Inclusion of closed subschemes.











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Let $X$ be a scheme and $Y$ and $Z$ closed subschemes. What does it mean for $Y$ to be contained in $Z$?




This question adresses the same question, but it covers only the affine case and it uses a different definition of a closed subscheme. So for the non-affine case, let $(Y, mathcal{O}_Y)$ and $(Z, mathcal{O}_Z)$ be two closed subschemas of a scheme $(X, mathcal{O}_X)$. Then $mathcal{O}_Y$ and $mathcal{O}_Z$ are quotients of the structure sheaf $mathcal{O}_X$ by a quasi-coherent sheaf of ideals, that is,
$$mathcal{O}_Y = mathcal{O}_X/mathcal{I}$$
and
$$mathcal{O}_Z = mathcal{O}_X/mathcal{J} , .$$
Then, to say that $Y subset Z$, we would like to say that $|Y| subset |X|$ as topological spaces, and that $mathcal{O}_Y$ is a sub sheaf of $mathcal{O}_Z$. I believe we can sharpen it a bit more by saying that the quasi-coherent sheaf $mathcal{J}$ is a subsheaf of $mathcal{I}$. Would this suffice?










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    Does this question answer your question? math.stackexchange.com/questions/2352513/…
    – Alfred Yerger
    Nov 18 at 19:47










  • @AlfredYerger It is interesting since it shows the equivalence of the two definitions, but still I would like a definition of $Y subset X$ just in terms of the one Eisenbud-Harris give for closed subschemes.
    – user313212
    Nov 18 at 20:14















up vote
2
down vote

favorite













Let $X$ be a scheme and $Y$ and $Z$ closed subschemes. What does it mean for $Y$ to be contained in $Z$?




This question adresses the same question, but it covers only the affine case and it uses a different definition of a closed subscheme. So for the non-affine case, let $(Y, mathcal{O}_Y)$ and $(Z, mathcal{O}_Z)$ be two closed subschemas of a scheme $(X, mathcal{O}_X)$. Then $mathcal{O}_Y$ and $mathcal{O}_Z$ are quotients of the structure sheaf $mathcal{O}_X$ by a quasi-coherent sheaf of ideals, that is,
$$mathcal{O}_Y = mathcal{O}_X/mathcal{I}$$
and
$$mathcal{O}_Z = mathcal{O}_X/mathcal{J} , .$$
Then, to say that $Y subset Z$, we would like to say that $|Y| subset |X|$ as topological spaces, and that $mathcal{O}_Y$ is a sub sheaf of $mathcal{O}_Z$. I believe we can sharpen it a bit more by saying that the quasi-coherent sheaf $mathcal{J}$ is a subsheaf of $mathcal{I}$. Would this suffice?










share|cite|improve this question


















  • 1




    Does this question answer your question? math.stackexchange.com/questions/2352513/…
    – Alfred Yerger
    Nov 18 at 19:47










  • @AlfredYerger It is interesting since it shows the equivalence of the two definitions, but still I would like a definition of $Y subset X$ just in terms of the one Eisenbud-Harris give for closed subschemes.
    – user313212
    Nov 18 at 20:14













up vote
2
down vote

favorite









up vote
2
down vote

favorite












Let $X$ be a scheme and $Y$ and $Z$ closed subschemes. What does it mean for $Y$ to be contained in $Z$?




