Definition of closed subschemeInclusion of closed subschemes.Questions on Reduced Induced Closed SubschemeA...
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Definition of closed subscheme
Inclusion of closed subschemes.Questions on Reduced Induced Closed SubschemeA particular closed subschemeWhy a closed subscheme give rise to a closed immersionClosed subsets and closed subschemesWhy the ideal defining a closed subscheme is unique?Why do we have two definitions of Cartier divisor?Why is a locally closed subscheme defined in this way?Regarding a sheaf of $mathcal O_X$-modules as a sheaf of $mathcal O_Z$-modules, where $Z$ is a closed subschemeHelp understanding closed subschemes and closed immersionsCondition for a morphism of schemes to factor through a closed subscheme
$begingroup$
Let $(X,mathcal O_X)$ be a scheme. I'm using the following definition from Görtz-Wedhorn. A closed subscheme of $(X,mathcal O_X)$ is a scheme $(Z,mathcal O_Z)$, where $icolon Zhookrightarrow X$ is a closed subset of $X$, and for which there exists an ideal sheaf $mathcal Isubsetmathcal O_X$ together with an isomorphism $mathcal O_X/mathcal Icong i_*mathcal O_Z$.
Elsewhere I've seen the definition that a closed subscheme is a scheme of the form $(Z,i^{-1}(mathcal O_X/mathcal I))$, where $Zsubset X$ is a closed subset and $mathcal Isubsetmathcal O_X$ is an ideal sheaf.
It seems like these definitions are equivalent. Certainly the first implies the second, since $i^{-1}i_*mathcal O_z=mathcal O_Z$. How do I show the other implication?
algebraic-geometry schemes
$endgroup$
add a comment |
$begingroup$
Let $(X,mathcal O_X)$ be a scheme. I'm using the following definition from Görtz-Wedhorn. A closed subscheme of $(X,mathcal O_X)$ is a scheme $(Z,mathcal O_Z)$, where $icolon Zhookrightarrow X$ is a closed subset of $X$, and for which there exists an ideal sheaf $mathcal Isubsetmathcal O_X$ together with an isomorphism $mathcal O_X/mathcal Icong i_*mathcal O_Z$.
Elsewhere I've seen the definition that a closed subscheme is a scheme of the form $(Z,i^{-1}(mathcal O_X/mathcal I))$, where $Zsubset X$ is a closed subset and $mathcal Isubsetmathcal O_X$ is an ideal sheaf.
It seems like these definitions are equivalent. Certainly the first implies the second, since $i^{-1}i_*mathcal O_z=mathcal O_Z$. How do I show the other implication?
algebraic-geometry schemes
$endgroup$
add a comment |
$begingroup$
Let $(X,mathcal O_X)$ be a scheme. I'm using the following definition from Görtz-Wedhorn. A closed subscheme of $(X,mathcal O_X)$ is a scheme $(Z,mathcal O_Z)$, where $icolon Zhookrightarrow X$ is a closed subset of $X$, and for which there exists an ideal sheaf $mathcal Isubsetmathcal O_X$ together with an isomorphism $mathcal O_X/mathcal Icong i_*mathcal O_Z$.
Elsewhere I've seen the definition that a closed subscheme is a scheme of the form $(Z,i^{-1}(mathcal O_X/mathcal I))$, where $Zsubset X$ is a closed subset and $mathcal Isubsetmathcal O_X$ is an ideal sheaf.
It seems like these definitions are equivalent. Certainly the first implies the second, since $i^{-1}i_*mathcal O_z=mathcal O_Z$. How do I show the other implication?
algebraic-geometry schemes
$endgroup$
Let $(X,mathcal O_X)$ be a scheme. I'm using the following definition from Görtz-Wedhorn. A closed subscheme of $(X,mathcal O_X)$ is a scheme $(Z,mathcal O_Z)$, where $icolon Zhookrightarrow X$ is a closed subset of $X$, and for which there exists an ideal sheaf $mathcal Isubsetmathcal O_X$ together with an isomorphism $mathcal O_X/mathcal Icong i_*mathcal O_Z$.
Elsewhere I've seen the definition that a closed subscheme is a scheme of the form $(Z,i^{-1}(mathcal O_X/mathcal I))$, where $Zsubset X$ is a closed subset and $mathcal Isubsetmathcal O_X$ is an ideal sheaf.
