A natural construction of finite 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. For $n \geq 1$ we can consider the closed subspace $Z_ n \subset U$ defined by the quasi-coherent sheaf of ideals $\mathcal{I}^{n + 1}$.
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 $Z \subset Z_1$ over $X$ described above.
The $n$th order infinitesimal neighbourhood of $Z$ in $X$ is the $n$th order thickening $Z \subset Z_ n$ over $X$ described above.
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
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 $X$.
For $n \geq 1$ the $n$th order infinitesimal neighbourhood $Z_ n$ of $Z$ in $X$ has the following universal property: Given any commutative diagram
where $T \subset T'$ is an $n$th order thickening over $X$, there exists a unique morphism $(a', a) : (T \subset T') \to (Z \subset Z_ n)$ of thickenings over $X$.
\[ \xymatrix{ Z \ar[d]_ i & T \ar[l]^ a \ar[d] \\ X & T' \ar[l]_ b } \]
\[ \xymatrix{ Z \ar[d]_ i & T \ar[l]^ a \ar[d] \\ X & T' \ar[l]_ b } \]
Proof. We will only prove (1). 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 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.
\[ \xymatrix{ Z \ar[r] \ar[d] & Z' \ar[d] \\ Z \ar[r] & X } \]
Proof. This is clear from the construction of $Z'$ above. $\square$
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)