Lemma 36.11.6. Let $X$ be a Noetherian scheme. Let $E$ in $D(\mathcal{O}_ X)$ be perfect. Then

1. $E$ is in $D^ b_{\textit{Coh}}(\mathcal{O}_ X)$,

2. if $L$ is in $D_{\textit{Coh}}(\mathcal{O}_ X)$ then $E \otimes _{\mathcal{O}_ X}^\mathbf {L} L$ and $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(E, L)$ are in $D_{\textit{Coh}}(\mathcal{O}_ X)$,

3. if $L$ is in $D^ b_{\textit{Coh}}(\mathcal{O}_ X)$ then $E \otimes _{\mathcal{O}_ X}^\mathbf {L} L$ and $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(E, L)$ are in $D^ b_{\textit{Coh}}(\mathcal{O}_ X)$,

4. if $L$ is in $D^+_{\textit{Coh}}(\mathcal{O}_ X)$ then $E \otimes _{\mathcal{O}_ X}^\mathbf {L} L$ and $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(E, L)$ are in $D^+_{\textit{Coh}}(\mathcal{O}_ X)$,

5. if $L$ is in $D^-_{\textit{Coh}}(\mathcal{O}_ X)$ then $E \otimes _{\mathcal{O}_ X}^\mathbf {L} L$ and $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(E, L)$ are in $D^-_{\textit{Coh}}(\mathcal{O}_ X)$.

Proof. Since $X$ is quasi-compact, each of these statements can be checked over the members of any open covering of $X$. Thus we may assume $E$ is represented by a bounded complex $\mathcal{E}^\bullet$ of finite free modules, see Cohomology, Lemma 20.49.3. In this case each of the statements is clear as both $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(E, L)$ and $E \otimes _{\mathcal{O}_ X}^\mathbf {L} L$ can be computed on the level of complexes using $\mathcal{E}^\bullet$, see Cohomology, Lemmas 20.46.9 and 20.26.9. Some details omitted. $\square$

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