This question adresses the same question, but it covers only the affine case and it uses a different definition of a closed subscheme. So for the non-affine case, let $(Y, mathcal{O}_Y)$ and $(Z, mathcal{O}_Z)$ be two closed subschemas of a scheme $(X, mathcal{O}_X)$. Then $mathcal{O}_Y$ and $mathcal{O}_Z$ are quotients of the structure sheaf $mathcal{O}_X$ by a quasi-coherent sheaf of ideals, that is,
$$mathcal{O}_Y = mathcal{O}_X/mathcal{I}$$
and
$$mathcal{O}_Z = mathcal{O}_X/mathcal{J} , .$$
Then, to say that $Y subset Z$, we would like to say that $|Y| subset |X|$ as topological spaces, and that $mathcal{O}_Y$ is a sub sheaf of $mathcal{O}_Z$. I believe we can sharpen it a bit more by saying that the quasi-coherent sheaf $mathcal{J}$ is a subsheaf of $mathcal{I}$. Would this suffice?










share|cite|improve this question














Let $X$ be a scheme and $Y$ and $Z$ closed subschemes. What does it mean for $Y$ to be contained in $Z$?




This question adresses the same question, but it covers only the affine case and it uses a different definition of a closed subscheme. So for the non-affine case, let $(Y, mathcal{O}_Y)$ and $(Z, mathcal{O}_Z)$ be two closed subschemas of a scheme $(X, mathcal{O}_X)$. Then $mathcal{O}_Y$ and $mathcal{O}_Z$ are quotients of the structure sheaf $mathcal{O}_X$ by a quasi-coherent sheaf of ideals, that is,
$$mathcal{O}_Y = mathcal{O}_X/mathcal{I}$$
and
$$mathcal{O}_Z = mathcal{O}_X/mathcal{J} , .$$
Then, to say that $Y subset Z$, we would like to say that $|Y| subset |X|$ as topological spaces, and that $mathcal{O}_Y$ is a sub sheaf of $mathcal{O}_Z$. I believe we can sharpen it a bit more by saying that the quasi-coherent sheaf $mathcal{J}$ is a subsheaf of $mathcal{I}$. Would this suffice?







algebraic-geometry sheaf-theory schemes






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asked Nov 18 at 19:35









user313212

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  • 1




    Does this question answer your question? math.stackexchange.com/questions/2352513/…
    – Alfred Yerger
    Nov 18 at 19:47










  • @AlfredYerger It is interesting since it shows the equivalence of the two definitions, but still I would like a definition of $Y subset X$ just in terms of the one Eisenbud-Harris give for closed subschemes.
    – user313212
    Nov 18 at 20:14














  • 1




    Does this question answer your question? math.stackexchange.com/questions/2352513/…
    – Alfred Yerger
    Nov 18 at 19:47










  • @AlfredYerger It is interesting since it shows the equivalence of the two definitions, but still I would like a definition of $Y subset X$ just in terms of the one Eisenbud-Harris give for closed subschemes.
    – user313212
    Nov 18 at 20:14








1




1




Does this question answer your question? math.stackexchange.com/questions/2352513/…
– Alfred Yerger
Nov 18 at 19:47




Does this question answer your question? math.stackexchange.com/questions/2352513/…
– Alfred Yerger
Nov 18 at 19:47












@AlfredYerger It is interesting since it shows the equivalence of the two definitions, but still I would like a definition of $Y subset X$ just in terms of the one Eisenbud-Harris give for closed subschemes.
– user313212
Nov 18 at 20:14




@AlfredYerger It is interesting since it shows the equivalence of the two definitions, but still I would like a definition of $Y subset X$ just in terms of the one Eisenbud-Harris give for closed subschemes.
– user313212
Nov 18 at 20:14










1 Answer
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For some mysterious reason the notion of closed subscheme is handled in a rather confusing way in much of the literature (egregiously and very untypically in Hartshorne).

Be that as it may, just remember that there is a bijective order reversing corrrespondence between closed subschemes $Ysubset X$ and quasi-coherent ideal sheaves $mathcal Jsubset mathcal O_X$.

The corresponding items are written $Y=V(mathcal J)$, resp. $mathcal J=mathcal I_Y$.

And then the answer to your question is given by the biblical simple equivalence:




$$Ysubset Ziff mathcal I_Y supset mathcal I_Z$$




Edit: a simple but edifying example.

