 '18 at 23:44
$begingroup$
@BadamBaplan: Hi. Thank you for the reference. You refering to Hagen Knaf's answer, don't you? The only not satisfying point there is that for localizations wrt a prime ideal $mathfrak p$ he assumes that there exist always a $x_{mathfrak p}in B_{mathfrak p}$ with $B_{mathfrak p}=x_{mathfrak p}A_{mathfrak p}$. Could you explain why this is true?
$endgroup$
– KarlPeter
Dec 7 '18 at 0:06
$begingroup$
@BadamBaplan: Hi. Thank you for the reference. You refering to Hagen Knaf's answer, don't you? The only not satisfying point there is that for localizations wrt a prime ideal $mathfrak p$ he assumes that there exist always a $x_{mathfrak p}in B_{mathfrak p}$ with $B_{mathfrak p}=x_{mathfrak p}A_{mathfrak p}$. Could you explain why this is true?
$endgroup$
– KarlPeter
Dec 7 '18 at 0:06
$begingroup$
Actually I much prefer the other answer there, which does not use the assumption that the integral closure is an f.g. $R$-module.
$endgroup$
– Badam Baplan
Dec 7 '18 at 1:11
$begingroup$
Actually I much prefer the other answer there, which does not use the assumption that the integral closure is an f.g. $R$-module.
$endgroup$
– Badam Baplan
Dec 7 '18 at 1:11
$begingroup$
The answer you mention is using the fact that f.g. flat modules are locally cyclic. My feeling is that that is not the right answer to this question, since it has nothing to do with the integral closure. That approach shows that a domain has no proper f.g. flat overrings, but the second approach shows that a domain has no proper integral flat overrings. The second result implies the first.
$endgroup$
– Badam Baplan
Dec 7 '18 at 1:21
$begingroup$
The answer you mention is using the fact that f.g. flat modules are locally cyclic. My feeling is that that is not the right answer to this question, since it has nothing to do with the integral closure. That approach shows that a domain has no proper f.g. flat overrings, but the second approach shows that a domain has no proper integral flat overrings. The second result implies the first.
$endgroup$
– Badam Baplan
Dec 7 '18 at 1:21
|
show 5 more comments
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$begingroup$
See math.stackexchange.com/q/482908/660
$endgroup$
– Pierre-Yves Gaillard
Dec 6 '18 at 14:09
$begingroup$
and math.stackexchange.com/questions/385364/…
$endgroup$
– Badam Baplan
Dec 6 '18 at 23:44
$begingroup$
@BadamBaplan: Hi. Thank you for the reference. You refering to Hagen Knaf's answer, don't you? The only not satisfying point there is that for localizations wrt a prime ideal $mathfrak p$ he assumes that there exist always a $x_{mathfrak p}in B_{mathfrak p}$ with $B_{mathfrak p}=x_{mathfrak p}A_{mathfrak p}$. Could you explain why this is true?
$endgroup$
– KarlPeter
Dec 7 '18 at 0:06
$begingroup$
Actually I much prefer the other answer there, which does not use the assumption that the integral closure is an f.g. $R$-module.
$endgroup$
– Badam Baplan
Dec 7 '18 at 1:11
$begingroup$
The answer you mention is using the fact that f.g. flat modules are locally cyclic. My feeling is that that is not the right answer to this question, since it has nothing to do with the integral closure. That approach shows that a domain has no proper f.g. flat overrings, but the second approach shows that a domain has no proper integral flat overrings. The second result implies the first.
$endgroup$
– Badam Baplan
Dec 7 '18 at 1:21