Restriction of a sheaf of modules
$begingroup$
Let $X$ be a scheme and $Y$ be a closed subscheme. For $mathcal{F}$ a sheaf of modules on $X$ to be the pushforward of a sheaf of modules on $Y$ via the inclusion $i: Y rightarrow X$ is necessary and sufficient that $mathcal{I}mathcal{F} = 0$, where $mathcal{I}$ is the ideal defining $Y$. When we consider sheaves on a topological space $Z$ we know that a sheaf $mathcal{F}$ such that $mathcal{F} vert_U = 0$, $U subset Z$ open, is isomorphic to $j_{ast} j^{-1} mathcal{F}$, where $j : Z - U rightarrow X$. I think that the same is not true for schemes, as the condition above seems stronger to me as $mathcal{I}mathcal{F}$ need not to be zero on $Y$ if it is on $X-Y$, but I am not able to find a counterexample. Any help?
sheaf-theory schemes
asked Jan 5 at 15:57
FedericoFederico
933413
$endgroup$
add a comment |
$begingroup$
Let $X$ be a scheme and $Y$ be a closed subscheme. For $mathcal{F}$ a sheaf of modules on $X$ to be the pushforward of a sheaf of modules on $Y$ via the inclusion $i: Y rightarrow X$ is necessary and sufficient that $mathcal{I}mathcal{F} = 0$, where $mathcal{I}$ is the ideal defining $Y$. When we consider sheaves on a topological space $Z$ we know that a sheaf $mathcal{F}$ such that $mathcal{F} vert_U = 0$, $U subset Z$ open, is isomorphic to $j_{ast} j^{-1} mathcal{F}$, where $j : Z - U rightarrow X$. I think that the same is not true for schemes, as the condition above seems stronger to me as $mathcal{I}mathcal{F}$ need not to be zero on $Y$ if it is on $X-Y$, but I am not able to find a counterexample. Any help?
sheaf-theory schemes
asked Jan 5 at 15:57
FedericoFederico
933413
$endgroup$
$begingroup$
Trying to clarify the question here: Is the precise statement of your question that you don't think $mathcal{I}mathcal{F}=0$ is equivalent to $mathcal{F}|_U=0$?
$endgroup$
– jgon
Jan 5 at 16:18
add a comment |
$begingroup$
Let $X$ be a scheme and $Y$ be a closed subscheme. For $mathcal{F}$ a sheaf of modules on $X$ to be the pushforward of a sheaf of modules on $Y$ via the inclusion $i: Y rightarrow X$ is necessary and sufficient that $mathcal{I}mathcal{F} = 0$, where $mathcal{I}$ is the ideal defining $Y$. When we consider sheaves on a topological space $Z$ we know that a sheaf $mathcal{F}$ such that $mathcal{F} vert_U = 0$, $U subset Z$ open, is isomorphic to $j_{ast} j^{-1} mathcal{F}$, where $j : Z - U rightarrow X$. I think that the same is not true for schemes, as the condition above seems stronger to me as $mathcal{I}mathcal{F}$ need not to be zero on $Y$ if it is on $X-Y$, but I am not able to find a counterexample. Any help?
sheaf-theory schemes
asked Jan 5 at 15:57
FedericoFederico
933413
$endgroup$
Let $X$ be a scheme and $Y$ be a closed subscheme. For $mathcal{F}$ a sheaf of modules on $X$ to be the pushforward of a sheaf of modules on $Y$ via the inclusion $i: Y rightarrow X$ is necessary and sufficient that $mathcal{I}mathcal{F} = 0$, where $mathcal{I}$ is the ideal defining $Y$. When we consider sheaves on a topological space $Z$ we know that a sheaf $mathcal{F}$ such that $mathcal{F} vert_U = 0$, $U subset Z$ open, is isomorphic to $j_{ast} j^{-1} mathcal{F}$, where $j : Z - U rightarrow X$. I think that the same is not true for schemes, as the condition above seems stronger to me as $mathcal{I}mathcal{F}$ need not to be zero on $Y$ if it is on $X-Y$, but I am not able to find a counterexample. Any help?
sheaf-theory schemes
sheaf-theory schemes
asked Jan 5 at 15:57
FedericoFederico
933413
asked Jan 5 at 15:57
FedericoFederico
933413
asked Jan 5 at 15:57
FedericoFederico
933413
asked Jan 5 at 15:57
FedericoFederico
933413
asked Jan 5 at 15:57
FedericoFederico
933413
933413
$begingroup$
Trying to clarify the question here: Is the precise statement of your question that you don't think $mathcal{I}mathcal{F}=0$ is equivalent to $mathcal{F}|_U=0$?
