53.4 Duality
In this section we work out the consequences of the very general material on dualizing complexes and duality for proper $1$-dimensional schemes over fields. If you are interested in the analogous discussion for higher dimension proper schemes over fields, see Duality for Schemes, Section 48.27.
Lemma 53.4.1. Let $X$ be a proper scheme of dimension $\leq 1$ over a field $k$. There exists a dualizing complex $\omega _ X^\bullet $ with the following properties
$H^ i(\omega _ X^\bullet )$ is nonzero only for $i = -1, 0$,
$\omega _ X = H^{-1}(\omega _ X^\bullet )$ is a coherent Cohen-Macaulay module whose support is the irreducible components of dimension $1$,
for $x \in X$ closed, the module $H^0(\omega _{X, x}^\bullet )$ is nonzero if and only if either
$\dim (\mathcal{O}_{X, x}) = 0$ or
$\dim (\mathcal{O}_{X, x}) = 1$ and $\mathcal{O}_{X, x}$ is not Cohen-Macaulay,
for $K \in D_\mathit{QCoh}(\mathcal{O}_ X)$ there are functorial isomorphisms1
\[ \mathop{\mathrm{Ext}}\nolimits ^ i_ X(K, \omega _ X^\bullet ) = \mathop{\mathrm{Hom}}\nolimits _ k(H^{-i}(X, K), k) \]
compatible with shifts and distinguished triangles,
there are functorial isomorphisms $\mathop{\mathrm{Hom}}\nolimits (\mathcal{F}, \omega _ X) = \mathop{\mathrm{Hom}}\nolimits _ k(H^1(X, \mathcal{F}), k)$ for $\mathcal{F}$ quasi-coherent on $X$,
if $X \to \mathop{\mathrm{Spec}}(k)$ is smooth of relative dimension $1$, then $\omega _ X \cong \Omega _{X/k}$.
Proof.
Denote $f : X \to \mathop{\mathrm{Spec}}(k)$ the structure morphism. We start with the relative dualizing complex
\[ \omega _ X^\bullet = \omega _{X/k}^\bullet = a(\mathcal{O}_{\mathop{\mathrm{Spec}}(k)}) \]
as described in Duality for Schemes, Remark 48.12.5. Then property (4) holds by construction as $a$ is the right adjoint for $f_* : D_\mathit{QCoh}(\mathcal{O}_ X) \to D(\mathcal{O}_{\mathop{\mathrm{Spec}}(k)})$. Since $f$ is proper we have $f^!(\mathcal{O}_{\mathop{\mathrm{Spec}}(k)}) = a(\mathcal{O}_{\mathop{\mathrm{Spec}}(k)})$ by definition, see Duality for Schemes, Section 48.16. Hence $\omega _ X^\bullet $ and $\omega _ X$ are as in Duality for Schemes, Example 48.22.1 and as in Duality for Schemes, Example 48.22.2. Parts (1) and (2) follow from Duality for Schemes, Lemma 48.22.4. For a closed point $x \in X$ we see that $\omega _{X, x}^\bullet $ is a normalized dualizing complex over $\mathcal{O}_{X, x}$, see Duality for Schemes, Lemma 48.21.1. Assertion (3) then follows from Dualizing Complexes, Lemma 47.20.2. Assertion (5) follows from Duality for Schemes, Lemma 48.22.5 for coherent $\mathcal{F}$ and in general by unwinding (4) for $K = \mathcal{F}[0]$ and $i = -1$. Assertion (6) follows from Duality for Schemes, Lemma 48.15.7.
$\square$
Lemma 53.4.2. Let $X$ be a proper scheme over a field $k$ which is Cohen-Macaulay and equidimensional of dimension $1$. The module $\omega _ X$ of Lemma 53.4.1 has the following properties
$\omega _ X$ is a dualizing module on $X$ (Duality for Schemes, Section 48.22),
$\omega _ X$ is a coherent Cohen-Macaulay module whose support is $X$,
there are functorial isomorphisms $\mathop{\mathrm{Ext}}\nolimits ^ i_ X(K, \omega _ X[1]) = \mathop{\mathrm{Hom}}\nolimits _ k(H^{-i}(X, K), k)$ compatible with shifts for $K \in D_\mathit{QCoh}(X)$,
there are functorial isomorphisms $\mathop{\mathrm{Ext}}\nolimits ^{1 + i}(\mathcal{F}, \omega _ X) = \mathop{\mathrm{Hom}}\nolimits _ k(H^{-i}(X, \mathcal{F}), k)$ for $\mathcal{F}$ quasi-coherent on $X$.
Proof.
Recall from the proof of Lemma 53.4.1 that $\omega _ X$ is as in Duality for Schemes, Example 48.22.1 and hence is a dualizing module. The other statements follow from Lemma 53.4.1 and the fact that $\omega _ X^\bullet = \omega _ X[1]$ as $X$ is Cohen-Macualay (Duality for Schemes, Lemma 48.23.1).
