The Stacks project

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. $\square$


Comments (2)

Comment #4275 by Bogdan on

I think it should be (not ).

There are also:

  • 1 comment(s) on Section 47.15: Dualizing complexes

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0A7E. Beware of the difference between the letter 'O' and the digit '0'.