$N/K$ is direct summand of $M/K$ implies $N$ is direct summand of $M$

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Let $Ksubset Nsubset M$ be $R-$submodules where $R$ is a commutative ring with unity. If $K$ is a direct summand of $M$ then show that $K$ is a direct summand of $N$. Further, if $N/K$ is a direct summand of $M/K$ then show that $N$ is a direct summand of $M$.

It is easy to show that

If $K$ is a direct summand of $M$ then show that $K$ is a direct summand of $N$.

I am stuck on the "Further,..." part. I don't see how it is related to the previous part. Here's my attempt.

Since $N/K$ is a direct summand of $M/K$ there exists an $R-$ submodule $A$ of $M/K$ such that $N/Koplus A=M/K$. Now $A$ is $N'/K$ for a $R-$submodule $N'$ of $M$ containing $K$. So we have $N/Koplus N'/K=M/K$.

I want to say something like $Noplus N'=M$ which is not true since $N'cap Nsupset Kneq (0)$. Set theoretically I believe that if I take $B=N'- K$ then maybe $M=Noplus B$ but $B$ is not a submodule.

commutative-algebra modules

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edited yesterday

asked yesterday

Landon Carter

7,13611540

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Let $Ksubset Nsubset M$ be $R-$submodules where $R$ is a commutative ring with unity. If $K$ is a direct summand of $M$ then show that $K$ is a direct summand of $N$. Further, if $N/K$ is a direct summand of $M/K$ then show that $N$ is a direct summand of $M$.

It is easy to show that

If $K$ is a direct summand of $M$ then show that $K$ is a direct summand of $N$.

I am stuck on the "Further,..." part. I don't see how it is related to the previous part. Here's my attempt.

Since $N/K$ is a direct summand of $M/K$ there exists an $R-$ submodule $A$ of $M/K$ such that $N/Koplus A=M/K$. Now $A$ is $N'/K$ for a $R-$submodule $N'$ of $M$ containing $K$. So we have $N/Koplus N'/K=M/K$.

I want to say something like $Noplus N'=M$ which is not true since $N'cap Nsupset Kneq (0)$. Set theoretically I believe that if I take $B=N'- K$ then maybe $M=Noplus B$ but $B$ is not a submodule.

commutative-algebra modules

share|cite|improve this question

edited yesterday

asked yesterday

Landon Carter

7,13611540

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Let $Ksubset Nsubset M$ be $R-$submodules where $R$ is a commutative ring with unity. If $K$ is a direct summand of $M$ then show that $K$ is a direct summand of $N$. Further, if $N/K$ is a direct summand of $M/K$ then show that $N$ is a direct summand of $M$.

It is easy to show that

If $K$ is a direct summand of $M$ then show that $K$ is a direct summand of $N$.

I am stuck on the "Further,..." part. I don't see how it is related to the previous part. Here's my attempt.

Since $N/K$ is a direct summand of $M/K$ there exists an $R-$ submodule $A$ of $M/K$ such that $N/Koplus A=M/K$. Now $A$ is $N'/K$ for a $R-$submodule $N'$ of $M$ containing $K$. So we have $N/Koplus N'/K=M/K$.

I want to say something like $Noplus N'=M$ which is not true since $N'cap Nsupset Kneq (0)$. Set theoretically I believe that if I take $B=N'- K$ then maybe $M=Noplus B$ but $B$ is not a submodule.

commutative-algebra modules

share|cite|improve this question

edited yesterday

asked yesterday

Landon Carter

7,13611540

Let $Ksubset Nsubset M$ be $R-$submodules where $R$ is a commutative ring with unity. If $K$ is a direct summand of $M$ then show that $K$ is a direct summand of $N$. Further, if $N/K$ is a direct summand of $M/K$ then show that $N$ is a direct summand of $M$.

It is easy to show that

If $K$ is a direct summand of $M$ then show that $K$ is a direct summand of $N$.

I am stuck on the "Further,..." part. I don't see how it is related to the previous part. Here's my attempt.

Since $N/K$ is a direct summand of $M/K$ there exists an $R-$ submodule $A$ of $M/K$ such that $N/Koplus A=M/K$. Now $A$ is $N'/K$ for a $R-$submodule $N'$ of $M$ containing $K$. So we have $N/Koplus N'/K=M/K$.

