global sections of locally free sheaf on projective space












4












$begingroup$


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











$endgroup$








  • 1




    $begingroup$
    Have a look at Base change for quasi-coherent sheaves.
    $endgroup$
    – abx
    Nov 27 '18 at 5:27
















4












$begingroup$


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











$endgroup$








  • 1




    $begingroup$
    Have a look at Base change for quasi-coherent sheaves.
    $endgroup$
    – abx
    Nov 27 '18 at 5:27














4












4








4


1



$begingroup$


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











$endgroup$




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




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 '18 at 22:30







mike

















asked Nov 26 '18 at 22:23









mikemike

263




263








  • 1




    $begingroup$
    Have a look at Base change for quasi-coherent sheaves.
    $endgroup$
    – abx
    Nov 27 '18 at 5:27














  • 1




    $begingroup$
    Have a look at Base change for quasi-coherent sheaves.
    $endgroup$
    – abx
    Nov 27 '18 at 5:27








1




1




$begingroup$
Have a look at Base change for quasi-coherent sheaves.
$endgroup$
– abx
Nov 27 '18 at 5:27




$begingroup$
Have a look at Base change for quasi-coherent sheaves.
$endgroup$
– abx
Nov 27 '18 at 5:27










1 Answer
1






active

oldest

votes


















8












$begingroup$

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









$endgroup$














    Your Answer








    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',
    autoActivateHeartbeat: false,
    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









    8












    $begingroup$

    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









    $endgroup$


















      8












      $begingroup$

      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









      $endgroup$
















        8












        8








        8





        $begingroup$

        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









        $endgroup$



        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 '18 at 0:01









        SashaSasha

        21.3k22756




        21.3k22756






























            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.




            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

            Wiesbaden

            Image
            Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
            Read more

            To store a contact into the json file from server.js file using a class in NodeJS

            Image
            0 I have created three files -contactPresentation.js, contactService.js, contacts.json. I will run the contactService.js file. And it will navigate to contactPresentation.js file that will only choose an option to either add or display a file. If add(), then add the data into add of service file.And the data will be added to the contacts.json file . And if display(), then display the display() of service file. And the display will take the data from contacts.json file. However, i know my code is flawed. Please correct it . I am also getting an error - this.contacts.push(new ContactService(userName, contactNumber, emailId)); ^ TypeError: Cannot read property 'push' of undefined at ContactService.add (e:NodeJSProjectcontactManager2contactService.js:57:23) at new C
            Read more

            Marschland

            Image
            Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
            Read more