Lemma 47.15.4. Let $A$ be a Noetherian ring. Let $F : D^ b_{\textit{Coh}}(A) \to D^ b_{\textit{Coh}}(A)$ be an $A$-linear equivalence of categories. Then $F(A)$ is an invertible object of $D(A)$.

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

$\mathop{\mathrm{Ext}}\nolimits ^ i_ A(\kappa , \kappa ) = \text{Ext}^ i_ A(F(\kappa ), F(\kappa )) = \mathop{\mathrm{Hom}}\nolimits _{D(A)}(F(\kappa ), F(\kappa )[i])$

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

$F(\kappa ) \to H^ b(F(\kappa ))[-b] \to H^ a(F(\kappa ))[-b] \to F(\kappa )[a - b]$

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 wich 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

$\mathop{\mathrm{Ext}}\nolimits ^ i_ A(A, E') = \text{Ext}^ i_ A(F(A), F(E')) = \mathop{\mathrm{Hom}}\nolimits _{D(A)}(F(A), E[-a + i])$

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 $E$ (Lemma 47.15.3) 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

$A_\mathfrak m = \mathop{\mathrm{Hom}}\nolimits _{D(A)}(A, A)_\mathfrak m = \mathop{\mathrm{Hom}}\nolimits _{D(A)}(F(A), F(A))_\mathfrak m = \mathop{\mathrm{Hom}}\nolimits _{A_\mathfrak m}(M, M)$

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. $\square$

Comment #4275 by Bogdan on

I think it should be $\operatorname{Ext}^i_{A}(k,k)=\operatorname{Ext}^i_{A}(F(k),F(k))=\operatorname{Hom}_{D(A)}(F(k),F(k)[i])$ (not $-i$).

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