Being finitely generated is a Morita invariant

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




Let $R$ and $S$ Morita equivalent via $F: Rtext-mathrmmod to Stext-mathrmmod$ 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 $M=sum_i in IN_i$ there exists a finite set $J subset I$ such that $M=sum_iin J N_i$.



Now I consider a family $N_i mid i in I$ of submodules of $F(M)$ such that $sum_i in IN_i=F(M)$. 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 $F(M)=F(sum_i in IA_i)$. Then, if I apply $G$ to both sides I have $M= sum_i in IA_i$. 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!







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  • 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!
    – robbis
    Aug 27 at 14:40










  • 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.
    – Randall
    Aug 27 at 14:49










  • Sorry @Randall I didn't see your answer. What do you mean?
    – robbis
    Sep 6 at 19:08










  • It boils down to the same argument given in the answer below.
    – Randall
    Sep 6 at 19:11














up vote
2
down vote

favorite












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




Let $R$ and $S$ Morita equivalent via $F: Rtext-mathrmmod to Stext-mathrmmod$ 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 $M=sum_i in IN_i$ there exists a finite set $J subset I$ such that $M=sum_iin J N_i$.



Now I consider a family $N_i mid i in I$ of submodules of $F(M)$ such that $sum_i in IN_i=F(M)$. 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 $F(M)=F(sum_i in IA_i)$. Then, if I apply $G$ to both sides I have $M= sum_i in IA_i$. 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!







share|cite|improve this question






















  • 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!
    – robbis
    Aug 27 at 14:40










  • 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.
    – Randall
    Aug 27 at 14:49










  • Sorry @Randall I didn't see your answer. What do you mean?
    – robbis
    Sep 6 at 19:08










  • It boils down to the same argument given in the answer below.
    – Randall
    Sep 6 at 19:11












up vote
2
down vote

favorite









up vote
2
down vote

favorite











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




Let $R$ and $S$ Morita equivalent via $F: Rtext-mathrmmod to Stext-mathrmmod$ 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 $M=sum_i in IN_i$ there exists a finite set $J subset I$ such that $M=sum_iin J N_i$.



Now I consider a family $N_i mid i in I$ of submodules of $F(M)$ such that $sum_i in IN_i=F(M)$. 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 $F(M)=F(sum_i in IA_i)$. Then, if I apply $G$ to both sides I have $M= sum_i in IA_i$. 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!







share|cite|improve this question














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




Let $R$ and $S$ Morita equivalent via $F: Rtext-mathrmmod to Stext-mathrmmod$ 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 $M=sum_i in IN_i$ there exists a finite set $J subset I$ such that $M=sum_iin J N_i$.



Now I consider a family $N_i mid i in I$ of submodules of $F(M)$ such that $sum_i in IN_i=F(M)$. 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 $F(M)=F(sum_i in IA_i)$. Then, if I apply $G$ to both sides I have $M= sum_i in IA_i$. 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!









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edited Aug 31 at 13:05









Jendrik Stelzner

7,63121037




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asked Aug 27 at 13:43









robbis

528




528











  • 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!
    – robbis
    Aug 27 at 14:40










  • 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.
    – Randall
    Aug 27 at 14:49










  • Sorry @Randall I didn't see your answer. What do you mean?
    – robbis
    Sep 6 at 19:08










  • It boils down to the same argument given in the answer below.
    – Randall
    Sep 6 at 19:11
















  • 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!
    – robbis
    Aug 27 at 14:40










  • 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.
    – Randall
    Aug 27 at 14:49










  • Sorry @Randall I didn't see your answer. What do you mean?
    – robbis
    Sep 6 at 19:08










  • It boils down to the same argument given in the answer below.
    – Randall
    Sep 6 at 19:11















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!
– robbis
Aug 27 at 14:40




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!
– robbis
Aug 27 at 14:40












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.
– Randall
Aug 27 at 14:49




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.
– Randall
Aug 27 at 14:49












Sorry @Randall I didn't see your answer. What do you mean?
– robbis
Sep 6 at 19:08




Sorry @Randall I didn't see your answer. What do you mean?
– robbis
Sep 6 at 19:08












It boils down to the same argument given in the answer below.
– Randall
Sep 6 at 19:11




It boils down to the same argument given in the answer below.
– Randall
Sep 6 at 19:11










1 Answer
1






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up vote
1
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accepted