I want to say something like $Noplus N'=M$ which is not true since $N'cap Nsupset Kneq (0)$. Set theoretically I believe that if I take $B=N'- K$ then maybe $M=Noplus B$ but $B$ is not a submodule.

commutative-algebra modules

share|cite|improve this question

edited yesterday

asked yesterday

Landon Carter

7,13611540

share|cite|improve this question

share|cite|improve this question

edited yesterday

edited yesterday

edited yesterday

asked yesterday

Landon Carter

7,13611540

asked yesterday

Landon Carter

7,13611540

asked yesterday

Landon Carter

7,13611540

7,13611540

add a comment | 

add a comment | 

1 Answer
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The statement is just incorrect: Consider the commutative ring $k[x, y]$ for $k$ any field. Let $M=k^3$ be a module over it with submodules $K=langle e_3rangle subset N=langle e_2, e_3rangle$. The action by $x e_2=e_3$, $ye_1=e_3$ and $xe_1=ye_2=0$. Then $M$ is indecomposable but $M/K$ decomposes into $N/K$ and the submodule spanned by the residue class of $e_1$.

share|cite|improve this answer

answered yesterday

Julian Kuelshammer

7,08632464

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Sorry but what happens to $xe_3$?
  – Landon Carter
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It is zero as well as $ye_3$.
  – Julian Kuelshammer
  yesterday
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote

The statement is just incorrect: Consider the commutative ring $k[x, y]$ for $k$ any field. Let $M=k^3$ be a module over it with submodules $K=langle e_3rangle subset N=langle e_2, e_3rangle$. The action by $x e_2=e_3$, $ye_1=e_3$ and $xe_1=ye_2=0$. Then $M$ is indecomposable but $M/K$ decomposes into $N/K$ and the submodule spanned by the residue class of $e_1$.

share|cite|improve this answer

answered yesterday

Julian Kuelshammer

7,08632464

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Sorry but what happens to $xe_3$?
  – Landon Carter
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It is zero as well as $ye_3$.
  – Julian Kuelshammer
  yesterday
  

  
  
  
  

add a comment | 

up vote
0
down vote

The statement is just incorrect: Consider the commutative ring $k[x, y]$ for $k$ any field. Let $M=k^3$ be a module over it with submodules $K=langle e_3rangle subset N=langle e_2, e_3rangle$. The action by $x e_2=e_3$, $ye_1=e_3$ and $xe_1=ye_2=0$. Then $M$ is indecomposable but $M/K$ decomposes into $N/K$ and the submodule spanned by the residue class of $e_1$.

share|cite|improve this answer

answered yesterday

Julian Kuelshammer

7,08632464

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Sorry but what happens to $xe_3$?
  – Landon Carter
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It is zero as well as $ye_3$.
  – Julian Kuelshammer
  yesterday
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

The statement is just incorrect: Consider the commutative ring $k[x, y]$ for $k$ any field. Let $M=k^3$ be a module over it with submodules $K=langle e_3rangle subset N=langle e_2, e_3rangle$. The action by $x e_2=e_3$, $ye_1=e_3$ and $xe_1=ye_2=0$. Then $M$ is indecomposable but $M/K$ decomposes into $N/K$ and the submodule spanned by the residue class of $e_1$.

share|cite|improve this answer

answered yesterday

Julian Kuelshammer

7,08632464

The statement is just incorrect: Consider the commutative ring $k[x, y]$ for $k$ any field. Let $M=k^3$ be a module over it with submodules $K=langle e_3rangle subset N=langle e_2, e_3rangle$. The action by $x e_2=e_3$, $ye_1=e_3$ and $xe_1=ye_2=0$. Then $M$ is indecomposable but $M/K$ decomposes into $N/K$ and the submodule spanned by the residue class of $e_1$.

share|cite|improve this answer

answered yesterday

Julian Kuelshammer

7,08632464

share|cite|improve this answer

share|cite|improve this answer

answered yesterday

Julian Kuelshammer

7,08632464

answered yesterday

Julian Kuelshammer

7,08632464

answered yesterday

Julian Kuelshammer

7,08632464

7,08632464

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Sorry but what happens to $xe_3$?
  – Landon Carter
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It is zero as well as $ye_3$.
  – Julian Kuelshammer
  yesterday
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Sorry but what happens to $xe_3$?
  – Landon Carter
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It is zero as well as $ye_3$.
  – Julian Kuelshammer
  yesterday
  

  
  
  
  

1

1

Sorry but what happens to $xe_3$?
– Landon Carter
yesterday

Sorry but what happens to $xe_3$?
– Landon Carter
yesterday

It is zero as well as $ye_3$.
– Julian Kuelshammer
yesterday

It is zero as well as $ye_3$.
– Julian Kuelshammer
yesterday

add a comment | 

 

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