Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
add comment

2 Answers

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 $k$.]

share|improve this answer
add comment

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.

share|improve this answer
    
Dear Robert, thanks for your response. I want to use the criterion that an invertible sheaf is generated by a collection of global sections iff those global sections don't simultaneously vanish at some point of $X$. If I take the sections $x^2, y^2, z^2$, at which point do they simultaneously vanish? I am calculating that, for example, $x^2$ vanishes at the prime $\mathfrak{p}$ iff $x^2 \in \mathfrak{p}$. i.e. iff $x \in \mathfrak{p}$. Similarly for $y^2, z^2$. Thus, they simultaneously vanish at $\mathfrak{p}$ iff $x,y,z \in \mathfrak{p}$, contradiction. –  David Mar 13 '12 at 14:43
    
Sorry, I meant a priori that $x^2$ vanishes at $\mathfrak{p}$ iff $x^4 \in \mathfrak{p}$. i.e. iff $x \in \mathfrak{p}$. Of course this is equivalent to what I wrote above. –  David Mar 13 '12 at 14:49
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.