The Stacks project

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 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

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. $\square$