The Stacks project

Proof. Pick an integer $n$ and consider the distinguished triangle

see Derived Categories, Remark 13.12.4. Since $L$ has finite injective dimension we see that $R\mathop{\mathrm{Hom}}\nolimits _ A(\tau _{\geq n + 1}K, L)$ has vanishing cohomology in degrees $\geq c - n$ for some constant $c$. Hence, given $i$, we see that $\mathop{\mathrm{Ext}}\nolimits ^ i_ A(K, L) \to \mathop{\mathrm{Ext}}\nolimits ^ i_ A(\tau _{\leq n}K, L)$ is an isomorphism for some $n \gg - i$. By Derived Categories of Schemes, Lemma 36.11.5 applied to $\tau _{\leq n}K$ and $L$ we see conclude that $\mathop{\mathrm{Ext}}\nolimits ^ i_ A(K, L)$ is a finite $A$-module for all $i$. Hence $R\mathop{\mathrm{Hom}}\nolimits _ A(K, L)$ is indeed an object of $D_{\textit{Coh}}(A)$. $\square$

Proof. Let $K$ be an object of $D_{\textit{Coh}}(A)$. From Lemma 47.15.2 we see $R\mathop{\mathrm{Hom}}\nolimits _ A(K, \omega _ A^\bullet )$ is an object of $D_{\textit{Coh}}(A)$. By More on Algebra, Lemma 15.98.2 and the assumptions on the dualizing complex we obtain a canonical isomorphism

Thus our functor has a quasi-inverse and the proof is complete. $\square$

Proof. Let $\mathfrak m \subset A$ be a maximal ideal with residue field $\kappa$. Consider the object $F(\kappa )$. Since $\kappa = \mathop{\mathrm{Hom}}\nolimits _{D(A)}(\kappa , \kappa )$ we find that all cohomology groups of $F(\kappa )$ are annihilated by $\mathfrak m$. We also see that

is zero for $i < 0$. Say $H^ a(F(\kappa )) \not= 0$ and $H^ b(F(\kappa )) \not= 0$ with $a$ minimal and $b$ maximal (so in particular $a \leq b$). Then there is a nonzero map

in $D(A)$ (nonzero because it induces a nonzero map on cohomology). This proves that $b = a$. We conclude that $F(\kappa ) = \kappa [-a]$.

Let $G$ be a quasi-inverse to our functor $F$. Arguing as above we find an integer $b$ such that $G(\kappa ) = \kappa [-b]$. On composing we find $a + b = 0$. Let $E$ be a finite $A$-module which is annihilated by a power of $\mathfrak m$. Arguing by induction on the length of $E$ we find that $G(E) = E'[-b]$ for some finite $A$-module $E'$ annihilated by a power of $\mathfrak m$. Then $E[-a] = F(E')$. Next, we consider the groups

The left hand side is nonzero if and only if $i = 0$ and then we get $E'$. Applying this with $E = E' = \kappa$ and using Nakayama's lemma this implies that $H^ j(F(A))_\mathfrak m$ is zero for $j > a$ and generated by $1$ element for $j = a$. On the other hand, if $H^ j(F(A))_\mathfrak m$ is not zero for some $j < a$, then there is a map $F(A) \to E[-a + i]$ for some $i < 0$ and some $E$ (More on Algebra, Lemma 15.65.7) which is a contradiction. Thus we see that $F(A)_\mathfrak m = M[-a]$ for some $A_\mathfrak m$-module $M$ generated by $1$ element. However, since

we see that $M \cong A_\mathfrak m$. We conclude that there exists an element $f \in A$, $f \not\in \mathfrak m$ such that $F(A)_ f$ is isomorphic to $A_ f[-a]$. This finishes the proof. $\square$

Proof. By Lemmas 47.15.3 and 47.15.4 the functor $K \mapsto R\mathop{\mathrm{Hom}}\nolimits _ A(R\mathop{\mathrm{Hom}}\nolimits _ A(K, \omega _ A^\bullet ), (\omega _ A')^\bullet )$ maps $A$ to an invertible object $L$. In other words, there is an isomorphism

Since $L$ has finite tor dimension, this means that we can apply More on Algebra, Lemma 15.98.2 to see that

is an isomorphism for $K$ in $D^ b_{\textit{Coh}}(A)$. In particular, setting $K = \omega _ A^\bullet$ finishes the proof. $\square$

