Let $i : (Z, \mathcal{O}_ Z) \to (X, \mathcal{O}_ X)$ be a morphism of ringed spaces such that $i$ is a homeomorphism onto a closed subset and such that $i^\sharp : \mathcal{O}_ X \to i_*\mathcal{O}_ Z$ is surjective. (For example a closed immersion of schemes.) Let $\mathcal{I} = \mathop{\mathrm{Ker}}(i^\sharp )$. For a sheaf of $\mathcal{O}_ X$-modules $\mathcal{F}$ the sheaf
a sheaf of $\mathcal{O}_ X$-modules annihilated by $\mathcal{I}$. Hence by Modules, Lemma 17.13.4 there is a sheaf of $\mathcal{O}_ Z$-modules, which we will denote $\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, \mathcal{F})$, such that
as $\mathcal{O}_ X$-modules. We spell out what this means.
Lemma 48.9.1. With notation as above. The functor $\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, -)$ is a right adjoint to the functor $i_* : \textit{Mod}(\mathcal{O}_ Z) \to \textit{Mod}(\mathcal{O}_ X)$. For $V \subset Z$ open we have where $U \subset X$ is an open whose intersection with $Z$ is $V$.
\[ \Gamma (V, \mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, \mathcal{F})) = \{ s \in \Gamma (U, \mathcal{F}) \mid \mathcal{I}s = 0\} \]
Proof. Let $\mathcal{G}$ be a sheaf of $\mathcal{O}_ Z$-modules. Then
The first equality by Modules, Lemma 17.22.3 and the second by the fully faithfulness of $i_*$, see Modules, Lemma 17.13.4. The description of sections is left to the reader. $\square$
The functor
is left exact and has a derived extension
Lemma 48.9.2. With notation as above. The functor $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, -)$ is the right adjoint of the functor $Ri_* : D(\mathcal{O}_ Z) \to D(\mathcal{O}_ X)$.
Proof. This is a consequence of the fact that $i_*$ and $\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, -)$ are adjoint functors by Lemma 48.9.1. See Derived Categories, Lemma 13.30.3. $\square$
Lemma 48.9.3. With notation as above. We have in $D(\mathcal{O}_ X)$ for all $K$ in $D(\mathcal{O}_ X)$.
\[ Ri_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, K) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(i_*\mathcal{O}_ Z, K) \]
Proof. This is immediate from the construction of the functor $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, -)$. $\square$
Lemma 48.9.4. With notation as above. For $M \in D(\mathcal{O}_ Z)$ we have in $D(\mathcal{O}_ Z)$ for all $K$ in $D(\mathcal{O}_ X)$.
\[ R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(Ri_*M, K) = Ri_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Z}(M, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, K)) \]
Proof. This is immediate from the construction of the functor $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, -)$ and the fact that if $\mathcal{K}^\bullet $ is a K-injective complex of $\mathcal{O}_ X$-modules, then $\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, \mathcal{K}^\bullet )$ is a K-injective complex of $\mathcal{O}_ Z$-modules, see Derived Categories, Lemma 13.31.9. $\square$
Lemma 48.9.5. Let $i : Z \to X$ be a pseudo-coherent closed immersion of schemes (any closed immersion if $X$ is locally Noetherian). Then
$R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, -)$ maps $D^+_\mathit{QCoh}(\mathcal{O}_ X)$ into $D^+_\mathit{QCoh}(\mathcal{O}_ Z)$, and
if $X = \mathop{\mathrm{Spec}}(A)$ and $Z = \mathop{\mathrm{Spec}}(B)$, then the diagram
is commutative.
\[ \xymatrix{ D^+(B) \ar[r] & D_\mathit{QCoh}^+(\mathcal{O}_ Z) \\ D^+(A) \ar[r] \ar[u]^{R\mathop{\mathrm{Hom}}\nolimits (B, -)} & D_\mathit{QCoh}^+(\mathcal{O}_ X) \ar[u]_{R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, -)} } \]
Proof. To explain the parenthetical remark, if $X$ is locally Noetherian, then $i$ is pseudo-coherent by More on Morphisms, Lemma 37.60.9.
