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









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    1 Answer
    1






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    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






























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