Is the intersection of two independent $\sigma$-algebras the trivial $\sigma$-algebra (i.e. $\{\emptyset,\Omega\}$)?
My attempt: yes. Let $P$ and $Q$ be two independent $\sigma$-algebras. Then for any set $A\in P\cap Q$, by independence, we have $$P(A\cap A)=P(A)P(A).$$ That is, $P(A)=1$ or $0$. Since only the whole space $\Omega$ has measure $1$ and empty set has measure $0$, we have $P\cap Q=\{\emptyset,\Omega\}$ (here i'm not so certain).
This is not true in general. Consider a probability space $(\Omega,\mathcal{F},\mathsf{P})$ and let $\mathcal{G}$ be the collection of $\mathsf{P}$-trivial events of $\mathcal{F}$, i.e. $$ \mathcal{G}=\{F\in \mathcal{F}:\mathsf{P}(F)\in \{0,1\}\}. $$ Then $\mathcal{G}$ is independent of any sub-$\sigma$-algebra of $\mathcal{F}$. In particular, it is independent of $\mathcal{F}$. So $\mathcal{G}\cap \mathcal{F}\ne \{\emptyset,\Omega\}$ unless $\mathcal{G}$ is a trivial $\sigma$-algebra.