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. $\square$

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