
Lemma 15.62.9. Let $R$ be a ring. Let $m \in \mathbf{Z}$. If $K^\bullet \oplus L^\bullet$ is $m$-pseudo-coherent (resp. pseudo-coherent) so are $K^\bullet$ and $L^\bullet$.

Proof. In this proof we drop the superscript ${}^\bullet$. Assume that $K \oplus L$ is $m$-pseudo-coherent. It is clear that $K, L \in D^{-}(R)$. Note that there is a distinguished triangle

$(K \oplus L, K \oplus L, L \oplus L[1]) = (K, K, 0) \oplus (L, L, L \oplus L[1])$

see Derived Categories, Lemma 13.4.9. By Lemma 15.62.2 we see that $L \oplus L[1]$ is $m$-pseudo-coherent. Hence also $L[1] \oplus L[2]$ is $m$-pseudo-coherent. By induction $L[n] \oplus L[n + 1]$ is $m$-pseudo-coherent. By Lemma 15.62.8 we see that $L[n]$ is $m$-pseudo-coherent for large $n$. Hence working backwards, using the distinguished triangles

$(L[n], L[n] \oplus L[n - 1], L[n - 1])$

we conclude that $L[n], L[n - 1], \ldots , L$ are $m$-pseudo-coherent as desired. The pseudo-coherent case follows from this and Lemma 15.62.5. $\square$

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).