Lemma 33.47.1. Let $k$ be a field. Let $X$ be a proper scheme over $k$. Let $\mathcal{L}$ be an ample invertible $\mathcal{O}_ X$-module. Let $Z \subset X$ be a closed subscheme. Then there exists an integer $n_0$ such that for all $n \geq n_0$ the kernel $V_ n$ of $\Gamma (X, \mathcal{L}^{\otimes n}) \to \Gamma (Z, \mathcal{L}^{\otimes n}|_ Z)$ generates $\mathcal{L}^{\otimes n}|_{X \setminus Z}$ and the canonical morphism

$X \setminus Z \longrightarrow \mathbf{P}(V_ n)$

is an immersion of schemes over $k$.

Proof. Let $\mathcal{I} \subset \mathcal{O}_ X$ be the quasi-coherent ideal sheaf of $Z$. Observe that via the inclusion $\mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n} \subset \mathcal{L}^{\otimes n}$ we have $V_ n = \Gamma (X, \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n})$. Choose $n_1$ such that for $n \geq n_1$ the sheaf $\mathcal{I} \otimes \mathcal{L}^{\otimes n}$ is globally generated, see Properties, Proposition 28.26.13. It follows that $V_ n$ generates $\mathcal{L}^{\otimes n}|_{X \setminus Z}$ for $n \geq n_1$.

For $n \geq n_1$ denote $\psi _ n : V_ n \to \Gamma (X \setminus Z, \mathcal{L}^{\otimes n}|_{X \setminus Z})$ the restriction map. We get a canonical morphism

$\varphi = \varphi _{\mathcal{L}^{\otimes n}|_{X \setminus Z}, \psi _ n} : X \setminus Z \longrightarrow \mathbf{P}(V_ n)$

by Constructions, Example 27.21.2. Choose $n_2$ such that for all $n \geq n_2$ the invertible sheaf $\mathcal{L}^{\otimes n}$ is very ample on $X$. We claim that $n_0 = n_1 + n_2$ works.

Proof of the claim. Say $n \geq n_0$ and write $n = n_1 + n'$. For $x \in X \setminus Z$ we can choose $s_1 \in V_1$ not vanishing at $x$. Set $V' = \Gamma (X, \mathcal{L}^{\otimes n'})$. By our choice of $n$ and $n'$ we see that the corresponding morphism $\varphi ' : X \to \mathbf{P}(V')$ is a closed immersion. Thus if we choose $s' \in \Gamma (X, \mathcal{L}^{\otimes n'})$ not vanishing at $x$, then $X_{s'} = (\varphi ')^{-1}(D_+(s'))$ (see Constructions, Lemma 27.14.1) is affine and $X_{s'} \to D_+(s')$ is a closed immersion. Then $s = s_1 \otimes s' \in V_ n$ does not vanish at $x$. If $D_+(s) \subset \mathbf{P}(V_ n)$ denotes the corresponding open affine space of our projective space, then $\varphi ^{-1}(D_+(s)) = X_ s \subset X \setminus Z$ (see reference above). The open $X_ s = X_{s'} \cap X_{s_1}$ is affine, see Properties, Lemma 28.26.4. Consider the ring map

$\text{Sym}(V)_{(s)} \longrightarrow \mathcal{O}_ X(X_ s)$

defining the morphism $X_ s \to D_+(s)$. Because $X_{s'} \to D_+(s')$ is a closed immersion, the images of the elements

$\frac{s_1 \otimes t'}{s_1 \otimes s'}$

where $t' \in V'$ generate the image of $\mathcal{O}_ X(X_{s'}) \to \mathcal{O}_ X(X_ s)$. Since $X_ s \to X_{s'}$ is an open immersion, this implies that $X_ s \to D_+(s)$ is an immersion of affine schemes (see below). Thus $\varphi _ n$ is an immersion by Morphisms, Lemma 29.3.5.

Let $a : A' \to A$ and $c : B \to A$ be ring maps such that $\mathop{\mathrm{Spec}}(a)$ is an immersion and $\mathop{\mathrm{Im}}(a) \subset \mathop{\mathrm{Im}}(c)$. Set $B' = A' \times _ A B$ with projections $b : B' \to B$ and $c' : B' \to A'$. By assumption $c'$ is surjective and hence $\mathop{\mathrm{Spec}}(c')$ is a closed immersion. Whence $\mathop{\mathrm{Spec}}(c') \circ \mathop{\mathrm{Spec}}(a)$ is an immersion (Schemes, Lemma 26.24.3). Then $\mathop{\mathrm{Spec}}(c)$ has to be an immersion because it factors the immersion $\mathop{\mathrm{Spec}}(c') \circ \mathop{\mathrm{Spec}}(a) = \mathop{\mathrm{Spec}}(b) \circ \mathop{\mathrm{Spec}}(c)$, see Morphisms, Lemma 29.3.1. $\square$

Comment #7839 by quasicompact on

There's a missing parenthesis at "$X_{s'} = (\varphi ')^{-1}(D_+(s')$"

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