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This is a simple question of notation. Let $k(p)$ be the residue field of a point $p$ on $\mathbb{P}^{N}$. How is defined (and where is) the sheaf $\mathcal{O}_{\mathbb{P}^{N}}(1) \otimes k(p)$? Also, what could mean the tensor $H^{0}(\mathbb{P}^{N},\mathcal{O}_{\mathbb{P}^{N}}(1)) \otimes \mathcal{O}_{\mathbb{P}^{N}}$?

Thank you.

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up vote 5 down vote accepted

a) The sheaf $\mathcal{O}_{\mathbb{P}^{N}}(1) \otimes k(p)$ is the sky-scraper sheaf on $\mathbb{P}^{N}$ whose only non-zero fiber is over $p$, the value of that fiber being the one-dimensional $k(p)$-vector space $L(p)^\ast $, where $L(p)\subset k^{n+1}$ is the line represented by the point $p$ and the asterisk means "dual vector space".
The notation $\mathcal{O}_{\mathbb{P}^{N}}(1) \otimes k(p)$ may also denote just the vector space $L(p)^\ast $ or may denote the sheaf with that fiber over the one-point scheme $\operatorname {Spec } (k(p))$ .

b) The notation $H^{0}(\mathbb{P}^{N},\mathcal{O}_{\mathbb{P}^{N}}(1)) \otimes \mathcal{O}_{\mathbb{P}^{N}}$ denotes the trivial vector bundle (= locally free sheaf) of rank $n+1$ with fiber $H^{0}(\mathbb{P}^{N}, \mathcal{O}_{\mathbb{P}^{N}}(1))$ .

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Let $\mathscr{F}$ be some $\mathscr{O}_X$-module. To say what Georges has expertly said in a different way (I hope):

(1) To me, $\mathscr{F} \otimes k(p)$ usually means "pull back $\mathscr{F}$ to the point $\operatorname{Spec} k(p)$". Since that's just a point, we can identify the sheaf with the module: it's $\mathscr{F}_p \otimes_{\mathscr{O}_{X, p}} k(p) = \mathscr{F}_p/\mathfrak{m}_p\mathscr{F}_p$.

(2) Working over some field $k$ (it could be a ring; I don't think it matters), the constant sheaf $\underline{k}$ with values in $k$ is a subsheaf of rings of $\mathscr{O}_X$; and the constant sheaf with values in $H^0(X, \mathscr{F})$ is a sheaf of modules over that, so we can tensor over $\underline{k}$ with $\mathscr{O}_X$. I think this is usually done in order to talk about global generation, ie, whether the multiplication map $H^0(X, \mathscr{F}) \otimes \mathscr{O}_X \to \mathscr{F}$ is surjective.

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(2) feels a little weird just because I haven't seen it anywhere, but it has the right fibers... – Hoot May 12 '14 at 15:44

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