Locally ringed space with noetherian stalks and a non-coherent structural sheaf












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I am looking for a locally ringed space the stalks of which are noetherian and such that the structural sheaf is not coherent over itself. Can you provide me an example of this?



Notice that one may not find such an example which is a scheme, since schemes with noetherian stalks are locally noetherian and any locally noetherian scheme has a coherent structural sheaf (over itself).










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    $begingroup$


    I am looking for a locally ringed space the stalks of which are noetherian and such that the structural sheaf is not coherent over itself. Can you provide me an example of this?



    Notice that one may not find such an example which is a scheme, since schemes with noetherian stalks are locally noetherian and any locally noetherian scheme has a coherent structural sheaf (over itself).










    share|cite|improve this question







    New contributor




    Gaussian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







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      4





      $begingroup$


      I am looking for a locally ringed space the stalks of which are noetherian and such that the structural sheaf is not coherent over itself. Can you provide me an example of this?



      Notice that one may not find such an example which is a scheme, since schemes with noetherian stalks are locally noetherian and any locally noetherian scheme has a coherent structural sheaf (over itself).










      share|cite|improve this question







      New contributor




      Gaussian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      I am looking for a locally ringed space the stalks of which are noetherian and such that the structural sheaf is not coherent over itself. Can you provide me an example of this?



      Notice that one may not find such an example which is a scheme, since schemes with noetherian stalks are locally noetherian and any locally noetherian scheme has a coherent structural sheaf (over itself).







      sheaf-theory coherent-sheaves locally-ringed-spaces






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          Welcome new contributor. It is best to think about these kinds of examples on one's own. Thus, please try this for yourself before reading the following.



          Let $(Y,Z)$ be any pair of a Noetherian scheme $Y$ and a nonempty closed subset $Z$ that is nowhere dense. For instance, let $Y$ be the following Spec of a DVR, $Y=text{Spec} k[t]_{langle trangle}$, and let $Z$ be the singleton set ${z}$ of the $k$-point of the unique maximal ideal. Denote by $i$ the inclusion continuous function from $Z$ to $Y$.



          Consider the sheaf of $mathcal{O}_{Y}$-modules $i_*i^{-1}mathcal{O}_{Y}$. For every open subset $U$ of $Y$ that does not intersect $Z$, the only section of this sheaf on $U$ is the zero section, and thus the stalk at every $yin U$ is the zero module. Also, for every $zin Z$, the stalk at $z$ equals $mathcal{O}_{Y,z}$.



          Now form the sheaf of $mathcal{O}_Y$-modules $mathcal{O}_X:=mathcal{O}_Y oplus left( i_*i^{-1}mathcal{O}_Ycdot epsilon right)$, where $epsilon$ is just a placeholder. Give this the unique structure of $mathcal{O}_Y$-algebra such that $epsiloncdot epsilon$ equals $0$ and such that the following natural inclusion map is a morphism of $mathcal{O}_Y$-algebras, $$mathcal{O}_Y hookrightarrow mathcal{O}_Y oplus left( i_*i^{-1}mathcal{O}_Ycdot epsilon right), fmapsto (f,0cdot epsilon).$$



          For every $yin Ysetminus Z$, the stalk $mathcal{O}_{X,y}$ equals $mathcal{O}_{Y,y}$ as an $mathcal{O}_{Y,y}$-algebra. Also, for every $zin Z$, the $mathcal{O}_{Y,z}$-algebra $mathcal{O}_{X,z}$ equals $mathcal{O}_{Y,z}oplus mathcal{O}_{Y,z}cdot epsilon$ with $epsiloncdot epsilon = 0$. In every case, the stalk is a Noetherian local ring.



          There is a natural retraction of the above algebra homomorphism, $$mathcal{O}_Xto mathcal{O}_Y, (f,gcdot epsilon) mapsto f.$$ Consider the $mathcal{O}_X$-module morphism from $mathcal{O}_X$ to itself that multiplies by the global section $(0,1cdot epsilon)$. Denote the kernel sheaf by $mathcal{K}$. Denote by $mathcal{K}'$ the associated $mathcal{O}_Y$-module $mathcal{K}otimes_{mathcal{O}_X}mathcal{O}_Y$.



          If $mathcal{K}$ is locally finitely generated as an $mathcal{O}_X$-module, then also $mathcal{K}'$ is locally finitely generated as an $mathcal{O}_Y$-module. However, $mathcal{K}'$ is the extension by zero of the structure sheaf on $Ysetminus Z$. This is not a finitely generated $mathcal{O}_X$-module. In fact, for every irreducible open neighborhood $U$ that intersects $Z$, the sections of $mathcal{K}'$ on $U$ are zero, yet they are nonzero for $Usetminus Z$. Thus, every map $mathcal{O}_U^{oplus n}to mathcal{K}'|_U$ is the zero map, yet the restriction of $mathcal{K}'|_U$ to $Usetminus Z$ is nonzero.






