TITLE: Is the Artinian property dual to the Noetherian property?
QUESTION [7 upvotes]: If $R$ is a ring and $M$ is a left $R$-module, we say that $M$ is Noetherian whenever it satisfies any of the equivalent conditions:
1N. Every ascending (under subspace inclusion) chain of submodules of $M$ stabilises,
2N. Every non-empty collection of submodules of $M$ has a maximal element under subspace inclusion,
3N. Every submodule of $M$ is finitely generated.
Whilst we say that $M$ is Artinian whenever it satisfies either of the equivalent conditions:
1A. Every descending (under subspace inclusion) chain of submodules of $M$ stabilises,
2A. Every non-empty collection of submodules of $M$ has a minimal element under subspace inclusion.
It seems to be that conditions 1N and 1A are dual, whilst conditions 2N and 2A are also dual. However I can't seem to think of a property of Artinian modules that seems obviously dual to condition 3N. 
My question is whether or not there does exist such a condition?

REPLY [4 votes]: My question is whether or not there does exist such a condition?

Yes. 3A should be 

Every quotient module of $M$ is finitely cogenerated.

Finite cogeneration is easy to understand if you first rephrase finite generation this way:

A module is finitely generated if for every chain of submodules $N_0\subseteq N_1\subseteq N_2\ldots$ such that $\cup_{i\in I} N_i=M$, there is necessarily a finite subset $F\subseteq I$ such that $\cup_{i\in F} N_i=M$.

Then

A module is finitely cogenerated if for every chain of submodules $N_0\supseteq N_1\supseteq N_2\ldots$ such that $\cap_{i\in I} N_i=\{0\}$, there is necessarily a finite subset $F\subset I$ such that $\cap_{i\in F} N_i=\{0\}$.