When does $(\square P \land \square Q) \to \square (P \land Q)$ hold? If all axioms of classical propositional calculus hold and we work in modal logic that is at least K (ie. extremely weak), it is trivial to show $\square(P \land Q) \to (\square P \land \square Q)$. What other modality axioms, if any, have to be added for $(\square P \land  \square  Q) \to \square (P \land Q)$ to hold?
 A: You don't need any other axiom, $(\square P \land  \square  Q) \to \square (P \land Q)$ is derivable in the system $\mathbf{K}$, i.e. in the propositional calculus (with modus ponens and closure under substitution) augmented with the necessitation rule (i.e. if $A$ is derivable then $\square A$ is derivable) and the distribution axiom (i.e. $\square (P \to Q) \to (\square P \to \square Q)$).
A formal proof in Hilbert system for $\mathbf{K}$ is the following: 


*

*$P \to (Q \to (P \land Q)) \qquad$ (axiom for conjunction in propositional calculus)

*$\square (P \to (Q \to (P \land Q))) \qquad$ (necessitation rule applied to 1.)

*$\square (P \to (Q \to (P \land Q))) \to (\square P \to \square (Q \to (P \land Q))) \qquad$ (distribution axiom)

*$\square P \to \square (Q \to (P \land Q)) \qquad$ (modus ponens of 3. and 2.)

*$\square (Q \to (P \land Q)) \to (\square Q \to \square (P \land Q)) \qquad$ (distribution axiom)

*$(\square P \to \square(Q \to (P \land Q)) ) \to \big( ( \square(Q \to (P \land Q)) \to (\square Q \to \square(P \land Q) )) \to ((\square P \land \square Q) \to \square(P \land Q) \big) \qquad (*)$ 

*$( \square(Q \to (P \land Q)) \to (\square Q \to \square(P \land Q) )) \to ((\square P \land \square Q) \to \square(P \land Q) \qquad$ (modus ponens of 6. and 4.)

*$(\square P \land \square Q) \to \square(P \land Q) \qquad$ (modus ponens of 7. and 5.)


where the formula in ($*$) is derivable because it is an instance of the tautology in propositional $(p \to q) \to \big( (q \to (r \to s)) \to ((p \land r) \to s)\big)$ obtained through the substitution 
\begin{equation}\begin{cases}
p \mapsto \square P \\ 
q \mapsto \square (Q \to (P \land Q)) \\ 
r \mapsto \square Q \\
s \mapsto \square (P \land Q) 
\end{cases}\end{equation}
Remind that, by the completeness theorem, any tautology in propositional calculus can be derived in a formal proof system such the Hilbert system.
For a reference, you can see "A New Introduction to Modal Logic" by Hughes and  Cresswell, p. 27.
