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The homogeneous polynomials in $\mathbb{C}[x_0,\cdots,x_n]$ can be considered as the global sections of a line bundle over $\mathbb{P}^n$ (the line bundle corresponding to Serre's twisting sheaf). In fact if $\mathcal{L}$ is the line bundle corresponding to $\mathcal{O}(1)$ then the homogeneous polynomials in degree $d$ are (isomorphic to) $H^{0}(\mathbb{P}^n, \mathcal{L}^{\otimes d})$. The line bundle $\mathcal{L}$ is globally generated since $\mathbb{P}^n$ can be covered by open sets $U_i = \{x = [x_0:\cdots:x_n] \in \mathbb{P}^n: x_i \not = 0\}$.

According to this example/theorem in Lazarsfeld (2.1.29):

http://books.google.dk/books?id=T87ftUcU_hEC&lpg=PA126&ots=dPxmwqbLJJ&dq=lazarsfeld%20positivity%20%22section%20ring%22&hl=da&pg=PA131#v=onepage&q&f=false

the multiplication map: $H^0(\mathbb{P}^n, \mathcal{L}^{\otimes a}) \otimes H^0(\mathbb{P}^n, \mathcal{L}^{\otimes b}) \rightarrow H^0(\mathbb{P}^n, \mathcal{L}^{\otimes (a+b)})$ is surjective for $a,b$ big enough. This seems wrong to me since it would mean that there are no irreducible homogeneous of arbitrarily high degree, which is not true if $n > 1$.

I'm a beginner in Algebraic Geometry so I would appreciate it if someone can explain to me what is it that I'm missing or misunderstanding about this example. Thanks.

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The $\otimes$ means all linear combinations of products. And clearly any monomial of degree $a+b$ is a product of monomials of degree $a$ and $b$ respectively. –  WimC Mar 15 '12 at 9:03
    
OH, right! That's it, I was clearly being stupid. Thank you for your answer. –  Nadim Rustom Mar 15 '12 at 9:07
    
@WimC Please consider promoting your comment to an answer, since it indeed answered the question. –  ˈjuː.zɚ79365 Jun 13 '13 at 4:04

1 Answer 1

The $\otimes$ means all linear combinations of products. And clearly any monomial of degree $a+b$ is a product of monomials of degree $a$ and $b$ respectively.

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