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Let $R$ be a semi-simple ring. All left modules are semi-simple. Then all right modules are semi-simple. So all modules are semi-simple on both sides?

Problem with a semisimple ring exampleMorita equivalence: Is $_mathrmEnd_R(P)P$ projective if $P_R$ is?Submodules of semi-simple modulesSimple Cogenerator for the category of left $R$-modulesLeft and right R-modulesMatrix Ring of a Semisimple RingUsing Exchange Lemma in a decompositionRings of finite uniform dimensionProof of $R$ is semi simple $iff $ every finitely generated modules over $R$ is semi-simple.category of semi-simple modules is not closed under extension?

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$begingroup$

Let $R$ be a semi-simple ring not necessarily commutative. Let $R-Mod$ be the category of left $R-$modules and $Mod-R$ be the category of right $R$ modules.

Then I have both $R-Mod$ and $Mod-R$ composed of semi-simple modules on left and right correspondingly. Suppose $M$ is $(R,R)$ bimodule on both-sides. Say $M_1R$ is one of the left simple module factor of $M$(i.e. $M_1R$ is one of the direct summand of $M$ as right $R-$module.) Now consider simple $ _RM_j$ as some direct summand of $M$ as left $R-$module.

$textbfQ:$ Is $M_1R$ necessarily agrees with some $ _R M_j$ for some $j$? In other words, if a module is left-semi simple and it is a bi-module over the same ring, then it is also right-semi simple with the same direct summand? Consider ring $R$ itself. Since $R$ is semi-simple, say it decomposes into $oplus_jI_j$ with $I_j$ simple. Pick out orthogonal idempotents $e_j$ as left module.(Is it even obvious to pick out orthogonal idempotents in general?) If it is possible to pick out orthogonal idempotents, then it is clear that $e_je_k=delta_jke_j$. Since left decomposition of $R$ by those orthogonal idempotents is also a right decomposition of $R$ by those orthogonal idempotents, decomposition in left and right for $R$ is the same.

abstract-algebra ring-theory

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asked Mar 29 at 17:06

user45765user45765

2,6932724

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0

0

$begingroup$

Let $R$ be a semi-simple ring not necessarily commutative. Let $R-Mod$ be the category of left $R-$modules and $Mod-R$ be the category of right $R$ modules.

Then I have both $R-Mod$ and $Mod-R$ composed of semi-simple modules on left and right correspondingly. Suppose $M$ is $(R,R)$ bimodule on both-sides. Say $M_1R$ is one of the left simple module factor of $M$(i.e. $M_1R$ is one of the direct summand of $M$ as right $R-$module.) Now consider simple $ _RM_j$ as some direct summand of $M$ as left $R-$module.

$textbfQ:$ Is $M_1R$ necessarily agrees with some $ _R M_j$ for some $j$? In other words, if a module is left-semi simple and it is a bi-module over the same ring, then it is also right-semi simple with the same direct summand? Consider ring $R$ itself. Since $R$ is semi-simple, say it decomposes into $oplus_jI_j$ with $I_j$ simple. Pick out orthogonal idempotents $e_j$ as left module.(Is it even obvious to pick out orthogonal idempotents in general?) If it is possible to pick out orthogonal idempotents, then it is clear that $e_je_k=delta_jke_j$. Since left decomposition of $R$ by those orthogonal idempotents is also a right decomposition of $R$ by those orthogonal idempotents, decomposition in left and right for $R$ is the same.

abstract-algebra ring-theory

share|cite|improve this question

asked Mar 29 at 17:06

user45765user45765

2,6932724

$endgroup$

add a comment | 

$begingroup$

Let $R$ be a semi-simple ring not necessarily commutative. Let $R-Mod$ be the category of left $R-$modules and $Mod-R$ be the category of right $R$ modules.

Then I have both $R-Mod$ and $Mod-R$ composed of semi-simple modules on left and right correspondingly. Suppose $M$ is $(R,R)$ bimodule on both-sides. Say $M_1R$ is one of the left simple module factor of $M$(i.e. $M_1R$ is one of the direct summand of $M$ as right $R-$module.) Now consider simple $ _RM_j$ as some direct summand of $M$ as left $R-$module.

