global sections of locally free sheaf on projective space











up vote
4
down vote

favorite
1












Let $mathcal{E}$ be a locally free sheaf on $mathbb{P}^n_A=mathbb{P}^ntimes_{Spec k} Spec A$, where $A$ is a finitely generated algebra over a field $k$. By a well known theorem (see e.g. Hartshorne's Algebraic Geometry, Thm 5.19) $H^0(mathbb{P}^n_A, mathcal{E})$ is a finitely generated $A$- module.




  1. Is it true that $H^0(mathbb{P}^n_A, mathcal{E})$ is a projective module?


Let $B$ be a finitely generated $k$-algebra, $f: Spec B to Spec A$ a morphism and $f^*mathcal{E}$ the pullback of $mathcal{E}$ to $Spec B$.




  1. Is it true that $H^0(mathbb{P}^n_A, mathcal{E})otimes _A B= H^0(mathbb{P}^n_B, f^*mathcal{E})$ ?


1 and 2 above are true when $mathcal {E}$ is a direct sum of line bundles of the form $mathcal O(n)$. I was wondering if they are true for general $mathcal{E}$.










share|cite|improve this question




















  • 1




    Have a look at Base change for quasi-coherent sheaves.
    – abx
    Nov 27 at 5:27















up vote
4
down vote

favorite
1












Let $mathcal{E}$ be a locally free sheaf on $mathbb{P}^n_A=mathbb{P}^ntimes_{Spec k} Spec A$, where $A$ is a finitely generated algebra over a field $k$. By a well known theorem (see e.g. Hartshorne's Algebraic Geometry, Thm 5.19) $H^0(mathbb{P}^n_A, mathcal{E})$ is a finitely generated $A$- module.




  1. Is it true that $H^0(mathbb{P}^n_A, mathcal{E})$ is a projective module?


Let $B$ be a finitely generated $k$-algebra, $f: Spec B to Spec A$ a morphism and $f^*mathcal{E}$ the pullback of $mathcal{E}$ to $Spec B$.




  1. Is it true that $H^0(mathbb{P}^n_A, mathcal{E})otimes _A B= H^0(mathbb{P}^n_B, f^*mathcal{E})$ ?


1 and 2 above are true when $mathcal {E}$ is a direct sum of line bundles of the form $mathcal O(n)$. I was wondering if they are true for general $mathcal{E}$.










share|cite|improve this question




















  • 1




    Have a look at Base change for quasi-coherent sheaves.
    – abx
    Nov 27 at 5:27













up vote
4
down vote

favorite
1









up vote
4
down vote

favorite
1






1





Let $mathcal{E}$ be a locally free sheaf on $mathbb{P}^n_A=mathbb{P}^ntimes_{Spec k} Spec A$, where $A$ is a finitely generated algebra over a field $k$. By a well known theorem (see e.g. Hartshorne's Algebraic Geometry, Thm 5.19) $H^0(mathbb{P}^n_A, mathcal{E})$ is a finitely generated $A$- module.




  1. Is it true that $H^0(mathbb{P}^n_A, mathcal{E})$ is a projective module?


Let $B$ be a finitely generated $k$-algebra, $f: Spec B to Spec A$ a morphism and $f^*mathcal{E}$ the pullback of $mathcal{E}$ to $Spec B$.




  1. Is it true that $H^0(mathbb{P}^n_A, mathcal{E})otimes _A B= H^0(mathbb{P}^n_B, f^*mathcal{E})$ ?


1 and 2 above are true when $mathcal {E}$ is a direct sum of line bundles of the form $mathcal O(n)$. I was wondering if they are true for general $mathcal{E}$.










share|cite|improve this question















Let $mathcal{E}$ be a locally free sheaf on $mathbb{P}^n_A=mathbb{P}^ntimes_{Spec k} Spec A$, where $A$ is a finitely generated algebra over a field $k$. By a well known theorem (see e.g. Hartshorne's Algebraic Geometry, Thm 5.19) $H^0(mathbb{P}^n_A, mathcal{E})$ is a finitely generated $A$- module.




  1. Is it true that $H^0(mathbb{P}^n_A, mathcal{E})$ is a projective module?


Let $B$ be a finitely generated $k$-algebra, $f: Spec B to Spec A$ a morphism and $f^*mathcal{E}$ the pullback of $mathcal{E}$ to $Spec B$.




