The inclusion $\bigcap aM_n\subset a\bigcap M_n$ for descending chains of modules

Question

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".

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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.

Answer 1

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 $M_n:=(a)^n$.

Answer 2

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.