Definition 76.12.1. Let $i : Z \to X$ be an immersion of algebraic spaces. The first order infinitesimal neighbourhood of $Z$ in $X$ is the first order thickening $Z \subset Z'$ over $X$ described above.
76.12 First order infinitesimal neighbourhood
A natural construction of first order thickenings is the following. Suppose that $i : Z \to X$ be an immersion of algebraic spaces. Choose an open subspace $U \subset X$ such that $i$ identifies $Z$ with a closed subspace $Z \subset U$ (see Morphisms of Spaces, Remark 67.12.4). Let $\mathcal{I} \subset \mathcal{O}_ U$ be the quasi-coherent sheaf of ideals defining $Z$ in $U$, see Morphisms of Spaces, Lemma 67.13.1. Then we can consider the closed subspace $Z' \subset U$ defined by the quasi-coherent sheaf of ideals $\mathcal{I}^2$.
This thickening has the following universal property (which will assuage any fears that the construction above depends on the choice of the open $U$).
Lemma 76.12.2. Let $i : Z \to X$ be an immersion of algebraic spaces. The first order infinitesimal neighbourhood $Z'$ of $Z$ in $X$ has the following universal property: Given any commutative diagram
\[ \xymatrix{ Z \ar[d]_ i & T \ar[l]^ a \ar[d] \\ X & T' \ar[l]_ b } \]
where $T \subset T'$ is a first order thickening over $X$, there exists a unique morphism $(a', a) : (T \subset T') \to (Z \subset Z')$ of thickenings over $X$.
Proof. Let $U \subset X$ be the open subspace used in the construction of $Z'$, i.e., an open such that $Z$ is identified with a closed subspace of $U$ cut out by the quasi-coherent sheaf of ideals $\mathcal{I}$. Since $|T| = |T'|$ we see that $|b|(|T'|) \subset |U|$. Hence we can think of $b$ as a morphism into $U$, see Properties of Spaces, Lemma 66.4.9. Let $\mathcal{J} \subset \mathcal{O}_{T'}$ be the square zero quasi-coherent sheaf of ideals cutting out $T$. By the commutativity of the diagram we have $b|_ T = i \circ a$ where $i : Z \to U$ is the closed immersion. We conclude that $b^\sharp (b^{-1}\mathcal{I}) \subset \mathcal{J}$ by Morphisms of Spaces, Lemma 67.13.1. As $T'$ is a first order thickening of $T$ we see that $\mathcal{J}^2 = 0$ hence $b^\sharp (b^{-1}(\mathcal{I}^2)) = 0$. By Morphisms of Spaces, Lemma 67.13.1 this implies that $b$ factors through $Z'$. Letting $a' : T' \to Z'$ be this factorization we win. $\square$
Lemma 76.12.3. Let $i : Z \to X$ be an immersion of algebraic spaces. Let $Z \subset Z'$ be the first order infinitesimal neighbourhood of $Z$ in $X$. Then the diagram
\[ \xymatrix{ Z \ar[r] \ar[d] & Z' \ar[d] \\ Z \ar[r] & X } \]
induces a map of conormal sheaves $\mathcal{C}_{Z/X} \to \mathcal{C}_{Z/Z'}$ by Lemma 76.5.3. This map is an isomorphism.
Proof. This is clear from the construction of $Z'$ above. $\square$
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