1

I am considering an alternative proof of Grothendieck's classification of vector bundles on $\mathbb{P}^1$. Given a vector bundle $E$ on $\mathbb{P}^1$ one can associate a graded module $\Gamma(E)$ over $k[x,y]$, then if I can say it is projective I am done(Seshadri's theorem).

To show this, we can reduce it locally(localising at a maximal ideal) and since it is regular it is enough to show that it is a maximal Cohen Macaulay module.

Can someone please help me by explaining how to show this? (one can use Hilbert-syzygy, but I don't know how to do this?)

Sidana
  • 21
  • 2
    Sorry, what is $\Gamma(\mathcal E)$ when $\mathcal E$ is $\mathcal O(1)$? Is it projective? – Z. M Feb 08 '24 at 13:36
  • It is projective, just the same ring but different grading(shifted towards left by 1) – Sidana Feb 08 '24 at 18:19
  • This might not be helpful, but there is an elementary proof of this theorem using only linear algebra.. https://mathoverflow.net/a/16450/613 – Deane Yang Feb 09 '24 at 19:38
  • I know this proof, but I am thinking of the other proof. Thanks – Sidana Feb 10 '24 at 03:44

1 Answers1

4

Consider the exact sequence, $0\to E(-1)\to E\to E_p\to 0$, where the first map is just multiplication by $x$. Then $E_p$ is just the skyscraper sheaf at $p=(0,1)$. Taking cohomologies you get $0\to \Gamma (E)\to\Gamma (E)\to \Gamma (E_p)$ exact. One checks that $\Gamma (E_p)$ is just a free $k[y]$ module and then it follows that $\Gamma (E) $ is Cohen- Macaulay.

Graded projective modules over a polynomial ring are free follows from Nakayama. Projective modules over $k[x,y]$ are free was originally proved by Seshadri long before Quillen-Suslin.

Mohan
  • 6,117