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I have a geometric line bundle $L$ on $\mathbb{P}^1 = \{[x_0:x_1]\}$. With respect to the standard affine cover $U_0 = \{x_0 \neq 0\}$ and $U_1 = \{x_1 \neq 0\}$, I have the transition function $[x_0:x_1] \mapsto (\frac{x_0}{x_1})^n$. If I'm not mistaken, the invertible sheaf of sections $\mathscr{L}$ associated to this line bundle should be \begin{align*} \mathscr{L}(U) = \{&(h_0: U \cap U_0 \to k, h_1: U \cap U_1 \to k) \mid \\ &h_0([x_0:x_1]) = (\frac{x_0}{x_1})^n h_1([x_0:x_1]) \text{ on } U \cap U_0 \cap U_1\} \\ \Gamma(\mathbb{P}^1, \mathscr{L}) = \{&(h_0 \in k[x_0,x_1,x_0^{-1}], h_1 \in k[x_0,x_1,x_1^{-1}]) \mid \frac{h_0}{x_0^n} = \frac{h_1}{x_1^n} \text{ on } U \cap U_0 \cap U_1\}. \end{align*}

By the classification of line bundles on projective space, the sheaf $\mathscr{L}$ is isomorphic to $\mathscr{O}_{\mathbb{P}^1}(m)$ for some $m \in \mathbb{Z}$. My question is: which $m$ is it?

My guess is that $\mathscr{L} \cong \mathscr{O}_{\mathbb{P}^1}(-n)$, but I can't seem to show this. In fact, the sheaf I wrote down doesn't even seem to be coherent since there is no bound on the degrees of $h_0$ and $h_1$, so I must be doing something wrong. Can someone help me out?

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  • $\begingroup$ More generally you can write down the transition functions for $\mathcal{O}(n)$ on $\mathbb{P}^d$ explicitly. $\endgroup$ Commented Aug 9, 2013 at 7:04

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Your guess $\mathcal L=\mathcal O_{\mathbb P^1}(-n)$ is correct, as is your description of the sections.
The sheaf $\mathcal L$ is coherent since locally it is isomorphic to $\mathcal O_{\mathbb P^1}$, a coherent sheaf, and by definition coherence can be checked locally.

The non-boundedness of the degrees of $h_0$ and $h_1$ is irrelevant:
I hope you are convinced that the structural sheaf $\mathcal O_{\mathbb P^1}$ is coherent.
But this does not prevent $\Gamma(U_0, \mathcal O_{\mathbb P^1})=\Gamma(\mathbb A^1, \mathcal O_{\mathbb P^1})$ from being equal to $k[x]$, the algebra of all polynomials over the base field $k$, without any condition of boundedness of their degree.

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  • $\begingroup$ Thank you so much for your help. I think I was confused about the "coherent" business because I somehow thought of $k[x]$ as a module over $k$ instead, so thanks again for clearing that up! $\endgroup$
    – JHF
    Commented Aug 9, 2013 at 16:06
  • $\begingroup$ Dear JHF, I'm happy that the question has now been cleared up. $\endgroup$ Commented Aug 9, 2013 at 21:19

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