$I$-torsion functor and $sqrt{I}$
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Let $R$ be a Noetherian ring and $I,Jsubseteq R$ ideals. For any $R$-module $M$, define $Gamma_I(M)={xin Mmid I^n x=0text{ for some }ninmathbb{N}}$. Suppose that $Gamma_I=Gamma_J$. I want to show that $sqrt{I}=sqrt{J}$.
My issue is that I don't know which $R$-module to consider. Picking any $rin sqrt{I}$, I think I want some quotient $R/A$ (where $A$ is an ideal involving $I$, or maybe $r$) such that $rinGamma_I(R/A)=Gamma_J(R/A)$, i.e. such that $(r+A)J^esubseteq A$ for some $e$.
abstract-algebra ring-theory commutative-algebra modules local-cohomology
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Let $R$ be a Noetherian ring and $I,Jsubseteq R$ ideals. For any $R$-module $M$, define $Gamma_I(M)={xin Mmid I^n x=0text{ for some }ninmathbb{N}}$. Suppose that $Gamma_I=Gamma_J$. I want to show that $sqrt{I}=sqrt{J}$.
My issue is that I don't know which $R$-module to consider. Picking any $rin sqrt{I}$, I think I want some quotient $R/A$ (where $A$ is an ideal involving $I$, or maybe $r$) such that $rinGamma_I(R/A)=Gamma_J(R/A)$, i.e. such that $(r+A)J^esubseteq A$ for some $e$.
abstract-algebra ring-theory commutative-algebra modules local-cohomology
New contributor
When you say $Gamma_I = Gamma_J$, do you mean that for all $R$-modules $M$, $Gamma_I(M) = Gamma_J(M)$?
– Badam Baplan
Nov 29 at 23:08
@BadamBaplan Yes.
– user234988
Nov 29 at 23:46
Sorry that should have been obvious :)
– Badam Baplan
Nov 29 at 23:51
add a comment |
up vote
1
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favorite
up vote
1
down vote
favorite
Let $R$ be a Noetherian ring and $I,Jsubseteq R$ ideals. For any $R$-module $M$, define $Gamma_I(M)={xin Mmid I^n x=0text{ for some }ninmathbb{N}}$. Suppose that $Gamma_I=Gamma_J$. I want to show that $sqrt{I}=sqrt{J}$.
My issue is that I don't know which $R$-module to consider. Picking any $rin sqrt{I}$, I think I want some quotient $R/A$ (where $A$ is an ideal involving $I$, or maybe $r$) such that $rinGamma_I(R/A)=Gamma_J(R/A)$, i.e. such that $(r+A)J^esubseteq A$ for some $e$.
abstract-algebra ring-theory commutative-algebra modules local-cohomology
New contributor
Let $R$ be a Noetherian ring and $I,Jsubseteq R$ ideals. For any $R$-module $M$, define $Gamma_I(M)={xin Mmid I^n x=0text{ for some }ninmathbb{N}}$. Suppose that $Gamma_I=Gamma_J$. I want to show that $sqrt{I}=sqrt{J}$.
My issue is that I don't know which $R$-module to consider. Picking any $rin sqrt{I}$, I think I want some quotient $R/A$ (where $A$ is an ideal involving $I$, or maybe $r$) such that $rinGamma_I(R/A)=Gamma_J(R/A)$, i.e. such that $(r+A)J^esubseteq A$ for some $e$.
abstract-algebra ring-theory commutative-algebra modules local-cohomology
abstract-algebra ring-theory commutative-algebra modules local-cohomology
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New contributor
edited 2 days ago
user26857
39.1k123882
39.1k123882
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asked Nov 29 at 21:39
user234988
62
62
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When you say $Gamma_I = Gamma_J$, do you mean that for all $R$-modules $M$, $Gamma_I(M) = Gamma_J(M)$?
– Badam Baplan
Nov 29 at 23:08
@BadamBaplan Yes.
– user234988
Nov 29 at 23:46
Sorry that should have been obvious :)
– Badam Baplan
Nov 29 at 23:51
add a comment |
When you say $Gamma_I = Gamma_J$, do you mean that for all $R$-modules $M$, $Gamma_I(M) = Gamma_J(M)$?
– Badam Baplan
Nov 29 at 23:08
@BadamBaplan Yes.
– user234988
Nov 29 at 23:46
Sorry that should have been obvious :)
– Badam Baplan
Nov 29 at 23:51
When you say $Gamma_I = Gamma_J$, do you mean that for all $R$-modules $M$, $Gamma_I(M) = Gamma_J(M)$?
– Badam Baplan
Nov 29 at 23:08
When you say $Gamma_I = Gamma_J$, do you mean that for all $R$-modules $M$, $Gamma_I(M) = Gamma_J(M)$?
– Badam Baplan
Nov 29 at 23:08
@BadamBaplan Yes.
– user234988
Nov 29 at 23:46
@BadamBaplan Yes.
– user234988
Nov 29 at 23:46
Sorry that should have been obvious :)
– Badam Baplan
Nov 29 at 23:51
Sorry that should have been obvious :)
– Badam Baplan
Nov 29 at 23:51
add a comment |
1 Answer
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First check that we can use the Noetherian hypothesis to rewrite the definition of the $I$-torsion functor as $Gamma_I(M) = {m in M mid I subseteq sqrt{text{Ann}_R(m)} }$. I think this well illuminates an approach to the problem by associated primes.