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            $begingroup$

            Welcome new contributor. It is best to think about these kinds of examples on one's own. Thus, please try this for yourself before reading the following.



            Let $(Y,Z)$ be any pair of a Noetherian scheme $Y$ and a nonempty closed subset $Z$ that is nowhere dense. For instance, let $Y$ be the following Spec of a DVR, $Y=text{Spec} k[t]_{langle trangle}$, and let $Z$ be the singleton set ${z}$ of the $k$-point of the unique maximal ideal. Denote by $i$ the inclusion continuous function from $Z$ to $Y$.



            Consider the sheaf of $mathcal{O}_{Y}$-modules $i_*i^{-1}mathcal{O}_{Y}$. For every open subset $U$ of $Y$ that does not intersect $Z$, the only section of this sheaf on $U$ is the zero section, and thus the stalk at every $yin U$ is the zero module. Also, for every $zin Z$, the stalk at $z$ equals $mathcal{O}_{Y,z}$.



            Now form the sheaf of $mathcal{O}_Y$-modules $mathcal{O}_X:=mathcal{O}_Y oplus left( i_*i^{-1}mathcal{O}_Ycdot epsilon right)$, where $epsilon$ is just a placeholder. Give this the unique structure of $mathcal{O}_Y$-algebra such that $epsiloncdot epsilon$ equals $0$ and such that the following natural inclusion map is a morphism of $mathcal{O}_Y$-algebras, $$mathcal{O}_Y hookrightarrow mathcal{O}_Y oplus left( i_*i^{-1}mathcal{O}_Ycdot epsilon right), fmapsto (f,0cdot epsilon).$$



            For every $yin Ysetminus Z$, the stalk $mathcal{O}_{X,y}$ equals $mathcal{O}_{Y,y}$ as an $mathcal{O}_{Y,y}$-algebra. Also, for every $zin Z$, the $mathcal{O}_{Y,z}$-algebra $mathcal{O}_{X,z}$ equals $mathcal{O}_{Y,z}oplus mathcal{O}_{Y,z}cdot epsilon$ with $epsiloncdot epsilon = 0$. In every case, the stalk is a Noetherian local ring.



            There is a natural retraction of the above algebra homomorphism, $$mathcal{O}_Xto mathcal{O}_Y, (f,gcdot epsilon) mapsto f.$$ Consider the $mathcal{O}_X$-module morphism from $mathcal{O}_X$ to itself that multiplies by the global section $(0,1cdot epsilon)$. Denote the kernel sheaf by $mathcal{K}$. Denote by $mathcal{K}'$ the associated $mathcal{O}_Y$-module $mathcal{K}otimes_{mathcal{O}_X}mathcal{O}_Y$.



            If $mathcal{K}$ is locally finitely generated as an $mathcal{O}_X$-module, then also $mathcal{K}'$ is locally finitely generated as an $mathcal{O}_Y$-module. However, $mathcal{K}'$ is the extension by zero of the structure sheaf on $Ysetminus Z$. This is not a finitely generated $mathcal{O}_X$-module. In fact, for every irreducible open neighborhood $U$ that intersects $Z$, the sections of $mathcal{K}'$ on $U$ are zero, yet they are nonzero for $Usetminus Z$. Thus, every map $mathcal{O}_U^{oplus n}to mathcal{K}'|_U$ is the zero map, yet the restriction of $mathcal{K}'|_U$ to $Usetminus Z$ is nonzero.






            share|cite|improve this answer











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              5












              $begingroup$

              Welcome new contributor. It is best to think about these kinds of examples on one's own. Thus, please try this for yourself before reading the following.



              Let $(Y,Z)$ be any pair of a Noetherian scheme $Y$ and a nonempty closed subset $Z$ that is nowhere dense. For instance, let $Y$ be the following Spec of a DVR, $Y=text{Spec} k[t]_{langle trangle}$, and let $Z$ be the singleton set ${z}$ of the $k$-point of the unique maximal ideal. Denote by $i$ the inclusion continuous function from $Z$ to $Y$.



              Consider the sheaf of $mathcal{O}_{Y}$-modules $i_*i^{-1}mathcal{O}_{Y}$. For every open subset $U$ of $Y$ that does not intersect $Z$, the only section of this sheaf on $U$ is the zero section, and thus the stalk at every $yin U$ is the zero module. Also, for every $zin Z$, the stalk at $z$ equals $mathcal{O}_{Y,z}$.