$\square$
Here is a sanity check for the dualizing complex.
Lemma 53.4.4. Let $X$ be a proper scheme of dimension $\leq 1$ over a field $k$. Let $\omega _ X^\bullet $ and $\omega _ X$ be as in Lemma 53.4.1.
If $X \to \mathop{\mathrm{Spec}}(k)$ factors as $X \to \mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k)$ for some field $k'$, then $\omega _ X^\bullet $ and $\omega _ X$ satisfy properties (4), (5), (6) with $k$ replaced with $k'$.
If $K/k$ is a field extension, then the pullback of $\omega _ X^\bullet $ and $\omega _ X$ to the base change $X_ K$ are as in Lemma 53.4.1 for the morphism $X_ K \to \mathop{\mathrm{Spec}}(K)$.
Proof.
Denote $f : X \to \mathop{\mathrm{Spec}}(k)$ the structure morphism. Assertion (1) really means that $\omega _ X^\bullet $ and $\omega _ X$ are as in Lemma 53.4.1 for the morphism $f' : X \to \mathop{\mathrm{Spec}}(k')$. In the proof of Lemma 53.4.1 we took $\omega _ X^\bullet = a(\mathcal{O}_{\mathop{\mathrm{Spec}}(k)})$ where $a$ be is the right adjoint of Duality for Schemes, Lemma 48.3.1 for $f$. Thus we have to show $a(\mathcal{O}_{\mathop{\mathrm{Spec}}(k)}) \cong a'(\mathcal{O}_{\mathop{\mathrm{Spec}}(k)})$ where $a'$ be is the right adjoint of Duality for Schemes, Lemma 48.3.1 for $f'$. Since $k' \subset H^0(X, \mathcal{O}_ X)$ we see that $k'/k$ is a finite extension (Cohomology of Schemes, Lemma 30.19.2). By uniqueness of adjoints we have $a = a' \circ b$ where $b$ is the right adjoint of Duality for Schemes, Lemma 48.3.1 for $g : \mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k)$. Another way to say this: we have $f^! = (f')^! \circ g^!$. Thus it suffices to show that $\mathop{\mathrm{Hom}}\nolimits _ k(k', k) \cong k'$ as $k'$-modules, see Duality for Schemes, Example 48.3.2. This holds because these are $k'$-vector spaces of the same dimension (namely dimension $1$).
Proof of (2). This holds because we have base change for $a$ by Duality for Schemes, Lemma 48.6.2. See discussion in Duality for Schemes, Remark 48.12.5.
$\square$
Lemma 53.4.5. Let $X$ be a proper scheme of dimension $\leq 1$ over a field $k$. Let $i : Y \to X$ be a closed immersion. Let $\omega _ X^\bullet $, $\omega _ X$, $\omega _ Y^\bullet $, $\omega _ Y$ be as in Lemma 53.4.1. Then
$\omega _ Y^\bullet = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Y, \omega _ X^\bullet )$,
$\omega _ Y = \mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Y, \omega _ X)$ and $i_*\omega _ Y = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(i_*\mathcal{O}_ Y, \omega _ X)$.
Proof.
Denote $g : Y \to \mathop{\mathrm{Spec}}(k)$ and $f : X \to \mathop{\mathrm{Spec}}(k)$ the structure morphisms. Then $g = f \circ i$. Denote $a, b, c$ the right adjoint of Duality for Schemes, Lemma 48.3.1 for $f, g, i$. Then $b = c \circ a$ by uniqueness of right adjoints and because $Rg_* = Rf_* \circ Ri_*$. In the proof of Lemma 53.4.1 we set $\omega _ X^\bullet = a(\mathcal{O}_{\mathop{\mathrm{Spec}}(k)})$ and $\omega _ Y^\bullet = b(\mathcal{O}_{\mathop{\mathrm{Spec}}(k)})$. Hence $\omega _ Y^\bullet = c(\omega _ X^\bullet )$ which implies (1) by Duality for Schemes, Lemma 48.9.7. Since $\omega _ X = H^{-1}(\omega _ X^\bullet )$ and $\omega _ Y = H^{-1}(\omega _ Y^\bullet )$ we conclude that $\omega _ Y = \mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Y, \omega _ X)$. This implies $i_*\omega _ Y = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(i_*\mathcal{O}_ Y, \omega _ X)$ by Duality for Schemes, Lemma 48.9.3.
