The Stacks project

Lemma 76.9.5. Let $S$ be a scheme. Let $X \subset X'$ be a thickening of algebraic spaces over $S$. If $X$ is (representable by) a scheme, then so is $X'$.

Proof. Note that $X'_{red} = X_{red}$. Hence if $X$ is a scheme, then $X'_{red}$ is a scheme. Thus the result follows from Limits of Spaces, Lemma 70.15.3. Below we give a direct proof for finite order thickenings which is the case most often used in practice. $\square$

Proof for finite order thickenings. It suffices to prove this when $X'$ is a first order thickening of $X$. By Properties of Spaces, Lemma 66.13.1 there is a largest open subspace of $X'$ which is a scheme. Thus we have to show that every point $x$ of $|X'| = |X|$ is contained in an open subspace of $X'$ which is a scheme. Using Lemma 76.9.3 we may replace $X \subset X'$ by $U \subset U'$ with $x \in U$ and $U$ an affine scheme. Hence we may assume that $X$ is affine. Thus we reduce to the case discussed in the next paragraph.

Assume $X \subset X'$ is a first order thickening where $X$ is an affine scheme. Set $A = \Gamma (X, \mathcal{O}_ X)$ and $A' = \Gamma (X', \mathcal{O}_{X'})$. By Lemma 76.9.4 the map $A \to A'$ is surjective. The kernel $I$ is an ideal of square zero. By Properties of Spaces, Lemma 66.33.1 we obtain a canonical morphism $f : X' \to \mathop{\mathrm{Spec}}(A')$ which fits into the following commutative diagram

\[ \xymatrix{ X \ar@{=}[d] \ar[r] & X' \ar[d]^ f \\ \mathop{\mathrm{Spec}}(A) \ar[r] & \mathop{\mathrm{Spec}}(A') } \]

Because the horizontal arrows are thickenings it is clear that $f$ is universally injective and surjective. Hence it suffices to show that $f$ is étale, since then Morphisms of Spaces, Lemma 67.51.2 will imply that $f$ is an isomorphism.

To prove that $f$ is étale choose an affine scheme $U'$ and an étale morphism $U' \to X'$. It suffices to show that $U' \to X' \to \mathop{\mathrm{Spec}}(A')$ is étale, see Properties of Spaces, Definition 66.16.2. Write $U' = \mathop{\mathrm{Spec}}(B')$. Set $U = X \times _{X'} U'$. Since $U$ is a closed subspace of $U'$, it is a closed subscheme, hence $U = \mathop{\mathrm{Spec}}(B)$ with $B' \to B$ surjective. Denote $J = \mathop{\mathrm{Ker}}(B' \to B)$ and note that $J = \Gamma (U, \mathcal{I})$ where $\mathcal{I} = \mathop{\mathrm{Ker}}(\mathcal{O}_{X'} \to \mathcal{O}_ X)$ on $X_{spaces, {\acute{e}tale}}$ as in the proof of Lemma 76.9.4. The morphism $U' \to X' \to \mathop{\mathrm{Spec}}(A')$ induces a commutative diagram

\[ \xymatrix{ 0 \ar[r] & J \ar[r] & B' \ar[r] & B \ar[r] & 0 \\ 0 \ar[r] & I \ar[r] \ar[u] & A' \ar[r] \ar[u] & A \ar[r] \ar[u] & 0 } \]

Now, since $\mathcal{I}$ is a quasi-coherent $\mathcal{O}_ X$-module we have $\mathcal{I} = (\widetilde I)^ a$, see Descent, Definition 35.8.2 for notation and Descent, Proposition 35.8.9 for why this is true. Hence we see that $J = I \otimes _ A B$. Finally, note that $A \to B$ is étale as $U \to X$ is étale as the base change of the étale morphism $U' \to X'$. We conclude that $A' \to B'$ is étale by Algebra, Lemma 10.143.11. $\square$


Comments (1)

Comment #9503 by Shizhang on

Proof for finite order thickenings, second paragraph, line 2: should be ``the map is surjective''.


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 05ZR. Beware of the difference between the letter 'O' and the digit '0'.