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?
Question: "Is there a natural ring structure on $Pic(P^1)$ such that the isomorphism between $Pic(P^1)$ and $Z$ becomes an isomorphism of rings?"
Answer: If $C$ is a non-singular algebraic curve there is an isomorphism
$$K_0(C) \cong \mathbb{Z}\oplus Pic(C)$$
where $K_0(C)$ is the Grothendieck group of finite rank locally free sheaves on $C$. Since $C$ is non-singular there is a product on $K_0(C)$ induced by the tensor product.
Example: If $C$ is the projective line it follows $Pic(C) \cong \mathbb{Z}$. The projective bundle formula proves that there is an isomorphism of rings
$$K_0(C) \cong \mathbb{Z}[t]/(t^2)\cong \mathbb{Z}\{1,t\}.$$
In general if $G$ is an abelian group you may define $F(G):=\mathbb{Z}\oplus G$ with multiplicative unit $1:=(1,0)$ and the following multiplication: $(m,g)*(n,h):=(mn, mh+ng)$. It follows $F(G)$ is a commutative unital ring and this construction is functorial and satisfies a certain universal property. If you do this construction with the projective line $C$ and $Pic(C)$ you get an isomorphism
$$F(Pic(C)) \cong K_0(C)$$
hence there is in some sense a "functorial way" to introduce a ring structure on an abelian extension $F(Pic(C))$ of $Pic(C)$ recovering the Grothendieck ring $K_0(C)$.