Definition 29.31.1. Let $i : Z \to X$ be an immersion. The conormal sheaf $\mathcal{C}_{Z/X}$ of $Z$ in $X$ or the conormal sheaf of $i$ is the quasi-coherent $\mathcal{O}_ Z$-module $\mathcal{I}/\mathcal{I}^2$ described above.
29.31 Conormal sheaf of an immersion
Let $i : Z \to X$ be a closed immersion. Let $\mathcal{I} \subset \mathcal{O}_ X$ be the corresponding quasi-coherent sheaf of ideals. Consider the short exact sequence
\[ 0 \to \mathcal{I}^2 \to \mathcal{I} \to \mathcal{I}/\mathcal{I}^2 \to 0 \]
of quasi-coherent sheaves on $X$. Since the sheaf $\mathcal{I}/\mathcal{I}^2$ is annihilated by $\mathcal{I}$ it corresponds to a sheaf on $Z$ by Lemma 29.4.1. This quasi-coherent $\mathcal{O}_ Z$-module is called the conormal sheaf of $Z$ in $X$ and is often simply denoted $\mathcal{I}/\mathcal{I}^2$ by the abuse of notation mentioned in Section 29.4.
In case $i : Z \to X$ is a (locally closed) immersion we define the conormal sheaf of $i$ as the conormal sheaf of the closed immersion $i : Z \to X \setminus \partial Z$, where $\partial Z = \overline{Z} \setminus Z$. It is often denoted $\mathcal{I}/\mathcal{I}^2$ where $\mathcal{I}$ is the ideal sheaf of the closed immersion $i : Z \to X \setminus \partial Z$.
In [IV Definition 16.1.2, EGA] this sheaf is denoted $\mathcal{N}_{Z/X}$. We will not follow this convention since we would like to reserve the notation $\mathcal{N}_{Z/X}$ for the normal sheaf of the immersion. It is defined as
\[ \mathcal{N}_{Z/X} = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Z}(\mathcal{C}_{Z/X}, \mathcal{O}_ Z) = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Z}(\mathcal{I}/\mathcal{I}^2, \mathcal{O}_ Z) \]
provided the conormal sheaf is of finite presentation (otherwise the normal sheaf may not even be quasi-coherent). We will come back to the normal sheaf later (insert future reference here).
Lemma 29.31.2. Let $i : Z \to X$ be an immersion. The conormal sheaf of $i$ has the following properties:
Let $U \subset X$ be any open subscheme such that $i$ factors as $Z \xrightarrow {i'} U \to X$ where $i'$ is a closed immersion. Let $\mathcal{I} = \mathop{\mathrm{Ker}}((i')^\sharp ) \subset \mathcal{O}_ U$. Then
For any affine open $\mathop{\mathrm{Spec}}(R) = U \subset X$ such that $Z \cap U = \mathop{\mathrm{Spec}}(R/I)$ there is a canonical isomorphism $\Gamma (Z \cap U, \mathcal{C}_{Z/X}) = I/I^2$.
\[ \mathcal{C}_{Z/X} = (i')^*\mathcal{I}\quad \text{and}\quad i'_*\mathcal{C}_{Z/X} = \mathcal{I}/\mathcal{I}^2 \]
Proof. Mostly clear from the definitions. Note that given a ring $R$ and an ideal $I$ of $R$ we have $I/I^2 = I \otimes _ R R/I$. Details omitted. $\square$
Lemma 29.31.3. Let be a commutative diagram in the category of schemes. Assume $i$, $i'$ immersions. There is a canonical map of $\mathcal{O}_ Z$-modules characterized by the following property: For every pair of affine opens $(\mathop{\mathrm{Spec}}(R) = U \subset X, \mathop{\mathrm{Spec}}(R') = U' \subset X')$ with $f(U) \subset U'$ such that $Z \cap U = \mathop{\mathrm{Spec}}(R/I)$ and $Z' \cap U' = \mathop{\mathrm{Spec}}(R'/I')$ the induced map is the one induced by the ring map $f^\sharp : R' \to R$ which has the property $f^\sharp (I') \subset I$.
