Lemma 81.13.4. In Situation 81.13.2 assume X quasi-compact. In (81.13.3.1) for all n large enough, there exists an m such that X_ n \to X_{n + m} factors through a closed immersion X \to X_{n + m}.
Proof. Let's look a bit more closely at the construction of X_ n and how it changes as we increase n. We have X_ n = \underline{\mathop{\mathrm{Spec}}}(\mathcal{A}_ n) where \mathcal{A}_ n is the equalizer of s_ n^\sharp and t_ n^\sharp going from g_{n , *}\mathcal{O}_{Y_ n} to h_{n, *}\mathcal{O}_{R_ n}. Here g_ n : Y_ n = X' \amalg Z_ n \to X and h_ n : R_ n = Y_ n \times _ X Y_ n \to X are the given morphisms. Let \mathcal{I} \subset \mathcal{O}_ X be the coherent sheaf of ideals corresponding to Z. Then
Similarly, we have a decomposition
As Z_ n \to X is a monomorphism, we see that X' \times _ X Z_ n = Z_ n \times _ X X' and that this identification is compatible with the two morphisms to X, with the two morphisms to X', and with the two morphisms to Z_ n. Denote f_ n : X' \times _ X Z_ n \to X the morphism to X. Denote
By the remarks above we find that
We have canonical maps
of coherent \mathcal{O}_ X-algebras. The statement of the lemma means that for n large enough there exists an m \geq 0 such that the image of \mathcal{A}_{n + m} \to \mathcal{A}_ n is isomorphic to \mathcal{O}_ X. This we may check étale locally on X. Hence by Properties of Spaces, Lemma 66.6.3 we may assume X is an affine Noetherian scheme.
Since X_ n \to X is an isomorphism over U we see that the kernel of \mathcal{O}_ X \to \mathcal{A}_ n is supported on |Z|. Since X is Noetherian, the sequence of kernels \mathcal{J}_ n = \mathop{\mathrm{Ker}}(\mathcal{O}_ X \to \mathcal{A}_ n) stabilizes (Cohomology of Spaces, Lemma 69.13.1). Say \mathcal{J}_{n_0} = \mathcal{J}_{n_0 + 1} = \ldots = \mathcal{J}. By Cohomology of Spaces, Lemma 69.13.2 we find that \mathcal{I}^ t \mathcal{J} = 0 for some t \geq 0. On the other hand, there is an \mathcal{O}_ X-algebra map \mathcal{A}_ n \to \mathcal{O}_ X/\mathcal{I}^ n and hence \mathcal{J} \subset \mathcal{I}^ n for all n. By Artin-Rees (Cohomology of Spaces, Lemma 69.13.3) we find that \mathcal{J} \cap \mathcal{I}^ n \subset \mathcal{I}^{n - c}\mathcal{J} for some c \geq 0 and all n \gg 0. We conclude that \mathcal{J} = 0.
Pick n \geq n_0 as in the previous paragraph. Then \mathcal{O}_ X \to \mathcal{A}_ n is injective. Hence it now suffices to find m \geq 0 such that the image of \mathcal{A}_{n + m} \to \mathcal{A}_ n is equal to the image of \mathcal{O}_ X. Observe that \mathcal{A}_ n sits in a short exact sequence
and similarly for \mathcal{A}_{n + m}. Hence it suffices to show
for some m \geq 0. To do this we may work étale locally on X and since X is Noetherian we may assume that X is a Noetherian affine scheme. Say X = \mathop{\mathrm{Spec}}(R) and \mathcal{I} corresponds to the ideal I \subset R. Let \mathcal{A} = \widetilde{A} for a finite R-algebra A. Let f_*\mathcal{O}_{X'} = \widetilde{B} for a finite R-algebra B. Then R \to A \subset B and these maps become isomorphisms on inverting any element of I.
Note that f_{n, *}\mathcal{O}_{X' \times _ X Z_ n} is equal to f_*(\mathcal{O}_{X'}/I^ n\mathcal{O}_{X'}) in the notation used in Cohomology of Spaces, Section 69.22. By Cohomology of Spaces, Lemma 69.22.4 we see that there exists a c \geq 0 such that
is contained in I^{n + m}B. On the other hand, as R \to B is finite and an isomorphism after inverting any element of I we see that I^{n + m}B \subset \mathop{\mathrm{Im}}(I^ n \to B) for m large enough (can be chosen independent of n). This finishes the proof as A \subset B. \square
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