$endgroup$
– jgon
Jan 5 at 16:18
add a comment |
$begingroup$
Trying to clarify the question here: Is the precise statement of your question that you don't think $mathcal{I}mathcal{F}=0$ is equivalent to $mathcal{F}|_U=0$?
$endgroup$
– jgon
Jan 5 at 16:18
$begingroup$
Trying to clarify the question here: Is the precise statement of your question that you don't think $mathcal{I}mathcal{F}=0$ is equivalent to $mathcal{F}|_U=0$?
$endgroup$
– jgon
Jan 5 at 16:18
$begingroup$
Trying to clarify the question here: Is the precise statement of your question that you don't think $mathcal{I}mathcal{F}=0$ is equivalent to $mathcal{F}|_U=0$?
$endgroup$
– jgon
Jan 5 at 16:18
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Working from the assumption that your question is:
Are the conditions $mathcal{I}mathcal{F}=0$ and $mathcal{F}|_U=0$ equivalent for a sheaf of $mathcal{O}_X$-modules?
Edit:
I was careless before. The correct answer is that they are not equivalent. $newcommandcalF{mathcal{F}}newcommandcalI{mathcal{I}}newcommandcalO{mathcal{O}}newcommandpp{mathfrak{p}}newcommandSpec{operatorname{Spec}}$
Counterexample:
Take the sky scraper sheaf $k[x]_{(x)}$ on $Bbb{A}^1$ at $(x)$. This is a sheaf of $calO_X$-modules in the obvious way, but $(x)k[x]_{(x)}$ is certainly nontrivial. (Note that this is not a quasicoherent sheaf though.) For a quasicoherent sheaf, take the module $k[x]/(x^2)$ over $k[x]$, which has support precisely $(x)$, but $(x)k[x]/(x^2)ne 0$.
However $calIcalF=0$ does imply $calF|_U=0$, so you are correct, $calIcalF$ is stronger.
Proof.
If $calIcalF=0$, but $calF|_Une 0$, then choose a point $xin U$ with $calF_xne 0$. Now restrict to an affine open nhood of $x$, $Spec R$ with $x=pp$, and $calI$ and $calF$ given by an ideal $Isubseteq R$ and $R$-module $M$ respectively on this open subset. Then $IM=0$. Since $x=ppnotin Z$, $Inotsubseteq pp$, but then localizing at $pp$, we have $I_pp = R_pp$, but then we get
$$0=(IM)_pp=I_pp M_pp = R_pp M_pp = M_pp ne 0,$$
contradiction. $blacksquare$
Another way to see this is the fact that $calIcalF = 0$ implies that $calI$ is a subsheaf of the annihilator sheaf of $calF$, so $Y$ contains the support of $M$ (the closed subscheme cut out by the annihilator sheaf). Thus $calF|_U=0$.
answered Jan 5 at 16:27
jgonjgon
16.4k32143
$endgroup$
$begingroup$
This was my question, thank you very much!
$endgroup$
– Federico
Jan 5 at 16:43
$begingroup$
Sorry, I thought I understood but now it comes a question: why can you say $mathcal{I} vert_Y = 0$? For example, if I consider the ideal $(x,y) subset mathbb{C}[x,y]$ then the pullback to $(0,0)$ is given by $(x,y) / (x,y)^2$ which means that the restriction can't be zero. Am I missing something?
$endgroup$
– Federico
Jan 5 at 16:50
$begingroup$
@Federico Sorry, you're right that was careless, let me edit.
$endgroup$
– jgon
Jan 5 at 17:04
$begingroup$
@Federico, Wow that was really bad. Fixed it. You were correct, and I've added the counterexample that you were looking for.
$endgroup$
– jgon
Jan 5 at 17:17
$begingroup$
Thank you very much for your help. However, I think that only the second counterexample is correct. The skyscraper sheaf $k[x]_{(x)}$ has $mathcal{O}$ structure given by the composition $mathbb{C}[x] rightarrow mathbb{C}$ and therefore it is actually true that $(x)k[x]_{(x)} = 0$. Am I right?
$endgroup$
– Federico
Jan 5 at 17:31
|
show 2 more comments
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$begingroup$
Working from the assumption that your question is:
Are the conditions $mathcal{I}mathcal{F}=0$ and $mathcal{F}|_U=0$ equivalent for a sheaf of $mathcal{O}_X$-modules?
Edit:
I was careless before. The correct answer is that they are not equivalent. $newcommandcalF{mathcal{F}}newcommandcalI{mathcal{I}}newcommandcalO{mathcal{O}}newcommandpp{mathfrak{p}}newcommandSpec{operatorname{Spec}}$
Counterexample:
Take the sky scraper sheaf $k[x]_{(x)}$ on $Bbb{A}^1$ at $(x)$. This is a sheaf of $calO_X$-modules in the obvious way, but $(x)k[x]_{(x)}$ is certainly nontrivial. (Note that this is not a quasicoherent sheaf though.) For a quasicoherent sheaf, take the module $k[x]/(x^2)$ over $k[x]$, which has support precisely $(x)$, but $(x)k[x]/(x^2)ne 0$.
However $calIcalF=0$ does imply $calF|_U=0$, so you are correct, $calIcalF$ is stronger.
Proof.
If $calIcalF=0$, but $calF|_Une 0$, then choose a point $xin U$ with $calF_xne 0$. Now restrict to an affine open nhood of $x$, $Spec R$ with $x=pp$, and $calI$ and $calF$ given by an ideal $Isubseteq R$ and $R$-module $M$ respectively on this open subset. Then $IM=0$. Since $x=ppnotin Z$, $Inotsubseteq pp$, but then localizing at $pp$, we have $I_pp = R_pp$, but then we get
$$0=(IM)_pp=I_pp M_pp = R_pp M_pp = M_pp ne 0,$$
contradiction. $blacksquare$
Another way to see this is the fact that $calIcalF = 0$ implies that $calI$ is a subsheaf of the annihilator sheaf of $calF$, so $Y$ contains the support of $M$ (the closed subscheme cut out by the annihilator sheaf). Thus $calF|_U=0$.
answered Jan 5 at 16:27
jgonjgon
16.4k32143
$endgroup$
$begingroup$
This was my question, thank you very much!
$endgroup$
– Federico
Jan 5 at 16:43
$begingroup$
Sorry, I thought I understood but now it comes a question: why can you say $mathcal{I} vert_Y = 0$? For example, if I consider the ideal $(x,y) subset mathbb{C}[x,y]$ then the pullback to $(0,0)$ is given by $(x,y) / (x,y)^2$ which means that the restriction can't be zero. Am I missing something?
$endgroup$
– Federico
Jan 5 at 16:50
$begingroup$
@Federico Sorry, you're right that was careless, let me edit.
$endgroup$
– jgon
Jan 5 at 17:04
$begingroup$
@Federico, Wow that was really bad. Fixed it. You were correct, and I've added the counterexample that you were looking for.
$endgroup$
– jgon
Jan 5 at 17:17
$begingroup$
Thank you very much for your help. However, I think that only the second counterexample is correct. The skyscraper sheaf $k[x]_{(x)}$ has $mathcal{O}$ structure given by the composition $mathbb{C}[x] rightarrow mathbb{C}$ and therefore it is actually true that $(x)k[x]_{(x)} = 0$. Am I right?
$endgroup$
– Federico
Jan 5 at 17:31
|
show 2 more comments
$begingroup$
Working from the assumption that your question is:
Are the conditions $mathcal{I}mathcal{F}=0$ and $mathcal{F}|_U=0$ equivalent for a sheaf of $mathcal{O}_X$-modules?
