# Why is the generator of the Picard group of $\mathbb{P}^1$ isomorphic to $\mathcal{O}(1)$?

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?

Edit: I managed to prove this fact.

• There is some serious confusion here. $\mathscr O(1)$ generates the Picard group. A sheaf (or line bundle) is not a group! Jun 8 at 17:49
• 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 $\mathcal{O}(1)$. Jun 8 at 18:35
• 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! Jun 8 at 19:21
• My argument is not Hartshorne style. The exponential sheaf sequence gives $H^1(X,\mathscr O^*) \cong H^2(X,\Bbb Z)$ for $X=\Bbb P^1$. Jun 8 at 19:22
• You may also wish to correct the title, too (it looks like you missed this in your last edit clarifying the question). Jun 8 at 19:33

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 $$U_x$$.
• Just a picky remark. By $(1,0)$ you mean the divisor $P=[1,0]$. Jun 9 at 0:14