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






share|cite|improve this answer




















  • So is it correct requiring that the $A_i$ are submodules of $M$? Can you tell me why the functor $G$ is exact? Thanks!
    – robbis
    Aug 28 at 15:51










  • Every equivalence of categories is exact.
    – Tashi Walde
    Aug 28 at 15:53










  • Being exact means that the functor transforms an exact sequence into another exact?
    – robbis
    Aug 28 at 15:57






  • 1




    Exactly ;) In particular it sends monos to monos.
    – Tashi Walde
    Aug 28 at 16:07










  • Thanks a lot for your help!
    – robbis
    Aug 28 at 16:11










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










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






share|cite|improve this answer




















  • So is it correct requiring that the $A_i$ are submodules of $M$? Can you tell me why the functor $G$ is exact? Thanks!
    – robbis
    Aug 28 at 15:51










  • Every equivalence of categories is exact.
    – Tashi Walde
    Aug 28 at 15:53










  • Being exact means that the functor transforms an exact sequence into another exact?
    – robbis
    Aug 28 at 15:57






  • 1




    Exactly ;) In particular it sends monos to monos.
    – Tashi Walde
    Aug 28 at 16:07










  • Thanks a lot for your help!
    – robbis
    Aug 28 at 16:11














up vote
1
down vote



accepted










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






share|cite|improve this answer




















  • So is it correct requiring that the $A_i$ are submodules of $M$? Can you tell me why the functor $G$ is exact? Thanks!
    – robbis
    Aug 28 at 15:51










  • Every equivalence of categories is exact.
    – Tashi Walde
    Aug 28 at 15:53










  • Being exact means that the functor transforms an exact sequence into another exact?
    – robbis
    Aug 28 at 15:57






  • 1




    Exactly ;) In particular it sends monos to monos.
    – Tashi Walde
    Aug 28 at 16:07










  • Thanks a lot for your help!
    – robbis
    Aug 28 at 16:11












up vote
1
down vote



accepted







up vote
1
down vote



accepted






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






share|cite|improve this answer












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







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 27 at 15:56









Tashi Walde

1,23410




1,23410











  • So is it correct requiring that the $A_i$ are submodules of $M$? Can you tell me why the functor $G$ is exact? Thanks!
    – robbis
    Aug 28 at 15:51










  • Every equivalence of categories is exact.
    – Tashi Walde
    Aug 28 at 15:53










  • Being exact means that the functor transforms an exact sequence into another exact?
    – robbis
    Aug 28 at 15:57






  • 1




    Exactly ;) In particular it sends monos to monos.
    – Tashi Walde
    Aug 28 at 16:07










  • Thanks a lot for your help!
    – robbis
    Aug 28 at 16:11
















  • So is it correct requiring that the $A_i$ are submodules of $M$? Can you tell me why the functor $G$ is exact? Thanks!
    – robbis
    Aug 28 at 15:51










  • Every equivalence of categories is exact.
    – Tashi Walde
    Aug 28 at 15:53










  • Being exact means that the functor transforms an exact sequence into another exact?
    – robbis
    Aug 28 at 15:57






  • 1




    Exactly ;) In particular it sends monos to monos.
    – Tashi Walde
    Aug 28 at 16:07










  • Thanks a lot for your help!
    – robbis
    Aug 28 at 16:11















So is it correct requiring that the $A_i$ are submodules of $M$? Can you tell me why the functor $G$ is exact? Thanks!
– robbis
Aug 28 at 15:51




So is it correct requiring that the $A_i$ are submodules of $M$? Can you tell me why the functor $G$ is exact? Thanks!
– robbis
Aug 28 at 15:51












Every equivalence of categories is exact.
– Tashi Walde
Aug 28 at 15:53




Every equivalence of categories is exact.
– Tashi Walde
Aug 28 at 15:53












Being exact means that the functor transforms an exact sequence into another exact?
– robbis
Aug 28 at 15:57




Being exact means that the functor transforms an exact sequence into another exact?
– robbis
Aug 28 at 15:57




1




1




Exactly ;) In particular it sends monos to monos.
– Tashi Walde
Aug 28 at 16:07




Exactly ;) In particular it sends monos to monos.
– Tashi Walde
Aug 28 at 16:07












Thanks a lot for your help!
– robbis
Aug 28 at 16:11




Thanks a lot for your help!
– robbis
Aug 28 at 16:11

















 

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