Edit:
I was careless before. The correct answer is that they are not equivalent. $newcommandcalF{mathcal{F}}newcommandcalI{mathcal{I}}newcommandcalO{mathcal{O}}newcommandpp{mathfrak{p}}newcommandSpec{operatorname{Spec}}$
Counterexample:
Take the sky scraper sheaf $k[x]_{(x)}$ on $Bbb{A}^1$ at $(x)$. This is a sheaf of $calO_X$-modules in the obvious way, but $(x)k[x]_{(x)}$ is certainly nontrivial. (Note that this is not a quasicoherent sheaf though.) For a quasicoherent sheaf, take the module $k[x]/(x^2)$ over $k[x]$, which has support precisely $(x)$, but $(x)k[x]/(x^2)ne 0$.
However $calIcalF=0$ does imply $calF|_U=0$, so you are correct, $calIcalF$ is stronger.
Proof.
If $calIcalF=0$, but $calF|_Une 0$, then choose a point $xin U$ with $calF_xne 0$. Now restrict to an affine open nhood of $x$, $Spec R$ with $x=pp$, and $calI$ and $calF$ given by an ideal $Isubseteq R$ and $R$-module $M$ respectively on this open subset. Then $IM=0$. Since $x=ppnotin Z$, $Inotsubseteq pp$, but then localizing at $pp$, we have $I_pp = R_pp$, but then we get
$$0=(IM)_pp=I_pp M_pp = R_pp M_pp = M_pp ne 0,$$
contradiction. $blacksquare$
Another way to see this is the fact that $calIcalF = 0$ implies that $calI$ is a subsheaf of the annihilator sheaf of $calF$, so $Y$ contains the support of $M$ (the closed subscheme cut out by the annihilator sheaf). Thus $calF|_U=0$.
answered Jan 5 at 16:27
jgonjgon
16.4k32143
$endgroup$
$begingroup$
This was my question, thank you very much!
$endgroup$
– Federico
Jan 5 at 16:43
$begingroup$
Sorry, I thought I understood but now it comes a question: why can you say $mathcal{I} vert_Y = 0$? For example, if I consider the ideal $(x,y) subset mathbb{C}[x,y]$ then the pullback to $(0,0)$ is given by $(x,y) / (x,y)^2$ which means that the restriction can't be zero. Am I missing something?
$endgroup$
– Federico
Jan 5 at 16:50
$begingroup$
@Federico Sorry, you're right that was careless, let me edit.
$endgroup$
– jgon
Jan 5 at 17:04
$begingroup$
@Federico, Wow that was really bad. Fixed it. You were correct, and I've added the counterexample that you were looking for.
$endgroup$
– jgon
Jan 5 at 17:17
$begingroup$
Thank you very much for your help. However, I think that only the second counterexample is correct. The skyscraper sheaf $k[x]_{(x)}$ has $mathcal{O}$ structure given by the composition $mathbb{C}[x] rightarrow mathbb{C}$ and therefore it is actually true that $(x)k[x]_{(x)} = 0$. Am I right?
$endgroup$
– Federico
Jan 5 at 17:31
|
show 2 more comments
$begingroup$
Working from the assumption that your question is:
Are the conditions $mathcal{I}mathcal{F}=0$ and $mathcal{F}|_U=0$ equivalent for a sheaf of $mathcal{O}_X$-modules?
Edit:
I was careless before. The correct answer is that they are not equivalent. $newcommandcalF{mathcal{F}}newcommandcalI{mathcal{I}}newcommandcalO{mathcal{O}}newcommandpp{mathfrak{p}}newcommandSpec{operatorname{Spec}}$
Counterexample:
Take the sky scraper sheaf $k[x]_{(x)}$ on $Bbb{A}^1$ at $(x)$. This is a sheaf of $calO_X$-modules in the obvious way, but $(x)k[x]_{(x)}$ is certainly nontrivial. (Note that this is not a quasicoherent sheaf though.) For a quasicoherent sheaf, take the module $k[x]/(x^2)$ over $k[x]$, which has support precisely $(x)$, but $(x)k[x]/(x^2)ne 0$.
However $calIcalF=0$ does imply $calF|_U=0$, so you are correct, $calIcalF$ is stronger.
Proof.
