The Stacks project

Lemma 29.38.7. Let $f : X \to S$ be a morphism of schemes. Let $\mathcal{L}$ be an invertible sheaf on $X$. Assume $f$ is quasi-compact. The following are equivalent

  1. $\mathcal{L}$ is relatively very ample on $X/S$,

  2. there exists an open covering $S = \bigcup V_ j$ such that $\mathcal{L}|_{f^{-1}(V_ j)}$ is relatively very ample on $f^{-1}(V_ j)/V_ j$ for all $j$,

  3. there exists a quasi-coherent sheaf of graded $\mathcal{O}_ S$-algebras $\mathcal{A}$ generated in degree $1$ over $\mathcal{O}_ S$ and a map of graded $\mathcal{O}_ X$-algebras $\psi : f^*\mathcal{A} \to \bigoplus _{n \geq 0} \mathcal{L}^{\otimes n}$ such that $f^*\mathcal{A}_1 \to \mathcal{L}$ is surjective and the associated morphism $r_{\mathcal{L}, \psi } : X \to \underline{\text{Proj}}_ S(\mathcal{A})$ is an immersion, and

  4. $f$ is quasi-separated, the canonical map $\psi : f^*f_*\mathcal{L} \to \mathcal{L}$ is surjective, and the associated map $r_{\mathcal{L}, \psi } : X \to \mathbf{P}(f_*\mathcal{L})$ is an immersion.

Proof. It is clear that (1) implies (2). It is also clear that (4) implies (1); the hypothesis of quasi-separation in (4) is used to guarantee that $f_*\mathcal{L}$ is quasi-coherent via Schemes, Lemma 26.24.1.

Assume (2). We will prove (4). Let $S = \bigcup V_ j$ be an open covering as in (2). Set $X_ j = f^{-1}(V_ j)$ and $f_ j : X_ j \to V_ j$ the restriction of $f$. We see that $f$ is separated by Lemma 29.38.6 (as being separated is local on the base). By assumption there exists a quasi-coherent $\mathcal{O}_{V_ j}$-module $\mathcal{E}_ j$ and an immersion $i_ j : X_ j \to \mathbf{P}(\mathcal{E}_ j)$ with $\mathcal{L}|_{X_ j} \cong i_ j^*\mathcal{O}_{\mathbf{P}(\mathcal{E}_ j)}(1)$. The morphism $i_ j$ corresponds to a surjection $f_ j^*\mathcal{E}_ j \to \mathcal{L}|_{X_ j}$, see Constructions, Section 27.21. This map is adjoint to a map $\mathcal{E}_ j \to f_*\mathcal{L}|_{V_ j}$ such that the composition

\[ f_ j^*\mathcal{E}_ j \to (f^*f_*\mathcal{L})|_{X_ j} \to \mathcal{L}|_{X_ j} \]

is surjective. We conclude that $\psi : f^*f_*\mathcal{L} \to \mathcal{L}$ is surjective. Let $r_{\mathcal{L}, \psi } : X \to \mathbf{P}(f_*\mathcal{L})$ be the associated morphism. We still have to show that $r_{\mathcal{L}, \psi }$ is an immersion; we urge the reader to prove this for themselves. The $\mathcal{O}_{V_ j}$-module map $\mathcal{E}_ j \to f_*\mathcal{L}|_{V_ j}$ determines a homomorphism on symmetric algebras, which in turn defines a morphism

\[ \mathbf{P}(f_*\mathcal{L}|_{V_ j}) \supset U_ j \longrightarrow \mathbf{P}(\mathcal{E}_ j) \]

where $U_ j$ is the open subscheme of Constructions, Lemma 27.18.1. The compatibility of $\psi $ with $\mathcal{E}_ j \to f_*\mathcal{L}|_{V_ j}$ shows that $r_{\mathcal{L}, \psi }(X_ j) \subset U_ j$ and that there is a factorization

\[ \xymatrix{ X_ j \ar[r]^-{r_{\mathcal{L}, \psi }} & U_ j \ar[r] & \mathbf{P}(\mathcal{E}_ j) } \]

We omit the verification. This shows that $r_{\mathcal{L}, \psi }$ is an immersion.

At this point we see that (1), (2) and (4) are equivalent. Clearly (4) implies (3). Assume (3). We will prove (1). Let $\mathcal{A}$ be a quasi-coherent sheaf of graded $\mathcal{O}_ S$-algebras generated in degree $1$ over $\mathcal{O}_ S$. Consider the map of graded $\mathcal{O}_ S$-algebras $\text{Sym}(\mathcal{A}_1) \to \mathcal{A}$. This is surjective by hypothesis and hence induces a closed immersion

\[ \underline{\text{Proj}}_ S(\mathcal{A}) \longrightarrow \mathbf{P}(\mathcal{A}_1) \]

which pulls back $\mathcal{O}(1)$ to $\mathcal{O}(1)$, see Constructions, Lemma 27.18.5. Hence it is clear that (3) implies (1). $\square$

Comments (7)

Comment #5799 by Luke on

The last step in proving (4) should be modified, since even in the case when base is a field there are examples in which is not surjective, therefore the second map defined only on an open. Lukely we can chose the open as the complement of , then show that image of map is contained in the chosen open.

Comment #5800 by on

@#5799: If is defined over a field and is relatively very ample on over , then it follows from the definition that there is a -vector space and a surjective module map . Sure this does not mean that is surjective, but we never use this in the argument. (The argument actually shows you can take to correspond to and then the surjectivity also holds on the base but it doesn't always need to be true.)

Comment #5813 by Luke on

@5800: But the proof implicitly assumes the surjectivity since you claim in the proof that there is a (linear) morphism . When we try to show (4) from (2), we cannot specify E to correspond to , because the E here comes from the definition of relatively very ample line bundle(def 29.38.1). Otherwise we will make a circular argument. My claim in previous comment is to remedy this by just admitting that the linear "morphism" is really a rational map and the image of morphism lies in the defined open.

Comment #5815 by on

Please tell me the exact step in the proof that is wrong. Thanks!

Comment #5816 by Luke on

In the 3rd paragraph, Using the maps we see that there is a factorization Here should be a rational map rather than a morphism.

Comment #5817 by on

Ah, yes, OK, thanks very much! I will fix this when I next go through all the comments.

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