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

$g_{n, *}\mathcal{O}_{Y_ n} = f_*\mathcal{O}_{X'} \times \mathcal{O}_ X/\mathcal{I}^ n$

Similarly, we have a decomposition

$R_ n = X' \times _ X X' \amalg X' \times _ X Z_ n \amalg Z_ n \times _ X X' \amalg Z_ n \times _ X Z_ n$

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

$\mathcal{A} = \text{Equalizer}( \xymatrix{ f_*\mathcal{O}_{X'} \ar@<1ex>[r] \ar@<-1ex>[r] & (f \times f)_*\mathcal{O}_{X' \times _ X X'} } )$

By the remarks above we find that

$\mathcal{A}_ n = \text{Equalizer}( \xymatrix{ \mathcal{A} \times \mathcal{O}_ X/\mathcal{I}^ n \ar@<1ex>[r] \ar@<-1ex>[r] & f_{n, *}\mathcal{O}_{X' \times _ X Z_ n} } )$

We have canonical maps

$\mathcal{O}_ X \to \ldots \to \mathcal{A}_3 \to \mathcal{A}_2 \to \mathcal{A}_1$

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

$0 \to \mathop{\mathrm{Ker}}(\mathcal{A} \to f_{n, *}\mathcal{O}_{X' \times _ X Z_ n}) \to \mathcal{A}_ n \to \mathcal{O}_ X/\mathcal{I}^ n \to 0$

and similarly for $\mathcal{A}_{n + m}$. Hence it suffices to show

$\mathop{\mathrm{Ker}}(\mathcal{A} \to f_{n + m, *}\mathcal{O}_{X' \times _ X Z_{n + m}}) \subset \mathop{\mathrm{Im}}(\mathcal{I}^ n \to \mathcal{A})$

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

$\mathop{\mathrm{Ker}}(B \to \Gamma (X, f_*(\mathcal{O}_{X'}/I^{n + m + c}\mathcal{O}_{X'}))$

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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