TITLE: Is the derived category of a commutative ring monoidal?
QUESTION [5 upvotes]: Let $A$ be a commutative ring, and consider the derived category $D(A)$.

Is this a symmetric monoidal category? We have an obvious product, that is $-\otimes^L_A - $, and it is clear that we have an associator, that is, an isomorphism $(M\otimes^L_A N)\otimes^L_A K \cong M\otimes^L_A (N\otimes^L_A K)$. However, for this to be a monoidal category, we need that the pentagon diagram (the two ways to get associativity with 4 terms) commute on the nose, and not just up to isomorphism. So, can we define the derived tensor product in a suitable way so that the pentagon diagram will commute on the nose?
Assuming the answer to 1 is yes, is this also a closed monoidal category? Again, we have an obvious candidate for internal hom, that is: $RHom_A(-,-)$. But we have a problem: morphisms in $D(A)$ are only $H^0(RHom_A(-,-))$, and it follows from the axioms of a symmetric monoidal category that the internal hom and the hom are isomorphic as sets. So, can this be fixed in any sensible way?

REPLY [8 votes]: Hovey has proved the following general result:

Theorem. If $\mathcal{M}$ is a (symmetric) monoidal model category, then $\operatorname{Ho} \mathcal{M}$ is a (symmetric) monoidal closed category under the derived tensor product.

The category of unbounded chain complex of $A$-modules is a symmetric monoidal model category where the weak equivalences are the quasi-isomorphisms and the monoidal product is the Koszul tensor product. Thus the derived category is a symmetric monoidal closed category under the derived tensor product. The details can be found in Chapter 4 of [Hovey, Model categories].