If $calIcalF=0$, but $calF|_Une 0$, then choose a point $xin U$ with $calF_xne 0$. Now restrict to an affine open nhood of $x$, $Spec R$ with $x=pp$, and $calI$ and $calF$ given by an ideal $Isubseteq R$ and $R$-module $M$ respectively on this open subset. Then $IM=0$. Since $x=ppnotin Z$, $Inotsubseteq pp$, but then localizing at $pp$, we have $I_pp = R_pp$, but then we get
$$0=(IM)_pp=I_pp M_pp = R_pp M_pp = M_pp ne 0,$$
contradiction. $blacksquare$
Another way to see this is the fact that $calIcalF = 0$ implies that $calI$ is a subsheaf of the annihilator sheaf of $calF$, so $Y$ contains the support of $M$ (the closed subscheme cut out by the annihilator sheaf). Thus $calF|_U=0$.
answered Jan 5 at 16:27
jgonjgon
16.4k32143
$endgroup$
Working from the assumption that your question is:
Are the conditions $mathcal{I}mathcal{F}=0$ and $mathcal{F}|_U=0$ equivalent for a sheaf of $mathcal{O}_X$-modules?
Edit:
I was careless before. The correct answer is that they are not equivalent. $newcommandcalF{mathcal{F}}newcommandcalI{mathcal{I}}newcommandcalO{mathcal{O}}newcommandpp{mathfrak{p}}newcommandSpec{operatorname{Spec}}$
Counterexample:
Take the sky scraper sheaf $k[x]_{(x)}$ on $Bbb{A}^1$ at $(x)$. This is a sheaf of $calO_X$-modules in the obvious way, but $(x)k[x]_{(x)}$ is certainly nontrivial. (Note that this is not a quasicoherent sheaf though.) For a quasicoherent sheaf, take the module $k[x]/(x^2)$ over $k[x]$, which has support precisely $(x)$, but $(x)k[x]/(x^2)ne 0$.
However $calIcalF=0$ does imply $calF|_U=0$, so you are correct, $calIcalF$ is stronger.
Proof.
If $calIcalF=0$, but $calF|_Une 0$, then choose a point $xin U$ with $calF_xne 0$. Now restrict to an affine open nhood of $x$, $Spec R$ with $x=pp$, and $calI$ and $calF$ given by an ideal $Isubseteq R$ and $R$-module $M$ respectively on this open subset. Then $IM=0$. Since $x=ppnotin Z$, $Inotsubseteq pp$, but then localizing at $pp$, we have $I_pp = R_pp$, but then we get
$$0=(IM)_pp=I_pp M_pp = R_pp M_pp = M_pp ne 0,$$
contradiction. $blacksquare$
Another way to see this is the fact that $calIcalF = 0$ implies that $calI$ is a subsheaf of the annihilator sheaf of $calF$, so $Y$ contains the support of $M$ (the closed subscheme cut out by the annihilator sheaf). Thus $calF|_U=0$.
answered Jan 5 at 16:27
jgonjgon
16.4k32143
edited Jan 5 at 17:23
answered Jan 5 at 16:27
jgonjgon
16.4k32143
answered Jan 5 at 16:27
jgonjgon
16.4k32143
answered Jan 5 at 16:27
jgonjgon
16.4k32143
16.4k32143
$begingroup$
This was my question, thank you very much!
$endgroup$
– Federico
Jan 5 at 16:43
$begingroup$
Sorry, I thought I understood but now it comes a question: why can you say $mathcal{I} vert_Y = 0$? For example, if I consider the ideal $(x,y) subset mathbb{C}[x,y]$ then the pullback to $(0,0)$ is given by $(x,y) / (x,y)^2$ which means that the restriction can't be zero. Am I missing something?
$endgroup$
– Federico
Jan 5 at 16:50
$begingroup$
@Federico Sorry, you're right that was careless, let me edit.
$endgroup$
– jgon
Jan 5 at 17:04
$begingroup$
@Federico, Wow that was really bad. Fixed it. You were correct, and I've added the counterexample that you were looking for.
$endgroup$
– jgon
Jan 5 at 17:17
$begingroup$
Thank you very much for your help. However, I think that only the second counterexample is correct. The skyscraper sheaf $k[x]_{(x)}$ has $mathcal{O}$ structure given by the composition $mathbb{C}[x] rightarrow mathbb{C}$ and therefore it is actually true that $(x)k[x]_{(x)} = 0$. Am I right?
