Example to show we may have $Ann_{R}(M_{1}) + Ann_{R}(M_{2}) \subset Ann_{R}(M_{1} \cap M_{2})$ If $M_{1}$ and $M_{2}$ are submodules of a left $R$-module $M$ for a ring $R$, we know that we always have $Ann_{R}(M_{1}) + Ann_{R}(M_{2}) \subseteq Ann_{R}(M_{1} \cap M_{2})$. So what's a good example where the inequality is strict?
 A: Take $ R = \mathbb{Z}$. Any finite $\mathbb{Z}$-module is isomorphic to $\mathbb{Z}_{d_1} \times \cdots \times \mathbb{Z}_{d_n}$, where $d_1 \ |\  d_2 \ | \cdots \ |\ d_n$ are positive integers. If $n \geq 2$, then we can take
$$M_1 = \mathbb{Z}_{d_1} \times \{0\} \times \cdots \times \{0\}$$ and  $$M_2 = \{0\} \times \mathbb{Z}_{d_2} \times \{0\} \times\cdots \times \{0\}$$ (or any two factors of the above factorization) Indeed, $Ann(M_i) = d_i \mathbb{Z}$, and $Ann(M_1 \cap M_2) = \mathbb{Z}$. Then
$$  Ann(M_1) + Ann(M_2) = d_1 \mathbb{Z} \subsetneq \mathbb{Z} = Ann(M_1 \cap M_2)  $$
A: This is a good example of how striving for extremes can yield a counterexample. The first thing that came to mind for me in this question was: well, annihilators of  nonzero left ideals in prime rings are always zero. And if I found two such ideals that had a trivial intersection, that would make the annihilator of the intersection the entire ring. So:
Take $R=M_2(F)$ for any field $F$, and $M_1=\begin{bmatrix}F&0\\F&0\end{bmatrix}$ and $M_2=\begin{bmatrix}0&F\\0&F\end{bmatrix}$.
Since $R$ is a prime ring, $Ann(M_1)$ and $Ann(M_2)$ are both $\{0\}$.
So in this case, 
$\{0\}=Ann_{R}(M_{1}) + Ann_{R}(M_{2}) \subseteq Ann_{R}(M_{1} \cap M_{2})=Ann_R(\{0\})=R$
