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.
$\square$

## 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Your email address will not be published.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.

## Comments (1)

Comment #7825 by Federico on