$textbfQ:$ Is $M_1R$ necessarily agrees with some $ _R M_j$ for some $j$? In other words, if a module is left-semi simple and it is a bi-module over the same ring, then it is also right-semi simple with the same direct summand? Consider ring $R$ itself. Since $R$ is semi-simple, say it decomposes into $oplus_jI_j$ with $I_j$ simple. Pick out orthogonal idempotents $e_j$ as left module.(Is it even obvious to pick out orthogonal idempotents in general?) If it is possible to pick out orthogonal idempotents, then it is clear that $e_je_k=delta_jke_j$. Since left decomposition of $R$ by those orthogonal idempotents is also a right decomposition of $R$ by those orthogonal idempotents, decomposition in left and right for $R$ is the same.

abstract-algebra ring-theory

share|cite|improve this question

asked Mar 29 at 17:06

user45765user45765

2,6932724

$endgroup$

share|cite|improve this question

asked Mar 29 at 17:06

user45765user45765

2,6932724

share|cite|improve this question

asked Mar 29 at 17:06

user45765user45765

2,6932724

asked Mar 29 at 17:06

user45765user45765

2,6932724

asked Mar 29 at 17:06

user45765user45765

2,6932724

1 Answer
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5

$begingroup$

Let $R=M_2(k)$, the $2times 2$ matrices over a field $k$. This is a classic
example of a semi-simple ring. A classic example of an $R$-bimodule is $R$ itself.
Consider the set $N$ of matrices of the form $$pmatrix*&*\0&0.$$
Then $N$ is a simple right $R$-submodule of $R$, but is certainly not a left $R$-submodule.

share|cite|improve this answer

answered Mar 29 at 17:21

Lord Shark the UnknownLord Shark the Unknown

108k1162135

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• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  You will have each left (or right) ideal generated by an idempotent. But in general $Re$ and $eR$ are completely different. @user45765
  $endgroup$
  – Lord Shark the Unknown
  Mar 29 at 17:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see. Thanks a lot.
  $endgroup$
  – user45765
  Mar 29 at 17:32
  

  
  
  
  

add a comment | 

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5

$begingroup$

Let $R=M_2(k)$, the $2times 2$ matrices over a field $k$. This is a classic
example of a semi-simple ring. A classic example of an $R$-bimodule is $R$ itself.
Consider the set $N$ of matrices of the form $$pmatrix*&*\0&0.$$
Then $N$ is a simple right $R$-submodule of $R$, but is certainly not a left $R$-submodule.

share|cite|improve this answer

answered Mar 29 at 17:21

Lord Shark the UnknownLord Shark the Unknown

108k1162135

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  You will have each left (or right) ideal generated by an idempotent. But in general $Re$ and $eR$ are completely different. @user45765
  $endgroup$
  – Lord Shark the Unknown
  Mar 29 at 17:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see. Thanks a lot.
  $endgroup$
  – user45765
  Mar 29 at 17:32
  

  
  
  
  

add a comment | 

5

$begingroup$

Let $R=M_2(k)$, the $2times 2$ matrices over a field $k$. This is a classic
example of a semi-simple ring. A classic example of an $R$-bimodule is $R$ itself.
Consider the set $N$ of matrices of the form $$pmatrix*&*\0&0.$$
Then $N$ is a simple right $R$-submodule of $R$, but is certainly not a left $R$-submodule.

share|cite|improve this answer

answered Mar 29 at 17:21

Lord Shark the UnknownLord Shark the Unknown

108k1162135

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  You will have each left (or right) ideal generated by an idempotent. But in general $Re$ and $eR$ are completely different. @user45765
  $endgroup$
  – Lord Shark the Unknown
  Mar 29 at 17:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see. Thanks a lot.
  $endgroup$
  – user45765
  Mar 29 at 17:32
  

  
  
  
  

add a comment | 

$begingroup$

Let $R=M_2(k)$, the $2times 2$ matrices over a field $k$. This is a classic
example of a semi-simple ring. A classic example of an $R$-bimodule is $R$ itself.
Consider the set $N$ of matrices of the form $$pmatrix*&*\0&0.$$
Then $N$ is a simple right $R$-submodule of $R$, but is certainly not a left $R$-submodule.

share|cite|improve this answer

answered Mar 29 at 17:21

Lord Shark the UnknownLord Shark the Unknown

108k1162135

$endgroup$

share|cite|improve this answer

answered Mar 29 at 17:21

Lord Shark the UnknownLord Shark the Unknown

108k1162135

answered Mar 29 at 17:21

Lord Shark the UnknownLord Shark the Unknown

108k1162135

answered Mar 29 at 17:21

Lord Shark the UnknownLord Shark the Unknown

108k1162135