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The set of isomorphism classes of holomorphic line bundles on a complex manifold $X$ is a group under tensor product. This group is called the Picard group and is denoted $\operatorname{Pic}(X)$.

We obtain the isomorphism $\operatorname{Pic}(X) \cong H^1(X, \mathcal{O}^*)$ by considering transition functions. By looking at the long exact cohomology sequence associated to the exponential sequence of sheaves we obtain

$$\dots \to H^1(X, \mathcal{O}) \to H^1(X, \mathcal{O}^*) \xrightarrow{c_1} H^2(X, \mathbb{Z}) \to H^2(X, \mathcal{O}) \to \dots$$

By Dolbeault's Theorem, $H^k(X, \mathcal{O}) \cong H^{0,k}_{\bar{\partial}}(X)$. So, if

  • $H^{0,1}_{\bar{\partial}}(X) = 0$, the map $c_1 : \operatorname{Pic}(X) \to H^2(X, \mathbb{Z})$ is injective,
  • $H^{0,2}_{\bar{\partial}}(X) = 0$, the map $c_1 : \operatorname{Pic}(X) \to H^2(X, \mathbb{Z})$ is surjective.

In the case of $X = \mathbb{CP}^1$, both conditions are met so $\operatorname{Pic}(X) \cong H^2(\mathbb{CP}^1, \mathbb{Z})$. Furthermore, $H^2(\mathbb{CP}^1, \mathbb{Z}) \cong \mathbb{Z}$ so $\operatorname{Pic}(X) \cong \mathbb{Z}$; this isomorphism is given by the degree of a line bundle. In addition to being a group, $\mathbb{Z}$ is a ring.

Is there a natural ring structure on $\operatorname{Pic}(\mathbb{CP}^1)$ such that the isomorphism between $\operatorname{Pic}(\mathbb{CP}^1)$ and $\mathbb{Z}$ becomes an isomorphism of rings?

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