Let $R$ be a commutative algebra over a field $k$. My problem set asks me to
Show that every $R$-module is an iterated colimit of $R$.
My idea
Let $A$ be a free $R$-module with basis $M$, and let $\pi :A \to M$ be the R-module homomorphism taking each basis element to the corresponding element of $M$. Clearly $A$ is surjective, so $M \cong A/\ker \pi$.
It is easy to see that $A$ is a coproduct of $\lvert M \rvert$ copies of $R$, so $A$ is certainly a colimit of $R$.
We would then like to show that $M$ is a colimit of $A$. This is where I get less confident. Let $U \subseteq A \times A$ be the subset $U = \{(a, b) \in U\times U\mid a - b\in \ker \pi\}$. It is easy to check that $U$ is a submodule of $A\times A$.
Define the maps $$ f:U \to A, (a,b) \mapsto \pi(a),\quad g:U\to A, (a,b) \mapsto\pi(b). $$
Then $A/\ker \pi$ is the coequaliser of $U \rightrightarrows A$.
So we have obtained $M$ using two colimits.
The problem
I am not sure what is meant by an "iterated colimit". In particular, I am uneasy about the way I have defined $U$, where I have basically just prescribed exactly what I want the quotient to be. This feels potentially problematic because the diagram $U \rightrightarrows A$ is not constructed entirely as a colimit of $R$ (i.e. $A$ but not $U$ is a colimit of $R$).
I'm not sure whether this is a valid concern, especially because I don't have a precise definition of "iterated colimit".
Is my solution valid?
An iterated colimit of copies of $R$ is an object that can be inductively constructed by starting from $R$ and repeatedly taking colimits (possibly infinitely many times although in this case that’s unnecessary); said another way, the subcategory of iterated colimits of copies of $R$ is the intersection of all subcategories containing $R$ and closed under colimits.
Your argument is incomplete because you haven't shown that $U$ can be expressed as an iterated colimit of copies of $R$. (In fact you don't need this; it suffices to replace $U$ with an epimorphism from a free module to $U$.)