Proof.
Since the squares in the diagram are cartesian and since the morphisms $S_ n \to S_{n + 1}$ are closed immersions, we see that the morphisms $i_ n$ are closed immersions too. In particular we may think of $X_ m$ as a closed subscheme of $X_ n$ for $m < n$. In fact $X_ m$ is the closed subscheme cut out by the quasi-coherent sheaf of ideals $I^ m\mathcal{O}_{X_ n}$. Moreover, the underlying topological spaces of the schemes $X_1, X_2, X_3, \ldots $ are all identified, hence we may (and do) think of sheaves $\mathcal{O}_{X_ n}$ as living on the same underlying topological space; similarly for coherent $\mathcal{O}_{X_ n}$-modules. Set
\[ \mathcal{F}_ n = \mathop{\mathrm{Ker}}(\mathcal{O}_{X_{n + 1}} \to \mathcal{O}_{X_ n}) \]
so that we obtain short exact sequences
\[ 0 \to \mathcal{F}_ n \to \mathcal{O}_{X_{n + 1}} \to \mathcal{O}_{X_ n} \to 0 \]
By the above we have $\mathcal{F}_ n = I^ n\mathcal{O}_{X_{n + 1}}$. It follows $\mathcal{F}_ n$ is a coherent sheaf on $X_{n + 1}$ annihilated by $I$, hence we may (and do) think of it as a coherent module $\mathcal{O}_{X_1}$-module. Observe that for $m > n$ the sheaf
\[ I^ n\mathcal{O}_{X_ m}/I^{n + 1}\mathcal{O}_{X_ m} \]
maps isomorphically to $\mathcal{F}_ n$ under the map $\mathcal{O}_{X_ m} \to \mathcal{O}_{X_{n + 1}}$. Hence given $n_1, n_2 \geq 0$ we can pick an $m > n_1 + n_2$ and consider the multiplication map
\[ I^{n_1}\mathcal{O}_{X_ m} \times I^{n_2}\mathcal{O}_{X_ m} \longrightarrow I^{n_1 + n_2}\mathcal{O}_{X_ m} \to \mathcal{F}_{n_1 + n_2} \]
This induces an $\mathcal{O}_{X_1}$-bilinear map
\[ \mathcal{F}_{n_1} \times \mathcal{F}_{n_2} \longrightarrow \mathcal{F}_{n_1 + n_2} \]
which in turn defines the structure of a graded $\mathcal{O}_{X_1}$-algebra on $\mathcal{F} = \bigoplus _{n \geq 0} \mathcal{F}_ n$.
Set $B = \bigoplus I^ n/I^{n + 1}$; this is a finitely generated graded $A/I$-algebra. Set $\mathcal{B} = (X_1 \to S_1)^*\widetilde{B}$. The discussion above provides us with a canonical surjection
\[ \mathcal{B} \longrightarrow \mathcal{F} \]
of graded $\mathcal{O}_{X_1}$-algebras. In particular we see that $\mathcal{F}$ is a finite type quasi-coherent graded $\mathcal{B}$-module. By Lemma 30.19.3 we can find an integer $d_0$ such that $H^1(X_1, \mathcal{F} \otimes \mathcal{L}^{\otimes d}) = 0$ for all $d \geq d_0$. Pick a $d \geq d_0$ such that there exist sections $s_{0, 1}, \ldots , s_{N, 1} \in \Gamma (X_1, \mathcal{L}_1^{\otimes d})$ which induce an immersion
\[ \psi _1 : X_1 \to \mathbf{P}^ N_{S_1} \]
over $S_1$, see Morphisms, Lemma 29.39.4. As $X_1$ is proper over $S_1$ we see that $\psi _1$ is a closed immersion, see Morphisms, Lemma 29.41.7 and Schemes, Lemma 26.10.4. We are going to “lift” $\psi _1$ to a compatible system of closed immersions of $X_ n$ into $\mathbf{P}^ N$.
Upon tensoring the short exact sequences of the first paragraph of the proof by $\mathcal{L}_{n + 1}^{\otimes d}$ we obtain short exact sequences
\[ 0 \to \mathcal{F}_ n \otimes \mathcal{L}_{n + 1}^{\otimes d} \to \mathcal{L}_{n + 1}^{\otimes d} \to \mathcal{L}_{n + 1}^{\otimes d} \to 0 \]
Using the isomorphisms $\varphi _ n$ we obtain isomorphisms $\mathcal{L}_{n + 1} \otimes \mathcal{O}_{X_ l} = \mathcal{L}_ l$ for $l \leq n$. Whence the sequence above becomes
\[ 0 \to \mathcal{F}_ n \otimes \mathcal{L}_1^{\otimes d} \to \mathcal{L}_{n + 1}^{\otimes d} \to \mathcal{L}_ n^{\otimes d} \to 0 \]
The vanishing of $H^1(X, \mathcal{F}_ n \otimes \mathcal{L}_1^{\otimes d})$ implies we can inductively lift $s_{0, 1}, \ldots , s_{N, 1} \in \Gamma (X_1, \mathcal{L}_1^{\otimes d})$ to sections $s_{0, n}, \ldots , s_{N, n} \in \Gamma (X_ n, \mathcal{L}_ n^{\otimes d})$. Thus we obtain a commutative diagram
\[ \xymatrix{ X_1 \ar[r]_{i_1} \ar[d]_{\psi _1} & X_2 \ar[r]_{i_2} \ar[d]_{\psi _2} & X_3 \ar[r] \ar[d]_{\psi _3} & \ldots \\ \mathbf{P}^ N_{S_1} \ar[r] & \mathbf{P}^ N_{S_2} \ar[r] & \mathbf{P}^ N_{S_3} \ar[r] & \ldots } \]
where $\psi _ n = \varphi _{(\mathcal{L}_ n, (s_{0, n}, \ldots , s_{N, n}))}$ in the notation of Constructions, Section 27.13. As the squares in the statement of the theorem are cartesian we see that the squares in the above diagram are cartesian. We win by applying Lemma 30.28.1.
$\square$
Comments (0)