              Now form the sheaf of $mathcal{O}_Y$-modules $mathcal{O}_X:=mathcal{O}_Y oplus left( i_*i^{-1}mathcal{O}_Ycdot epsilon right)$, where $epsilon$ is just a placeholder. Give this the unique structure of $mathcal{O}_Y$-algebra such that $epsiloncdot epsilon$ equals $0$ and such that the following natural inclusion map is a morphism of $mathcal{O}_Y$-algebras, $$mathcal{O}_Y hookrightarrow mathcal{O}_Y oplus left( i_*i^{-1}mathcal{O}_Ycdot epsilon right), fmapsto (f,0cdot epsilon).$$



              For every $yin Ysetminus Z$, the stalk $mathcal{O}_{X,y}$ equals $mathcal{O}_{Y,y}$ as an $mathcal{O}_{Y,y}$-algebra. Also, for every $zin Z$, the $mathcal{O}_{Y,z}$-algebra $mathcal{O}_{X,z}$ equals $mathcal{O}_{Y,z}oplus mathcal{O}_{Y,z}cdot epsilon$ with $epsiloncdot epsilon = 0$. In every case, the stalk is a Noetherian local ring.



              There is a natural retraction of the above algebra homomorphism, $$mathcal{O}_Xto mathcal{O}_Y, (f,gcdot epsilon) mapsto f.$$ Consider the $mathcal{O}_X$-module morphism from $mathcal{O}_X$ to itself that multiplies by the global section $(0,1cdot epsilon)$. Denote the kernel sheaf by $mathcal{K}$. Denote by $mathcal{K}'$ the associated $mathcal{O}_Y$-module $mathcal{K}otimes_{mathcal{O}_X}mathcal{O}_Y$.



              If $mathcal{K}$ is locally finitely generated as an $mathcal{O}_X$-module, then also $mathcal{K}'$ is locally finitely generated as an $mathcal{O}_Y$-module. However, $mathcal{K}'$ is the extension by zero of the structure sheaf on $Ysetminus Z$. This is not a finitely generated $mathcal{O}_X$-module. In fact, for every irreducible open neighborhood $U$ that intersects $Z$, the sections of $mathcal{K}'$ on $U$ are zero, yet they are nonzero for $Usetminus Z$. Thus, every map $mathcal{O}_U^{oplus n}to mathcal{K}'|_U$ is the zero map, yet the restriction of $mathcal{K}'|_U$ to $Usetminus Z$ is nonzero.






              share|cite|improve this answer











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                5












                5








                5





                $begingroup$

                Welcome new contributor. It is best to think about these kinds of examples on one's own. Thus, please try this for yourself before reading the following.



                Let $(Y,Z)$ be any pair of a Noetherian scheme $Y$ and a nonempty closed subset $Z$ that is nowhere dense. For instance, let $Y$ be the following Spec of a DVR, $Y=text{Spec} k[t]_{langle trangle}$, and let $Z$ be the singleton set ${z}$ of the $k$-point of the unique maximal ideal. Denote by $i$ the inclusion continuous function from $Z$ to $Y$.



                Consider the sheaf of $mathcal{O}_{Y}$-modules $i_*i^{-1}mathcal{O}_{Y}$. For every open subset $U$ of $Y$ that does not intersect $Z$, the only section of this sheaf on $U$ is the zero section, and thus the stalk at every $yin U$ is the zero module. Also, for every $zin Z$, the stalk at $z$ equals $mathcal{O}_{Y,z}$.



                Now form the sheaf of $mathcal{O}_Y$-modules $mathcal{O}_X:=mathcal{O}_Y oplus left( i_*i^{-1}mathcal{O}_Ycdot epsilon right)$, where $epsilon$ is just a placeholder. Give this the unique structure of $mathcal{O}_Y$-algebra such that $epsiloncdot epsilon$ equals $0$ and such that the following natural inclusion map is a morphism of $mathcal{O}_Y$-algebras, $$mathcal{O}_Y hookrightarrow mathcal{O}_Y oplus left( i_*i^{-1}mathcal{O}_Ycdot epsilon right), fmapsto (f,0cdot epsilon).$$



                For every $yin Ysetminus Z$, the stalk $mathcal{O}_{X,y}$ equals $mathcal{O}_{Y,y}$ as an $mathcal{O}_{Y,y}$-algebra. Also, for every $zin Z$, the $mathcal{O}_{Y,z}$-algebra $mathcal{O}_{X,z}$ equals $mathcal{O}_{Y,z}oplus mathcal{O}_{Y,z}cdot epsilon$ with $epsiloncdot epsilon = 0$. In every case, the stalk is a Noetherian local ring.



