I'm working on a problem that says if $I$ is finitely generated ideal and $M$ is a flat module, then $(0:_M I)=(0:_R I)M$. I have reduced the problem into proving
$$\bigcap_{\mathrm{finite}} (0:_R a_i)M= (0:_R I)M$$
where $I$ is generated by all the $a_i$. I believe this is extremely simple, but writing down the definitions does not help.
Since $I$ is generated by finitely many elements $a_{1},\dots,a_{n}$, we have $(0:_{M}I)=\bigcap\limits_{i=1}^{n}(0:_{M}a_{i})$ and $(0:_{R}I)=\bigcap\limits_{i=1}^{n}(0:_{R}a_{i})$, thus we can reduce the problem to the case that $I=(a)$ is a principal ideal.
We have a short exact sequence $$0\longrightarrow (0:_{R}a)\longrightarrow R\overset{x\mapsto xa}{\longrightarrow} Ra\longrightarrow 0.$$ By the flatness of $M$, the sequence$$0\longrightarrow (0:_{R}a)M\longrightarrow M\overset{m\mapsto ma}{\longrightarrow} Ma$$ is also exact. Hence $(0:_{R}a)M$ is isomorphic to the kernel of $M\overset{m\mapsto ma}{\longrightarrow} Ma$, which is $(0:_{M}a)$.