Lemma 15.79.3. Let $R$ be a Noetherian regular ring of dimension $d < \infty $. Let $K, L \in D^-(R)$. Assume there exists an $k$ such that $H^ i(K) = 0$ for $i \leq k$ and $H^ i(L) = 0$ for $i \geq k - d + 1$. Then $\mathop{\mathrm{Hom}}\nolimits _{D(R)}(K, L) = 0$.

**Proof.**
Let $K^\bullet $ be a bounded above complex representing $K$, say $K^ i = 0$ for $i \geq n + 1$. After replacing $K^\bullet $ by $\tau _{\geq k + 1}K^\bullet $ we may assume $K^ i = 0$ for $i \leq k$. Then we may use the distinguished triangle

to see it suffices to prove the lemma for $K^ n[-n]$ and $\sigma _{\leq n - 1}K^\bullet $. By induction on $n$, we conclude that it suffices to prove the lemma in case $K$ is represented by the complex $M[-m]$ for some $R$-module $M$ and some $m \geq k + 1$. Since $R$ has global dimension $d$ by Algebra, Lemma 10.110.8 we see that $M$ has a projective resolution $0 \to P_ d \to \ldots \to P_0 \to M \to 0$. Then the complex $P^\bullet $ having $P_ i$ in degree $m - i$ is a bounded complex of projectives representing $M[-m]$. On the other hand, we can choose a complex $L^\bullet $ representing $L$ with $L^ i = 0$ for $i \geq k - d + 1$. Hence any map of complexes $P^\bullet \to L^\bullet $ is zero. This implies the lemma by Derived Categories, Lemma 13.19.8. $\square$

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