An MO user had a question about the “if” part of the following claim:
The question got shut down rapidly as “not research level”. But I thought it deserved an answer. Anyone who has a simpler proof should feel free to contact me (my email should be readily available through my MO user page).
The idea of proof is to establish an adjunction ; see the proposition below. Since we also have and right adjoints are unique up to canonical isomorphism, we get an isomorphism
But , being itself left adjoint to , preserves colimits, so preserves colimits. The fact that it preserves coequalizers means is projective (i.e., is right exact). To see is finitely generated, write as the (filtered) colimit of the system of its finitely generated submodules and inclusions between them. Then the comparison map
is an isomorphism. In particular, some element on the left, represented by an element of some , gets mapped to . That is to say, there is some whose composition with the inclusion is . This forces to be a monomorphism of into the f.g. submodule , i.e., and are mutually submodules of one another, with the inverse of the inclusion , forcing to be finitely generated.
Given that the canonical map is an isomorphism, there is an adjunction .
For this we need a unit and a counit. The counit is given by the evaluation map . The unit is given by the composite
and the proof is completed by verifying the triangular equations for an adjunction, viz. that the following composites are identity maps:
We leave this verification to the reader.