Inclusion of closed subschemes.











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













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 18 at 19:35









user313212

287520




287520








  • 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
1






active

oldest

votes

















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























    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004021%2finclusion-of-closed-subschemes%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    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










      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






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004021%2finclusion-of-closed-subschemes%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Quarter-circle Tiles

            build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

            Mont Emei