Let $A$ be an Artin algebra and $\text{mod}\,A$ the category of finite length modules. Further, let $X\_0 \longrightarrow X\_1 \longrightarrow X\_2 \longrightarrow ...$ and $Y\_0 \longleftarrow Y\_1 \longleftarrow Y\_2 \longleftarrow ...$ be chains of morphisms in $\text{mod}\,A$ such that $\underset{\longrightarrow}{\lim} X\_i \cong \underset{\longleftarrow}{\lim} Y\_j$. What can we say about the module $Z:=\underset{\longrightarrow}{\lim} X\_i \cong \underset{\longleftarrow}{\lim} Y\_j$? Does $Z$ have any nice properties? Does someone know an example, where $Z$ is not of finite length?



Let A be an Artin algebra. What happens if the limit and inverse limit are the same in mod A?
Let me show that every such $Z$ has finite length.


First, note that being of the form $\varinjlim X\_i$ (with the $X\_i$ of finite length) is the same as being countably generated over $A$. So let us start with $Z=\varprojlim Y\_j$ (with the $Y\_j$ of finite length) and prove that $Z$ cannot be countably generated unless it has finite length.


Endow $A$ and the $Y\_j$'s with the discrete topology, and $Z$ with the limit topology. Then $Z$ is a pro-discrete (hence complete metrizable) space, and a topological $A$-module. For each $j$, put $U\_j:=\ker(Z\to Y\_j)$: these are submodules of $Z$ forming a basis of neighborhoods of zero.


**Claim.** For a submodule $E$ of $Z$, the following conditions are equivalent, and imply that $E$ is closed:


(1) $E$ is discrete;  

(2) $U\_j\cap E=\{0\}$ for large $j$;  

(3) $E$ has finite length.


Indeed, $(1)\Leftrightarrow(2)$ is clear. (2) implies that $E\to Y\_j$ is injective, hence (3); conversely, the $U\_j\cap E$ form a decreasing sequence of submodules of $E$, with zero intersection, so (3) implies (2). Finally, if $E$ is discrete it is a locally closed subgroup of $Z$, hence closed.


Now assume that $Z$ is not of finite length. Then it is not discrete (by the above), i.e. $0$ is not isolated. It follows that every discrete (i.e. finite type) submodule has empty interior in $Z$. By Baire's theorem, so does every countable union of such submodules. In other words, $Z$ is not countably generated.


