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Assume $\mathbb{P}^1_k$ is the projective line over an algebraically closed field. In Hartshorne, Chapter II.6, Corollary 6.17, Hartshorne claims that the generator of $\text{Pic}(\mathbb{P}^1_k)$ is isomorphic to $\mathcal{O}(1)$. Why?

I understand that in the case of $\mathbb{P}^1_k$, we have an isomorphism of Weil and Cartier divisors, as well as an isomorphism of Cartier divisors and invertible sheaves. I understand that these isomorphisms descend onto isomorphism of $Cl$, $CaCl$ and $Pic$. I understand that $Cl(\mathbb{P}^1_k)$ is isomorphic to $\mathbb{Z}$ and is generated by the class of a hyperplane. I know how to compute the corresponding Cartier divisor, but am struggling to show that any sheaf $\mathcal{O}(D)$ generated by a representative of $\text{CaCl}(\mathbb{P}^1_k)$ having $\text{deg}(D) = 1$ is isomorphic to $\mathcal{O}(1)$.

Any help?

Thanks in advance!

Edit: I managed to prove this fact.

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    $\begingroup$ There is some serious confusion here. $\mathscr O(1)$ generates the Picard group. A sheaf (or line bundle) is not a group! $\endgroup$
    – Ted Shifrin
    Commented Jun 8, 2022 at 17:49
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    $\begingroup$ I can't believe the legendary Theodore Shifrin replied to my stack post! While I do not protest the fact that I am generically confused, I meant to ask why is it true that all sheaves in the isomorphism class of $\mathcal{O}(D)$ for $D\in\text{Cl}(\mathbb{P}^1_k)$ with $\text{deg}(D) = 1$ are isomorphic to $\mathcal{O}(1)$. $\endgroup$
    – kindasorta
    Commented Jun 8, 2022 at 18:35
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    $\begingroup$ If you've managed to prove it, then you should write up your solution and accept the answer so others can benefit from what you've learned! $\endgroup$
    – Stahl
    Commented Jun 8, 2022 at 19:21
  • $\begingroup$ My argument is not Hartshorne style. The exponential sheaf sequence gives $H^1(X,\mathscr O^*) \cong H^2(X,\Bbb Z)$ for $X=\Bbb P^1$. $\endgroup$
    – Ted Shifrin
    Commented Jun 8, 2022 at 19:22
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    $\begingroup$ You may also wish to correct the title, too (it looks like you missed this in your last edit clarifying the question). $\endgroup$
    – KReiser
    Commented Jun 8, 2022 at 19:33

1 Answer 1

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Hartshorne already proves in II.6.13.c that if $D_1\sim D_2$ are two Cartier divisors, then $\mathcal{L}(D_1)$ is isomorphic to $\mathcal{L}(D_2)$. Therefore, in order to prove that the generator of the class of $\text{CaCl}(\mathbb{P}^1_k)$ is isomorphic to $\mathcal{O}(1)$, it is enough to show that any invertible sheaf in the linear equivalence class of the generator of $\text{CaCl}(\mathbb{P}^1_k)$ is isomorphic to $\mathcal{O}(1)$.

Let's consider the representative corresponding to the Weil divisor $D = 1\cdot (1,0)$. The corresponding Cartier divisor is given by $\{1, U_x\}, \{x/y,U_y\}$, which corresponds to the sheaf $\mathcal{L}(D)$, given by the trivialization $\mathcal{L}(D)|_{U_x} = \mathcal{O}_{\mathbb{P}^1_k}(U_x)$ and $\mathcal{L}(D)|_{U_y} = \mathcal{O}_{\mathbb{P}^1_k}(U_y)\cdot \dfrac{y}{x}$.

To see that the sheaf $\mathcal{L}(D)$ is isomorphic to the sheaf $\mathcal{O}(1)$, consider the sheaf morphism, which over the open $U$ sends the section $s\in \mathcal{L}(D)(U)$ to $x\cdot s\in \mathcal{O}(1)$. Injectivity follows from the fact that $\mathbb{P}^1_k$ is integral. Surjectivity can be checked locally on $U_x$ and $U_y$. For example, for $U_y$ we have $\dfrac{y}{x}\mapsto y$ and $1\mapsto x$, which generate $\mathcal{O}(1)|_{U_y}$, and similarly for $U_x$.

Our sheaf morphism is injective and locally surjective and therefore an isomorphism, which completes the proof.

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    $\begingroup$ Just a picky remark. By $(1,0)$ you mean the divisor $P=[1,0]$. $\endgroup$
    – Ted Shifrin
    Commented Jun 9, 2022 at 0:14

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