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!
category-theory modules
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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!
category-theory modules
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!
â 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.
â Randall
Aug 27 at 14:49
Sorry @Randall I didn't see your answer. What do you mean?
â robbis
Sep 6 at 19:08
It boils down to the same argument given in the answer below.
â Randall
Sep 6 at 19:11
add a comment |Â
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!
category-theory modules
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!
category-theory modules
edited Aug 31 at 13:05
Jendrik Stelzner
7,63121037
7,63121037
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!
â 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.
â Randall
Aug 27 at 14:49
Sorry @Randall I didn't see your answer. What do you mean?
â robbis
Sep 6 at 19:08
It boils down to the same argument given in the answer below.
â Randall
Sep 6 at 19:11
add a comment |Â
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!
â 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.
â Randall
Aug 27 at 14:49
Sorry @Randall I didn't see your answer. What do you mean?
â robbis
Sep 6 at 19:08
It boils down to the same argument given in the answer below.
â 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!
â 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!
â 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.
â 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.
â Randall
Aug 27 at 14:49
Sorry @Randall I didn't see your answer. What do you mean?
â robbis
Sep 6 at 19:08
Sorry @Randall I didn't see your answer. What do you mean?
â robbis
Sep 6 at 19:08
It boils down to the same argument given in the answer below.
â Randall
Sep 6 at 19:11
It boils down to the same argument given in the answer below.
â Randall
Sep 6 at 19:11
add a comment |Â
1 Answer
1
active
oldest
votes
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 $M$
So is it correct requiring that the $A_i$ are submodules of $M$? Can you tell me why the functor $G$ is exact? Thanks!
â robbis
Aug 28 at 15:51
Every equivalence of categories is exact.
â Tashi Walde
Aug 28 at 15:53
Being exact means that the functor transforms an exact sequence into another exact?
â robbis
Aug 28 at 15:57
1
Exactly ;) In particular it sends monos to monos.
â Tashi Walde
Aug 28 at 16:07
Thanks a lot for your help!
â robbis
Aug 28 at 16:11
add a comment |Â
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 $M$
So is it correct requiring that the $A_i$ are submodules of $M$? Can you tell me why the functor $G$ is exact? Thanks!
â robbis
Aug 28 at 15:51
Every equivalence of categories is exact.
â Tashi Walde
Aug 28 at 15:53
Being exact means that the functor transforms an exact sequence into another exact?
â robbis
Aug 28 at 15:57
1
Exactly ;) In particular it sends monos to monos.
â Tashi Walde
Aug 28 at 16:07
Thanks a lot for your help!
â robbis
Aug 28 at 16:11
add a comment |Â
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 $M$
So is it correct requiring that the $A_i$ are submodules of $M$? Can you tell me why the functor $G$ is exact? Thanks!
â robbis
Aug 28 at 15:51
Every equivalence of categories is exact.
â Tashi Walde
Aug 28 at 15:53
Being exact means that the functor transforms an exact sequence into another exact?
â robbis
Aug 28 at 15:57
1
Exactly ;) In particular it sends monos to monos.
â Tashi Walde
Aug 28 at 16:07
Thanks a lot for your help!
â robbis
Aug 28 at 16:11
add a comment |Â
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 $M$
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 $M$
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!
â robbis
Aug 28 at 15:51
Every equivalence of categories is exact.
â Tashi Walde
Aug 28 at 15:53
Being exact means that the functor transforms an exact sequence into another exact?
â robbis
Aug 28 at 15:57
1
Exactly ;) In particular it sends monos to monos.
â Tashi Walde
Aug 28 at 16:07
Thanks a lot for your help!
â robbis
Aug 28 at 16:11
add a comment |Â
So is it correct requiring that the $A_i$ are submodules of $M$? Can you tell me why the functor $G$ is exact? Thanks!
â robbis
Aug 28 at 15:51
Every equivalence of categories is exact.
â Tashi Walde
Aug 28 at 15:53
Being exact means that the functor transforms an exact sequence into another exact?
â robbis
Aug 28 at 15:57
1
Exactly ;) In particular it sends monos to monos.
â Tashi Walde
Aug 28 at 16:07
Thanks a lot for your help!
â 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!
â 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!
â robbis
Aug 28 at 15:51
Every equivalence of categories is exact.
â Tashi Walde
Aug 28 at 15:53
Every equivalence of categories is exact.
â Tashi Walde
Aug 28 at 15:53
Being exact means that the functor transforms an exact sequence into another exact?
â robbis
Aug 28 at 15:57
Being exact means that the functor transforms an exact sequence into another exact?
â robbis
Aug 28 at 15:57
1
1
Exactly ;) In particular it sends monos to monos.
â Tashi Walde
Aug 28 at 16:07
Exactly ;) In particular it sends monos to monos.
â Tashi Walde
Aug 28 at 16:07
Thanks a lot for your help!
â robbis
Aug 28 at 16:11
Thanks a lot for your help!
â robbis
Aug 28 at 16:11
add a comment |Â
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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!
â 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.
â Randall
Aug 27 at 14:49
Sorry @Randall I didn't see your answer. What do you mean?
â robbis
Sep 6 at 19:08
It boils down to the same argument given in the answer below.
â Randall
Sep 6 at 19:11