Proving that $\mathcal{O}(m) \otimes_{\mathcal{O_{\mathbb P^{1}}}} \mathcal{O}(n) = \mathcal{O}(m+n)$

Question

Is there a way to prove $$ \mathcal{O}(m) \otimes_{\mathcal{O_{\mathbb P^{1}}}} \mathcal{O}(n) = \mathcal{O}(m+n)$$ for all $m,n \in \mathbb Z$ without resorting to the cases where both are positive, both are negative and one is positive and one is negative?

Answer 1

Over $\mathbb{P}^N$, being $k$ positive, negative or zero, $\mathcal{O}(k)$ is given by the cocycles $(z^k_j/z^k_i)$, with respect to the standard open covering $U_i = \{z_i \neq 0\}$. Since the tensor product corresponds to the actual product of the cocyles we get that $\mathcal{O}(n) \otimes \mathcal{O}(m)$ is given by the cocycles $(z^n_j/z^n_i \cdot z^m_j/z^m_i) = (z^{n+m}_j/z^{n+m}_i)$ and therefore $\mathcal{O}(n) \otimes \mathcal{O}(m) \simeq \mathcal{O}(n+m)$.