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

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?

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Do you have a definition/characterization of what $\mathcal O(n)$ is which is not by cases? –  Mariano Suárez-Alvarez Nov 10 '11 at 18:11
Unfortunately no. I was hoping someone else might. What motivated me to ask this question was proving the one positive, one negative part. –  Abelsh Nov 10 '11 at 18:43
This boils down to the simple fact that the degree of polynomails is additive under multiplication. –  Andrea Mori Nov 10 '11 at 20:23

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 $\mathcal{O}(n) \otimes \mathcal{O}(m) \simeq \mathcal{O}(n+m)$.