\[ \xymatrix{ Z \ar[r]_ i \ar[d]_ f & X \ar[d]^ g \\ Z' \ar[r]^{i'} & X' } \]
\[ f^*\mathcal{C}_{Z'/X'} \longrightarrow \mathcal{C}_{Z/X} \]
\[ \Gamma (Z' \cap U', \mathcal{C}_{Z'/X'}) = I'/I'^2 \longrightarrow I/I^2 = \Gamma (Z \cap U, \mathcal{C}_{Z/X}) \]
Proof. Let $\partial Z' = \overline{Z'} \setminus Z'$ and $\partial Z = \overline{Z} \setminus Z$. These are closed subsets of $X'$ and of $X$. Replacing $X'$ by $X' \setminus \partial Z'$ and $X$ by $X \setminus \big (g^{-1}(\partial Z') \cup \partial Z\big )$ we see that we may assume that $i$ and $i'$ are closed immersions.
The fact that $g \circ i$ factors through $i'$ implies that $g^*\mathcal{I}'$ maps into $\mathcal{I}$ under the canonical map $g^*\mathcal{I}' \to \mathcal{O}_ X$, see Schemes, Lemmas 26.4.6 and 26.4.7. Hence we get an induced map of quasi-coherent sheaves $g^*(\mathcal{I}'/(\mathcal{I}')^2) \to \mathcal{I}/\mathcal{I}^2$. Pulling back by $i$ gives $i^*g^*(\mathcal{I}'/(\mathcal{I}')^2) \to i^*(\mathcal{I}/\mathcal{I}^2)$. Note that $i^*(\mathcal{I}/\mathcal{I}^2) = \mathcal{C}_{Z/X}$. On the other hand, $i^*g^*(\mathcal{I}'/(\mathcal{I}')^2) = f^*(i')^*(\mathcal{I}'/(\mathcal{I}')^2) = f^*\mathcal{C}_{Z'/X'}$. This gives the desired map.
Checking that the map is locally described as the given map $I'/(I')^2 \to I/I^2$ is a matter of unwinding the definitions and is omitted. Another observation is that given any $x \in i(Z)$ there do exist affine open neighbourhoods $U$, $U'$ with $f(U) \subset U'$ and $Z \cap U$ as well as $U' \cap Z'$ closed such that $x \in U$. Proof omitted. Hence the requirement of the lemma indeed characterizes the map (and could have been used to define it). $\square$
Lemma 29.31.4. Let be a fibre product diagram in the category of schemes with $i$, $i'$ immersions. Then the canonical map $f^*\mathcal{C}_{Z'/X'} \to \mathcal{C}_{Z/X}$ of Lemma 29.31.3 is surjective. If $g$ is flat, then it is an isomorphism.
\[ \xymatrix{ Z \ar[r]_ i \ar[d]_ f & X \ar[d]^ g \\ Z' \ar[r]^{i'} & X' } \]
Proof. Let $R' \to R$ be a ring map, and $I' \subset R'$ an ideal. Set $I = I'R$. Then $I'/(I')^2 \otimes _{R'} R \to I/I^2$ is surjective. If $R' \to R$ is flat, then $I = I' \otimes _{R'} R$ and $I^2 = (I')^2 \otimes _{R'} R$ and we see the map is an isomorphism. $\square$
Lemma 29.31.5. Let $Z \to Y \to X$ be immersions of schemes. Then there is a canonical exact sequence where the maps come from Lemma 29.31.3 and $i : Z \to Y$ is the first morphism.
\[ i^*\mathcal{C}_{Y/X} \to \mathcal{C}_{Z/X} \to \mathcal{C}_{Z/Y} \to 0 \]
Proof. Via Lemma 29.31.3 this translates into the following algebra fact. Suppose that $C \to B \to A$ are surjective ring maps. Let $I = \mathop{\mathrm{Ker}}(B \to A)$, $J = \mathop{\mathrm{Ker}}(C \to A)$ and $K = \mathop{\mathrm{Ker}}(C \to B)$. Then there is an exact sequence
\[ K/K^2 \otimes _ B A \to J/J^2 \to I/I^2 \to 0. \]
This follows immediately from the observation that $I = J/K$. $\square$
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