Let $\mathcal{O}_{\Bbb P^1}$ be the structure sheaf of the projective line $X=\Bbb P^1_k$ over some field $k$ (algebraically closed of characteristic $0$). What is a good (and, preferably, easy) way to compute the cohomology group $H^1(\Bbb P^1,\mathcal{O}_X)$?
I thought about taking the affine open cover consisting of $U_0=\{x_0\neq 0\}$ and $U_1=\{x_1\neq 0\}$ and using the exact skyscraper sequence $$0\to\mathcal{O}_X(-1)\to\mathcal{O}_X\to k_p\to 0,$$ where $p=(1:0)$. Here $\varphi\in\mathcal{O}_X(-1)(U)$ is mapped to $x_0\varphi\in\mathcal{O}_X(U)$, while $\mathcal{O}_X(U_0)\ni\psi\mapsto \psi(p)\in k_p(U_0)=k$.
Then we get a long exact cohomology sequence $$0\to \Gamma(\mathcal{O}_X(-1))\to\Gamma(\mathcal{O}_X)\to\Gamma(k_p)\to H^1(\mathcal{O}_X(-1))\to H^1(\mathcal{O}_X)\to H^1(k_p)\to 0.$$ Using $\Gamma(\mathcal{O}_X(-1))=0$, $\Gamma(\mathcal{O}_X)=k=\Gamma(k_p)$, and $H^1(k_p)$ should be zero, too, since by definition $C^1(k_p)=\prod_{i<j}k_p(U_i\cap U_j)=k_p(U_0\cap U_1)=0$. This simplifies the sequence a bit, but I don't know how to proceed from here.
Is there another way to compute the cohomology group (which should be $0$), maybe even without the long exact sequence? If yes, I could maybe use this to also prove $H^1(\Bbb P^1,\mathcal{O}_X(n))=\{x_0^ix_1^j:i+j=n,i<0,j<0\}$.
Edit: Just to be more precise, I'm talking about a variety point-of-view here, not scheme-theoretic, although I hope this shouldn't make much difference :)
Thanks a lot!
As the comments above suggest, it is proven in Hartshorne III, Theorem 5.1 that $H^n( \mathcal{O}_{\mathbb{P}_{k}^{n}}(m))$ is the $k$-vector space with basis equal to the monomials $x_0^{l_0}...x_n^{l_n}$ where $l_i$ are all negative and sum to $-m$. This of course implies that $H^1(\mathcal{O}_{\mathbb{P}_{k}^{n}}) = H^1(\mathcal{O}_{\mathbb{P}_{k}^{n}}(0)) = 0$.
The trick of the calculation is to compute the cohomology of $\bigoplus_{m} \mathcal{O}_{\mathbb{P}_{k}^{n}}(m)$ while keeping track of the grading.