$I$-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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  • $begingroup$
    When you say $Gamma_I = Gamma_J$, do you mean that for all $R$-modules $M$, $Gamma_I(M) = Gamma_J(M)$?
    $endgroup$
    – Badam Baplan
    Nov 29 '18 at 23:08










  • $begingroup$
    @BadamBaplan Yes.
    $endgroup$
    – user234988
    Nov 29 '18 at 23:46












  • $begingroup$
    Sorry that should have been obvious :)
    $endgroup$
    – Badam Baplan
    Nov 29 '18 at 23:51
















1












$begingroup$



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







share|cite|improve this question











$endgroup$












  • $begingroup$
    When you say $Gamma_I = Gamma_J$, do you mean that for all $R$-modules $M$, $Gamma_I(M) = Gamma_J(M)$?
    $endgroup$
    – Badam Baplan
    Nov 29 '18 at 23:08










  • $begingroup$
    @BadamBaplan Yes.
    $endgroup$
    – user234988
    Nov 29 '18 at 23:46












  • $begingroup$
    Sorry that should have been obvious :)
    $endgroup$
    – Badam Baplan
    Nov 29 '18 at 23:51














1












1








1


1



$begingroup$



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







share|cite|improve this question











$endgroup$





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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '18 at 17:40









user26857

39.3k124183




39.3k124183










asked Nov 29 '18 at 21:39









user234988user234988

62




62












  • $begingroup$
    When you say $Gamma_I = Gamma_J$, do you mean that for all $R$-modules $M$, $Gamma_I(M) = Gamma_J(M)$?
    $endgroup$
    – Badam Baplan
    Nov 29 '18 at 23:08










  • $begingroup$
    @BadamBaplan Yes.
    $endgroup$
    – user234988
    Nov 29 '18 at 23:46












  • $begingroup$
    Sorry that should have been obvious :)
    $endgroup$
    – Badam Baplan
    Nov 29 '18 at 23:51


















  • $begingroup$
    When you say $Gamma_I = Gamma_J$, do you mean that for all $R$-modules $M$, $Gamma_I(M) = Gamma_J(M)$?
    $endgroup$
    – Badam Baplan
    Nov 29 '18 at 23:08










  • $begingroup$
    @BadamBaplan Yes.
    $endgroup$
    – user234988
    Nov 29 '18 at 23:46












  • $begingroup$
    Sorry that should have been obvious :)
    $endgroup$
    – Badam Baplan
    Nov 29 '18 at 23:51
















$begingroup$
When you say $Gamma_I = Gamma_J$, do you mean that for all $R$-modules $M$, $Gamma_I(M) = Gamma_J(M)$?
$endgroup$
– Badam Baplan
Nov 29 '18 at 23:08




$begingroup$
When you say $Gamma_I = Gamma_J$, do you mean that for all $R$-modules $M$, $Gamma_I(M) = Gamma_J(M)$?
$endgroup$
– Badam Baplan
Nov 29 '18 at 23:08












$begingroup$
@BadamBaplan Yes.
$endgroup$
– user234988
Nov 29 '18 at 23:46






$begingroup$
@BadamBaplan Yes.
$endgroup$
– user234988
Nov 29 '18 at 23:46














$begingroup$
Sorry that should have been obvious :)
$endgroup$
– Badam Baplan
Nov 29 '18 at 23:51




$begingroup$
Sorry that should have been obvious :)
$endgroup$
– Badam Baplan
Nov 29 '18 at 23:51










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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    $begingroup$

    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.







    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      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.







      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        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.







        share|cite|improve this answer









        $endgroup$



        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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        share|cite|improve this answer



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        answered Nov 30 '18 at 0:44









        Badam BaplanBadam Baplan

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