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Definition. [Hartshorne] If $X$ is any scheme over $Y$, an invertible sheaf $\mathcal{L}$ is very ample relative to $Y$, if there is an imersion $i:X \to \mathbb{P}_Y^r$ for some $r$ such that $i^\ast(\mathcal{O}(1)) \simeq \mathcal{L}$.

My question is: what is the right way (interpret "right way" as you wish) to think about very ample sheaves? In particular, why is the word "ample" being used? What is it that I have an ample amount of? Degree 1 elements?

In the simple case when $Y=\text{Spec}(A)$ is affine then $i^\ast(\mathcal{O}(1))$ is just $\mathcal{O}(1)$ as defined on $\text{Proj} A[x_0,\ldots x_r]$, that is, it's she sheafification of the degree 1 part of the polynomial ring $A[x_0,\ldots,x_n]$. So it seems like the more general definition is just meant to generalize this phenomenon. Is this true? If so, why is it worth generalizing? What's special about degree 1 elements? The only thing I can think of is that the polynomial ring is generated as an $A$-algebra by its degree 1 elements.

As you can tell, my question is not very well formed, so feel free to add anything you think is relevant. I am also happy to expand on anything I've written here.

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up vote 8 down vote accepted

$\newcommand{P}{\mathbb{P}}\newcommand{E}{\mathscr{E}}\newcommand{O}{\mathscr{O}}\newcommand{L}{\mathscr{L}}$Let me first briefly summarize a construction from EGA (II, 4.2). Let $X$ and $Y$ be schemes, $q : X \to Y$ a morphism, $\E$ a quasi-coherent $\O_Y$-module, and $\P(\E)$ the projective bundle defined by $\E$. There is a bijection between the $Y$-morphisms from $X$ to $\P(\E)$ and equivalence classes of pairs $(\L, \varphi)$ of invertible $\O_X$-modules $\L$ and surjective homomorphisms $\varphi : q^*(\E) \to \L$, under the relation where $(\L, \varphi)$ and $(\L', \varphi')$ are identified if there is an isomorphism $\tau : \L \stackrel{\sim}{\to} \L'$ such that $\varphi' = \tau \circ \varphi$. This correspondence is such that a morphism $r : X \to \P(\E)$ corresponds to the invertible $\O_X$-module $r^*(\O_{\P(\E)}(1))$; see EGA for the details. In the particular case $\E = \O_Y^{n+1}$, $P = \P(\O_Y^{n+1}) = \P^n_Y$, it follows that morphisms $r : X \to \P^n_Y$ are in bijection with invertible $\O_X$-modules $\L$ and surjective homomorphisms $\varphi : q^*(\O_Y^{n+1}) \to \L$; but $q^*(\O_Y^{n+1}) = \O_X^{n+1}$ and surjective homomorphisms $\O_X^{n+1} \to \L$ are in bijection with surjective homomorphisms $\O_{X,x}^{n+1} \to \L_x \stackrel{\sim}{\to} \O_{X,x}$, which are again in bijection with tuples $(s_0, \ldots, s_n)$ of global sections of $\L$ that generate $\L$ (i.e. have no common zeros).

Now an invertible $\O_X$-module $\L$ is called very ample for $q$ if there exists a quasi-coherent $\O_Y$-module $\E$ and an immersion $i : X \to P = \P(\E)$ such that $\L$ is isomorphic to $i^*(\O_P(1))$. One immediately sees that this is equivalent to the condition that there exists a quasi-coherent $\O_Y$-module $\E$ and a surjective homomorphism $\varphi : q^*(\E) \to \L$ such that the corresponding morphism $X \to P = \P(\E)$ is an immersion. As we saw above, in the case $\E = \O_Y^{n+1}$, this means that $\L$ is globally generated by $n+1$ sections. Basically, the term very ample is referring to the global sections: roughly speaking, $\L$ is very ample if there are "enough" global sections to define an immersion into projective space.

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Thanks for the answer. In general, is it true that very ample sheaves are generated by global sections? If so, what else is needed for the converse to hold. That is: generated by global sections + X = very ample. What is X? – Derek Allums Feb 27 '13 at 19:46
Taking Hartshorne's definition ($\E = \O_Y^{n+1}$), $\L$ is very ample iff it is globally generated by global sections $s_0, \ldots, s_n$ such that the corresponding morphism $X \to \P^n_Y$ is an immersion. – Adeel Feb 27 '13 at 21:35
Awesome, thanks @Adeel. – Derek Allums Feb 27 '13 at 22:17

Intuitively and for my answer, I am working over an algebraically closed field $k$.

I personally think about invertible sheaves as sheaves of functions on a scheme, where $s\in\mathcal L(U)$ can be evaluated at a point $P\in U$ in the sense that $s(P)$ is the image of $s$ under the morphism $\mathcal L(U) \to \mathcal L_P \cong \mathcal O_{X,P} \twoheadrightarrow \mathcal O_{X,P}/\mathfrak m_P=k(P)=k$. Of course, this depends on the local isomorphism we choose, but if I choose the same local isomorphism for $s_0,\ldots,s_n\in\mathcal L(U)$, then $[s_0(P):\ldots:s_n(P)]\in\mathbb P_k^n$ is well-defined as a point in projective space.

Hence intuitively, $\mathcal L$ is very ample if these functions can serve as coordinates, i.e. I have enough functions to distinguish points - and, in fact, this is close to an alternative characterization, see Proposition II.7.3 in Hartshorne (page 152): The morphism $\varphi:X\to\mathbb P^n$ corresponding to $\mathcal L$ and a choice of global sections $s_0,\ldots,s_n\in\mathcal L(X)$ is a closed immersion if and only if it separates points and tangent vectors, i.e.

  1. For closed points $P,Q\in X$ with $P\ne Q$ there exists $s\in V:=\langle s_0,\ldots,s_n\rangle$ with $s(P)=0$ and $s(Q)\ne 0$, using my above notation.
  2. For each closed point $P\in X$, the set $\{ s\in V \mid s(P)=0 \}$ spans the $k$-vector space $\mathfrak m_P\mathcal L_P/\mathfrak m_P^2\mathcal L_P$.

The second condition is a bit more subtle, but I found the geometric intuition given at the end of this blog post quite satisfying.

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Thanks for the answer. I'll come back to it once I get to section 7 in a few weeks. Until then: your answer sounds (roughly) like "very ample means enough global generators." Would you agree with this? – Derek Allums Feb 27 '13 at 19:51
Oh I totally agree with that, and even more so, Adeel's answer is beautiful. It carries the modern spirit of algebraic geometry, and it answers your question in perfect generality. Mine comes from a more classical angle, from where it is easier to quickly develop some intuition - but keep in mind the downside; I have forced $Y$ to be of the form $\mathrm{Spec}(k)$ with $k$ algebraically closed. – Jesko Hüttenhain Feb 27 '13 at 20:53
True, but since I'm a beginning, I do like relating more general constructions and ideas to elementary examples when possible, and your answer definitely helps with that. – Derek Allums Feb 27 '13 at 22:18

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