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 Mar 8 awarded Popular Question Jan 8 awarded Popular Question Nov 4 awarded Popular Question Nov 3 awarded Nice Question Nov 9 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 $D=\{0\}$. Nov 9 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 $(\eta_\alpha)\neq 0$. Nov 9 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 $\psi_{\alpha\alpha'}$. Nov 9 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 $\beta$, $s_D|_{U_\beta}=\eta_\alpha^{-1}|_{U_\beta}$. Nov 9 awarded Commentator Nov 9 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? Nov 9 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 $\mathbb{P}^1$? Nov 5 asked Global sections of the line bundle $\mathcal{O}(D)$ Nov 1 revised (Weil divisors : Cartier divisors) = (p-Cycles : ? ) added 70 characters in body Nov 1 revised (Weil divisors : Cartier divisors) = (p-Cycles : ? ) added 70 characters in body Nov 1 asked (Weil divisors : Cartier divisors) = (p-Cycles : ? ) Oct 29 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). Oct 27 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 $x$) Oct 27 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) Oct 27 accepted Generically non degenerate quadratic forms on a scheme Oct 26 awarded Supporter