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There is somewhat of symmetry in the definition of flat module and pure short exact sequence which can be made precise as follows.

Let $\mathcal{R}$ be the class of all right $R$-modules, $\mathcal{S}$ be the class of all short exact sequences of left $R$-modules.

For $\mathcal{A}\subseteq\mathcal{R}, \mathcal{B}\subseteq\mathcal{S} $, define a binary relation as $$ \mathcal{A}\perp \mathcal{B}:\Leftrightarrow ~\forall M\in\mathcal{A},\forall\mathcal{E}\in\mathcal{B},M\otimes_R \mathcal{E}~~\text{is still exact}.$$ which induces an (antitone) Galois connection as

$$\mathcal{A}^\perp=\{\mathcal{E}\in\mathcal{B}|\mathcal{A}\perp\{\mathcal{E}\}\}$$ $$^\perp\mathcal{B}=\{M\in\mathcal{A}|\{M\} \perp\mathcal{B}\}$$

Observe that $$^\perp\mathcal{S}=\{\text{flat modules}\}$$ $$\mathcal{R}^\perp=\{\text{pure short exact sequences}\}$$

My question is what are the closure operators corresponding to this Galois connection? i.e. how to describe $$^\perp(\mathcal{A}^\perp), (^\perp\mathcal{B})^\perp$$ explicitly?

An audacious and rough guess is that $$^\perp(\mathcal{A}^\perp)=\{\text{filtered colimits of modules in}~\mathcal{A}\},$$ $$(^\perp\mathcal{B})^\perp=\{\text{filtered colimits of s.e.s. in}~\mathcal{B}\}$$ which is based on the fact that the left hand side are all closed under filtered colimits due to the exactness of taking filtered colimits.

The guess is to sharp to be true because it easily implies two non-trivial results:

  • (Lazard’s theorem) Every flat module is a filtered colimit of finitely generated free modules. (Vice versa)
  • Every pure exact sequence is a filtered colimit of split exact sequences. (Vice versa)

So it may need some adjustments such as adding the direct summands, etc., as the right hand side may be too small.

Could someone give me some guidance or insights?

Edit: A less audacious guess is

$^\perp(\mathcal{A}^\perp)$= the smallest full subcategory containing $\mathcal{A}$ which is closed under taking (finite) coproducts, filtered colimits, direct summands, and contains a generator.

Now it can’t imply the Lazard’s theorem (still, I used it to formulate the guess). And there is a dual statement lurking in the s.e.s. side, but I’m not sure what’s the counterpart of $R$ in the category of s.e.s..

The question is now posted here on MathOverflow, just to clarify.

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  • $\begingroup$ I don't know the answer, but note that for any $\mathcal{A}$, even $\mathcal{A}=\emptyset$, $^\perp(\mathcal{A}^\perp)$ contains all flat modules and for any $\mathcal{B}$, $(^\perp\mathcal{B})^\perp$ contains all pure short exact sequences, so your "audacious guess" would at least need to be adjusted to take account of that. $\endgroup$ May 2 at 8:27
  • $\begingroup$ @JeremyRickard: Thanks for pointing out that! I guess the flat modules and pure exact sequences behave like trivial elements (or isotopic vectors) in this context and any closure should add them inside rudely. This indeed feels unnatural since it’s unlikely to construct flat module out of nowhere. $\endgroup$ May 2 at 8:39
  • $\begingroup$ I’ve improved the guess a little in the first equality. Some more observation: “closed” subclass w.r.t. this Galois connection (i.e. of the form $^\perp\mathcal{B}$) should be closed under taking coproducts. A cute use of the snake lemma shows that it’s also closed under taking direct summands. So maybe we could adjust the guess as $^\perp(\mathcal{A}^\perp)$= the smallest full subcategory containing $\mathcal{A}$ which is closed under taking coproducts, filtered colimits, direct summands, and contains a generator. Now the RHS contains all the flat modules. $\endgroup$ May 2 at 9:31
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    $\begingroup$ Sorry the bounty didn't help. You could try posting on MO? $\endgroup$ Jun 19 at 6:08
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    $\begingroup$ @CalvinKhor Yes I posted it on MO yesterday, but didn’t get any response 😢. Thanks for your concern, though. $\endgroup$ Jun 19 at 8:11

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