Proof. Let $\omega _ A^\bullet \to I^\bullet$ be a quasi-isomorphism with $I^\bullet$ a bounded complex of injectives. Then $S^{-1}I^\bullet$ is a bounded complex of injective $B = S^{-1}A$-modules (Lemma 47.3.8) representing $\omega _ A^\bullet \otimes _ A B$. Thus $\omega _ A^\bullet \otimes _ A B$ has finite injective dimension. Since $H^ i(\omega _ A^\bullet \otimes _ A B) = H^ i(\omega _ A^\bullet ) \otimes _ A B$ by flatness of $A \to B$ we see that $\omega _ A^\bullet \otimes _ A B$ has finite cohomology modules. Finally, the map

is a quasi-isomorphism as formation of internal hom commutes with flat base change in this case, see More on Algebra, Lemma 15.99.2. $\square$

Proof. Consider the double complex

The associated total complex is quasi-isomorphic to $\omega _ A^\bullet$ for example by Descent, Remark 35.3.10 or by Derived Categories of Schemes, Lemma 36.9.4. By assumption the complexes $(\omega _ A^\bullet )_{f_ i}$ have finite injective dimension as complexes of $A_{f_ i}$-modules. This implies that each of the complexes $(\omega _ A^\bullet )_{f_{i_0} \ldots f_{i_ p}}$, $p > 0$ has finite injective dimension over $A_{f_{i_0} \ldots f_{i_ p}}$, see Lemma 47.3.8. This in turn implies that each of the complexes $(\omega _ A^\bullet )_{f_{i_0} \ldots f_{i_ p}}$, $p > 0$ has finite injective dimension over $A$, see Lemma 47.3.2. Hence $\omega _ A^\bullet$ has finite injective dimension as a complex of $A$-modules (as it can be represented by a complex endowed with a finite filtration whose graded parts have finite injective dimension). Since $H^ n(\omega _ A^\bullet )_{f_ i}$ is a finite $A_{f_ i}$ module for each $i$ we see that $H^ i(\omega _ A^\bullet )$ is a finite $A$-module, see Algebra, Lemma 10.23.2. Finally, the (derived) base change of the map $A \to R\mathop{\mathrm{Hom}}\nolimits _ A(\omega _ A^\bullet , \omega _ A^\bullet )$ to $A_{f_ i}$ is the map $A_{f_ i} \to R\mathop{\mathrm{Hom}}\nolimits _ A((\omega _ A^\bullet )_{f_ i}, (\omega _ A^\bullet )_{f_ i})$ by More on Algebra, Lemma 15.99.2. Hence we deduce that $A \to R\mathop{\mathrm{Hom}}\nolimits _ A(\omega _ A^\bullet , \omega _ A^\bullet )$ is an isomorphism and the proof is complete. $\square$

Proof. Let $\omega _ A^\bullet \to I^\bullet$ be a quasi-isomorphism with $I^\bullet$ a bounded complex of injectives. Then $\mathop{\mathrm{Hom}}\nolimits _ A(B, I^\bullet )$ is a bounded complex of injective $B$-modules (Lemma 47.3.4) representing $R\mathop{\mathrm{Hom}}\nolimits (B, \omega _ A^\bullet )$. Thus $R\mathop{\mathrm{Hom}}\nolimits (B, \omega _ A^\bullet )$ has finite injective dimension. By Lemma 47.13.4 it is an object of $D_{\textit{Coh}}(B)$. Finally, we compute

and for $n \not= 0$ we compute

which proves the last property of a dualizing complex. In the displayed equations, the first equality holds by Lemma 47.13.1 and the second equality holds by Lemma 47.15.3. $\square$

Proof. Special case of Lemma 47.15.8. $\square$

Proof. Set $B = A[x]$ and $\omega _ B^\bullet = \omega _ A^\bullet \otimes _ A B$. It follows from Lemma 47.3.10 and More on Algebra, Lemma 15.69.5 that $\omega _ B^\bullet$ has finite injective dimension. Since $H^ i(\omega _ B^\bullet ) = H^ i(\omega _ A^\bullet ) \otimes _ A B$ by flatness of $A \to B$ we see that $\omega _ A^\bullet \otimes _ A B$ has finite cohomology modules. Finally, the map

is a quasi-isomorphism as formation of internal hom commutes with flat base change in this case, see More on Algebra, Lemma 15.99.2. $\square$

Proof. This follows from a combination of Lemmas 47.15.6, 47.15.9, and 47.15.10. $\square$

Proof. This is true because $R\mathop{\mathrm{Hom}}\nolimits _ A(\kappa , \omega _ A^\bullet )$ is a dualizing complex over $\kappa$ (Lemma 47.15.9), because dualizing complexes over $\kappa$ are unique up to shifts (Lemma 47.15.5), and because $\kappa$ is a dualizing complex over $\kappa$. $\square$