$endgroup$
– Federico
Jan 5 at 17:31
|
show 2 more comments
$begingroup$
This was my question, thank you very much!
$endgroup$
– Federico
Jan 5 at 16:43
$begingroup$
Sorry, I thought I understood but now it comes a question: why can you say $mathcal{I} vert_Y = 0$? For example, if I consider the ideal $(x,y) subset mathbb{C}[x,y]$ then the pullback to $(0,0)$ is given by $(x,y) / (x,y)^2$ which means that the restriction can't be zero. Am I missing something?
$endgroup$
– Federico
Jan 5 at 16:50
$begingroup$
@Federico Sorry, you're right that was careless, let me edit.
$endgroup$
– jgon
Jan 5 at 17:04
$begingroup$
@Federico, Wow that was really bad. Fixed it. You were correct, and I've added the counterexample that you were looking for.
$endgroup$
– jgon
Jan 5 at 17:17
$begingroup$
Thank you very much for your help. However, I think that only the second counterexample is correct. The skyscraper sheaf $k[x]_{(x)}$ has $mathcal{O}$ structure given by the composition $mathbb{C}[x] rightarrow mathbb{C}$ and therefore it is actually true that $(x)k[x]_{(x)} = 0$. Am I right?
$endgroup$
– Federico
Jan 5 at 17:31
$begingroup$
This was my question, thank you very much!
$endgroup$
– Federico
Jan 5 at 16:43
$begingroup$
This was my question, thank you very much!
$endgroup$
– Federico
Jan 5 at 16:43
$begingroup$
Sorry, I thought I understood but now it comes a question: why can you say $mathcal{I} vert_Y = 0$? For example, if I consider the ideal $(x,y) subset mathbb{C}[x,y]$ then the pullback to $(0,0)$ is given by $(x,y) / (x,y)^2$ which means that the restriction can't be zero. Am I missing something?
$endgroup$
– Federico
Jan 5 at 16:50
$begingroup$
Sorry, I thought I understood but now it comes a question: why can you say $mathcal{I} vert_Y = 0$? For example, if I consider the ideal $(x,y) subset mathbb{C}[x,y]$ then the pullback to $(0,0)$ is given by $(x,y) / (x,y)^2$ which means that the restriction can't be zero. Am I missing something?
$endgroup$
– Federico
Jan 5 at 16:50
$begingroup$
@Federico Sorry, you're right that was careless, let me edit.
$endgroup$
– jgon
Jan 5 at 17:04
$begingroup$
@Federico Sorry, you're right that was careless, let me edit.
$endgroup$
– jgon
Jan 5 at 17:04
$begingroup$
@Federico, Wow that was really bad. Fixed it. You were correct, and I've added the counterexample that you were looking for.
$endgroup$
– jgon
Jan 5 at 17:17
$begingroup$
@Federico, Wow that was really bad. Fixed it. You were correct, and I've added the counterexample that you were looking for.
$endgroup$
– jgon
Jan 5 at 17:17
$begingroup$
Thank you very much for your help. However, I think that only the second counterexample is correct. The skyscraper sheaf $k[x]_{(x)}$ has $mathcal{O}$ structure given by the composition $mathbb{C}[x] rightarrow mathbb{C}$ and therefore it is actually true that $(x)k[x]_{(x)} = 0$. Am I right?
$endgroup$
– Federico
Jan 5 at 17:31
$begingroup$
Thank you very much for your help. However, I think that only the second counterexample is correct. The skyscraper sheaf $k[x]_{(x)}$ has $mathcal{O}$ structure given by the composition $mathbb{C}[x] rightarrow mathbb{C}$ and therefore it is actually true that $(x)k[x]_{(x)} = 0$. Am I right?
$endgroup$
– Federico
Jan 5 at 17:31
|
show 2 more comments
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$begingroup$
Trying to clarify the question here: Is the precise statement of your question that you don't think $mathcal{I}mathcal{F}=0$ is equivalent to $mathcal{F}|_U=0$?
$endgroup$
– jgon
Jan 5 at 16:18