Let $(A,mathfrak m)$ be a discrete valuation ring, $X=operatorname {Spec}A={M,eta}$ its associated affine scheme and $U={eta}$ its non trivial open subset.



a) The only non zero ideals of $A$ are the powers $mathfrak m^n$ and the corresponding quasi-coherent sheaves of ideals $mathcal I_n$ define the closed subschemes $Y_n=V(mathcal I_n)subset X$.

These $Y_n$'s are the infinitesimal neighbourhoods of the closed reduced point $Y_1={M}$ and they constitute the collection of all the closed subschemes $subsetneq X$.

Notice that $Y_n$ is non reduced for $ngt 1$ and that $Y_nsubsetneq Y_m$ for $n lt m$.



b) However $mathcal O_X$ has another sheaf of ideals $mathcal J$ defined by $mathcal J(X)=A$ and $mathcal J(U)={0}$.

This sheaf is not quasi-coherent and thus does not correspond to a closed subscheme of $X$.






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    up vote
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    accepted










    For some mysterious reason the notion of closed subscheme is handled in a rather confusing way in much of the literature (egregiously and very untypically in Hartshorne).

    Be that as it may, just remember that there is a bijective order reversing corrrespondence between closed subschemes $Ysubset X$ and quasi-coherent ideal sheaves $mathcal Jsubset mathcal O_X$.

    The corresponding items are written $Y=V(mathcal J)$, resp. $mathcal J=mathcal I_Y$.

    And then the answer to your question is given by the biblical simple equivalence:




    $$Ysubset Ziff mathcal I_Y supset mathcal I_Z$$




    Edit: a simple but edifying example.

    Let $(A,mathfrak m)$ be a discrete valuation ring, $X=operatorname {Spec}A={M,eta}$ its associated affine scheme and $U={eta}$ its non trivial open subset.



    a) The only non zero ideals of $A$ are the powers $mathfrak m^n$ and the corresponding quasi-coherent sheaves of ideals $mathcal I_n$ define the closed subschemes $Y_n=V(mathcal I_n)subset X$.

    These $Y_n$'s are the infinitesimal neighbourhoods of the closed reduced point $Y_1={M}$ and they constitute the collection of all the closed subschemes $subsetneq X$.

    Notice that $Y_n$ is non reduced for $ngt 1$ and that $Y_nsubsetneq Y_m$ for $n lt m$.



    b) However $mathcal O_X$ has another sheaf of ideals $mathcal J$ defined by $mathcal J(X)=A$ and $mathcal J(U)={0}$.

    This sheaf is not quasi-coherent and thus does not correspond to a closed subscheme of $X$.






    share|cite|improve this answer



























      up vote
      4
      down vote



      accepted










      For some mysterious reason the notion of closed subscheme is handled in a rather confusing way in much of the literature (egregiously and very untypically in Hartshorne).

      Be that as it may, just remember that there is a bijective order reversing corrrespondence between closed subschemes $Ysubset X$ and quasi-coherent ideal sheaves $mathcal Jsubset mathcal O_X$.

      The corresponding items are written $Y=V(mathcal J)$, resp. $mathcal J=mathcal I_Y$.

      And then the answer to your question is given by the biblical simple equivalence:




      $$Ysubset Ziff mathcal I_Y supset mathcal I_Z$$




      Edit: a simple but edifying example.

      Let $(A,mathfrak m)$ be a discrete valuation ring, $X=operatorname {Spec}A={M,eta}$ its associated affine scheme and $U={eta}$ its non trivial open subset.



      a) The only non zero ideals of $A$ are the powers $mathfrak m^n$ and the corresponding quasi-coherent sheaves of ideals $mathcal I_n$ define the closed subschemes $Y_n=V(mathcal I_n)subset X$.

      These $Y_n$'s are the infinitesimal neighbourhoods of the closed reduced point $Y_1={M}$ and they constitute the collection of all the closed subschemes $subsetneq X$.

