First of all I am sorry if this question is duplicate. I tried to search it but didn't succeed.
Let $X$ be a blowing up of projective plane $\mathbb{P}_k^2$ at one point and $L$ be an exceptional divisor. How can one compute cohomologies of $\mathcal{O}_X(-L)$? I am trying to check that they are all zero.
The general theory suggests that from the exact sequence
$$0\rightarrow\mathcal{O}_X(-L)\rightarrow\mathcal{O}_X\rightarrow\mathcal{O}_L\rightarrow0$$
we have the exact sequence of cohomologies
$$0\rightarrow H^0(X,\mathcal{O}_X(-L))\rightarrow H^0(X,\mathcal{O}_X)\rightarrow H^0(X, \mathcal{O}_L)\rightarrow H^1(X,\mathcal{O}_X(-L))\rightarrow H^1(X,\mathcal{O}_X)\rightarrow H^1(X, \mathcal{O}_L)\rightarrow...$$
and from the known fact $H^1(X,\mathcal{O}_X)=0$ we obtain the exact sequence
$$0\rightarrow H^0(X,\mathcal{O}_X(-L))\rightarrow H^0(X,\mathcal{O}_X)\rightarrow H^0(X, \mathcal{O}_L)\rightarrow H^1(X,\mathcal{O}_X(-L))\rightarrow0$$
in which $H^0(X,\mathcal{O}_X)=k$ and $H^0(X, \mathcal{O}_L)=k$. I guess that the map from $H^0(X,\mathcal{O}_X)$ to $H^0(X, \mathcal{O}_L)$ is injective which means that $H^0(X,\mathcal{O}_X(-L))=H^1(X,\mathcal{O}_X(-L))=0$ but I have some difficulties to justify this fact.
How can I finish these computations? (I mean using this approach)