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 Nov9 comment Global sections of the line bundle $\mathcal{O}(D)$ Errata. Everything that I wrote in the last comments is wrong! For example, $1/x_0$ as a rational function on $\mathbb{P}^1$ of course has degree zero and divisor $-\{0\}+\infty$ which is different from $D=\{0\}$. Nov9 comment Global sections of the line bundle $\mathcal{O}(D)$ Errata. I think as $\eta_\alpha\in\mathcal{K}(X)$ we must take one of the $\eta_\alpha$'s such that $(\eta_\alpha)\neq 0$. Nov9 comment Global sections of the line bundle $\mathcal{O}(D)$ And if you picked a different $\alpha$, say $\alpha'$, $s_D$ would be multiplied by the global invertible function $\psi_{\alpha\alpha'}$. Nov9 comment Global sections of the line bundle $\mathcal{O}(D)$ Conjecture: $s_D$ is in fact determined only up to global invertible functions $\lambda\in\mathcal{O}_X^{\;*}(X)$, and, in terms of the $\eta_\alpha$'s, $s_D$ is obtained as follows: fix an $\alpha$ and consider $\eta_\alpha$ as a global rational function on $X$, then, for any $\beta$, $s_D|_{U_\beta}=\eta_\alpha^{-1}|_{U_\beta}$. Nov9 awarded Commentator Nov9 comment Global sections of the line bundle $\mathcal{O}(D)$ Well, in this case we can take $\eta_0=x_0$ and $\eta_1=1$ (the constant function $1$ on $U_1$). So $\psi_{01}=x_0$ on $U_{01}=\mathbb{C}\setminus\{0\}$. So I guess it's $s_D=1/x_0$, thought of as a global rational function on $\mathbb{P}^1$. As I remarked in the question, $s_D|_{U_\alpha}$ cannot be equal to $\eta_\alpha^{-1}$ for every $\alpha$: indeed $s_D|_{U_1}=1/x_0\neq 1=\eta_1^{-1}$. Is that all right? Nov9 comment Global sections of the line bundle $\mathcal{O}(D)$ I realize maybe there can't be an explicit description of $s_D$ in terms of $\eta_\alpha$, because in fact the $\eta_alpha$'s are determined up to a coboundary for $\mathcal{O}_X^{\;*}$. For $X=\mathbb{P}^1$ and $D=\{0\}$, how can we describe $s_D$ in terms of homogeneus coordinates $[x_0:x_1]$ on $\mathbb{P}^1$? Nov5 asked Global sections of the line bundle $\mathcal{O}(D)$ Nov1 revised (Weil divisors : Cartier divisors) = (p-Cycles : ? ) added 70 characters in body Nov1 revised (Weil divisors : Cartier divisors) = (p-Cycles : ? ) added 70 characters in body Nov1 asked (Weil divisors : Cartier divisors) = (p-Cycles : ? ) Oct29 comment Generically non degenerate quadratic forms on a scheme Wait. I was talking about the map of bundles not the corresponding map of sheaves. Take for example an inclusion $\mathcal{O}(-1)\to\mathcal{O}$ on $\mathbb{P}^1$: it's an inclusion of sheaves (it's injective on stalks), but it is not an injective map of line bundles (it's not injective on fibers). Oct27 comment Generically non degenerate quadratic forms on a scheme (when I say the map is zero at a point $x$, I mean on the fiber at $x$ not at the stalk at $x$) Oct27 comment Generically non degenerate quadratic forms on a scheme I think non-degenerate means $\det(V)\to\det(V^*)\otimes\ell^{n}$ is never zero on $X$, and generically non-degenerate means it can be zero at most along a divisor. Right? (assume $X$ irreducible) Oct27 accepted Generically non degenerate quadratic forms on a scheme Oct26 awarded Supporter Oct26 comment Sum of inverse squares of denominators Even if this answer is ok, I think I should accept coffeemath's answer instead, because it's more complete. Oct26 accepted Sum of inverse squares of denominators Oct26 comment Sum of inverse squares of denominators Great. Thank you! Oct26 revised Sum of inverse squares of denominators deleted 5 characters in body