**Proof.**
The implication (2) $\Rightarrow $ (1) is immediate. Assume (1) holds. First we choose a K-flat complex $K^\bullet $ with flat terms representing $K$, see Lemma 15.58.12. For any $R$-module $M$ the cohomology of

\[ K^{n - 1} \otimes _ R M \to K^ n \otimes _ R M \to K^{n + 1} \otimes _ R M \]

computes $H^ n(K \otimes _ R^\mathbf {L} M)$. This is always zero for $n < a$. Hence if we apply Lemma 15.65.2 to the complex $\ldots \to K^{a - 1} \to K^ a \to K^{a + 1}$ we conclude that $N = \mathop{\mathrm{Coker}}(K^{a - 1} \to K^ a)$ is a flat $R$-module. We set

\[ E^\bullet = \tau _{\geq a}K^\bullet = (\ldots \to 0 \to N \to K^{a + 1} \to \ldots ) \]

The kernel $L^\bullet $ of $K^\bullet \to E^\bullet $ is the complex

\[ L^\bullet = (\ldots \to K^{a - 1} \to I \to 0 \to \ldots ) \]

where $I \subset K^ a$ is the image of $K^{a - 1} \to K^ a$. Since we have the short exact sequence $0 \to I \to K^ a \to N \to 0$ we see that $I$ is a flat $R$-module. Thus $L^\bullet $ is a bounded above complex of flat modules, hence K-flat by Lemma 15.58.9. It follows that $E^\bullet $ is K-flat by Lemma 15.58.8.
$\square$

## Comments (0)

There are also: