I want to show that $$0\to \mathcal O_{\mathbb{P}_k^1}\to \mathcal K\to \mathcal K/\mathcal O_{\mathbb{P}_k^1}\to 0$$ is a flasque resolution of $\mathcal O_{\mathbb{P}_k^1}$ with $k$ infinite, but not necessarily algebraically closed. Where $\mathcal K$ is the total ring of fractions.

I was able to show that $\mathcal K$ is flasque, and by extension, the presheaf $U\mapsto \mathcal K(U)/\mathcal O_{\mathbb{P}_k^1}(U)$ is 'flasque' as a presheaf. However, being a quotient, we aren't guaranteed that this is a sheaf. I tried to show that it was, but I kept running into troubles. So, instead, I tried to show flasqueness of the sheafification, but again, I couldn't lift any of the maps. Additionally, $\mathcal O_{\mathbb{P}_k^1}$ isn't flasque; so, I'm stuck.

Any advice is greatly appreciated! Thank you all in advance :).

  • 2
    $\begingroup$ What can you say about $H^1(U,\mathcal{O})$ if $U$ is any open subset of $\mathbb{P}^1$ ? $\endgroup$ – Roland Dec 3 '18 at 20:40
  • $\begingroup$ @Roland I assume that you'd like me to see it's 0 somehow, which would then give me what I want, but after thinking for a few minutes, I don't see why this ought to be 0. $\endgroup$ – Laarz Dec 3 '18 at 21:01
  • 1
    $\begingroup$ Yes that is my point. There is two cases to be treated separately : $U=\mathbb{P}^1$ and $U\neq\mathbb{P}^1$. For the first one, it is a classical computation. For the second, use that $U$ is affine. $\endgroup$ – Roland Dec 3 '18 at 21:04
  • $\begingroup$ Ahhh! Well, now I feel silly. Thank you so much!! $\endgroup$ – Laarz Dec 3 '18 at 21:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.