Hint about what module to apply the functor to:
I would look at $Gamma_I(R/I) = Gamma_J(R/I)$.
More details, if it's helpful:
In particular, pick a prime $mathfrak{p}$ minimal over $I$, so that $mathfrak{p}$ is an associate prime of $R/I$. There exists an element $m in R / I$ such that $text{Ann}_R(m) = mathfrak{p}$, hence $I subseteq mathfrak{p} = sqrt{text{Ann}_R(m)}$ and $m in Gamma_I(R/I)$. By assumption, $m in Gamma_J(R/I)$, too. Thus $J subseteq mathfrak{p}$ as well. It follows easily that $V(I) subseteq V(J)$. A symmetric argument shows the opposite inclusion.
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
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up vote
2
down vote
First check that we can use the Noetherian hypothesis to rewrite the definition of the $I$-torsion functor as $Gamma_I(M) = {m in M mid I subseteq sqrt{text{Ann}_R(m)} }$. I think this well illuminates an approach to the problem by associated primes.
Hint about what module to apply the functor to:
I would look at $Gamma_I(R/I) = Gamma_J(R/I)$.
More details, if it's helpful:
In particular, pick a prime $mathfrak{p}$ minimal over $I$, so that $mathfrak{p}$ is an associate prime of $R/I$. There exists an element $m in R / I$ such that $text{Ann}_R(m) = mathfrak{p}$, hence $I subseteq mathfrak{p} = sqrt{text{Ann}_R(m)}$ and $m in Gamma_I(R/I)$. By assumption, $m in Gamma_J(R/I)$, too. Thus $J subseteq mathfrak{p}$ as well. It follows easily that $V(I) subseteq V(J)$. A symmetric argument shows the opposite inclusion.
add a comment |
up vote
2
down vote
First check that we can use the Noetherian hypothesis to rewrite the definition of the $I$-torsion functor as $Gamma_I(M) = {m in M mid I subseteq sqrt{text{Ann}_R(m)} }$. I think this well illuminates an approach to the problem by associated primes.
Hint about what module to apply the functor to:
I would look at $Gamma_I(R/I) = Gamma_J(R/I)$.
More details, if it's helpful:
In particular, pick a prime $mathfrak{p}$ minimal over $I$, so that $mathfrak{p}$ is an associate prime of $R/I$. There exists an element $m in R / I$ such that $text{Ann}_R(m) = mathfrak{p}$, hence $I subseteq mathfrak{p} = sqrt{text{Ann}_R(m)}$ and $m in Gamma_I(R/I)$. By assumption, $m in Gamma_J(R/I)$, too. Thus $J subseteq mathfrak{p}$ as well. It follows easily that $V(I) subseteq V(J)$. A symmetric argument shows the opposite inclusion.
add a comment |
up vote
2
down vote
up vote
2
down vote
First check that we can use the Noetherian hypothesis to rewrite the definition of the $I$-torsion functor as $Gamma_I(M) = {m in M mid I subseteq sqrt{text{Ann}_R(m)} }$. I think this well illuminates an approach to the problem by associated primes.
Hint about what module to apply the functor to:
I would look at $Gamma_I(R/I) = Gamma_J(R/I)$.
More details, if it's helpful:
In particular, pick a prime $mathfrak{p}$ minimal over $I$, so that $mathfrak{p}$ is an associate prime of $R/I$. There exists an element $m in R / I$ such that $text{Ann}_R(m) = mathfrak{p}$, hence $I subseteq mathfrak{p} = sqrt{text{Ann}_R(m)}$ and $m in Gamma_I(R/I)$. By assumption, $m in Gamma_J(R/I)$, too. Thus $J subseteq mathfrak{p}$ as well. It follows easily that $V(I) subseteq V(J)$. A symmetric argument shows the opposite inclusion.
First check that we can use the Noetherian hypothesis to rewrite the definition of the $I$-torsion functor as $Gamma_I(M) = {m in M mid I subseteq sqrt{text{Ann}_R(m)} }$. I think this well illuminates an approach to the problem by associated primes.
Hint about what module to apply the functor to:
I would look at $Gamma_I(R/I) = Gamma_J(R/I)$.
More details, if it's helpful:
In particular, pick a prime $mathfrak{p}$ minimal over $I$, so that $mathfrak{p}$ is an associate prime of $R/I$. There exists an element $m in R / I$ such that $text{Ann}_R(m) = mathfrak{p}$, hence $I subseteq mathfrak{p} = sqrt{text{Ann}_R(m)}$ and $m in Gamma_I(R/I)$. By assumption, $m in Gamma_J(R/I)$, too. Thus $J subseteq mathfrak{p}$ as well. It follows easily that $V(I) subseteq V(J)$. A symmetric argument shows the opposite inclusion.
answered Nov 30 at 0:44
Badam Baplan
4,151722
4,151722
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When you say $Gamma_I = Gamma_J$, do you mean that for all $R$-modules $M$, $Gamma_I(M) = Gamma_J(M)$?
– Badam Baplan
Nov 29 at 23:08
@BadamBaplan Yes.
– user234988
Nov 29 at 23:46
Sorry that should have been obvious :)
– Badam Baplan
Nov 29 at 23:51