The inclusion $\bigcap aM_n\subset a\bigcap M_n$ for descending chains of modules Let $A$ be a commutative ring with one, let 
$$
M=M_0>M_1>\cdots
$$ 
be a descending chain of finitely generated $A$-modules, and let $a$ be in $A$.

Does the inclusion $$\bigcap_{n\in\mathbb N}aM_n\subset a\left(\bigcap_{n\in\mathbb N}M_n\right)$$ always hold?

Here of course "$\subset$" means "is a not necessarily proper subset of".

Let us show that the answer is Yes if $A$ is a principal ideal domain.
We can assume $a\ne0$.
Let $(x_n)_{n\in\mathbb N}$ be a sequence of elements of $M$ such that $ax_n=ax_0$ for all $n$. 
It suffices to find an element $x$ in the intersection $I$ of the $M_n$ such that $ax=ax_0$. 
We have $M_n=T_n\oplus F_n$, with $T_n$ torsion and $F_n$ free. The $(T_n)_{n\in\mathbb N}$ forming a weakly decreasing sequence of artinian modules, we can assume $T_n=T\subset I$ for all $n$, and it suffices to prove $x_0\in I$.
Writing $x_n=t_n+f_n$ (obvious notation), it is enough to verify $f_0\in I$.
Our equation $ax_n=ax_0$ becomes the system 
$$
at_n=at_0,\quad af_n=af_0.
$$ 
This implies $f_0\in  f_n+T\subset M_n$ for all $n$, and thus $f_0\in I$, as was to be shown.
 A: The answer is No.
As pointed out by user26857 in a comment, a counterexample is given by Example 2.4 in 
Anderson, D., Matijevic, J., and Nichols, W. (1976). The Krull intersection theorem. II. Pacific Journal of Mathematics, 66(1), 15-22.
Let $K$ be a field and let $A$ be the commutative unital $K$-algebra generated by the symbols $a,x,y_1,y_2,\dots$ subject to the relations 
$$
x=ay_1=a^2y_2=a^3y_3=\cdots
$$ 
We have
$$
x\in\bigcap_{n\in\mathbb N}\ (a)^n=\bigcap_{n\in\mathbb N}\ a\ (a)^n,\quad x\notin a\,\bigcap_{n\in\mathbb N}\ (a)^n,
$$ 
so that we get our counterexample by setting $M:=A$ and $M_n:=(a)^n$.
A: Here is a noetherian counterexample:
Let $K$ be a field and $A$ the commutative $K$-algebra with one generated by the symbols $a$ and $b$ subject to the relations $ab^2=ab$ and $a^2=0$. 
We clearly have $ab=ab^2=ab^3=\cdots$
The set 
$$
\{a,ab\}\cup\{b^n\ |\ n\ge0\}
$$ 
is a $K$-basis of $A$. This implies in particular
$$
A=K[b]\oplus Ka\oplus Kab,
$$ 
and we get 
$$
a\ \bigcap_{n\in\mathbb N}\ (b^n)=a\,(ab)=(0),
$$
$$
\bigcap_{n\in\mathbb N}\ a\,(b^n)=(ab)\ne(0).
$$
This answer shows that any counterexample which is a $K$-algebra must have at least two generators.
