I'm studying Morita theorem. I have the following property:

Let $R$ and $S$ Morita equivalent via $F: R\text{-}\mathrm{mod} \to S\text{-}\mathrm{mod}$ and suppose $M$ is an $R$-module. Then $M$ is finitely generated if and only if $F(M)$ is finitely generated.

I use the following characterization of finitely generated modules: $M$ is finitely generated if and only if, given a family $\{N_i \mid i \in I\}$ of submodules of $M$ such that $M=\sum_{i \in I}N_i$ there exists a finite set $J \subset I$ such that $M=\sum_{i\in J} N_i$.

Now I consider a family $\{N_i \mid i \in I\}$ of submodules of $F(M)$ such that $\sum_{i \in I}N_i=F(M)$. There exist $A_i$ such that $F(A_i)=N_i$ for every $i \in I$. The author says that $A_i$ are $R$-modules. So $F(M)=F(\sum_{i \in I}A_i)$. Then, if I apply $G$ to both sides I have $M= \sum_{i \in I}A_i$. I use the fact that $M$ is finitely generated to find a subset $J$ of $I$ and I can conclude that $F(M)$ is finitely generated.

My doubt is when he says that $A_i$ is an $R$-module. It is true, but I need the $A_i$ to be submodules of $M$ right? So I can apply the characterization of finitely generated modules. Is there a way to prove that the $A_i$ are submodules of $M$ knowing that the $N_i$ are submodules of $F(M)$ and that $N_i=F(A_i)$? Thanks!

  • $\begingroup$ If the proof is correct, is there a way to prove that the $A_i$ are submodules of $M$? or it isn't necessary? Thanks! $\endgroup$ – robbis Aug 27 '18 at 14:40
  • $\begingroup$ It may be easier to think of f.g. as quotient of a free module of finite rank. Then you can think in terms of exactness and not sub-objects. $\endgroup$ – Randall Aug 27 '18 at 14:49
  • $\begingroup$ Sorry @Randall I didn't see your answer. What do you mean? $\endgroup$ – robbis Sep 6 '18 at 19:08
  • $\begingroup$ It boils down to the same argument given in the answer below. $\endgroup$ – Randall Sep 6 '18 at 19:11

Let $G$ denote an inverse of $F$ with $G(N_i) = A_i$. Then $G$, being exact, sends the monomorphism $N_i\hookrightarrow F(M)$ to the monomorphism $A_i\hookrightarrow G(F(M))\cong M$. Hence you can identify $A_i$ with a submodule of $M$

  • $\begingroup$ So is it correct requiring that the $A_i$ are submodules of $M$? Can you tell me why the functor $G$ is exact? Thanks! $\endgroup$ – robbis Aug 28 '18 at 15:51
  • $\begingroup$ Every equivalence of categories is exact. $\endgroup$ – Tashi Walde Aug 28 '18 at 15:53
  • $\begingroup$ Being exact means that the functor transforms an exact sequence into another exact? $\endgroup$ – robbis Aug 28 '18 at 15:57
  • 1
    $\begingroup$ Exactly ;) In particular it sends monos to monos. $\endgroup$ – Tashi Walde Aug 28 '18 at 16:07
  • $\begingroup$ Thanks a lot for your help! $\endgroup$ – robbis Aug 28 '18 at 16:11

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