Lemma 15.74.17. Let R = \mathop{\mathrm{colim}}\nolimits _{i \in I} R_ i be a filtered colimit of rings.
Given a perfect K in D(R) there exists an i \in I and a perfect K_ i in D(R_ i) such that K \cong K_ i \otimes _{R_ i}^\mathbf {L} R in D(R).
Given 0 \in I and K_0, L_0 \in D(R_0) with K_0 perfect, we have
\mathop{\mathrm{Hom}}\nolimits _{D(R)}(K_0 \otimes _{R_0}^\mathbf {L} R, L_0 \otimes _{R_0}^\mathbf {L} R) = \mathop{\mathrm{colim}}\nolimits _{i \geq 0} \mathop{\mathrm{Hom}}\nolimits _{D(R_ i)}(K_0 \otimes _{R_0}^\mathbf {L} R_ i, L_0 \otimes _{R_0}^\mathbf {L} R_ i)
In other words, the triangulated category of perfect complexes over R is the colimit of the triangulated categories of perfect complexes over R_ i.
Proof.
We will use the results of Algebra, Lemmas 10.127.5 and 10.127.6 without further mention. These lemmas in particular say that the category of finitely presented R-modules is the colimit of the categories of finitely presented R_ i-modules. Since finite projective modules can be characterized as summands of finite free modules (Algebra, Lemma 10.78.2) we see that the same is true for the category of finite projective modules. This proves (1) by our definition of perfect objects of D(R).
To prove (2) we may represent K_0 by a bounded complex K_0^\bullet of finite projective R_0-modules. We may represent L_0 by a K-flat complex L_0^\bullet (Lemma 15.59.10). Then we have
\mathop{\mathrm{Hom}}\nolimits _{D(R)}(K_0 \otimes _{R_0}^\mathbf {L} R, L_0 \otimes _{R_0}^\mathbf {L} R) = \mathop{\mathrm{Hom}}\nolimits _{K(R)}(K_0^\bullet \otimes _{R_0} R, L_0^\bullet \otimes _{R_0} R)
by Derived Categories, Lemma 13.19.8. Similarly for the \mathop{\mathrm{Hom}}\nolimits with R replaced by R_ i. Since in the right hand side only a finite number of terms are involved, since
\mathop{\mathrm{Hom}}\nolimits _ R(K_0^ p \otimes _{R_0} R, L_0^ q \otimes _{R_0} R) = \mathop{\mathrm{colim}}\nolimits _{i \geq 0} \mathop{\mathrm{Hom}}\nolimits _{R_ i}(K_0^ p \otimes _{R_0} R_ i, L_0^ q \otimes _{R_0} R_ i)
by the lemmas cited at the beginning of the proof, and since filtered colimits are exact (Algebra, Lemma 10.8.8) we conclude that (2) holds as well.
\square
Comments (0)
There are also: