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category of semi-simple modules is not closed under extension?

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$begingroup$

Let $R$ be a ring and $mathcalC$ be the category of left semi-simple $R-$modules.

$textbfQ:$ What is the example of $mathcalC$ is not closed under extension?(i.e. Given $A,B$ semi-simple modules, there is $X$ not a semi-simple module s.t. $0to Ato Cto Bto 0$ is $R$ left module exact sequence.) I have thought of $Z_6$ over $Z_6$. $Z_6cong Z_2times Z_3$. Now $0to Z_2to Z_2times Z_3to Z_3to 0$ is an extension of $Z_2,Z_3$ as $Z_6$ module. I do not see other obvious possible extensions.

Ref. Auslander, Reiten, Representation Theory of Artin Algebras.

abstract-algebra ring-theory

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asked 23 hours ago

user45765user45765

2,6792724

$endgroup$

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  $begingroup$
  What examples of not semi-simple modules do you know?
  $endgroup$
  – Alex B.
  23 hours ago
  

  
  
  
  


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  @AlexB. That is the part that I have not seen so far. I knew some basic semi-simple rings as product of fields. Then modules over semi-simple rings are semi-simple. So I think I basically have no idea about non semi simples modules.
  $endgroup$
  – user45765
  23 hours ago
  

  
  
  
  


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  $begingroup$
  The case $R = mathbbZ$ is already instructive. Try figuring out what the simple and semisimple modules are.
  $endgroup$
  – Qiaochu Yuan
  21 hours ago
  

  
  
  
  


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  $begingroup$
  @QiaochuYuan All simple modules over $Z$ are of form $Z/(p)$ with $p$ prime. Semi-simple ones are built up from products of $Z/(p)$. So I guess I could consider $0to (2)to Z/(4)to Z/(2)to 0$ where first arrow is by canonical injection and second arrow is by standard projection. Clearly $Z/(4)$ is not semi-simple or it will be $Z/(2)times Z/(2)$. Is this correct? So in order to look for not closed under extensions, one has to look for non-semi simple rings? Maybe this is dumb, are you implying over semi-simple rings, the extension property is closed.
  $endgroup$
  – user45765
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @QiaochuYuan I think over semi-simple rings, the extension property is closed by $Ext$ group applied to any module being 0 as every module is both projective and injective over semi-simple rings. So extensions are always trivial. So semi-simpleness is closed under extension.
  $endgroup$
  – user45765
  9 hours ago
  

  
  
  
  

 | 
show 1 more comment

0

$begingroup$

Let $R$ be a ring and $mathcalC$ be the category of left semi-simple $R-$modules.

$textbfQ:$ What is the example of $mathcalC$ is not closed under extension?(i.e. Given $A,B$ semi-simple modules, there is $X$ not a semi-simple module s.t. $0to Ato Cto Bto 0$ is $R$ left module exact sequence.) I have thought of $Z_6$ over $Z_6$. $Z_6cong Z_2times Z_3$. Now $0to Z_2to Z_2times Z_3to Z_3to 0$ is an extension of $Z_2,Z_3$ as $Z_6$ module. I do not see other obvious possible extensions.

Ref. Auslander, Reiten, Representation Theory of Artin Algebras.

abstract-algebra ring-theory

share|cite|improve this question

asked 23 hours ago

user45765user45765

2,6792724

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What examples of not semi-simple modules do you know?
  $endgroup$
  – Alex B.
  23 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AlexB. That is the part that I have not seen so far. I knew some basic semi-simple rings as product of fields. Then modules over semi-simple rings are semi-simple. So I think I basically have no idea about non semi simples modules.
  $endgroup$
  – user45765
  23 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The case $R = mathbbZ$ is already instructive. Try figuring out what the simple and semisimple modules are.
  $endgroup$
  – Qiaochu Yuan
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @QiaochuYuan All simple modules over $Z$ are of form $Z/(p)$ with $p$ prime. Semi-simple ones are built up from products of $Z/(p)$. So I guess I could consider $0to (2)to Z/(4)to Z/(2)to 0$ where first arrow is by canonical injection and second arrow is by standard projection. Clearly $Z/(4)$ is not semi-simple or it will be $Z/(2)times Z/(2)$. Is this correct? So in order to look for not closed under extensions, one has to look for non-semi simple rings? Maybe this is dumb, are you implying over semi-simple rings, the extension property is closed.
  $endgroup$
  – user45765
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @QiaochuYuan I think over semi-simple rings, the extension property is closed by $Ext$ group applied to any module being 0 as every module is both projective and injective over semi-simple rings. So extensions are always trivial. So semi-simpleness is closed under extension.
  $endgroup$
  – user45765
  9 hours ago
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

Let $R$ be a ring and $mathcalC$ be the category of left semi-simple $R-$modules.

$textbfQ:$ What is the example of $mathcalC$ is not closed under extension?(i.e. Given $A,B$ semi-simple modules, there is $X$ not a semi-simple module s.t. $0to Ato Cto Bto 0$ is $R$ left module exact sequence.) I have thought of $Z_6$ over $Z_6$. $Z_6cong Z_2times Z_3$. Now $0to Z_2to Z_2times Z_3to Z_3to 0$ is an extension of $Z_2,Z_3$ as $Z_6$ module. I do not see other obvious possible extensions.

Ref. Auslander, Reiten, Representation Theory of Artin Algebras.

abstract-algebra ring-theory

share|cite|improve this question

asked 23 hours ago

user45765user45765

2,6792724

$endgroup$

share|cite|improve this question

asked 23 hours ago

user45765user45765

2,6792724

share|cite|improve this question

asked 23 hours ago

user45765user45765

2,6792724

asked 23 hours ago

user45765user45765

2,6792724

asked 23 hours ago

user45765user45765

2,6792724