      Notice that $Y_n$ is non reduced for $ngt 1$ and that $Y_nsubsetneq Y_m$ for $n lt m$.



      b) However $mathcal O_X$ has another sheaf of ideals $mathcal J$ defined by $mathcal J(X)=A$ and $mathcal J(U)={0}$.

      This sheaf is not quasi-coherent and thus does not correspond to a closed subscheme of $X$.






      share|cite|improve this answer

























        up vote
        4
        down vote



        accepted







        up vote
        4
        down vote



        accepted






        For some mysterious reason the notion of closed subscheme is handled in a rather confusing way in much of the literature (egregiously and very untypically in Hartshorne).

        Be that as it may, just remember that there is a bijective order reversing corrrespondence between closed subschemes $Ysubset X$ and quasi-coherent ideal sheaves $mathcal Jsubset mathcal O_X$.

        The corresponding items are written $Y=V(mathcal J)$, resp. $mathcal J=mathcal I_Y$.

        And then the answer to your question is given by the biblical simple equivalence:




        $$Ysubset Ziff mathcal I_Y supset mathcal I_Z$$




        Edit: a simple but edifying example.

        Let $(A,mathfrak m)$ be a discrete valuation ring, $X=operatorname {Spec}A={M,eta}$ its associated affine scheme and $U={eta}$ its non trivial open subset.



        a) The only non zero ideals of $A$ are the powers $mathfrak m^n$ and the corresponding quasi-coherent sheaves of ideals $mathcal I_n$ define the closed subschemes $Y_n=V(mathcal I_n)subset X$.

        These $Y_n$'s are the infinitesimal neighbourhoods of the closed reduced point $Y_1={M}$ and they constitute the collection of all the closed subschemes $subsetneq X$.

        Notice that $Y_n$ is non reduced for $ngt 1$ and that $Y_nsubsetneq Y_m$ for $n lt m$.



        b) However $mathcal O_X$ has another sheaf of ideals $mathcal J$ defined by $mathcal J(X)=A$ and $mathcal J(U)={0}$.

        This sheaf is not quasi-coherent and thus does not correspond to a closed subscheme of $X$.






        share|cite|improve this answer














        For some mysterious reason the notion of closed subscheme is handled in a rather confusing way in much of the literature (egregiously and very untypically in Hartshorne).

        Be that as it may, just remember that there is a bijective order reversing corrrespondence between closed subschemes $Ysubset X$ and quasi-coherent ideal sheaves $mathcal Jsubset mathcal O_X$.

        The corresponding items are written $Y=V(mathcal J)$, resp. $mathcal J=mathcal I_Y$.

        And then the answer to your question is given by the biblical simple equivalence:




        $$Ysubset Ziff mathcal I_Y supset mathcal I_Z$$




        Edit: a simple but edifying example.

        Let $(A,mathfrak m)$ be a discrete valuation ring, $X=operatorname {Spec}A={M,eta}$ its associated affine scheme and $U={eta}$ its non trivial open subset.



        a) The only non zero ideals of $A$ are the powers $mathfrak m^n$ and the corresponding quasi-coherent sheaves of ideals $mathcal I_n$ define the closed subschemes $Y_n=V(mathcal I_n)subset X$.

        These $Y_n$'s are the infinitesimal neighbourhoods of the closed reduced point $Y_1={M}$ and they constitute the collection of all the closed subschemes $subsetneq X$.

        Notice that $Y_n$ is non reduced for $ngt 1$ and that $Y_nsubsetneq Y_m$ for $n lt m$.



        b) However $mathcal O_X$ has another sheaf of ideals $mathcal J$ defined by $mathcal J(X)=A$ and $mathcal J(U)={0}$.

        This sheaf is not quasi-coherent and thus does not correspond to a closed subscheme of $X$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 19 at 16:22

























        answered Nov 18 at 23:27









        Georges Elencwajg

        118k7180325




        118k7180325






























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