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seen Dec 27 '12 at 14:04

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 : ? )
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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
Oct
26
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.
Oct
26
accepted Sum of inverse squares of denominators
Oct
26
comment Sum of inverse squares of denominators
Great. Thank you!
Oct
26
revised Sum of inverse squares of denominators
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