Interchanging the tensor product with infinite product Let $R$ be a $k$-algebra (not necessary commutative) and let $\mathbf{D}(R)$ be its derived category (right modules). I'm interested in the class of objects $V$ of $\mathbf{D}(R^{op})$ having the following property (P):
For any (infinite) product $\prod_{i\in I} M_{i}\in \mathbf{D}(R)$ the natural map
$$ [\prod_{i\in I} M_{i}]\otimes^{L}_{R} V\longrightarrow \prod_{i\in I}[ M_{i}\otimes^{L}_{R} V]$$ is an isomorphism in  $\mathbf{D}(k)$.
I have two questions.


*

*Does every perfect complex $V$ satisfy the property (P) ?


*If the answer to the first question is positive, what is the biggest class of objects of $\mathbf{D}(R)$ having the property (P)? Does it have a name ?

PS: $R^{op}$ is the ring $R$ but with opposite multiplication.
Thank you.
Ed.
 A: The class of objects with property (P) is a thick subcategory of $\mathbf{D}(R^{op})$ (i.e., a triangulated subcategory closed under taking direct summands), and contains $R$, so it contains all perfect complexes, since the category of perfect complexes is the thick subcategory generated by $R$.
I think that the class of objects with property (P) is the thick subcategory generated by finitely generated flat left $R$-modules. In particular, if $R$ is left noetherian, so that finitely generated flat modules are projective, then this is just the perfect complexes.
Proof: If $N$ is a finitely generated flat module then, by flatness, $N$ has property (P) if
$$\left[\prod_{i\in I}M_i\right]\otimes_RN\to\prod_{i\in I}\left[M_i\otimes_RN\right]$$ 
for every collection of right modules $\{M_i\}$, and this is true since $N$ is finitely generated. So the class of objects with property (P) certainly contains the thick subcategory generated by finitely generated flat modules.
Now suppose that $V$ has property (P). For every collection $\{M_i\vert i\in \mathbb{Z}\}$ of right modules, the direct product $\prod_{i\in\mathbb{Z}}M_i[i]$ coincides with the direct sum $\bigoplus_{i\in\mathbb{Z}}M_i[i]$, and so by property (P) the map
$$\bigoplus_{i\in\mathbb{Z}}\left[M_i[i]\otimes^{\mathbf{L}}_RV\right]\to
\prod_{i\in\mathbb{Z}}\left[M_i[i]\otimes^{\mathbf{L}}_RV\right]$$
is an isomorphism. In particular (taking $M_i=R$ for all $i$) $V$ has bounded cohomology, and (taking $M_i$ where possible with $H^0\!\!\left(M_i[i]\otimes^{\mathbf{L}}_RV\right)\neq0$) $V$ has finite flat dimension.
So up to a shift $V$ is isomorphic to a complex
$$\dots\to0\to V^0\to V^1\to\dots\to V^{n-1}\to V^n\to0\to\dots\tag{$*$}$$
with $V^0$ flat and $V^i$ projective for $i>0$.
If $n=0$, so that $V=V^0$ is just a flat module, then property (P) for a countable collection of copies of $R$ gives that
$$\left[\prod_{i\in\mathbb{Z}}R\right]\otimes_RV^0\to\prod_{i\in\mathbb{Z}}V^0$$
is an isomorphism, which is only true when $V^0$ is finitely generated.
If $n>0$, the same argument, together with right exactness of tensor product, shows that the cokernel of $V^{n-1}\to V^n$ must be finitely generated. Taking $P$ to be a finitely generated projective module mapping onto this cokernel, and lifting to a map $P\to V^n$, we get a map of complexes $P[-n]\to V$ whose mapping cone also has property (P). But the mapping cone is isomorphic to a shorter complex of the form $(*)$, so we are done by induction on $n$. 