                There is a natural retraction of the above algebra homomorphism, $$mathcal{O}_Xto mathcal{O}_Y, (f,gcdot epsilon) mapsto f.$$ Consider the $mathcal{O}_X$-module morphism from $mathcal{O}_X$ to itself that multiplies by the global section $(0,1cdot epsilon)$. Denote the kernel sheaf by $mathcal{K}$. Denote by $mathcal{K}'$ the associated $mathcal{O}_Y$-module $mathcal{K}otimes_{mathcal{O}_X}mathcal{O}_Y$.



                If $mathcal{K}$ is locally finitely generated as an $mathcal{O}_X$-module, then also $mathcal{K}'$ is locally finitely generated as an $mathcal{O}_Y$-module. However, $mathcal{K}'$ is the extension by zero of the structure sheaf on $Ysetminus Z$. This is not a finitely generated $mathcal{O}_X$-module. In fact, for every irreducible open neighborhood $U$ that intersects $Z$, the sections of $mathcal{K}'$ on $U$ are zero, yet they are nonzero for $Usetminus Z$. Thus, every map $mathcal{O}_U^{oplus n}to mathcal{K}'|_U$ is the zero map, yet the restriction of $mathcal{K}'|_U$ to $Usetminus Z$ is nonzero.






                share|cite|improve this answer











                $endgroup$



                Welcome new contributor. It is best to think about these kinds of examples on one's own. Thus, please try this for yourself before reading the following.



                Let $(Y,Z)$ be any pair of a Noetherian scheme $Y$ and a nonempty closed subset $Z$ that is nowhere dense. For instance, let $Y$ be the following Spec of a DVR, $Y=text{Spec} k[t]_{langle trangle}$, and let $Z$ be the singleton set ${z}$ of the $k$-point of the unique maximal ideal. Denote by $i$ the inclusion continuous function from $Z$ to $Y$.



                Consider the sheaf of $mathcal{O}_{Y}$-modules $i_*i^{-1}mathcal{O}_{Y}$. For every open subset $U$ of $Y$ that does not intersect $Z$, the only section of this sheaf on $U$ is the zero section, and thus the stalk at every $yin U$ is the zero module. Also, for every $zin Z$, the stalk at $z$ equals $mathcal{O}_{Y,z}$.



                Now form the sheaf of $mathcal{O}_Y$-modules $mathcal{O}_X:=mathcal{O}_Y oplus left( i_*i^{-1}mathcal{O}_Ycdot epsilon right)$, where $epsilon$ is just a placeholder. Give this the unique structure of $mathcal{O}_Y$-algebra such that $epsiloncdot epsilon$ equals $0$ and such that the following natural inclusion map is a morphism of $mathcal{O}_Y$-algebras, $$mathcal{O}_Y hookrightarrow mathcal{O}_Y oplus left( i_*i^{-1}mathcal{O}_Ycdot epsilon right), fmapsto (f,0cdot epsilon).$$



                For every $yin Ysetminus Z$, the stalk $mathcal{O}_{X,y}$ equals $mathcal{O}_{Y,y}$ as an $mathcal{O}_{Y,y}$-algebra. Also, for every $zin Z$, the $mathcal{O}_{Y,z}$-algebra $mathcal{O}_{X,z}$ equals $mathcal{O}_{Y,z}oplus mathcal{O}_{Y,z}cdot epsilon$ with $epsiloncdot epsilon = 0$. In every case, the stalk is a Noetherian local ring.



                There is a natural retraction of the above algebra homomorphism, $$mathcal{O}_Xto mathcal{O}_Y, (f,gcdot epsilon) mapsto f.$$ Consider the $mathcal{O}_X$-module morphism from $mathcal{O}_X$ to itself that multiplies by the global section $(0,1cdot epsilon)$. Denote the kernel sheaf by $mathcal{K}$. Denote by $mathcal{K}'$ the associated $mathcal{O}_Y$-module $mathcal{K}otimes_{mathcal{O}_X}mathcal{O}_Y$.



                If $mathcal{K}$ is locally finitely generated as an $mathcal{O}_X$-module, then also $mathcal{K}'$ is locally finitely generated as an $mathcal{O}_Y$-module. However, $mathcal{K}'$ is the extension by zero of the structure sheaf on $Ysetminus Z$. This is not a finitely generated $mathcal{O}_X$-module. In fact, for every irreducible open neighborhood $U$ that intersects $Z$, the sections of $mathcal{K}'$ on $U$ are zero, yet they are nonzero for $Usetminus Z$. Thus, every map $mathcal{O}_U^{oplus n}to mathcal{K}'|_U$ is the zero map, yet the restriction of $mathcal{K}'|_U$ to $Usetminus Z$ is nonzero.







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