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?
Homological Algebra – Are All Short Exact Sequences Direct Limits of Short Exact Sequences?
exact-sequencehomological-algebralimits-colimitsmodulesrepresentation-theory
Related Solutions
One has the exact sequence $$0\to K\to R^{(I)}\to M\to 0$$ We can assume w.l.o.g. that $K\subseteq R^{(I)}$, $M=R^{(I)}/K$. Let's consider now the set $$\mathcal{S}=\{(N,S):|S|<\infty,\;\;N\subseteq K\cap R^S, N\text{ is finitely generated}\}$$ with the order relation $(N,S)\leq (N',S') \iff N\subseteq N',S\subseteq S'$. This is a directed poset, in fact $N+N'$ is f.g. if $N,N'$ are, $S\cup S'$ is finite if $S,S'$ are. So by construction $\forall (N,S)\in \mathcal{S}$ the module $R^S/N$ is finitely presented. One verifies that $$\displaystyle \varinjlim_{(N,S)\in\mathcal{S}}R^S/N\cong R^{(I)}/K.$$
It is not true that sequentially small modules are always finitely presented. For instance, let $A$ be an ordered abelian group with the property that every countable set of positive elements has a positive strict lower bound. Let $R$ be a valuation ring with value group $A$ and let $k$ be the residue field of $R$. Then $k$ is not finitely presented as an $R$-module, since the maximal ideal of $R$ is not finitely generated (indeed, it is not even countably generated, by our assumption on $A$). However, I claim $k$ is sequentially small.
Indeed, suppose $$M_0\to M_1\to M_2\to\dots$$ is a sequential diagram of $R$-modules with colimit $M$. Since $k$ is finitely generated, if a homomorphism $f:k\to M_n$ becomes $0$ after mapping into $B$, then it becomes $0$ after mapping into $M_m$ for some $m>n$, since we just need the images of the finitely many generators of $k$ to become $0$. So, the canonical map $\operatorname{colim}\operatorname{Hom}(k,M_n)\to\operatorname{Hom}(k,M)$ is injective.
To show it is surjective, let $f:k\to M$ be a homomorphism; we wish to show $f$ can be lifted to $M_n$ for some $n$. First, $f(1)$ can be lifted to to an element $x\in M_i$ for some $i$. Let $x_n$ be the image of $x$ in $M_n$ for each $n\geq i$, and let $I_n\subseteq R$ be the annihilator of $x_n$. If $I_n$ contains the maximal ideal for some $n$, then we can lift $f$ to a homomorphism $k\to M_n$ by mapping $1$ to $x_n$ and we're done, so we may assume $I_n$ does not contain the maximal ideal for each $n$. Then for each $n$, we can pick some $a_n>0$ in $A$ such that every element of $I_n$ has valuation at least $a_n$. By our choice of $A$, there exists $a>0$ such that $a<a_n$ for all $n$. If $r\in R$ has valuation $a$, then, the image of $ax$ in $M_n$ is nonzero for all $n$, and thus the image of $ax$ in $M$ is nonzero. But the image of $ax$ in $M$ is $af(1)=0$ since $a$ annihilates $k$. This is a contradiction.
Best Answer
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.