It seems like these definitions are equivalent. Certainly the first implies the second, since $i^{-1}i_*mathcal O_z=mathcal O_Z$. How do I show the other implication?
algebraic-geometry schemes
algebraic-geometry schemes
asked Jul 9 '17 at 14:17
user363520
add a comment |
add a comment |
1 Answer
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$begingroup$
Suppose you have a closed subscheme $(Z,mathcal{O}_Z)$ according to the second definition. Let $mathcal{J}$ be the ideal sheaf on $X$ defined by saying $finmathcal{J}(U)$ iff for each $pin Ucap Z$, $f_pinmathcal{I}_p$. That is, $mathcal{J}$ consists of sections of $mathcal{O}_X$ which are locally in $mathcal{I}$ at every point of $Z$. We then have $i^{-1}(mathcal{O}_X/mathcal{J})cong i^{-1}(mathcal{O}_X/mathcal{I})= mathcal{O}_Z$ since $mathcal{I}$ and $mathcal{J}$ have the same stalks at points of $Z$, so $(Z,mathcal{O}_Z)$ also satisfies the second definition with respect to $mathcal{J}$. I claim that $(Z,mathcal{O}_Z)$ satisfies the first definition with respect to $mathcal{J}$.
The isomorphism $F:i^{-1}(mathcal{O}_X/mathcal{J})tomathcal{O}_Z$ induces a homomorphism $G:mathcal{O}_X/mathcal{J}to i_*mathcal{O}_Z$. We wish to show $G$ is an isomorphism. To prove this, we look at stalks. If $pin Z$, then the stalk of $i_*mathcal{O}_Z=i_*i^{-1}(mathcal{O}_X/mathcal{J})$ at $p$ is just the stalk of $mathcal{O}_X/mathcal{J}$ at $p$, so $G$ is an isomorphism on the stalks at $p$. If $pin Xsetminus Z$, then the stalk of $i_*mathcal{O}_Z$ at $p$ is $0$, so it suffices to show the stalk of $mathcal{O}_X/mathcal{J}$ at $p$ is $0$. But this is immediate from the definition of $mathcal{J}$: $mathcal{J}$ is all of $mathcal{O}_X$ on $Xsetminus Z$, since the test for $f$ to be an element of $mathcal{J}(U)$ is vacuous if $U$ is disjoint from $Z$.
Note that it is not necessarily true that $(Z,mathcal{O}_Z)$ satisfies the first definition with respect to $mathcal{I}$. For instance, if $Z=emptyset$, then the second definition is satisfied for any ideal $mathcal{I}$ at all, but the first definition can only be satisfied if $mathcal{I}=mathcal{O}_X$. More generally, the first definition can only be satisfied if $mathcal{I}$ is all of $mathcal{O}_X$ on $Xsetminus Z$, but the second definition doesn't care what $mathcal{I}$ is outside a neighborhood of $Z$.
answered Jul 9 '17 at 19:14
Eric WofseyEric Wofsey
192k14217351
$endgroup$
$begingroup$
Thanks! I'm following you until the start of the 2nd paragraph. I guess $G$ is the composite $mathcal O_X/mathcal Jto i_*i^{-1}(mathcal O_X/mathcal J)to i_*mathcal O_Z$. I'm having trouble seeing what the first morphism does on stalks.
$endgroup$
– user363520
Jul 9 '17 at 20:10
$begingroup$
Let $mathcal{F}$ be any sheaf on $X$. The stalk of $mathcal{F}$ at a point $pin Z$ is the direct limit of $mathcal{F}(U)$ over open neighborhoods $U$ of $p$. The stalk of $i_*i^{-1}(mathcal{F})$ at $p$ is the direct limit of $i_*i^{-1}(mathcal{F})(U)=i^{-1}(mathcal{F})(Zcap U)$. But $i^{-1}(mathcal{F})(Zcap U)$ is just the direct limit of $mathcal{F}(V)$ where $V$ ranges over open neighborhoods of $Zcap U$.
$endgroup$
– Eric Wofsey
Jul 9 '17 at 20:31
$begingroup$
So an element of the stalk of $i_*i^{-1}(mathcal{F})$ is represented by an element of $mathcal{F}(V)$ where $V$ is an arbitrarily small neighborhood of $Zcap U$ where $U$ is an arbitrarily small neighborhood of $p$. As $U$ varies over all open neighborhoods of $p$, $V$ will vary over all open neighborhoods of $p$ as well, and so the stalk of $i_*i^{-1}(mathcal{F})$ is the same as the stalk of $mathcal{F}$, via the canonical map.
$endgroup$
– Eric Wofsey
Jul 9 '17 at 20:34
add a comment |
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$begingroup$
Suppose you have a closed subscheme $(Z,mathcal{O}_Z)$ according to the second definition. Let $mathcal{J}$ be the ideal sheaf on $X$ defined by saying $finmathcal{J}(U)$ iff for each $pin Ucap Z$, $f_pinmathcal{I}_p$. That is, $mathcal{J}$ consists of sections of $mathcal{O}_X$ which are locally in $mathcal{I}$ at every point of $Z$. We then have $i^{-1}(mathcal{O}_X/mathcal{J})cong i^{-1}(mathcal{O}_X/mathcal{I})= mathcal{O}_Z$ since $mathcal{I}$ and $mathcal{J}$ have the same stalks at points of $Z$, so $(Z,mathcal{O}_Z)$ also satisfies the second definition with respect to $mathcal{J}$. I claim that $(Z,mathcal{O}_Z)$ satisfies the first definition with respect to $mathcal{J}$.
The isomorphism $F:i^{-1}(mathcal{O}_X/mathcal{J})tomathcal{O}_Z$ induces a homomorphism $G:mathcal{O}_X/mathcal{J}to i_*mathcal{O}_Z$. We wish to show $G$ is an isomorphism. To prove this, we look at stalks. If $pin Z$, then the stalk of $i_*mathcal{O}_Z=i_*i^{-1}(mathcal{O}_X/mathcal{J})$ at $p$ is just the stalk of $mathcal{O}_X/mathcal{J}$ at $p$, so $G$ is an isomorphism on the stalks at $p$. If $pin Xsetminus Z$, then the stalk of $i_*mathcal{O}_Z$ at $p$ is $0$, so it suffices to show the stalk of $mathcal{O}_X/mathcal{J}$ at $p$ is $0$. But this is immediate from the definition of $mathcal{J}$: $mathcal{J}$ is all of $mathcal{O}_X$ on $Xsetminus Z$, since the test for $f$ to be an element of $mathcal{J}(U)$ is vacuous if $U$ is disjoint from $Z$.
Note that it is not necessarily true that $(Z,mathcal{O}_Z)$ satisfies the first definition with respect to $mathcal{I}$. For instance, if $Z=emptyset$, then the second definition is satisfied for any ideal $mathcal{I}$ at all, but the first definition can only be satisfied if $mathcal{I}=mathcal{O}_X$. More generally, the first definition can only be satisfied if $mathcal{I}$ is all of $mathcal{O}_X$ on $Xsetminus Z$, but the second definition doesn't care what $mathcal{I}$ is outside a neighborhood of $Z$.
answered Jul 9 '17 at 19:14
Eric WofseyEric Wofsey
192k14217351
$endgroup$
$begingroup$
Thanks! I'm following you until the start of the 2nd paragraph. I guess $G$ is the composite $mathcal O_X/mathcal Jto i_*i^{-1}(mathcal O_X/mathcal J)to i_*mathcal O_Z$. I'm having trouble seeing what the first morphism does on stalks.
$endgroup$
– user363520
Jul 9 '17 at 20:10
$begingroup$
Let $mathcal{F}$ be any sheaf on $X$. The stalk of $mathcal{F}$ at a point $pin Z$ is the direct limit of $mathcal{F}(U)$ over open neighborhoods $U$ of $p$. The stalk of $i_*i^{-1}(mathcal{F})$ at $p$ is the direct limit of $i_*i^{-1}(mathcal{F})(U)=i^{-1}(mathcal{F})(Zcap U)$. But $i^{-1}(mathcal{F})(Zcap U)$ is just the direct limit of $mathcal{F}(V)$ where $V$ ranges over open neighborhoods of $Zcap U$.
