Is $\frac{\bigcap_{k=1}^\infty M_k}{\bigcup_{k=1}^\infty N_k}$ a direct limit? Suppose we have a sequence of $R$-modules $$N_1\subseteq N_2\subseteq\cdots \subseteq N_k\subseteq N_{k+1}\subseteq \cdots\subseteq M_{k+1}\subseteq M_k\subseteq \cdots\subseteq M_2\subseteq M_1.$$ 
My question is: is it possible to make the following set $$\bigg\{P_k:=\frac{M_k}{N_k}\bigg|k\geq 1\bigg\}$$ with some ordering and with some $R$-modules homomorphsims to a directed set so that $\frac{\bigcap_{k=1}^\infty M_k}{\bigcup_{k=1}^\infty N_k}$ is direct limit?
Of course, if all $M_k$ are equals to some fixed $M$, then for $i,j\in \Bbb N$ with $i\leq j (\text{ordering w.r.t. natural order of }\Bbb N)$ we have a $R$-module homomorphism $\varphi^i_j:P_i\longrightarrow P_j$ defined by $$\varphi^i_j:x+N_i\longmapsto x+N_j\text{ for all }x\in M.$$ And in this case, $\frac{\bigcap_{k=1}^\infty M_k}{\bigcup_{k=1}^\infty N_k}=\frac{M}{\bigcup_{k=1}^\infty N_k}$ is obviously the direct limit of $\{P_i,\varphi^i_j|i\leq j\}$.
Any help will be appreciated.
 A: For simplicity, let's consider the case where all the $N_k$ are zero, so the question becomes whether $\bigcap_{k=1}^\infty M_k$ can be a direct limit of some sequence with entries $M_k$.
[The only obvious maps are the inclusion maps, and then the intersection is the inverse limit rather than the direct limit, but of course that doesn't prove that there is no clever choice of order and maps such that the direct limit is isomorphic to the intersection.]
Any countable direct limit of countably generated modules is countably generated, so if we can find a descending sequence of countably generated modules whose intersection is not countably generated, then the intersection can't be isomorphic to a direct limit of any sequence with entries $M_k$.
There may be easier ways to do this, but here's one.
Let $k$ be a field, and $R$ the ring of polynomials over $k$ in $X\cup Y$, where $X=\{x_n\mid n\in\mathbb{N}\}$ is a countable set of indeterminates and $Y=\{y_\alpha\mid\alpha\in A\}$ is an uncountable set of indeterminates, modulo the ideal generated by
$$\{x_ix_j\mid i.j\in\mathbb{N}\}\cup\{x_iy_\alpha-x_jy_\alpha\mid i\in\mathbb{N},\alpha\in A\}.$$
[So in $R$, the elements $x_0,x_1,\dots$ are all different, but, for example, $x_0y_\alpha^2 y_\beta=x_1y_\alpha^2 y_\beta=x_2y_\alpha^2 y_\beta=\dots$ and we'll write it as $xy_\alpha^2y_\beta$.]
Let $M_k$ be the ideal of $R$ generated by $\{x_n\mid n>k\}$. Then $M_k$ is countably generated, but the intersection of the $M_k$ (which consists of the elements of the form $xf$, where $f$ is a polynomial in the $y_\alpha$ with zero constant term) is not countably generated.
