Tensoring short exact sequence being exact for all finitely presented modules implies pure exact(No direct limit proof) A short exact sequence of $R-$modules $0\to A\to B\to C\to 0$ is pure exact if the sequence resulted from any $R-$module tensoring the short exact sequence remain exact. 
This is a 3.42 problem in Rotman, Homological algebra. 
The question is to show that if for any finitely presented $R-$module $M$, $0\to M\otimes A\to M\otimes B\to M\otimes C\to 0$ is exact, then $0\to A\to B\to C\to 0$ is pure exact(i.e. $C$ is flat). 
Hint from the problem says: That an element lies in $Ker(M\otimes A\to M\otimes B)$ involves only finitely many elements of $A$. I think this hint is used to prove the other direction. 
It is clear that one only needs to show the injectivity remains intact. The following is what I did.
Assume $0\to A\to B\to C\to 0$ is not pure exact. So we must have a $R-$module $M$ such that $M\otimes A\to M\otimes B$ is not injective. Now we can shrink $M$ to finitely generated submodule by considering an element $\sum_i m_i\otimes a_i$ of $M\otimes A$ being sent to 0 as this element is only a finite combination of elements of $M$ and $A$. Consider $0\to K\to F\to M\to 0$ exact sequence where $M$ is finitely generated, $F$ is finite rank free module and $K$ is the module generated by relations in $M$. Define $K_0=(0)$, take any element $k_1$ from $K-K_0$. Define $K_1=(k_1)$. Define $K_i=(K_{i-1},k_{i-1})$ by taking any element $k_{i-1}$ from $K_i-K_{i-1}$. So clearly we get a chain $K_0\subset K_1\dots\subset K$.
One also obtain a chain of maps as well $F/K_0\to F/K_1\to\dots F/K$ and the map is clearly well defined. Even I have finite set of $m_i$ from $\sum_i m_i\otimes a_i=0\in M\otimes B$. 
It is not clear that I can impose only finitely many relations(i.e. it suffices to consider one of $F/K_i$ where $i$ is large enough to contain all relevant information).
Please give me a proof without direct limit or even mentioning direct limit.
 A: There's a short proof that relies on two key facts:


*

*Every $R$-module is a directed colimit of finitely presented modules

*Directed colimits are exact


(directed colimits are sometimes called direct limits)
Thus, writing an arbitrary module $M$ as a directed colimit
$$ M = \operatorname{colim}_j M_j $$
we have a directed system of exact sequences
$$ 0 \to M_j \otimes A \to M_j \otimes B \to M_j \otimes C \to 0 $$
and taking the colimit gives an exact sequence
$$ 0 \to M \otimes A \to M \otimes B \to M \otimes C \to 0 $$

The proof you are considering is, I think, basically repeating parts of the proof of these two key facts. I think the key idea you're overlooking is that you don't need to pick your $K_i$ to preserve all relevant information — instead, for each specific element you want to prove zero in $M \otimes A$, you merely need to take the specific relations needed to calculate that specific element is zero. Your choice of $k_i$ should be tailored specifically to that calculation, rather than being arbitrary choices.
