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Consequence of epimorphism from Noetherian $R$-module

Preimage of a module under a module homomorphism is a submodule?Example of Artinian module that is not NoetherianProof of $M$ Noetherian if and only if all submodules are finitely generatedA question about a proof of Noetherian modules and exact sequencesA question on Artinian and Noetherian rings.ACC on chains of finitely generated submodules$A$ noetherian, $A$-endomorphism not injective for all invariant submodules is nilpotent$R$ Noetherian $implies$ Every finitely generated $R$-module is NoetherianCommutative Noetherian prime ideals finitely generatedWhat can be said of the lattices of submodules of a noetherian/finitely generated module?Ascending / descending chain condition on graded modules.

1

$begingroup$

Let $R,S$ be a commutative rings with $1_R,1_S$ respectively. In the most commutative algebra one can find the following proposition.

Proposition. Let $φ:Rtwoheadrightarrow S$ be a ring epimorphism. Then, if $R$ is Noetherian/Artinian, $S$ is also Noetherian/Artinian.

I was wondering if this is still valid for $R$-modules.

That is, let $R$ be a commutative ring with $1_R$, $M,N$ two $R$-modules and $φ:M twoheadrightarrow N$ a module epimorphism. If $R$-module $M$ is Noetherian/Artinian, then $R$-module $N$ is Noetherian/Artinian.

Proof. Trying to follow the steps of the proposition above, we consider an arbitrary ascending chain of $R$-submodules of $N$,
$$N_1subseteq N_2 subseteq N_3 subseteq dotsb.$$
We set $M_i=φ^-1(N_i), forall i=1,2,3,dots.$ The preimage of $R$-module is again an $R$-module. So, taking the preimages on the chain above, we can constract the ascending chain of $R$-submodules of $M$,
$$M_1subseteq M_2 subseteq M_3 subseteq dotsb.$$
But $M$ is Noetherian $R$-module, so there is an index $min Bbb Z^+$ s.t.
beginalignat*2
M_m quad = & quad M_k, && forall kgeq m iff\
φ^-1 (N_m) quad = & quad φ^-1 (N_k), && forall kgeq m implies \
φ(φ^-1 (N_m)) quad = & quad φ(φ^-1 (N_k)), && forall kgeq m iff \
N_m quad = & quad N_k, && forall kgeq m
endalignat*
and $``iff"$ is valid because $φ$ is surjective.

Is this proof correct and complete?

Thank you.

proof-verification commutative-algebra modules noetherian artinian

share|cite|improve this question

edited 12 hours ago

user26857

39.3k124183

asked yesterday

ChrisChris

860411

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Looks ok to me.
  $endgroup$
  – jgon
  21 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $R,S$ be a commutative rings with $1_R,1_S$ respectively. In the most commutative algebra one can find the following proposition.

Proposition. Let $φ:Rtwoheadrightarrow S$ be a ring epimorphism. Then, if $R$ is Noetherian/Artinian, $S$ is also Noetherian/Artinian.

I was wondering if this is still valid for $R$-modules.

That is, let $R$ be a commutative ring with $1_R$, $M,N$ two $R$-modules and $φ:M twoheadrightarrow N$ a module epimorphism. If $R$-module $M$ is Noetherian/Artinian, then $R$-module $N$ is Noetherian/Artinian.

Proof. Trying to follow the steps of the proposition above, we consider an arbitrary ascending chain of $R$-submodules of $N$,
$$N_1subseteq N_2 subseteq N_3 subseteq dotsb.$$
We set $M_i=φ^-1(N_i), forall i=1,2,3,dots.$ The preimage of $R$-module is again an $R$-module. So, taking the preimages on the chain above, we can constract the ascending chain of $R$-submodules of $M$,
$$M_1subseteq M_2 subseteq M_3 subseteq dotsb.$$
But $M$ is Noetherian $R$-module, so there is an index $min Bbb Z^+$ s.t.
beginalignat*2
M_m quad = & quad M_k, && forall kgeq m iff\
φ^-1 (N_m) quad = & quad φ^-1 (N_k), && forall kgeq m implies \
φ(φ^-1 (N_m)) quad = & quad φ(φ^-1 (N_k)), && forall kgeq m iff \
N_m quad = & quad N_k, && forall kgeq m
endalignat*
and $``iff"$ is valid because $φ$ is surjective.

Is this proof correct and complete?

Thank you.

proof-verification commutative-algebra modules noetherian artinian

share|cite|improve this question

edited 12 hours ago

user26857

39.3k124183

asked yesterday

ChrisChris

860411

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Looks ok to me.
  $endgroup$
  – jgon
  21 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Let $R,S$ be a commutative rings with $1_R,1_S$ respectively. In the most commutative algebra one can find the following proposition.

Proposition. Let $φ:Rtwoheadrightarrow S$ be a ring epimorphism. Then, if $R$ is Noetherian/Artinian, $S$ is also Noetherian/Artinian.

I was wondering if this is still valid for $R$-modules.

That is, let $R$ be a commutative ring with $1_R$, $M,N$ two $R$-modules and $φ:M twoheadrightarrow N$ a module epimorphism. If $R$-module $M$ is Noetherian/Artinian, then $R$-module $N$ is Noetherian/Artinian.

Proof. Trying to follow the steps of the proposition above, we consider an arbitrary ascending chain of $R$-submodules of $N$,
$$N_1subseteq N_2 subseteq N_3 subseteq dotsb.$$
We set $M_i=φ^-1(N_i), forall i=1,2,3,dots.$ The preimage of $R$-module is again an $R$-module. So, taking the preimages on the chain above, we can constract the ascending chain of $R$-submodules of $M$,
$$M_1subseteq M_2 subseteq M_3 subseteq dotsb.$$
But $M$ is Noetherian $R$-module, so there is an index $min Bbb Z^+$ s.t.
beginalignat*2
M_m quad = & quad M_k, && forall kgeq m iff\
φ^-1 (N_m) quad = & quad φ^-1 (N_k), && forall kgeq m implies \
φ(φ^-1 (N_m)) quad = & quad φ(φ^-1 (N_k)), && forall kgeq m iff \
N_m quad = & quad N_k, && forall kgeq m
endalignat*
and $``iff"$ is valid because $φ$ is surjective.

Is this proof correct and complete?

Thank you.

proof-verification commutative-algebra modules noetherian artinian

share|cite|improve this question

edited 12 hours ago

user26857

39.3k124183

asked yesterday

ChrisChris

860411

$endgroup$

share|cite|improve this question

edited 12 hours ago

user26857

39.3k124183

asked yesterday

ChrisChris

860411

share|cite|improve this question

edited 12 hours ago

user26857

39.3k124183

asked yesterday

ChrisChris

860411

edited 12 hours ago

user26857

39.3k124183

edited 12 hours ago

user26857

39.3k124183

asked yesterday

ChrisChris

860411

asked yesterday

ChrisChris

860411

1 Answer
1

active

oldest

votes

2

$begingroup$

Yes it's correct.

I guess most commutative algebra courses/books show that:

Proposition.
If $0 rightarrow M' rightarrow M rightarrow M'' rightarrow 0$ is an exact sequence of $R$-modules, then:

$M$ is artinian/noetherian iff $M',M''$ are artinian/noetherian.

So you might want to try proving this result.

Your proposition is the special case $0 rightarrow kervarphi rightarrow M rightarrow N rightarrow 0$.

Note that being an epimorphism is equivalent to being surjective for modules, but this is not the case for rings, as $mathbb Ztomathbb Q$ is epi - I'm not sure if your version for ring epimorphisms is true, although I don't have a counter-example on hand.

share|cite|improve this answer

edited 12 hours ago

user26857

39.3k124183

answered 14 hours ago

lushlush

607112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think for ring epimorphisms the Noetherianity does not behave well. See here.
  $endgroup$
  – user26857
  12 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Btw, when I read the question I thought the OP refers to ring epimorphisms in the category of commutative rings.
  $endgroup$
  – user26857
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I thought of commutative rings in my asnwer as well.
  $endgroup$
  – lush
  11 hours ago
  

  
  
  
  

add a comment | 

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2

$begingroup$

Yes it's correct.

I guess most commutative algebra courses/books show that:

Proposition.
If $0 rightarrow M' rightarrow M rightarrow M'' rightarrow 0$ is an exact sequence of $R$-modules, then:

$M$ is artinian/noetherian iff $M',M''$ are artinian/noetherian.

So you might want to try proving this result.

Your proposition is the special case $0 rightarrow kervarphi rightarrow M rightarrow N rightarrow 0$.

Note that being an epimorphism is equivalent to being surjective for modules, but this is not the case for rings, as $mathbb Ztomathbb Q$ is epi - I'm not sure if your version for ring epimorphisms is true, although I don't have a counter-example on hand.

share|cite|improve this answer

edited 12 hours ago

user26857

39.3k124183

answered 14 hours ago

lushlush

607112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think for ring epimorphisms the Noetherianity does not behave well. See here.
  $endgroup$
  – user26857
  12 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Btw, when I read the question I thought the OP refers to ring epimorphisms in the category of commutative rings.
  $endgroup$
  – user26857
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I thought of commutative rings in my asnwer as well.
  $endgroup$
  – lush
  11 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Yes it's correct.

I guess most commutative algebra courses/books show that:

Proposition.
If $0 rightarrow M' rightarrow M rightarrow M'' rightarrow 0$ is an exact sequence of $R$-modules, then:

$M$ is artinian/noetherian iff $M',M''$ are artinian/noetherian.

So you might want to try proving this result.

Your proposition is the special case $0 rightarrow kervarphi rightarrow M rightarrow N rightarrow 0$.

Note that being an epimorphism is equivalent to being surjective for modules, but this is not the case for rings, as $mathbb Ztomathbb Q$ is epi - I'm not sure if your version for ring epimorphisms is true, although I don't have a counter-example on hand.

share|cite|improve this answer

edited 12 hours ago

user26857

39.3k124183

answered 14 hours ago

lushlush

607112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think for ring epimorphisms the Noetherianity does not behave well. See here.
  $endgroup$
  – user26857
  12 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Btw, when I read the question I thought the OP refers to ring epimorphisms in the category of commutative rings.
  $endgroup$
  – user26857
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I thought of commutative rings in my asnwer as well.
  $endgroup$
  – lush
  11 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Yes it's correct.

I guess most commutative algebra courses/books show that:

Proposition.
If $0 rightarrow M' rightarrow M rightarrow M'' rightarrow 0$ is an exact sequence of $R$-modules, then:

$M$ is artinian/noetherian iff $M',M''$ are artinian/noetherian.

So you might want to try proving this result.

Your proposition is the special case $0 rightarrow kervarphi rightarrow M rightarrow N rightarrow 0$.

Note that being an epimorphism is equivalent to being surjective for modules, but this is not the case for rings, as $mathbb Ztomathbb Q$ is epi - I'm not sure if your version for ring epimorphisms is true, although I don't have a counter-example on hand.

share|cite|improve this answer

edited 12 hours ago

user26857

39.3k124183

answered 14 hours ago

lushlush

607112

$endgroup$

share|cite|improve this answer

edited 12 hours ago

user26857

39.3k124183

answered 14 hours ago

lushlush

607112

edited 12 hours ago

user26857

39.3k124183

edited 12 hours ago

user26857

39.3k124183

answered 14 hours ago

lushlush

607112

answered 14 hours ago

lushlush

607112