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 $E \to f_\star L$ is not surjective, therefore the second map defined only on an open. Lukely we can chose the open as the complement of $P(coker\ E_j \to f_\star L)$, then show that image of map $r$ is contained in the chosen open.

Comment #5800 by on

@#5799: If $X$ is defined over a field $k$ and $\mathcal{L}$ is relatively very ample on $X$ over $k$, then it follows from the definition that there is a $k$-vector space $E$ and a surjective module map $E \otimes_k \mathcal{O}_X \to \mathcal{L}$. Sure this does not mean that $E \to f_*\mathcal{L}$ is surjective, but we never use this in the argument. (The argument actually shows you can take $E$ to correspond to $f_*\mathcal{L}$ 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 $P(f_\star L) \to P(E)$. When we try to show (4) from (2), we cannot specify E to correspond to $f_\star L$, 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 $r$ 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 $\mathcal{E}_ j \to (f_*\mathcal{L})|_{X_ j}$ we see that there is a factorization Here $\mathbf{P}(f_*\mathcal{L})|_{V_ j} \to \mathbf{P}(\mathcal{E}_ j)$ 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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