$\square$
Lemma 53.4.6. Let $X$ be a proper scheme over a field $k$ which is Gorenstein, reduced, and equidimensional of dimension $1$. Let $i : Y \to X$ be a reduced closed subscheme equidimensional of dimension $1$. Let $j : Z \to X$ be the scheme theoretic closure of $X \setminus Y$. Then
$Y$ and $Z$ are Cohen-Macaulay,
if $\mathcal{I} \subset \mathcal{O}_ X$, resp. $\mathcal{J} \subset \mathcal{O}_ X$ is the ideal sheaf of $Y$, resp. $Z$ in $X$, then
\[ \mathcal{I} = i_*\mathcal{I}' \quad \text{and}\quad \mathcal{J} = j_*\mathcal{J}' \]
where $\mathcal{I}' \subset \mathcal{O}_ Z$, resp. $\mathcal{J}' \subset \mathcal{O}_ Y$ is the ideal sheaf of $Y \cap Z$ in $Z$, resp. $Y$,
$\omega _ Y = \mathcal{J}'(i^*\omega _ X)$ and $i_*(\omega _ Y) = \mathcal{J}\omega _ X$,
$\omega _ Z = \mathcal{I}'(i^*\omega _ X)$ and $i_*(\omega _ Z) = \mathcal{I}\omega _ X$,
we have the following short exact sequences
\begin{align*} 0 \to \omega _ X \to i_*i^*\omega _ X \oplus j_*j^*\omega _ X \to \mathcal{O}_{Y \cap Z} \to 0 \\ 0 \to i_*\omega _ Y \to \omega _ X \to j_*j^*\omega _ X \to 0 \\ 0 \to j_*\omega _ Z \to \omega _ X \to i_*i^*\omega _ X \to 0 \\ 0 \to i_*\omega _ Y \oplus j_*\omega _ Z \to \omega _ X \to \mathcal{O}_{Y \cap Z} \to 0 \\ 0 \to \omega _ Y \to i^*\omega _ X \to \mathcal{O}_{Y \cap Z} \to 0 \\ 0 \to \omega _ Z \to j^*\omega _ X \to \mathcal{O}_{Y \cap Z} \to 0 \end{align*}
Here $\omega _ X$, $\omega _ Y$, $\omega _ Z$ are as in Lemma 53.4.1.
Proof.
A reduced $1$-dimensional Noetherian scheme is Cohen-Macaulay, so (1) is true. Since $X$ is reduced, we see that $X = Y \cup Z$ scheme theoretically. With notation as in Morphisms, Lemma 29.4.6 and by the statement of that lemma we have a short exact sequence
\[ 0 \to \mathcal{O}_ X \to \mathcal{O}_ Y \oplus \mathcal{O}_ Z \to \mathcal{O}_{Y \cap Z} \to 0 \]
Since $\mathcal{J} = \mathop{\mathrm{Ker}}(\mathcal{O}_ X \to \mathcal{O}_ Z)$, $\mathcal{J}' = \mathop{\mathrm{Ker}}(\mathcal{O}_ Y \to \mathcal{O}_{Y \cap Z})$, $\mathcal{I} = \mathop{\mathrm{Ker}}(\mathcal{O}_ X \to \mathcal{O}_ Y)$, and $\mathcal{I}' = \mathop{\mathrm{Ker}}(\mathcal{O}_ Z \to \mathcal{O}_{Y \cap Z})$ a diagram chase implies (2). Observe that $\mathcal{I} + \mathcal{J}$ is the ideal sheaf of $Y \cap Z$ and that $\mathcal{I} \cap \mathcal{J} = 0$. Hence we have the following exact sequences
\begin{align*} 0 \to \mathcal{O}_ X \to \mathcal{O}_ Y \oplus \mathcal{O}_ Z \to \mathcal{O}_{Y \cap Z} \to 0 \\ 0 \to \mathcal{J} \to \mathcal{O}_ X \to \mathcal{O}_ Z \to 0 \\ 0 \to \mathcal{I} \to \mathcal{O}_ X \to \mathcal{O}_ Y \to 0 \\ 0 \to \mathcal{J} \oplus \mathcal{I} \to \mathcal{O}_ X \to \mathcal{O}_{Y \cap Z} \to 0 \\ 0 \to \mathcal{J}' \to \mathcal{O}_ Y \to \mathcal{O}_{Y \cap Z} \to 0 \\ 0 \to \mathcal{I}' \to \mathcal{O}_ Z \to \mathcal{O}_{Y \cap Z} \to 0 \end{align*}
Since $X$ is Gorenstein $\omega _ X$ is an invertible $\mathcal{O}_ X$-module (Duality for Schemes, Lemma 48.24.4). Since $Y \cap Z$ has dimension $0$ we have $\omega _ X|_{Y \cap Z} \cong \mathcal{O}_{Y \cap Z}$. Thus if we prove (3) and (4), then we obtain the short exact sequences of the lemma by tensoring the above short exact sequence with the invertible module $\omega _ X$. By symmetry it suffices to prove (3) and by (2) it suffices to prove $i_*(\omega _ Y) = \mathcal{J}\omega _ X$.
We have $i_*\omega _ Y = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(i_*\mathcal{O}_ Y, \omega _ X)$ by Lemma 53.4.5. Again using that $\omega _ X$ is invertible we finally conclude that it suffices to show $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{O}_ X/\mathcal{I}, \mathcal{O}_ X)$ maps isomorphically to $\mathcal{J}$ by evaluation at $1$. In other words, that $\mathcal{J}$ is the annihilator of $\mathcal{I}$. This follows from the above.
$\square$
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