Is every short exact sequence the direct limit of short exact sequences between finitely presented modules?

Question

Let $R$ be a unitary, associative ring. It is well-known that every $R$-module is the direct limit (filtered colimit) of finitely presented $R$-modules. Is it also true that every short exact sequence of $R$-modules is the direct limit of short exact sequences between finitely presented $R$-modules?

Answer 1

This is false. Intuitively, the problem is that given a short exact sequence $0\to A\to B\to C\to 0$, you can write each of $A$ and $B$ as direct limits of finitely presented modules $(A_i)$ and $(B_i)$, but you can't necessarily do so in such a way that the $A_i$ map injectively to the $B_i$ and so can be part of a short exact sequence.

Here is a counterexample. Let $k$ be a field and $$R=k[x,y_0,y_1,y_2,\dots]/(xy_0,xy_1,xy_2,\dots).$$ I claim first that if $M$ is a finitely presented $R$-module and $a\in M$ is annihilated by $y_n$ for all $n$, then $a\in xM$. To prove this, consider any finite presentation for $M$. There is then some $y_n$ which does not appear in any of the relations in the presentation; let $S\subset R$ be the subring generated by $x$ and the $y_i$ for $i\neq n$. Then using the same presentation over $S$, we get an $S$-module $N$ such that $M=N\otimes_{k[x]}k[x,y_n]/(xy_n)$. It follows that $M$ can be identified with $N\oplus N/xN\oplus N/xN\oplus\dots$, with $y_n$ acting by shifting the direct summands forward by one (since $k[x,y_n]/(xy_n)$ has such a direct sum decomposition as a $k[x]$-module, with the summands being generated by the powers of $y_n$). In particular, any element of $M$ that is annihilated by $y_n$ must be in the first summand $N$ and be divisible by $x$. In particular, then, our element $a\in M$ which is annihilated by $y_n$ is divisible by $x$, as desired.

Now I claim that the short exact sequence $$0\to (x)\to R\to R/(x)\to 0$$ of $R$-modules cannot be written as a direct limit of short exact sequences of finitely presented modules. If it could, there would exist a short exact sequence $$0\to K\to M\to M/K\to 0$$ of finitely presented modules together with a homomorphism $f:M\to R$ inducing a map of short exact sequences and an element $a\in M$ such that $f(a)=1$ and $xa\in K$. Now note that $xa$ is annihilated by every $y_n$ and thus must be divisible by $x$ in $K$ since $K$ is finitely presented. But $f(K)\subseteq (x)$, so this would imply $f(xa)=x$ is divisible by $x^2$ in $R$, which is false.