Let $K$ be an object of $D^+_\mathit{QCoh}(\mathcal{O}_ X)$. To prove (1), by Morphisms, Lemma 29.4.1 it suffices to show that $i_*$ applied to $H^ n(R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, K))$ produces a quasi-coherent module on $X$. By Lemma 48.9.3 this means we have to show that $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(i_*\mathcal{O}_ Z, K)$ is in $D_\mathit{QCoh}(\mathcal{O}_ X)$. Since $i$ is pseudo-coherent the sheaf $\mathcal{O}_ Z$ is a pseudo-coherent $\mathcal{O}_ X$-module. Hence the result follows from Derived Categories of Schemes, Lemma 36.10.8.
Assume $X = \mathop{\mathrm{Spec}}(A)$ and $Z = \mathop{\mathrm{Spec}}(B)$ as in (2). Let $I^\bullet $ be a bounded below complex of injective $A$-modules representing an object $K$ of $D^+(A)$. Then we know that $R\mathop{\mathrm{Hom}}\nolimits (B, K) = \mathop{\mathrm{Hom}}\nolimits _ A(B, I^\bullet )$ viewed as a complex of $B$-modules. Choose a quasi-isomorphism
where $\mathcal{I}^\bullet $ is a bounded below complex of injective $\mathcal{O}_ X$-modules. It follows from the description of the functor $\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, -)$ in Lemma 48.9.1 that there is a map
Observe that $\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, \mathcal{I}^\bullet )$ represents $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, \widetilde{K})$. Applying the universal property of the $\widetilde{\ }$ functor we obtain a map
in $D(\mathcal{O}_ Z)$. We may check that this map is an isomorphism in $D(\mathcal{O}_ Z)$ after applying $i_*$. However, once we apply $i_*$ we obtain the isomorphism of Derived Categories of Schemes, Lemma 36.10.8 via the identification of Lemma 48.9.3. $\square$
Lemma 48.9.6. Let $i : Z \to X$ be a closed immersion of schemes. Assume $X$ is a locally Noetherian. Then $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, -)$ maps $D^+_{\textit{Coh}}(\mathcal{O}_ X)$ into $D^+_{\textit{Coh}}(\mathcal{O}_ Z)$.
Proof. The question is local on $X$, hence we may assume that $X$ is affine. Say $X = \mathop{\mathrm{Spec}}(A)$ and $Z = \mathop{\mathrm{Spec}}(B)$ with $A$ Noetherian and $A \to B$ surjective. In this case, we can apply Lemma 48.9.5 to translate the question into algebra. The corresponding algebra result is a consequence of Dualizing Complexes, Lemma 47.13.4. $\square$
Lemma 48.9.7. Let $X$ be a quasi-compact and quasi-separated scheme. Let $i : Z \to X$ be a pseudo-coherent closed immersion (if $X$ is Noetherian, then any closed immersion is pseudo-coherent). Let $a : D_\mathit{QCoh}(\mathcal{O}_ X) \to D_\mathit{QCoh}(\mathcal{O}_ Z)$ be the right adjoint to $Ri_*$. Then there is a functorial isomorphism for $K \in D_\mathit{QCoh}^+(\mathcal{O}_ X)$.
\[ a(K) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, K) \]
Proof. (The parenthetical statement follows from More on Morphisms, Lemma 37.60.9.) By Lemma 48.9.2 the functor $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, -)$ is a right adjoint to $Ri_* : D(\mathcal{O}_ Z) \to D(\mathcal{O}_ X)$. Moreover, by Lemma 48.9.5 and Lemma 48.3.5 both $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, -)$ and $a$ map $D_\mathit{QCoh}^+(\mathcal{O}_ X)$ into $D_\mathit{QCoh}^+(\mathcal{O}_ Z)$. Hence we obtain the isomorphism by uniqueness of adjoint functors. $\square$
Example 48.9.8. If $i : Z \to X$ is closed immersion of Noetherian schemes, then the diagram is commutative for $K \in D_\mathit{QCoh}^+(\mathcal{O}_ X)$. Here the horizontal equality sign is Lemma 48.9.3 and the lower horizontal arrow is induced by the map $\mathcal{O}_ X \to i_*\mathcal{O}_ Z$. The commutativity of the diagram is a consequence of Lemma 48.9.7.
\[ \xymatrix{ i_*a(K) \ar[rr]_-{\text{Tr}_{i, K}} \ar@{=}[d] & & K \ar@{=}[d] \\ i_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, K) \ar@{=}[r] & R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(i_*\mathcal{O}_ Z, K) \ar[r] & K } \]
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