Support of a graded module of a ring concentrated in non-negative dimensions
$begingroup$
I wanted to prove the following equivalence.
Consider $R$ a graded commutative Noetherian ring such that $R^{<0}=0$ and $M$ a graded, finitely generated $R$-module. Then $M^i=0$ for $i gg 0$ if and only if $text{Supp}_R M subset mathcal{V}(R^{geq 1})$. Here $text{Supp}_R M = { mathcal{P} in text{Spec}(R) : M_{mathcal{P}}neq 0 }$ is the support of $M$, there are other definitions but this is the easiest.
For the context of the exercise see corollary 4.25 of the following notes
https://arxiv.org/pdf/1107.4815.pdf.
The implication $Rightarrow$ is easy: take a prime ideal $mathcal{P}$ and suppose $R^{geq 1}notsubset mathcal{P}$, then this means there exists an element $r in R setminus mathcal{P}$ of positive degree, thus $r^km=0$ for any $m in M$ and $k in mathbb{N}$ such that $k|r|$ is big enough. This implies that $M_{mathcal{P}}=0$, thus $mathcal{P}$ cannot be in the support of $M$ and we deduce the inclusion $text{Supp}_R M subset mathcal{V}(R^{geq 1})$.
The other implication is the difficult part. I tried a direct approach: using the fact that $R^{geq 1}$ is in the support of $M$ I get that there exists $m in M$ such that for any $rin{R^0}$ $rm neq 0$ but from this I cannot deduce that $M^i=0$ for $i$ big enough. I tried other approaches but he fact is that the starting assumption gives information on how $R$ interact on $M$ while I want to prove $M^i=0$ for $igg0$ which is a property of the additive structure of the module. So I suppose that I cannot conclude using only the definitions but I have to use some result like Nakayama lemma.
noetherian graded-rings graded-modules
asked Dec 4 '18 at 11:18
N.B.N.B.
689313
$endgroup$
add a comment |
$begingroup$
I wanted to prove the following equivalence.
Consider $R$ a graded commutative Noetherian ring such that $R^{<0}=0$ and $M$ a graded, finitely generated $R$-module. Then $M^i=0$ for $i gg 0$ if and only if $text{Supp}_R M subset mathcal{V}(R^{geq 1})$. Here $text{Supp}_R M = { mathcal{P} in text{Spec}(R) : M_{mathcal{P}}neq 0 }$ is the support of $M$, there are other definitions but this is the easiest.
For the context of the exercise see corollary 4.25 of the following notes
https://arxiv.org/pdf/1107.4815.pdf.
The implication $Rightarrow$ is easy: take a prime ideal $mathcal{P}$ and suppose $R^{geq 1}notsubset mathcal{P}$, then this means there exists an element $r in R setminus mathcal{P}$ of positive degree, thus $r^km=0$ for any $m in M$ and $k in mathbb{N}$ such that $k|r|$ is big enough. This implies that $M_{mathcal{P}}=0$, thus $mathcal{P}$ cannot be in the support of $M$ and we deduce the inclusion $text{Supp}_R M subset mathcal{V}(R^{geq 1})$.
The other implication is the difficult part. I tried a direct approach: using the fact that $R^{geq 1}$ is in the support of $M$ I get that there exists $m in M$ such that for any $rin{R^0}$ $rm neq 0$ but from this I cannot deduce that $M^i=0$ for $i$ big enough. I tried other approaches but he fact is that the starting assumption gives information on how $R$ interact on $M$ while I want to prove $M^i=0$ for $igg0$ which is a property of the additive structure of the module. So I suppose that I cannot conclude using only the definitions but I have to use some result like Nakayama lemma.
noetherian graded-rings graded-modules
asked Dec 4 '18 at 11:18
N.B.N.B.
689313
$endgroup$
add a comment |
$begingroup$
I wanted to prove the following equivalence.
