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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.

user45765
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    Remark: another way to prove it is to use the facts that 1) every module is the direct limit of finitely presented modules and 2) direct limits are exact, if you already know these results. – Stahl Jun 06 '17 at 02:59
  • @Stahl Yes. I am aware that I can use direct limits and direct limits preserve exactness. However, I can trying to match two descriptions. The result from direct limits should be equivalent to reduction to finite relations somehow which I did not figure out. – user45765 Jun 06 '17 at 03:01
  • @Stahl. I think I figured out why I can assume there is finite relations enforced. I have posted my answer. Can you check whether I am heading towards the right direction? – user45765 Jun 06 '17 at 15:26

1 Answers1

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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.

  • Thanks. I knew this proof which simplifies my life a lot. I want to prove this problem without direct limit as there is no indication saying that direct limit is necessary. – user45765 Jun 06 '17 at 15:47
  • @user45765: I've finished my addendum since you commented. –  Jun 06 '17 at 15:49
  • That is the part I am stuck with. I do not see a way to choose a finite subset of that relations. Reduction from any modules to finitely generated modules is easy. Further reduction from finitely generated to finitely presented is not what I cannot see. Say $M=(x_1,x_2)$. Then $F$ is of rank 2. However $K$ may have infinitely many relations due to $r_{i1}x_i+r_{i2}x_2=0$ for $i\in N$. – user45765 Jun 06 '17 at 15:52
  • The choice of relations is the part which I do not see why it has finitely generated. Would you mind elaborate the finitely generation of the kernel part a bit? – user45765 Jun 06 '17 at 15:53
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    @user45765: You're missing the point: your goal is not to find a finite set of relations that present present $M$ as a quotient of $F$, your goal is to find a finite set of relations that can be chained to show $\sum_i m_i\otimes a_i \equiv 0$. You only care about this one element, not the whole module. –  Jun 06 '17 at 15:55
  • I think I see your point. Thanks. – user45765 Jun 06 '17 at 16:01