The Stacks project

Lemma 76.43.1. Let $A$ be a Noetherian ring complete with respect to an ideal $I$. Write $S = \mathop{\mathrm{Spec}}(A)$ and $S_ n = \mathop{\mathrm{Spec}}(A/I^ n)$. Let $X \to S$ be a morphism of algebraic spaces that is separated and of finite type. For $n \geq 1$ we set $X_ n = X \times _ S S_ n$. Suppose given a commutative diagram

\[ \xymatrix{ Z_1 \ar[r] \ar[d] & Z_2 \ar[r] \ar[d] & Z_3 \ar[r] \ar[d] & \ldots \\ X_1 \ar[r]^{i_1} & X_2 \ar[r]^{i_2} & X_3 \ar[r] & \ldots } \]

of algebraic spaces with cartesian squares. Assume that

  1. $Z_1 \to X_1$ is a closed immersion, and

  2. $Z_1 \to S_1$ is proper.

Then there exists a closed immersion of algebraic spaces $Z \to X$ such that $Z_ n = Z \times _ S S_ n$ for all $n \geq 1$. Moreover, $Z$ is proper over $S$.

Proof. Let's write $j_ n : Z_ n \to X_ n$ for the vertical morphisms. As the squares in the statement are cartesian we see that the base change of $j_ n$ to $X_1$ is $j_1$. Thus Limits of Spaces, Lemma 70.15.5 shows that $j_ n$ is a closed immersion. Set $\mathcal{F}_ n = j_{n, *}\mathcal{O}_{Z_ n}$, so that $j_ n^\sharp $ is a surjection $\mathcal{O}_{X_ n} \to \mathcal{F}_ n$. Again using that the squares are cartesian we see that the pullback of $\mathcal{F}_{n + 1}$ to $X_ n$ is $\mathcal{F}_ n$. Hence Grothendieck's existence theorem, as reformulated in Remark 76.42.12, tells us there exists a map $\mathcal{O}_ X \to \mathcal{F}$ of coherent $\mathcal{O}_ X$-modules whose restriction to $X_ n$ recovers $\mathcal{O}_{X_ n} \to \mathcal{F}_ n$. Moreover, the support of $\mathcal{F}$ is proper over $S$. As the completion functor is exact (Lemma 76.42.3) we see that $\mathcal{O}_ X \to \mathcal{F}$ is surjective. Thus $\mathcal{F} = \mathcal{O}_ X/\mathcal{J}$ for some quasi-coherent sheaf of ideals $\mathcal{J}$. Setting $Z = V(\mathcal{J})$ finishes the proof. $\square$


Comments (0)


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Your email address will not be published.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 08BG. Beware of the difference between the letter 'O' and the digit '0'.