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.
Best Answer
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.