Let $X$ be a scheme. $\mathcal K $ be the sheaf of total quotients of $X$. $\mathcal K^*$ be the sheaf corresponding sheaf of invertible elements of $\mathcal K$. So $\mathcal O_X^*$ is a subsheaf $\mathcal K^*$.

An Cartier Divisor on $X$ is by definition a global section of the sheaf ${\mathcal K^*} / {\mathcal O_X^*}$.

Now I want to prove that giving an global section of ${\mathcal K^*} / {\mathcal O_X^*}$ is same as providing the following data: $\mathrm{collection}{(U_i,f_i)}$ such that $f_i \in \Gamma (U_i, \mathcal K^*)$, where ${U_i}'s$ covers X and $f_i/f_j \in \Gamma(U_i \cap U_j, \mathcal O_X^*)$.

  • $\begingroup$ What've you tried out? $\endgroup$ – Ehsan M. Kermani Dec 19 '13 at 19:41
  • 5
    $\begingroup$ Repeat the construction of a quotient sheaf. $\endgroup$ – Martin Brandenburg Dec 19 '13 at 19:47
  • $\begingroup$ I have done it. I have the answer. I will write it soon. I thought, I should write down the formal proof here. $\endgroup$ – Babai Dec 19 '13 at 20:02
  • 1
    $\begingroup$ Basically one has to use the fact: A global section of ${\mathcal K^*} / {\mathcal O_X^*}$ is a function from $s: X\rightarrow {\prod_{p\in X} \widehat{(\frac{\mathcal K^*} {\mathcal O_X^*})}}_p$ where $\widehat{(\frac{\mathcal K^*} {\mathcal O_X^*})}$ is just the presheaf such that for any $x\in X$ (i) $s(x)=\widehat{(\frac{\mathcal K^*} {\mathcal O_X^*})}_x$ and (ii) For any $x\in X$ there exists an open set $U_x$ of X containing $x$ and an element $t\in\widehat{\frac{\mathcal K^*} {\mathcal O_X^*}}(U) $such that for any $q\in U$ , $t|_q=s(q)$ $\endgroup$ – Babai Dec 19 '13 at 20:02
  • 1
    $\begingroup$ The sheaf $\mathcal K$ is the sheaf of rational functions on X. Its relation to total fields of quotients is more subtle than one might think, even if "one" is Grothendieck... $\endgroup$ – Georges Elencwajg Dec 20 '13 at 7:58

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.