Cohomology groups of a non-degenerate algebraic variety.

Question

Let $X\subset\mathbb{P}^{n}$ be an algebraic variety. Let us suppose that $X$ is non-degenerate (it is not contained in any hyperplane of $\mathbb{P}^{n}$). I have read that (at least for curves) the natural map $$ H^{0}(\mathbb{P}^{n},\mathcal{O}_{\mathbb{P}^{n}}(1))\rightarrow H^{0}(X,\mathcal{O}_{X}(1)) $$ is injective, i.e. $H^{0}(\mathbb{P}^{n},\mathcal{I}_{X,\mathbb{P}^{n}}(1))=0$, where $\mathcal{I}_{X,\mathbb{P}^{n}}$ is the sheaf of modules induced by the ideal of $X\subset\mathbb{P}^{n}$. I don't know why this is true. I would appreciate if you could tell me why it holds or if you could recommend me a book where this is shown.

Answer 1

This is just because $H^{0}(\mathbb{P}^{n},\mathcal{O}_{\mathbb{P}^{n}}(1))$ is given by linear functions, so if one were in the kernel of $H^{0}(\mathbb{P}^{n},\mathcal{O}_{\mathbb{P}^{n}}(1))\rightarrow H^{0}(X,\mathcal{O}_{X}(1))$ it would precisely mean that it vanishes on $X$, so $X$ would be contained in the corresponding hyperplane.