The Stacks project

Lemma 15.69.14. Let $A$ be a ring. Let $K \in D(A)$ be perfect. Then $K^\vee = R\mathop{\mathrm{Hom}}\nolimits _ A(K, A)$ is a perfect complex and $K = (K^\vee )^\vee $. There are functorial isomorphisms

\[ K^\vee \otimes _ A^\mathbf {L} L = R\mathop{\mathrm{Hom}}\nolimits _ A(K, L) \quad \text{and}\quad H^0(K^\vee \otimes _ A^\mathbf {L} L) = \mathop{\mathrm{Ext}}\nolimits _ A^0(K, L) \]

for $L \in D(A)$.

Proof. We can represent $K$ by a complex $K^\bullet $ of finite projective $A$-modules. By Lemma 15.68.2 the object $K^\vee $ is represented by the complex $E^\bullet = \mathop{\mathrm{Hom}}\nolimits ^\bullet (K^\bullet , A)$. Note that $E^ n = \mathop{\mathrm{Hom}}\nolimits _ A(K^{-n}, A)$ and the differentials of $E^\bullet $ are the transpose of the differentials of $K^\bullet $. Thus the formula $(K^\vee )^\vee = K$ is clear from the fact that the double dual of a finite projective module is itself.

The second equality follows from the first by Lemma 15.68.1 and Derived Categories, Lemma 13.19.8 as well as the definition of Ext groups, see Derived Categories, Section 13.27. Let us prove the first equality.

Let $L^\bullet $ be a complex of $A$-modules representing $L$. The object on the left of the first equality is represented by $\text{Tot}(E^\bullet \otimes _ A L^\bullet )$. The object on the right of the first equality sign is represented by the complex $\mathop{\mathrm{Hom}}\nolimits ^\bullet (K^\bullet , L^\bullet )$ by the same lemma as before. Thus the equality follows from the fact that

\[ \mathop{\mathrm{Hom}}\nolimits _ A(K^ n, A) \otimes _ A L^ m = \mathop{\mathrm{Hom}}\nolimits _ A(K^ n, L^ m) \]

for all $n, m$ because $K^ n$ is finite projective. To be a bit more precise we define the map on the level of complexes

\[ \text{Tot}(E^\bullet \otimes _ A L^\bullet ) = \text{Tot}(\mathop{\mathrm{Hom}}\nolimits ^\bullet (A, L^\bullet ) \otimes _ A \mathop{\mathrm{Hom}}\nolimits ^\bullet (K^\bullet , A)) \longrightarrow \mathop{\mathrm{Hom}}\nolimits ^\bullet (K^\bullet , L^\bullet ) \]

using Lemma 15.67.2 and then the statement above shows this is an isomorphism of complexes. $\square$


Comments (0)

There are also:

  • 2 comment(s) on Section 15.69: Perfect 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Your email address will not be published.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 07VI. Beware of the difference between the letter 'O' and the digit '0'.