  1. Is it true that $H^0(mathbb{P}^n_A, mathcal{E})otimes _A B= H^0(mathbb{P}^n_B, f^*mathcal{E})$ ?


1 and 2 above are true when $mathcal {E}$ is a direct sum of line bundles of the form $mathcal O(n)$. I was wondering if they are true for general $mathcal{E}$.







ag.algebraic-geometry vector-bundles






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 26 at 22:30

























asked Nov 26 at 22:23









mike

263




263








  • 1




    Have a look at Base change for quasi-coherent sheaves.
    – abx
    Nov 27 at 5:27














  • 1




    Have a look at Base change for quasi-coherent sheaves.
    – abx
    Nov 27 at 5:27








1




1




Have a look at Base change for quasi-coherent sheaves.
– abx
Nov 27 at 5:27




Have a look at Base change for quasi-coherent sheaves.
– abx
Nov 27 at 5:27










1 Answer
1






active

oldest

votes

















up vote
8
down vote



accepted










The answer to both questions is negative, see counterexamples below.



1) Let $A = k[x,y,z]$, $n = 1$. Note that
$$
H^1(mathbb{P}^n_A,O(-2)) cong A.
$$

Consider the extension
$$
0 to O(-2) to E to O oplus O oplus O to 0
$$

whose extension class is $(x,y,z)$. Then the cohomolopgy exact sequence
$$
0 to H^0(mathbb{P}^n_A,E) to A oplus A oplus A stackrel{(x,y,z)}to A
$$

shows that $H^0(mathbb{P}^n_A,E)$ is reflexive but not locally free (hence not projective). In fact, this is the simplest example of a reflexive non-locally free sheaf.



2) Take $A = k[x,y,z,w]$ and define $E$ as the extension
$$
0 to O(-2) to E to O oplus O oplus O oplus O to 0
$$

whose extension class is $(x,y,z,w)$. Let $B = k$ with the morphism $A to B$ defined by $x,y,z,w mapsto 0$. Then $f^*E cong O(-2) oplus O oplus O oplus O oplus O$, hence
$$
H^0(mathbb{P}^n_B,f^*E) = B oplus B oplus B oplus B.
$$

On the other hand, tensoring
$$
0 to H^0(mathbb{P}^n_A,E) to A oplus A oplus A oplus A stackrel{(x,y,z,w)}to A to B to 0
$$

by $B$ (over $A$), we deduce
$$
H^0(mathbb{P}^n_A,E) otimes_A B cong Tor_2^A(B,B) cong B^{oplus 6}.
$$






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: "504"
    };
    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%2fmathoverflow.net%2fquestions%2f316285%2fglobal-sections-of-locally-free-sheaf-on-projective-space%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
    8
    down vote



    accepted










    The answer to both questions is negative, see counterexamples below.



    1) Let $A = k[x,y,z]$, $n = 1$. Note that
    $$
    H^1(mathbb{P}^n_A,O(-2)) cong A.
    $$

    Consider the extension
    $$
    0 to O(-2) to E to O oplus O oplus O to 0
    $$

    whose extension class is $(x,y,z)$. Then the cohomolopgy exact sequence
    $$
    0 to H^0(mathbb{P}^n_A,E) to A oplus A oplus A stackrel{(x,y,z)}to A
    $$

    shows that $H^0(mathbb{P}^n_A,E)$ is reflexive but not locally free (hence not projective). In fact, this is the simplest example of a reflexive non-locally free sheaf.



    2) Take $A = k[x,y,z,w]$ and define $E$ as the extension
    $$
    0 to O(-2) to E to O oplus O oplus O oplus O to 0
    $$

    whose extension class is $(x,y,z,w)$. Let $B = k$ with the morphism $A to B$ defined by $x,y,z,w mapsto 0$. Then $f^*E cong O(-2) oplus O oplus O oplus O oplus O$, hence
    $$
    H^0(mathbb{P}^n_B,f^*E) = B oplus B oplus B oplus B.
    $$

    On the other hand, tensoring
    $$
    0 to H^0(mathbb{P}^n_A,E) to A oplus A oplus A oplus A stackrel{(x,y,z,w)}to A to B to 0
    $$

    by $B$ (over $A$), we deduce
    $$
    H^0(mathbb{P}^n_A,E) otimes_A B cong Tor_2^A(B,B) cong B^{oplus 6}.
    $$






    share|cite|improve this answer

























      up vote
      8
      down vote



      accepted










      The answer to both questions is negative, see counterexamples below.



      1) Let $A = k[x,y,z]$, $n = 1$. Note that
      $$
      H^1(mathbb{P}^n_A,O(-2)) cong A.
      $$

      Consider the extension
      $$
      0 to O(-2) to E to O oplus O oplus O to 0
      $$

      whose extension class is $(x,y,z)$. Then the cohomolopgy exact sequence
      $$
      0 to H^0(mathbb{P}^n_A,E) to A oplus A oplus A stackrel{(x,y,z)}to A
      $$

      shows that $H^0(mathbb{P}^n_A,E)$ is reflexive but not locally free (hence not projective). In fact, this is the simplest example of a reflexive non-locally free sheaf.



      2) Take $A = k[x,y,z,w]$ and define $E$ as the extension
      $$
      0 to O(-2) to E to O oplus O oplus O oplus O to 0
      $$

      whose extension class is $(x,y,z,w)$. Let $B = k$ with the morphism $A to B$ defined by $x,y,z,w mapsto 0$. Then $f^*E cong O(-2) oplus O oplus O oplus O oplus O$, hence
      $$
      H^0(mathbb{P}^n_B,f^*E) = B oplus B oplus B oplus B.
      $$

      On the other hand, tensoring
      $$
      0 to H^0(mathbb{P}^n_A,E) to A oplus A oplus A oplus A stackrel{(x,y,z,w)}to A to B to 0
      $$

      by $B$ (over $A$), we deduce
      $$
      H^0(mathbb{P}^n_A,E) otimes_A B cong Tor_2^A(B,B) cong B^{oplus 6}.
      $$






      share|cite|improve this answer























        up vote
        8
        down vote



        accepted







        up vote
        8
        down vote



        accepted






        The answer to both questions is negative, see counterexamples below.



        1) Let $A = k[x,y,z]$, $n = 1$. Note that
        $$
        H^1(mathbb{P}^n_A,O(-2)) cong A.
        $$

        Consider the extension
        $$
        0 to O(-2) to E to O oplus O oplus O to 0
        $$

        whose extension class is $(x,y,z)$. Then the cohomolopgy exact sequence
        $$
        0 to H^0(mathbb{P}^n_A,E) to A oplus A oplus A stackrel{(x,y,z)}to A
        $$

        shows that $H^0(mathbb{P}^n_A,E)$ is reflexive but not locally free (hence not projective). In fact, this is the simplest example of a reflexive non-locally free sheaf.



        2) Take $A = k[x,y,z,w]$ and define $E$ as the extension
        $$
        0 to O(-2) to E to O oplus O oplus O oplus O to 0
        $$

        whose extension class is $(x,y,z,w)$. Let $B = k$ with the morphism $A to B$ defined by $x,y,z,w mapsto 0$. Then $f^*E cong O(-2) oplus O oplus O oplus O oplus O$, hence
        $$
        H^0(mathbb{P}^n_B,f^*E) = B oplus B oplus B oplus B.
        $$

        On the other hand, tensoring
        $$
        0 to H^0(mathbb{P}^n_A,E) to A oplus A oplus A oplus A stackrel{(x,y,z,w)}to A to B to 0
        $$

        by $B$ (over $A$), we deduce
        $$
        H^0(mathbb{P}^n_A,E) otimes_A B cong Tor_2^A(B,B) cong B^{oplus 6}.
        $$






        share|cite|improve this answer












        The answer to both questions is negative, see counterexamples below.



        1) Let $A = k[x,y,z]$, $n = 1$. Note that
        $$
        H^1(mathbb{P}^n_A,O(-2)) cong A.
        $$

        Consider the extension
        $$
        0 to O(-2) to E to O oplus O oplus O to 0
        $$

        whose extension class is $(x,y,z)$. Then the cohomolopgy exact sequence
        $$
        0 to H^0(mathbb{P}^n_A,E) to A oplus A oplus A stackrel{(x,y,z)}to A
        $$

        shows that $H^0(mathbb{P}^n_A,E)$ is reflexive but not locally free (hence not projective). In fact, this is the simplest example of a reflexive non-locally free sheaf.



        2) Take $A = k[x,y,z,w]$ and define $E$ as the extension
        $$
        0 to O(-2) to E to O oplus O oplus O oplus O to 0
        $$

        whose extension class is $(x,y,z,w)$. Let $B = k$ with the morphism $A to B$ defined by $x,y,z,w mapsto 0$. Then $f^*E cong O(-2) oplus O oplus O oplus O oplus O$, hence
        $$
        H^0(mathbb{P}^n_B,f^*E) = B oplus B oplus B oplus B.
        $$

        On the other hand, tensoring
        $$
        0 to H^0(mathbb{P}^n_A,E) to A oplus A oplus A oplus A stackrel{(x,y,z,w)}to A to B to 0
        $$

        by $B$ (over $A$), we deduce
        $$
        H^0(mathbb{P}^n_A,E) otimes_A B cong Tor_2^A(B,B) cong B^{oplus 6}.
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 27 at 0:01









        Sasha

        19.9k22652




        19.9k22652






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to MathOverflow!


            • 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%2fmathoverflow.net%2fquestions%2f316285%2fglobal-sections-of-locally-free-sheaf-on-projective-space%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

            "Incorrect syntax near the keyword 'ON'. (on update cascade, on delete cascade,)

            Alcedinidae

            Origin of the phrase “under your belt”?