Lemma 48.14.1. As above, let $X$ be a scheme and let $D \subset X$ be an effective Cartier divisor. There is a canonical isomorphism $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ D, \mathcal{O}_ X) = \mathcal{N}[-1]$ in $D(\mathcal{O}_ D)$.
48.14 Right adjoint of pushforward for effective Cartier divisors
Let $X$ be a scheme and let $i : D \to X$ be the inclusion of an effective Cartier divisor. Denote $\mathcal{N} = i^*\mathcal{O}_ X(D)$ the normal sheaf of $i$, see Morphisms, Section 29.31 and Divisors, Section 31.13. Recall that $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ D, -)$ denotes the right adjoint to $i_* : D(\mathcal{O}_ D) \to D(\mathcal{O}_ X)$ and has the property $i_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ D, -) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(i_*\mathcal{O}_ D, -)$, see Section 48.9.
Proof. Equivalently, we are saying that $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ D, \mathcal{O}_ X)$ has a unique nonzero cohomology sheaf in degree $1$ and that this sheaf is isomorphic to $\mathcal{N}$. Since $i_*$ is exact and fully faithful, it suffices to prove that $i_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ D, \mathcal{O}_ X)$ is isomorphic to $i_*\mathcal{N}[-1]$. We have $i_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ D, \mathcal{O}_ X) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(i_*\mathcal{O}_ D, \mathcal{O}_ X)$ by Lemma 48.9.3. We have a resolution
\[ 0 \to \mathcal{I} \to \mathcal{O}_ X \to i_*\mathcal{O}_ D \to 0 \]
where $\mathcal{I}$ is the ideal sheaf of $D$ which we can use to compute. Since $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{O}_ X, \mathcal{O}_ X) = \mathcal{O}_ X$ and $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{I}, \mathcal{O}_ X) = \mathcal{O}_ X(D)$ by a local computation, we see that
\[ R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(i_*\mathcal{O}_ D, \mathcal{O}_ X) = (\mathcal{O}_ X \to \mathcal{O}_ X(D)) \]
where on the right hand side we have $\mathcal{O}_ X$ in degree $0$ and $\mathcal{O}_ X(D)$ in degree $1$. The result follows from the short exact sequence
\[ 0 \to \mathcal{O}_ X \to \mathcal{O}_ X(D) \to i_*\mathcal{N} \to 0 \]
coming from the fact that $D$ is the zero scheme of the canonical section of $\mathcal{O}_ X(D)$ and from the fact that $\mathcal{N} = i^*\mathcal{O}_ X(D)$. $\square$
For every object $K$ of $D(\mathcal{O}_ X)$ there is a canonical map
48.14.1.1
\begin{equation} \label{duality-equation-map-effective-Cartier} Li^*K \otimes _{\mathcal{O}_ D}^\mathbf {L} R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ D, \mathcal{O}_ X) \longrightarrow R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ D, K) \end{equation}
in $D(\mathcal{O}_ D)$ functorial in $K$ and compatible with distinguished triangles. Namely, this map is adjoint to a map
\[ i_*(Li^*K \otimes ^\mathbf {L}_{\mathcal{O}_ D} R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ D, \mathcal{O}_ X)) = K \otimes ^\mathbf {L}_{\mathcal{O}_ X} R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(i_*\mathcal{O}_ D, \mathcal{O}_ X) \longrightarrow K \]
where the equality is Cohomology, Lemma 20.54.4 and the arrow comes from the canonical map $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(i_*\mathcal{O}_ D, \mathcal{O}_ X) \to \mathcal{O}_ X$ induced by $\mathcal{O}_ X \to i_*\mathcal{O}_ D$.
If $K \in D_\mathit{QCoh}(\mathcal{O}_ X)$, then (48.14.1.1) is equal to (48.8.0.1) via the identification $a(K) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ D, K)$ of Lemma 48.9.7. If $K \in D_\mathit{QCoh}(\mathcal{O}_ X)$ and $X$ is Noetherian, then the following lemma is a special case of Lemma 48.13.3.
Lemma 48.14.2. As above, let $X$ be a scheme and let $D \subset X$ be an effective Cartier divisor. Then (48.14.1.1) combined with Lemma 48.14.1 defines an isomorphism functorial in $K$ in $D(\mathcal{O}_ X)$.
\[ Li^*K \otimes _{\mathcal{O}_ D}^\mathbf {L} \mathcal{N}[-1] \longrightarrow R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ D, K) \]
Proof. Since $i_*$ is exact and fully faithful on modules, to prove the map is an isomorphism, it suffices to show that it is an isomorphism after applying $i_*$. We will use the short exact sequences $0 \to \mathcal{I} \to \mathcal{O}_ X \to i_*\mathcal{O}_ D \to 0$ and $0 \to \mathcal{O}_ X \to \mathcal{O}_ X(D) \to i_*\mathcal{N} \to 0$ used in the proof of Lemma 48.14.1 without further mention. By Cohomology, Lemma 20.54.4 which was used to define the map (48.14.1.1) the left hand side becomes
\[ K \otimes _{\mathcal{O}_ X}^\mathbf {L} i_*\mathcal{N}[-1] = K \otimes _{\mathcal{O}_ X}^\mathbf {L} (\mathcal{O}_ X \to \mathcal{O}_ X(D)) \]
The right hand side becomes
\begin{align*} R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(i_*\mathcal{O}_ D, K) & = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}((\mathcal{I} \to \mathcal{O}_ X), K) \\ & = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}((\mathcal{I} \to \mathcal{O}_ X), \mathcal{O}_ X) \otimes _{\mathcal{O}_ X}^\mathbf {L} K \end{align*}
the final equality by Cohomology, Lemma 20.50.5. Since the map comes from the isomorphism
\[ R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}((\mathcal{I} \to \mathcal{O}_ X), \mathcal{O}_ X) = (\mathcal{O}_ X \to \mathcal{O}_ X(D)) \]
the lemma is clear. $\square$
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