Lemma 37.2.2. Let $X$ be a scheme over a base $S$. Consider a short exact sequence
\[ 0 \to \mathcal{I} \to \mathcal{A} \to \mathcal{O}_ X \to 0 \]
of sheaves on $X$ where $\mathcal{A}$ is a sheaf of $f^{-1}\mathcal{O}_ S$-algebras, $\mathcal{A} \to \mathcal{O}_ X$ is a surjection of sheaves of $f^{-1}\mathcal{O}_ S$-algebras, and $\mathcal{I}$ is its kernel. If
$\mathcal{I}$ is an ideal of square zero in $\mathcal{A}$, and
$\mathcal{I}$ is quasi-coherent as an $\mathcal{O}_ X$-module
then $X' = (X, \mathcal{A})$ is a scheme and $X \to X'$ is a first order thickening over $S$. Moreover, any first order thickening over $S$ is of this form.
Proof.
It is clear that $X'$ is a locally ringed space. Let $U = \mathop{\mathrm{Spec}}(B)$ be an affine open of $X$. Set $A = \Gamma (U, \mathcal{A})$. Note that since $H^1(U, \mathcal{I}) = 0$ (see Cohomology of Schemes, Lemma 30.2.2) the map $A \to B$ is surjective. By assumption the kernel $I = \mathcal{I}(U)$ is an ideal of square zero in the ring $A$. By Schemes, Lemma 26.6.4 there is a canonical morphism of locally ringed spaces
\[ (U, \mathcal{A}|_ U) \longrightarrow \mathop{\mathrm{Spec}}(A) \]
coming from the map $B \to \Gamma (U, \mathcal{A})$. Since this morphism fits into the commutative diagram
\[ \xymatrix{ (U, \mathcal{O}_ X|_ U) \ar[d] \ar[r] & \mathop{\mathrm{Spec}}(B) \ar[d] \\ (U, \mathcal{A}|_ U) \ar[r] & \mathop{\mathrm{Spec}}(A) } \]
we see that it is a homeomorphism on underlying topological spaces. Thus to see that it is an isomorphism, it suffices to check it induces an isomorphism on the local rings. For $u \in U$ corresponding to the prime $\mathfrak p \subset A$ we obtain a commutative diagram of short exact sequences
\[ \xymatrix{ 0 \ar[r] & I_{\mathfrak p} \ar[r] \ar[d] & A_{\mathfrak p} \ar[r] \ar[d] & B_{\mathfrak p} \ar[r] \ar[d] & 0 \\ 0 \ar[r] & \mathcal{I}_ u \ar[r] & \mathcal{A}_ u \ar[r] & \mathcal{O}_{X, u} \ar[r] & 0. } \]
The left and right vertical arrows are isomorphisms because $\mathcal{I}$ and $\mathcal{O}_ X$ are quasi-coherent sheaves. Hence also the middle map is an isomorphism. Hence every point of $X' = (X, \mathcal{A})$ has an affine neighbourhood and $X'$ is a scheme as desired.
$\square$
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