Lemma 15.83.7. Let $R = \mathop{\mathrm{colim}}\nolimits _{i \in I} R_ i$ be a filtered colimit of rings. Let $0 \in I$ and $R_0 \to A_0$ be a flat ring map of finite presentation. For $i \geq 0$ set $A_ i = R_ i \otimes _{R_0} A_0$ and set $A = R \otimes _{R_0} A_0$.

Given an $R$-perfect $K$ in $D(A)$ there exists an $i \in I$ and an $R_ i$-perfect $K_ i$ in $D(A_ i)$ such that $K \cong K_ i \otimes _{A_ i}^\mathbf {L} A$ in $D(A)$.

Given $K_0, L_0 \in D(A_0)$ with $K_0$ pseudo-coherent and $L_0$ finite tor dimension over $R_0$, then we have

\[ \mathop{\mathrm{Hom}}\nolimits _{D(A)}(K_0 \otimes _{A_0}^\mathbf {L} A, L_0 \otimes _{A_0}^\mathbf {L} A) = \mathop{\mathrm{colim}}\nolimits _{i \geq 0} \mathop{\mathrm{Hom}}\nolimits _{D(A_ i)}(K_0 \otimes _{A_0}^\mathbf {L} A_ i, L_0 \otimes _{A_0}^\mathbf {L} A_ i) \]

In particular, the triangulated category of $R$-perfect complexes over $A$ is the colimit of the triangulated categories of $R_ i$-perfect complexes over $A_ i$.

**Proof.**
By Algebra, Lemma 10.127.6 the category of finitely presented $A$-modules is the colimit of the categories of finitely presented $A_ i$-modules. Given this, Algebra, Lemma 10.168.1 tells us that category of $R$-flat, finitely presented $A$-modules is the colimit of the categories of $R_ i$-flat, finitely presented $A_ i$-modules. Thus the characterization in Lemma 15.83.4 proves that (1) is true.

To prove (2) we choose $P_0^\bullet $ representing $K_0$ and $F_0^\bullet $ representing $L_0$ as in Lemma 15.83.6. Then $E_0^\bullet = \mathop{\mathrm{Hom}}\nolimits ^\bullet (P_0^\bullet , F_0^\bullet )$ satisfies

\[ H^0(E_0^\bullet \otimes _{R_0} R_ i) = \mathop{\mathrm{Hom}}\nolimits _{D(A_ i)}(K_0 \otimes _{A_0}^\mathbf {L} A_ i, L_0 \otimes _{A_0}^\mathbf {L} A_ i) \]

and

\[ H^0(E_0^\bullet \otimes _{R_0} R) = \mathop{\mathrm{Hom}}\nolimits _{D(A)}(K_0 \otimes _{A_0}^\mathbf {L} A, L_0 \otimes _{A_0}^\mathbf {L} A) \]

by the lemma. Thus the result because tensor product commutes with colimits and filtered colimits are exact (Algebra, Lemma 10.8.8).
$\square$

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