$endgroup$
– Eric Wofsey
Jul 9 '17 at 20:31
$begingroup$
So an element of the stalk of $i_*i^{-1}(mathcal{F})$ is represented by an element of $mathcal{F}(V)$ where $V$ is an arbitrarily small neighborhood of $Zcap U$ where $U$ is an arbitrarily small neighborhood of $p$. As $U$ varies over all open neighborhoods of $p$, $V$ will vary over all open neighborhoods of $p$ as well, and so the stalk of $i_*i^{-1}(mathcal{F})$ is the same as the stalk of $mathcal{F}$, via the canonical map.
$endgroup$
– Eric Wofsey
Jul 9 '17 at 20:34
add a comment |
$begingroup$
Suppose you have a closed subscheme $(Z,mathcal{O}_Z)$ according to the second definition. Let $mathcal{J}$ be the ideal sheaf on $X$ defined by saying $finmathcal{J}(U)$ iff for each $pin Ucap Z$, $f_pinmathcal{I}_p$. That is, $mathcal{J}$ consists of sections of $mathcal{O}_X$ which are locally in $mathcal{I}$ at every point of $Z$. We then have $i^{-1}(mathcal{O}_X/mathcal{J})cong i^{-1}(mathcal{O}_X/mathcal{I})= mathcal{O}_Z$ since $mathcal{I}$ and $mathcal{J}$ have the same stalks at points of $Z$, so $(Z,mathcal{O}_Z)$ also satisfies the second definition with respect to $mathcal{J}$. I claim that $(Z,mathcal{O}_Z)$ satisfies the first definition with respect to $mathcal{J}$.
The isomorphism $F:i^{-1}(mathcal{O}_X/mathcal{J})tomathcal{O}_Z$ induces a homomorphism $G:mathcal{O}_X/mathcal{J}to i_*mathcal{O}_Z$. We wish to show $G$ is an isomorphism. To prove this, we look at stalks. If $pin Z$, then the stalk of $i_*mathcal{O}_Z=i_*i^{-1}(mathcal{O}_X/mathcal{J})$ at $p$ is just the stalk of $mathcal{O}_X/mathcal{J}$ at $p$, so $G$ is an isomorphism on the stalks at $p$. If $pin Xsetminus Z$, then the stalk of $i_*mathcal{O}_Z$ at $p$ is $0$, so it suffices to show the stalk of $mathcal{O}_X/mathcal{J}$ at $p$ is $0$. But this is immediate from the definition of $mathcal{J}$: $mathcal{J}$ is all of $mathcal{O}_X$ on $Xsetminus Z$, since the test for $f$ to be an element of $mathcal{J}(U)$ is vacuous if $U$ is disjoint from $Z$.
Note that it is not necessarily true that $(Z,mathcal{O}_Z)$ satisfies the first definition with respect to $mathcal{I}$. For instance, if $Z=emptyset$, then the second definition is satisfied for any ideal $mathcal{I}$ at all, but the first definition can only be satisfied if $mathcal{I}=mathcal{O}_X$. More generally, the first definition can only be satisfied if $mathcal{I}$ is all of $mathcal{O}_X$ on $Xsetminus Z$, but the second definition doesn't care what $mathcal{I}$ is outside a neighborhood of $Z$.
answered Jul 9 '17 at 19:14
Eric WofseyEric Wofsey
192k14217351
$endgroup$
$begingroup$
Thanks! I'm following you until the start of the 2nd paragraph. I guess $G$ is the composite $mathcal O_X/mathcal Jto i_*i^{-1}(mathcal O_X/mathcal J)to i_*mathcal O_Z$. I'm having trouble seeing what the first morphism does on stalks.
$endgroup$
– user363520
Jul 9 '17 at 20:10
$begingroup$
Let $mathcal{F}$ be any sheaf on $X$. The stalk of $mathcal{F}$ at a point $pin Z$ is the direct limit of $mathcal{F}(U)$ over open neighborhoods $U$ of $p$. The stalk of $i_*i^{-1}(mathcal{F})$ at $p$ is the direct limit of $i_*i^{-1}(mathcal{F})(U)=i^{-1}(mathcal{F})(Zcap U)$. But $i^{-1}(mathcal{F})(Zcap U)$ is just the direct limit of $mathcal{F}(V)$ where $V$ ranges over open neighborhoods of $Zcap U$.
$endgroup$
– Eric Wofsey
Jul 9 '17 at 20:31
$begingroup$
So an element of the stalk of $i_*i^{-1}(mathcal{F})$ is represented by an element of $mathcal{F}(V)$ where $V$ is an arbitrarily small neighborhood of $Zcap U$ where $U$ is an arbitrarily small neighborhood of $p$. As $U$ varies over all open neighborhoods of $p$, $V$ will vary over all open neighborhoods of $p$ as well, and so the stalk of $i_*i^{-1}(mathcal{F})$ is the same as the stalk of $mathcal{F}$, via the canonical map.
$endgroup$
– Eric Wofsey
Jul 9 '17 at 20:34
add a comment |
$begingroup$
Suppose you have a closed subscheme $(Z,mathcal{O}_Z)$ according to the second definition. Let $mathcal{J}$ be the ideal sheaf on $X$ defined by saying $finmathcal{J}(U)$ iff for each $pin Ucap Z$, $f_pinmathcal{I}_p$. That is, $mathcal{J}$ consists of sections of $mathcal{O}_X$ which are locally in $mathcal{I}$ at every point of $Z$. We then have $i^{-1}(mathcal{O}_X/mathcal{J})cong i^{-1}(mathcal{O}_X/mathcal{I})= mathcal{O}_Z$ since $mathcal{I}$ and $mathcal{J}$ have the same stalks at points of $Z$, so $(Z,mathcal{O}_Z)$ also satisfies the second definition with respect to $mathcal{J}$. I claim that $(Z,mathcal{O}_Z)$ satisfies the first definition with respect to $mathcal{J}$.
The isomorphism $F:i^{-1}(mathcal{O}_X/mathcal{J})tomathcal{O}_Z$ induces a homomorphism $G:mathcal{O}_X/mathcal{J}to i_*mathcal{O}_Z$. We wish to show $G$ is an isomorphism. To prove this, we look at stalks. If $pin Z$, then the stalk of $i_*mathcal{O}_Z=i_*i^{-1}(mathcal{O}_X/mathcal{J})$ at $p$ is just the stalk of $mathcal{O}_X/mathcal{J}$ at $p$, so $G$ is an isomorphism on the stalks at $p$. If $pin Xsetminus Z$, then the stalk of $i_*mathcal{O}_Z$ at $p$ is $0$, so it suffices to show the stalk of $mathcal{O}_X/mathcal{J}$ at $p$ is $0$. But this is immediate from the definition of $mathcal{J}$: $mathcal{J}$ is all of $mathcal{O}_X$ on $Xsetminus Z$, since the test for $f$ to be an element of $mathcal{J}(U)$ is vacuous if $U$ is disjoint from $Z$.
Note that it is not necessarily true that $(Z,mathcal{O}_Z)$ satisfies the first definition with respect to $mathcal{I}$. For instance, if $Z=emptyset$, then the second definition is satisfied for any ideal $mathcal{I}$ at all, but the first definition can only be satisfied if $mathcal{I}=mathcal{O}_X$. More generally, the first definition can only be satisfied if $mathcal{I}$ is all of $mathcal{O}_X$ on $Xsetminus Z$, but the second definition doesn't care what $mathcal{I}$ is outside a neighborhood of $Z$.
answered Jul 9 '17 at 19:14
Eric WofseyEric Wofsey
192k14217351
$endgroup$
Suppose you have a closed subscheme $(Z,mathcal{O}_Z)$ according to the second definition. Let $mathcal{J}$ be the ideal sheaf on $X$ defined by saying $finmathcal{J}(U)$ iff for each $pin Ucap Z$, $f_pinmathcal{I}_p$. That is, $mathcal{J}$ consists of sections of $mathcal{O}_X$ which are locally in $mathcal{I}$ at every point of $Z$. We then have $i^{-1}(mathcal{O}_X/mathcal{J})cong i^{-1}(mathcal{O}_X/mathcal{I})= mathcal{O}_Z$ since $mathcal{I}$ and $mathcal{J}$ have the same stalks at points of $Z$, so $(Z,mathcal{O}_Z)$ also satisfies the second definition with respect to $mathcal{J}$. I claim that $(Z,mathcal{O}_Z)$ satisfies the first definition with respect to $mathcal{J}$.
The isomorphism $F:i^{-1}(mathcal{O}_X/mathcal{J})tomathcal{O}_Z$ induces a homomorphism $G:mathcal{O}_X/mathcal{J}to i_*mathcal{O}_Z$. We wish to show $G$ is an isomorphism. To prove this, we look at stalks. If $pin Z$, then the stalk of $i_*mathcal{O}_Z=i_*i^{-1}(mathcal{O}_X/mathcal{J})$ at $p$ is just the stalk of $mathcal{O}_X/mathcal{J}$ at $p$, so $G$ is an isomorphism on the stalks at $p$. If $pin Xsetminus Z$, then the stalk of $i_*mathcal{O}_Z$ at $p$ is $0$, so it suffices to show the stalk of $mathcal{O}_X/mathcal{J}$ at $p$ is $0$. But this is immediate from the definition of $mathcal{J}$: $mathcal{J}$ is all of $mathcal{O}_X$ on $Xsetminus Z$, since the test for $f$ to be an element of $mathcal{J}(U)$ is vacuous if $U$ is disjoint from $Z$.
Note that it is not necessarily true that $(Z,mathcal{O}_Z)$ satisfies the first definition with respect to $mathcal{I}$. For instance, if $Z=emptyset$, then the second definition is satisfied for any ideal $mathcal{I}$ at all, but the first definition can only be satisfied if $mathcal{I}=mathcal{O}_X$. More generally, the first definition can only be satisfied if $mathcal{I}$ is all of $mathcal{O}_X$ on $Xsetminus Z$, but the second definition doesn't care what $mathcal{I}$ is outside a neighborhood of $Z$.
answered Jul 9 '17 at 19:14
Eric WofseyEric Wofsey
192k14217351
answered Jul 9 '17 at 19:14
Eric WofseyEric Wofsey
192k14217351
answered Jul 9 '17 at 19:14
Eric WofseyEric Wofsey
192k14217351
answered Jul 9 '17 at 19:14
Eric WofseyEric Wofsey
192k14217351
192k14217351
$begingroup$
Thanks! I'm following you until the start of the 2nd paragraph. I guess $G$ is the composite $mathcal O_X/mathcal Jto i_*i^{-1}(mathcal O_X/mathcal J)to i_*mathcal O_Z$. I'm having trouble seeing what the first morphism does on stalks.
$endgroup$
– user363520
Jul 9 '17 at 20:10
$begingroup$
Let $mathcal{F}$ be any sheaf on $X$. The stalk of $mathcal{F}$ at a point $pin Z$ is the direct limit of $mathcal{F}(U)$ over open neighborhoods $U$ of $p$. The stalk of $i_*i^{-1}(mathcal{F})$ at $p$ is the direct limit of $i_*i^{-1}(mathcal{F})(U)=i^{-1}(mathcal{F})(Zcap U)$. But $i^{-1}(mathcal{F})(Zcap U)$ is just the direct limit of $mathcal{F}(V)$ where $V$ ranges over open neighborhoods of $Zcap U$.
$endgroup$
– Eric Wofsey
Jul 9 '17 at 20:31
$begingroup$
So an element of the stalk of $i_*i^{-1}(mathcal{F})$ is represented by an element of $mathcal{F}(V)$ where $V$ is an arbitrarily small neighborhood of $Zcap U$ where $U$ is an arbitrarily small neighborhood of $p$. As $U$ varies over all open neighborhoods of $p$, $V$ will vary over all open neighborhoods of $p$ as well, and so the stalk of $i_*i^{-1}(mathcal{F})$ is the same as the stalk of $mathcal{F}$, via the canonical map.
$endgroup$
– Eric Wofsey
Jul 9 '17 at 20:34
add a comment |
$begingroup$
Thanks! I'm following you until the start of the 2nd paragraph. I guess $G$ is the composite $mathcal O_X/mathcal Jto i_*i^{-1}(mathcal O_X/mathcal J)to i_*mathcal O_Z$. I'm having trouble seeing what the first morphism does on stalks.
$endgroup$
– user363520
Jul 9 '17 at 20:10
$begingroup$
Let $mathcal{F}$ be any sheaf on $X$. The stalk of $mathcal{F}$ at a point $pin Z$ is the direct limit of $mathcal{F}(U)$ over open neighborhoods $U$ of $p$. The stalk of $i_*i^{-1}(mathcal{F})$ at $p$ is the direct limit of $i_*i^{-1}(mathcal{F})(U)=i^{-1}(mathcal{F})(Zcap U)$. But $i^{-1}(mathcal{F})(Zcap U)$ is just the direct limit of $mathcal{F}(V)$ where $V$ ranges over open neighborhoods of $Zcap U$.
