# Cartier Divisor corresponds on $X$ to $(U_i,f_i)$

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 $f_i/f_j \in \Gamma(U_i \cap U_j, \mathcal O_X^*)$.

• What've you tried out? – Ehsan M. Kermani Dec 19 '13 at 19:41
• Repeat the construction of a quotient sheaf. – Martin Brandenburg Dec 19 '13 at 19:47
• I have done it. I have the answer. I will write it soon. I thought, I should write down the formal proof here. – Babai Dec 19 '13 at 20:02
• 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 $q\in U$ , $t|_q=s(q)$ – Babai Dec 19 '13 at 20:02
• 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... – Georges Elencwajg Dec 20 '13 at 7:58