global sections of ideal sheaf quotient $\mathcal{I}/\mathcal{I^{2}}$ Let $X=\mathbb{P}_{k}^{2}$, $P$ a point in $X$ with its ideal sheaf $\mathcal{I}$. Then I would like to know what is $H^{0}(X, \mathcal{I}/\mathcal{I}^{2})$.
My first thought was to consider the exact sequence
$$0\to \mathcal{I}^{2}\to \mathcal{I}\to \mathcal{I}/\mathcal{I}^{2}\to 0.$$
Taking the long exact sequence,
$$0\to H^{0}(X,\mathcal{I}^{2}) \to H^{0}(X,\mathcal{I})\to H^{0}(X,\mathcal{I}/\mathcal{I}^{2}) \to H^{1}(X,\mathcal{I}^{2})\to \cdots.$$
The first two $H^{0}$ are zero since they're powers of the ideal sheaf, then makes the fourth arrow an injective map. Hence what I only know is that dimension is bounded up by $h^{1}(X,\mathcal{I}^{2})$. Then what should I do next?
 A: Tabes Bridges' comment pretty much answers the question but I want to elaborate on some of the facts. Let's first consider the affine situation. Take an affine scheme $\operatorname{Spec} A$ and pick your closed point $\operatorname{Spec} k \to \operatorname{Spec} A$ corresponding to a maximal ideal $\mathfrak{m}$. Associated to such a point is a quasicoherent sheaf $\mathcal{I}$ of ideals, which is just the quasicoherent sheaf corresponding to $\mathfrak{m}$.
Now the quotient $\mathcal{I} / \mathcal{I}^2$ of a quasicoherent sheaf of ideals is often called the conormal sheaf because, geometrically, it identifies with the dual of the normal bundle of the closed subscheme. As the closed subscheme is just a point, this is geometrically the same as the cotangent space at that point, and indeed $\mathcal{I} / \mathcal{I}^2$ is also the cotangent sheaf at the closed point $[\mathfrak{m}]$.
When we say 'at the closed point' what do we mean? I claim that $\mathcal{I} / \mathcal{I}^2$ is the skyscraper sheaf supported on the closed point $[\mathfrak{m}]$ with value $\mathfrak{m} / \mathfrak{m}^2$. I don't actually know what the best way of seeing this (and I'd be happy if someone can remark on this), though the following commutative algebra lemma suffices in many cases:
Lemma. Let $M$ be a finitely generated $A$-module, and let $I$ be an ideal of $A$. Then $\operatorname{Supp}(M / IM) = V(I) \cap \operatorname{Supp} M$.
Now, turning to the global situation, nothing really changes. All the behaviour of the cotangent bundle should be 'local', i.e. concentrated around a small open neighbourhood of whatever closed point you consider. Therefore, it remains the case when $X$ is not affine that $\mathcal{I} / \mathcal{I}^2$ will be a skyscraper sheaf. In particular, the global sections may be identified with $\mathfrak{m} / \mathfrak{m}^2$ after choosing an open neighbourhood $\operatorname{Spec} A$ of $P$ such that $P$ identifies with $[\mathfrak{m}]$.
Remark: The module $\mathfrak{m} / \mathfrak{m}^2$ in fact has the structure of a vector space over the field $k = A / \mathfrak{m}$. Let's consider the example $X = \mathbb{P}^2_k$ that you're interested in, and assume that $k$ is algebraically closed*. Locally, this variety looks like $\operatorname{Spec} k[x,y]$, and a closed point corresponds to a maximal ideal $(x - a,y - b)$. A direct computation shows that $\mathfrak{m} / \mathfrak{m}^2$ must then be $2$-dimensional as a $k$-vector space. This is in line with the geometric interpretation: $\mathbb{P}^2$, as say a manifold, is $2$-dimensional after all.
*: Without this assumption there are additional maximal ideals, and the observation needs to be corrected. Thanks to Georges Elencwajg for noting this.
