Definition 37.5.1. Let $i : Z \to X$ be an immersion of schemes. The *first order infinitesimal neighbourhood* of $Z$ in $X$ is the first order thickening $Z \subset Z'$ over $X$ described above.

## 37.5 First order infinitesimal neighbourhood

A natural construction of first order thickenings is the following. Suppose that $i : Z \to X$ be an immersion of schemes. Choose an open subscheme $U \subset X$ such that $i$ identifies $Z$ with a closed subscheme $Z \subset U$. Let $\mathcal{I} \subset \mathcal{O}_ U$ be the quasi-coherent sheaf of ideals defining $Z$ in $U$. Then we can consider the closed subscheme $Z' \subset U$ 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 37.5.2. Let $i : Z \to X$ be an immersion of schemes. The first order infinitesimal neighbourhood $Z'$ of $Z$ in $X$ has the following universal property: Given any commutative diagram

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 used in the construction of $Z'$, i.e., an open such that $Z$ is identified with a closed subscheme 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 $U$. Let $\mathcal{J} \subset \mathcal{O}_{T'}$ be the ideal cutting out $T$. Since $b(T) \subset Z$ by the diagram above we see that $b^\sharp (b^{-1}\mathcal{I}) \subset \mathcal{J}$. 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 Schemes, Lemma 26.4.6 this implies that $b$ factors through $Z'$. Denote $a' : T' \to Z'$ this factorization and everything is clear.
$\square$

Lemma 37.5.3. Let $i : Z \to X$ be an immersion of schemes. Let $Z \subset Z'$ be the first order infinitesimal neighbourhood of $Z$ in $X$. Then the diagram

induces a map of conormal sheaves $\mathcal{C}_{Z/X} \to \mathcal{C}_{Z/Z'}$ by Morphisms, Lemma 29.31.3. This map is an isomorphism.

**Proof.**
This is clear from the construction of $Z'$ above.
$\square$

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## Comments (2)

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