Proof.
Assume (2). Consider the object $R\mathop{\mathrm{Hom}}\nolimits _ R(M, R)$ and the composition map
\[ R\mathop{\mathrm{Hom}}\nolimits (M, R) \otimes _ R^\mathbf {L} M \to R \]
Checking locally we see that this is an isomorphism; we omit the details. Because $D(R)$ is symmetric monoidal we see that $M$ is invertible.
Assume (1). Observe that an invertible object of a monoidal category has a left dual, namely, its inverse. Thus $M$ is perfect by Lemma 15.126.3. Consider a prime ideal $\mathfrak p \subset R$ with residue field $\kappa $. Then we see that $M \otimes _ R^\mathbf {L} \kappa $ is an invertible object of $D(\kappa )$. Clearly this implies that $\dim H^ i(M \otimes _ R^\mathbf {L} \kappa )$ is nonzero exactly for one $i$ and equal to $1$ in that case. By Lemma 15.75.7 this gives (2).
In the proof above we have seen that (a) holds. Let $U_ n \subset \mathop{\mathrm{Spec}}(R)$ be the union of the opens of the form $D(f)$ such that $M_ f \cong R_ f[-n]$. Clearly, $U_ n \cap U_{n'} = \emptyset $ if $n \not= n'$. If $M$ has tor amplitude in $[a, b]$, then $U_ n = \emptyset $ if $n \not\in [a, b]$. Hence we see that we have a product decomposition $R = \prod _{a \leq n \leq b} R_ n$ as in (d) such that $U_ n$ corresponds to $\mathop{\mathrm{Spec}}(R_ n)$, see Algebra, Lemma 10.24.3. Since $D(R) = \prod _{a \leq n \leq b} D(R_ n)$ and similarly for the category of modules parts (b), (c), and (d) follow immediately.
$\square$
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