$endgroup$
– Eric Wofsey
Jul 9 '17 at 20:31
$begingroup$
So an element of the stalk of $i_*i^{-1}(mathcal{F})$ is represented by an element of $mathcal{F}(V)$ where $V$ is an arbitrarily small neighborhood of $Zcap U$ where $U$ is an arbitrarily small neighborhood of $p$. As $U$ varies over all open neighborhoods of $p$, $V$ will vary over all open neighborhoods of $p$ as well, and so the stalk of $i_*i^{-1}(mathcal{F})$ is the same as the stalk of $mathcal{F}$, via the canonical map.
$endgroup$
– Eric Wofsey
Jul 9 '17 at 20:34
$begingroup$
Thanks! I'm following you until the start of the 2nd paragraph. I guess $G$ is the composite $mathcal O_X/mathcal Jto i_*i^{-1}(mathcal O_X/mathcal J)to i_*mathcal O_Z$. I'm having trouble seeing what the first morphism does on stalks.
$endgroup$
– user363520
Jul 9 '17 at 20:10
$begingroup$
Thanks! I'm following you until the start of the 2nd paragraph. I guess $G$ is the composite $mathcal O_X/mathcal Jto i_*i^{-1}(mathcal O_X/mathcal J)to i_*mathcal O_Z$. I'm having trouble seeing what the first morphism does on stalks.
$endgroup$
– user363520
Jul 9 '17 at 20:10
$begingroup$
Let $mathcal{F}$ be any sheaf on $X$. The stalk of $mathcal{F}$ at a point $pin Z$ is the direct limit of $mathcal{F}(U)$ over open neighborhoods $U$ of $p$. The stalk of $i_*i^{-1}(mathcal{F})$ at $p$ is the direct limit of $i_*i^{-1}(mathcal{F})(U)=i^{-1}(mathcal{F})(Zcap U)$. But $i^{-1}(mathcal{F})(Zcap U)$ is just the direct limit of $mathcal{F}(V)$ where $V$ ranges over open neighborhoods of $Zcap U$.
$endgroup$
– Eric Wofsey
Jul 9 '17 at 20:31
$begingroup$
Let $mathcal{F}$ be any sheaf on $X$. The stalk of $mathcal{F}$ at a point $pin Z$ is the direct limit of $mathcal{F}(U)$ over open neighborhoods $U$ of $p$. The stalk of $i_*i^{-1}(mathcal{F})$ at $p$ is the direct limit of $i_*i^{-1}(mathcal{F})(U)=i^{-1}(mathcal{F})(Zcap U)$. But $i^{-1}(mathcal{F})(Zcap U)$ is just the direct limit of $mathcal{F}(V)$ where $V$ ranges over open neighborhoods of $Zcap U$.
$endgroup$
– Eric Wofsey
Jul 9 '17 at 20:31
$begingroup$
So an element of the stalk of $i_*i^{-1}(mathcal{F})$ is represented by an element of $mathcal{F}(V)$ where $V$ is an arbitrarily small neighborhood of $Zcap U$ where $U$ is an arbitrarily small neighborhood of $p$. As $U$ varies over all open neighborhoods of $p$, $V$ will vary over all open neighborhoods of $p$ as well, and so the stalk of $i_*i^{-1}(mathcal{F})$ is the same as the stalk of $mathcal{F}$, via the canonical map.
$endgroup$
– Eric Wofsey
Jul 9 '17 at 20:34
$begingroup$
So an element of the stalk of $i_*i^{-1}(mathcal{F})$ is represented by an element of $mathcal{F}(V)$ where $V$ is an arbitrarily small neighborhood of $Zcap U$ where $U$ is an arbitrarily small neighborhood of $p$. As $U$ varies over all open neighborhoods of $p$, $V$ will vary over all open neighborhoods of $p$ as well, and so the stalk of $i_*i^{-1}(mathcal{F})$ is the same as the stalk of $mathcal{F}$, via the canonical map.
$endgroup$
– Eric Wofsey
Jul 9 '17 at 20:34
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