# Isn't the category of abelian groups obviously Grothendieck?

In Peter Freyd's Abelian Categories, it is mentioned in passing that the category of abelian groups (more generally, $$R$$-modules for a ring $$R$$) satisfy the axiom AB5:

For each linearly ordered family $$\{S_i\}_I$$ in the lattice of subobjects of an object $$S$$, and $$T$$ is any subobject of $$S$$, then we have $$T \cap \bigcup S_i = \bigcup (T \cap S_i).$$

Isn't this completely obvious? Let $$r \in RHS$$. Then $$r \in LHS$$. Conversely if $$r \in LHS$$ then $$r \in RHS$$. (The only non-triviality is that the union of a family of submodules is a module in the first place, and this holds because the family is linearly ordered.)

I am confused because here https://ncatlab.org/nlab/show/Grothendieck+category#Kiersz they link to a 9 page paper that proves that the category of $$R$$-modules satisfies what they say is an equivalent condition to AB5 ('small filtered colimits are exact').

• This is obvious for the case at hand because $\cap$ and $\bigcup$ for goups or $R$-,odules happen to be simply the set-theoretic intersection and union of he underlying set (and same underlying set means same module as long as we speak about submodules of a fixed module) – Hagen von Eitzen Dec 15 '18 at 4:38
• Thank you! May i ask then why in that link I gave (ncatlab.org/nlab/show/Grothendieck+category#Kiersz), under the section of examples they say "for $R$ a commutative ring, its category of modules $R$-Mod is a Grothendieck category. (see e.g Kiersz 06, prop. 4 for the proof that filtered colimits here are exact.)"? I mean to ask, is it really the case that their version of AB5 is that much harder to prove than the one I gave? – SSF Dec 15 '18 at 4:41