Lemma 15.75.5. Let $(R, \mathfrak m, \kappa )$ be a local ring. Let $K \in D(R)$ be pseudo-coherent. Set $d_ i = \dim _\kappa H^ i(K \otimes _ R^\mathbf {L} \kappa )$. Then $d_ i < \infty$ and for some $b \in \mathbf{Z}$ we have $d_ i = 0$ for $i > b$. Then there exists a complex

$\ldots \to R^{\oplus d_{b - 2}} \to R^{\oplus d_{b - 1}} \to R^{\oplus d_ b} \to 0 \to \ldots$

representing $K$ in $D(R)$. Moreover, this complex is unique up to isomorphism(!).

Proof. Observe that $K \otimes _ R^\mathbf {L} \kappa$ is pseudo-coherent as an object of $D(\kappa )$, see Lemma 15.64.12. Hence the cohomology spaces are finite dimensional and vanish above some cutoff. Every object of $D(\kappa )$ is isomorphic in $D(\kappa )$ to a complex $E^\bullet$ with zero differentials. In particular $E^ i \cong \kappa ^{\oplus d_ i}$ is finite free. Applying Lemma 15.75.4 we obtain the existence.

If we have two complexes $F^\bullet$ and $G^\bullet$ with $F^ i$ and $G^ i$ free of rank $d_ i$ representing $K$. Then we may choose a map of complexes $\beta : F^\bullet \to G^\bullet$ representing the isomorphism $F^\bullet \cong K \cong G^\bullet$, see Derived Categories, Lemma 13.19.8. The induced map of complexes $\beta \otimes 1 : F^\bullet \otimes _ R^\mathbf {L} \kappa \to G^\bullet \otimes _ R^\mathbf {L} \kappa$ must be an isomorphism of complexes as the differentials in $F^\bullet \otimes _ R^\mathbf {L} \kappa$ and $G^\bullet \otimes _ R^\mathbf {L} \kappa$ are zero. Thus $\beta ^ i : F^ i \to G^ i$ is a map of finite free $R$-modules whose reduction modulo $\mathfrak m$ is an isomorphism. Hence $\beta ^ i$ is an isomorphism and we win. $\square$

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.

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 0BCC. Beware of the difference between the letter 'O' and the digit '0'.