The Stacks project

Lemma 81.13.4. In Situation 81.13.2 assume $X$ quasi-compact. In ( 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 $A \subset B$. $\square$

Comments (0)

There are also:

  • 3 comment(s) on Section 81.13: Coequalizers and glueing

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0AGK. Beware of the difference between the letter 'O' and the digit '0'.