Lemma 15.59.9. Let $R$ be a ring. Let $K^\bullet$ be a complex of $R$-modules. If $K^\bullet \otimes _ R M$ is acyclic for all finitely presented $R$-modules $M$, then $K^\bullet$ is K-flat.

Proof. We will use repeatedly that tensor product commute with colimits (Algebra, Lemma 10.12.9). Thus we see that $K^\bullet \otimes _ R M$ is acyclic for any $R$-module $M$, because any $R$-module is a filtered colimit of finitely presented $R$-modules $M$, see Algebra, Lemma 10.11.3. Let $M^\bullet$ be an acyclic complex of $R$-modules. We have to show that $\text{Tot}(M^\bullet \otimes _ R K^\bullet )$ is acyclic. Since $M^\bullet = \mathop{\mathrm{colim}}\nolimits \tau _{\leq n} M^\bullet$ (termwise colimit) we have

$\text{Tot}(M^\bullet \otimes _ R K^\bullet ) = \mathop{\mathrm{colim}}\nolimits \text{Tot}(\tau _{\leq n} M^\bullet \otimes _ R K^\bullet )$

with truncations as in Homology, Section 12.15. As filtered colimits are exact (Algebra, Lemma 10.8.8) we may replace $M^\bullet$ by $\tau _{\leq n}M^\bullet$ and assume that $M^\bullet$ is bounded above. In the bounded above case, we can write $M^\bullet = \mathop{\mathrm{colim}}\nolimits \sigma _{\geq -n} M^\bullet$ where the complexes $\sigma _{\geq -n} M^\bullet$ are bounded but possibly no longer acyclic. Arguing as above we reduce to the case where $M^\bullet$ is a bounded complex. Finally, for a bounded complex $M^ a \to \ldots \to M^ b$ we can argue by induction on the length $b - a$ of the complex. The case $b - a = 1$ we have seen above. For $b - a > 1$ we consider the split short exact sequence of complexes

$0 \to \sigma _{\geq a + 1}M^\bullet \to M^\bullet \to M^ a[-a] \to 0$

and we apply Lemma 15.58.4 to do the induction step. Some details omitted. $\square$

There are also:

• 4 comment(s) on Section 15.59: Derived tensor product

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).