Does every sheaf embed into a quasicoherent sheaf?
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Question. Let $X$ be a scheme. Let $mathcalE$ be a sheaf of $mathcalO_X$-modules. Is there always a quasicoherent sheaf $mathcalE'$ together with a monomorphism $mathcalE to mathcalE'$?
Remark. The coherator yields a way to find a quasicoherent sheaf together with a morphism to $mathcalE$. But I'm interested in finding a quasicoherent sheaf together with a monomorphism from $mathcalE$.
Motivation. There is a way to set up the theory of sheaf cohomology for quasicoherent sheaves without injective or flabby resolutions. If any sheaf of modules would embed into a quasicoherent one, we might be able to extend this development to arbitrary (not necessarily quasicoherent) sheaves of modules.
ag.algebraic-geometry sheaf-cohomology quasi-coherent-modules
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Question. Let $X$ be a scheme. Let $mathcalE$ be a sheaf of $mathcalO_X$-modules. Is there always a quasicoherent sheaf $mathcalE'$ together with a monomorphism $mathcalE to mathcalE'$?
Remark. The coherator yields a way to find a quasicoherent sheaf together with a morphism to $mathcalE$. But I'm interested in finding a quasicoherent sheaf together with a monomorphism from $mathcalE$.
Motivation. There is a way to set up the theory of sheaf cohomology for quasicoherent sheaves without injective or flabby resolutions. If any sheaf of modules would embed into a quasicoherent one, we might be able to extend this development to arbitrary (not necessarily quasicoherent) sheaves of modules.
ag.algebraic-geometry sheaf-cohomology quasi-coherent-modules
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Question. Let $X$ be a scheme. Let $mathcalE$ be a sheaf of $mathcalO_X$-modules. Is there always a quasicoherent sheaf $mathcalE'$ together with a monomorphism $mathcalE to mathcalE'$?
Remark. The coherator yields a way to find a quasicoherent sheaf together with a morphism to $mathcalE$. But I'm interested in finding a quasicoherent sheaf together with a monomorphism from $mathcalE$.
Motivation. There is a way to set up the theory of sheaf cohomology for quasicoherent sheaves without injective or flabby resolutions. If any sheaf of modules would embed into a quasicoherent one, we might be able to extend this development to arbitrary (not necessarily quasicoherent) sheaves of modules.
ag.algebraic-geometry sheaf-cohomology quasi-coherent-modules
Question. Let $X$ be a scheme. Let $mathcalE$ be a sheaf of $mathcalO_X$-modules. Is there always a quasicoherent sheaf $mathcalE'$ together with a monomorphism $mathcalE to mathcalE'$?
Remark. The coherator yields a way to find a quasicoherent sheaf together with a morphism to $mathcalE$. But I'm interested in finding a quasicoherent sheaf together with a monomorphism from $mathcalE$.
Motivation. There is a way to set up the theory of sheaf cohomology for quasicoherent sheaves without injective or flabby resolutions. If any sheaf of modules would embed into a quasicoherent one, we might be able to extend this development to arbitrary (not necessarily quasicoherent) sheaves of modules.
ag.algebraic-geometry sheaf-cohomology quasi-coherent-modules
ag.algebraic-geometry sheaf-cohomology quasi-coherent-modules
asked 3 hours ago
Ingo Blechschmidt
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That already fails for $X$ equal to $textSpec R$, where $R$ is a DVR with generic point $eta = textSpec K$. Since there are only two nonempty open subsets of $X$, namely all of $X$ and $eta$, there is a straightforward equivalence between the category of $mathcalO_X$-modules and the category of triples $(M,V,phi)$ of an $R$-module $M$, a $K$-module $V$, and an $R$-module homomorphism $$phi:Mto V.$$ This is quasi-coherent if and only if $phi$ induces an isomorphism $$Motimes_R K xrightarrowcong V,$$ i.e., the $mathcalO_X$-module is equivalent to $$(M,Motimes_R K,iota_M).$$ In particular, consider the $mathcalO_X$-module $$(R,0,0).$$
For every $mathcalO_X$-module homomorphism of this $mathcalO_X$-module to a quasi-coherent $mathcalO_X$-module, $$(psi_R,psi_eta):(R,0,0) to (M,Motimes_R K,iota_M),$$ the composite $iota_Mcirc psi_R$ equals $0$. Thus, the image $psi_R(R)$ is contained in the torsion submodule of $M$. Every torsion quotient of $R$ has nonzero kernel. Thus, $(psi_R,psi_eta)$ is not a monomorphism.
