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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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?
$begingroup$
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
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
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
asked Mar 29 at 17:06
user45765user45765
2,6932724
$endgroup$
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
abstract-algebra ring-theory
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
asked Mar 29 at 17:06
user45765user45765
2,6932724
asked Mar 29 at 17:06
user45765user45765
2,6932724
2,6932724
add a comment |
add a comment |
1 Answer
1
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$begingroup$
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.
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 |
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1 Answer
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1 Answer
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$begingroup$
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.
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.
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.
answered Mar 29 at 17:21
Lord Shark the UnknownLord Shark the Unknown
108k1162135
$endgroup$
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.
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
answered Mar 29 at 17:21
Lord Shark the UnknownLord Shark the Unknown
108k1162135
108k1162135
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 |
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
2
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
$begingroup$
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
$begingroup$
I see. Thanks a lot.
$endgroup$
– user45765
Mar 29 at 17:32
add a comment |
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