Consider $R$ a graded commutative Noetherian ring such that $R^{<0}=0$ and $M$ a graded, finitely generated $R$-module. Then $M^i=0$ for $i gg 0$ if and only if $text{Supp}_R M subset mathcal{V}(R^{geq 1})$. Here $text{Supp}_R M = { mathcal{P} in text{Spec}(R) : M_{mathcal{P}}neq 0 }$ is the support of $M$, there are other definitions but this is the easiest.
For the context of the exercise see corollary 4.25 of the following notes
https://arxiv.org/pdf/1107.4815.pdf.
The implication $Rightarrow$ is easy: take a prime ideal $mathcal{P}$ and suppose $R^{geq 1}notsubset mathcal{P}$, then this means there exists an element $r in R setminus mathcal{P}$ of positive degree, thus $r^km=0$ for any $m in M$ and $k in mathbb{N}$ such that $k|r|$ is big enough. This implies that $M_{mathcal{P}}=0$, thus $mathcal{P}$ cannot be in the support of $M$ and we deduce the inclusion $text{Supp}_R M subset mathcal{V}(R^{geq 1})$.
The other implication is the difficult part. I tried a direct approach: using the fact that $R^{geq 1}$ is in the support of $M$ I get that there exists $m in M$ such that for any $rin{R^0}$ $rm neq 0$ but from this I cannot deduce that $M^i=0$ for $i$ big enough. I tried other approaches but he fact is that the starting assumption gives information on how $R$ interact on $M$ while I want to prove $M^i=0$ for $igg0$ which is a property of the additive structure of the module. So I suppose that I cannot conclude using only the definitions but I have to use some result like Nakayama lemma.
noetherian graded-rings graded-modules
asked Dec 4 '18 at 11:18
N.B.N.B.
689313
$endgroup$
I wanted to prove the following equivalence.
Consider $R$ a graded commutative Noetherian ring such that $R^{<0}=0$ and $M$ a graded, finitely generated $R$-module. Then $M^i=0$ for $i gg 0$ if and only if $text{Supp}_R M subset mathcal{V}(R^{geq 1})$. Here $text{Supp}_R M = { mathcal{P} in text{Spec}(R) : M_{mathcal{P}}neq 0 }$ is the support of $M$, there are other definitions but this is the easiest.
For the context of the exercise see corollary 4.25 of the following notes
https://arxiv.org/pdf/1107.4815.pdf.
The implication $Rightarrow$ is easy: take a prime ideal $mathcal{P}$ and suppose $R^{geq 1}notsubset mathcal{P}$, then this means there exists an element $r in R setminus mathcal{P}$ of positive degree, thus $r^km=0$ for any $m in M$ and $k in mathbb{N}$ such that $k|r|$ is big enough. This implies that $M_{mathcal{P}}=0$, thus $mathcal{P}$ cannot be in the support of $M$ and we deduce the inclusion $text{Supp}_R M subset mathcal{V}(R^{geq 1})$.
The other implication is the difficult part. I tried a direct approach: using the fact that $R^{geq 1}$ is in the support of $M$ I get that there exists $m in M$ such that for any $rin{R^0}$ $rm neq 0$ but from this I cannot deduce that $M^i=0$ for $i$ big enough. I tried other approaches but he fact is that the starting assumption gives information on how $R$ interact on $M$ while I want to prove $M^i=0$ for $igg0$ which is a property of the additive structure of the module. So I suppose that I cannot conclude using only the definitions but I have to use some result like Nakayama lemma.
noetherian graded-rings graded-modules
noetherian graded-rings graded-modules
asked Dec 4 '18 at 11:18
N.B.N.B.
689313
asked Dec 4 '18 at 11:18
N.B.N.B.
689313
asked Dec 4 '18 at 11:18
N.B.N.B.
689313
asked Dec 4 '18 at 11:18
N.B.N.B.
689313
asked Dec 4 '18 at 11:18
N.B.N.B.
689313
689313
add a comment |
add a comment |
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