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
That already fails for $X$ equal to $textSpec R$, where $R$ is a DVR with generic point $eta = textSpec K$. Since there are only two nonempty open subsets of $X$, namely all of $X$ and $eta$, there is a straightforward equivalence between the category of $mathcalO_X$-modules and the category of triples $(M,V,phi)$ of an $R$-module $M$, a $K$-module $V$, and an $R$-module homomorphism $$phi:Mto V.$$ This is quasi-coherent if and only if $phi$ induces an isomorphism $$Motimes_R K xrightarrowcong V,$$ i.e., the $mathcalO_X$-module is equivalent to $$(M,Motimes_R K,iota_M).$$ In particular, consider the $mathcalO_X$-module $$(R,0,0).$$
For every $mathcalO_X$-module homomorphism of this $mathcalO_X$-module to a quasi-coherent $mathcalO_X$-module, $$(psi_R,psi_eta):(R,0,0) to (M,Motimes_R K,iota_M),$$ the composite $iota_Mcirc psi_R$ equals $0$. Thus, the image $psi_R(R)$ is contained in the torsion submodule of $M$. Every torsion quotient of $R$ has nonzero kernel. Thus, $(psi_R,psi_eta)$ is not a monomorphism.
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That already fails for $X$ equal to $textSpec R$, where $R$ is a DVR with generic point $eta = textSpec K$. Since there are only two nonempty open subsets of $X$, namely all of $X$ and $eta$, there is a straightforward equivalence between the category of $mathcalO_X$-modules and the category of triples $(M,V,phi)$ of an $R$-module $M$, a $K$-module $V$, and an $R$-module homomorphism $$phi:Mto V.$$ This is quasi-coherent if and only if $phi$ induces an isomorphism $$Motimes_R K xrightarrowcong V,$$ i.e., the $mathcalO_X$-module is equivalent to $$(M,Motimes_R K,iota_M).$$ In particular, consider the $mathcalO_X$-module $$(R,0,0).$$
For every $mathcalO_X$-module homomorphism of this $mathcalO_X$-module to a quasi-coherent $mathcalO_X$-module, $$(psi_R,psi_eta):(R,0,0) to (M,Motimes_R K,iota_M),$$ the composite $iota_Mcirc psi_R$ equals $0$. Thus, the image $psi_R(R)$ is contained in the torsion submodule of $M$. Every torsion quotient of $R$ has nonzero kernel. Thus, $(psi_R,psi_eta)$ is not a monomorphism.
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up vote
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That already fails for $X$ equal to $textSpec R$, where $R$ is a DVR with generic point $eta = textSpec K$. Since there are only two nonempty open subsets of $X$, namely all of $X$ and $eta$, there is a straightforward equivalence between the category of $mathcalO_X$-modules and the category of triples $(M,V,phi)$ of an $R$-module $M$, a $K$-module $V$, and an $R$-module homomorphism $$phi:Mto V.$$ This is quasi-coherent if and only if $phi$ induces an isomorphism $$Motimes_R K xrightarrowcong V,$$ i.e., the $mathcalO_X$-module is equivalent to $$(M,Motimes_R K,iota_M).$$ In particular, consider the $mathcalO_X$-module $$(R,0,0).$$
For every $mathcalO_X$-module homomorphism of this $mathcalO_X$-module to a quasi-coherent $mathcalO_X$-module, $$(psi_R,psi_eta):(R,0,0) to (M,Motimes_R K,iota_M),$$ the composite $iota_Mcirc psi_R$ equals $0$. Thus, the image $psi_R(R)$ is contained in the torsion submodule of $M$. Every torsion quotient of $R$ has nonzero kernel. Thus, $(psi_R,psi_eta)$ is not a monomorphism.
That already fails for $X$ equal to $textSpec R$, where $R$ is a DVR with generic point $eta = textSpec K$. Since there are only two nonempty open subsets of $X$, namely all of $X$ and $eta$, there is a straightforward equivalence between the category of $mathcalO_X$-modules and the category of triples $(M,V,phi)$ of an $R$-module $M$, a $K$-module $V$, and an $R$-module homomorphism $$phi:Mto V.$$ This is quasi-coherent if and only if $phi$ induces an isomorphism $$Motimes_R K xrightarrowcong V,$$ i.e., the $mathcalO_X$-module is equivalent to $$(M,Motimes_R K,iota_M).$$ In particular, consider the $mathcalO_X$-module $$(R,0,0).$$
For every $mathcalO_X$-module homomorphism of this $mathcalO_X$-module to a quasi-coherent $mathcalO_X$-module, $$(psi_R,psi_eta):(R,0,0) to (M,Motimes_R K,iota_M),$$ the composite $iota_Mcirc psi_R$ equals $0$. Thus, the image $psi_R(R)$ is contained in the torsion submodule of $M$. Every torsion quotient of $R$ has nonzero kernel. Thus, $(psi_R,psi_eta)$ is not a monomorphism.
edited 2 hours ago
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Jason Starr
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