Projective morphism, generated by global sections basic question I have a very dumb question.  Let $X = \mathbb{P}^2_k = Proj(k[x,y,z])$ where $k$ is algebraically closed.  We have an invertible sheaf $\mathcal{O}(2)$ on $X$.  Its space of global sections contains the elements $x^2, y^2, z^2, xy, yz, xz$.  
It seems to me (by my calculations), however, that $\mathcal{O}(2)$ is generated by $x^2, y^2$, and $z^2$.  Meaning, these 3 global sections generate the stalks at each point of $X$.  I'm suspicious, though.  Is this true?
David
 A: One should not confuse the two different statements :
a) The sections $x^2,y^2,z^2$ generate the $k$-vector space $\Gamma(\mathbb P^2_k,\mathcal O(2))$, which happens to be false.
b) The sections $x^2,y^2,z^2$ generate the line bundle $O(2)$, which happens to be true.  
a) The first statement probably needs no further eleboration since the $k$-vector space $\Gamma(\mathbb P^2_k,\mathcal O(2))$ is the 6-dimensional space $k[x,y,z]_2$ of homogeneous polynomials of degree 2, which obviously cannot be generated by only 3 vectors.  
b) And what does the second statement even mean? It means that that the fibre of $\mathcal O_{\mathbb P^2_k}(2)$ at any point $P\in \mathbb P^2_k$ is generated by the values of the three sections at $P$.
Since the fibres of at any $P\in \mathbb P^2_k$ are 1-dimensional, the condition just states that the three sections do not vanish simultaneously at $P$.
 And this is indeed the case since  at any $P\in \mathbb P^2_k$ it is impossible to have simultaneously $x^2=y^2=z^2=0$  .   
[It is also vital to carefully distinguish between the stalk  $\mathcal (O(2))_P$, which is a free module of rank one over $\mathcal O_P$ and thus an infinite-dimensional vector space over $k$, and the fibre $\mathcal O(2)_P\otimes_{O_P} k$ which is a 1-dimensional vector space over $k$.]
A: The thing is, with $x^2$, $y^2$ and $z^2$, there's no way of generating $xy$, $xz$ and $yz$; that's the problem. That's why these necessarily have to be part of the generating set.
