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Let $X$ be a quasi-compact scheme. Let $\mathcal L$ be an invertible sheaf on $X$. We say $\mathcal L$ is ample if for any finitely generated quasi-coherent sheaf $\mathcal F$ on $X$, there exists some $N\ge 1$ such that $\mathcal F\otimes \mathcal L^{\otimes n}$ is generated by its global sections for all $n\ge N$.

I have three questions.

$1.$ Both $\mathcal F$ and $\mathcal L$ are finitely generated. Using the quasi-compactness hypothesis, it seems we should be able to show that "generated by global sections" in the above statement can be replaced by "generated by a finite number of global sections." Is this true, and if so, how does one prove it?

I apologize if this is very easy -- I am getting tangled up in the definitions. Here is my attempt:

Suppose $\mathcal F\otimes \mathcal L^{\otimes n}$ is generated by its global sections for some $n$. Let this set be $\{s_\alpha\}$. Then for each open affine $U$, $\mathcal F\otimes \mathcal L^{\otimes n}|_U$ is generated by its global sections $\{s_\alpha|_U\}$. So it suffices to reduce to a finite affine cover, because if we show it for this cover, we can collect their respective finite generating sets into one large, but finite, set.

Since $\mathcal F$ and $\mathcal L$ are finitely generated, so is $\mathcal F\otimes \mathcal L^{\otimes n}$. So we may take a finite open cover by affine open subsets $U_i$ that are finitely generated as $\mathcal O_X$ modules, using quasi-compactness. Then, since the $\mathcal F\otimes \mathcal L^{\otimes n}(U_i)$ are finitely generated, a finite number of $\{s_\alpha|_{U_i}\}$ suffice to generate all the stalks.

The last sentence is just a guess. I don't think it's correct, but it also seems like the only way to proceed.

$2.$ Consider the following theorem

Let $f:X\rightarrow Y$ be a proper morphism of locally Noetherian schemes. Let $\mathcal L$ be an invertible sheaf on $X$. Let us fix a point $y\in Y$ and let $\phi: X \times _Y \operatorname{Spec} \mathcal O_{Y,y} \rightarrow X$ be the canonical morphism. Then if $\phi^* \mathcal L$ is ample, there exists an open neighborhood $V$ of $y$ such that $\mathcal L_{f^{-1}(V)}$ is ample.

The proof begins by saying that if we replace $\mathcal L$ by a tensor power, we may assume $\phi^*\mathcal L$ is very ample and generated by its global sections. Why is this?

(A very ample sheaf is a sheaf that is the pullback under some immersion $i:X\rightarrow \mathbb P^d_A$ of $\mathcal O_{P^d_A}(1).$)

$3.$ It was shown earlier that some tensor power of $\mathcal L$ is very ample. Is it true that every very ample sheaf is generated by its global sections? Wikipedia says this is true, but it is not transparent from the description in my book.

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For number one, I'm not saying this messes anything up, but you should be very careful about the difference between "finitely generated" and "generated by a finite number of global sections." You say "Then, since the $\mathcal F\otimes \mathcal L^{\otimes n}(U_i)$ are finitely generated, a finite number of $\{s_\alpha|_{U_i}\}$ ..." But finitely generated means that there is a sheaf surjection $\mathcal{O}_{U_i}\to\mathcal{F}\otimes L^n|_{U_i}$. It isn't clear that a finite number of global $\mathcal{F}\otimes L^n(U_i)$ will be a generating set. –  Matt Jul 4 '13 at 22:24
    
@Matt Yes. What I was trying (poorly) to get across is that I think that is true, but I can't prove it. Or maybe it's false. I don't know. –  Potato Jul 4 '13 at 22:28
    
Oops. That should be sheaf surjection from $\mathcal{O}_{U_i}^m$, but I think you got that. –  Matt Jul 4 '13 at 22:32

1 Answer 1

up vote 1 down vote accepted
+200

I will attempt to answer your questions in reverse order.

Q3: Consider an open affine $U\subset X$, we will show that if $\mathcal{L}$ is a very ample sheaf on $X$ then $\Gamma(X, \mathcal{L}) \rightarrow \Gamma(U, \mathcal{L})$ is surjective. Notice that shrinking $U$ if necessary, we may assume that $U \subset \mathbb{A}^{d} \cap X$, where $\mathbb{A}^{d}$ is a standard affine open (or more appropriately a hyperplane section) in $\mathbb{P}^{d}$. ( $U$ corresponds to the non-vanishing locus of a function on $X$, lift that function to $\mathbb{P}^{d}$).
Now take any section $s \in \Gamma(U,\mathcal{L})$ it lift to a section of $\mathcal{A}^{d}$ (because $U$ is a closed set in an affine space), this section further lifts to $\mathbb{P}^{d}$ (because $\mathcal{O}(1)$ is ample (this is a computation)) and then consider the restriction of this section to $X$.

Q2: Since $\mathcal{G} =\phi^{\ast}\mathcal{L}$ is ample, we can after replacing it by some tensor power assume that $\mathcal{G}^{\otimes n}$ it is generated by global sections $\{ s_{0}, \ldots, s_{N} \}$ which in turn gives a morphism to $X \rightarrow \mathbb{P}^{\mathcal{G}}= Proj(\oplus_{m} \Gamma(X,\mathcal{G}^{mn})$ (this is a standard fact you can find in Hartshorne or better Mumford explains this in Curves on an algebraic surface). Naturally $\mathcal{G}$ is the pullback of $\mathcal{O}(1)$ on $\mathbb{P}^{\mathcal{G}}$.

Q1: I agree with your approach. Recall that in algebra when we say an ideal is invertible we implicitly mean that some finite combination is identity. The only thing that I will be cautious about is if we do not assume $X$ is Noetherian and(or) separated scheme then there might be some pathological counterexamples. I cannot think of something right now... Probably it will still work in that case.

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For $1$, the issue is that we don't know that $\mathcal O_X^n(U)\rightarrow \mathcal F\otimes \mathcal L^n(U)$ is surjective (because this could fail to be surjective, even when we know $O_X^n\rightarrow \mathcal F\otimes \mathcal L^n$ is as a sheaf morphism). So we don't know that the $s_i|U$ can generate all of the generators of $\mathcal F\otimes \mathcal L^n(U)$. –  Potato Jul 9 '13 at 21:05
    
For $3$, you seem to be assuming that $X$ is $\mathbb P^d$ (you speak of taking one of the standard affine neighborhoods.) Is this permissible, or am I misunderstanding something? –  Potato Jul 9 '13 at 21:19
    
I don't follow your work for $2$ at all. The proof begins by taking powers of $\mathcal L$, not $\phi^*\mathcal L$. –  Potato Jul 9 '13 at 21:21
    
The functoriality of $\phi^{\ast}$ will imply that $\phi^{\ast}(\mathcal{L}^{\otimes n}) = (\phi^{\ast} \mathcal{L})^{n}$. –  DBS Jul 9 '13 at 22:03
    
I am sorry if it sounds little confusing, but what I am saying in 2 is that $X$ is a closed subscheme of $\mathbb{P}^{d}$. If you take a sufficiently small open affine set in $X$, then this affine set is contained in the intersection of $X$ with one of the standard affine open subsets of $\mathbb{P}^{d}$, so it will become a closed subscheme inside affine space. –  DBS Jul 9 '13 at 22:06

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