Assume $p$ is a prime number and $q = p^2$. Denote by $A$ the ring $\mathbb{Z} / q \mathbb{Z}$.
Consider a finite type module $M$ over $A$ whith cardinality $q^N$ where $N$ is an integer, $N>0$.
Assume we have two $A$-submodules $S$ and $T$ of $M$ of cardinality $q^s$ and $q^t$.
If $s + t > N$, can we assert that $S \cap T \neq \{ 0 \}$ ? If this is the case, can we find a lower bound for the cardinality of $S \cap T$ ?
(I think the answer to the first question is positive when $q = p$ since we are dealing with vector spaces and we can write the dimension of $S \cap T$ in terms of the dimension of $S$, $T$ and $S + T$).
Consider the module homomorphism
$$f:S\oplus T\to M$$ $$f(x,y)=x+y$$
Denote by $\Delta=\{(x,-x)\ |\ x\in S\cap T\}$ and note that $f(x,y)=0$ if and only if $x=-y$ and so $f(x,y)=0$ if and only if $(x,y)\in\Delta$. It follows that
$$Im(f)\simeq (S\oplus T)/\Delta$$
Now since every finite $A$-module has to have cardinality of the form $p^x$ then $|S\cap T|=|\Delta|=p^D$ which leads to
$$|Im(f)|=|S\oplus T|/|\Delta|=q^{s+t}/p^D$$
and so $p^D=q^{s+t}/|Im(f)|$. So it reaches its minimum when $Im(f)$ is the biggest, i.e. $Im(f)=M$ (or equivalently when $S,T$ generate $M$) and in that case $D=2(s+t-N)$ is the lowest possible value. Or equivalently $|S\cap T|\geq q^{s+t-N}$.
This also shows that if $s+t>N$ then $S\cap T\neq 0$.