Dimension of $End(V)$ with $V$ countable dimension irreducible module over a complex algebra
Let $A$ be a $mathbb{C}$-algebra and $V$ be an irreducible $A$-module with countable dimension. What is the dimension of $End(V)$ as $A$-module? Note that every endomorphism must be injective and surjective, because of the irreducibility of $V$. My claim (or at least my hope) is that $End(V)$ has countable dimension, but I can't see a proof of this fact.
linear-algebra abstract-algebra cardinals division-algebras
asked Dec 22 '17 at 12:51
Giuseppe Bargagnati
1,094514
|
show 7 more comments
Let $A$ be a $mathbb{C}$-algebra and $V$ be an irreducible $A$-module with countable dimension. What is the dimension of $End(V)$ as $A$-module? Note that every endomorphism must be injective and surjective, because of the irreducibility of $V$. My claim (or at least my hope) is that $End(V)$ has countable dimension, but I can't see a proof of this fact.
linear-algebra abstract-algebra cardinals division-algebras
asked Dec 22 '17 at 12:51
Giuseppe Bargagnati
1,094514
I made some mistakes I've just edited; however I mean irreducible module, that is a module with no bilateral submodules except himself and $(0)$.
– Giuseppe Bargagnati
Dec 22 '17 at 14:35
It is much clearer now, thanks for the edits.
– rschwieb
Dec 22 '17 at 15:12
1
Without loss of generality you can assume $V$ is a faithful $A$ module and think about right primitive rings and apply the Jacobson density theorem.
– rschwieb
Dec 22 '17 at 15:26
2
What is “dimension” “as $A$-module”? Also, what is “countable” - infinite or not necessarily?
– Dap
Dec 29 '17 at 7:37
2
How is $End(V)$ an $A$-module?
– Dap
Dec 30 '17 at 14:09
|
show 7 more comments
Let $A$ be a $mathbb{C}$-algebra and $V$ be an irreducible $A$-module with countable dimension. What is the dimension of $End(V)$ as $A$-module? Note that every endomorphism must be injective and surjective, because of the irreducibility of $V$. My claim (or at least my hope) is that $End(V)$ has countable dimension, but I can't see a proof of this fact.
linear-algebra abstract-algebra cardinals division-algebras
asked Dec 22 '17 at 12:51
Giuseppe Bargagnati
1,094514
Let $A$ be a $mathbb{C}$-algebra and $V$ be an irreducible $A$-module with countable dimension. What is the dimension of $End(V)$ as $A$-module? Note that every endomorphism must be injective and surjective, because of the irreducibility of $V$. My claim (or at least my hope) is that $End(V)$ has countable dimension, but I can't see a proof of this fact.
linear-algebra abstract-algebra cardinals division-algebras
linear-algebra abstract-algebra cardinals division-algebras
asked Dec 22 '17 at 12:51
Giuseppe Bargagnati
1,094514
asked Dec 22 '17 at 12:51
Giuseppe Bargagnati
1,094514
edited Dec 22 '17 at 14:33
asked Dec 22 '17 at 12:51
Giuseppe Bargagnati
1,094514
asked Dec 22 '17 at 12:51
Giuseppe Bargagnati
1,094514
asked Dec 22 '17 at 12:51
Giuseppe Bargagnati
1,094514
1,094514
I made some mistakes I've just edited; however I mean irreducible module, that is a module with no bilateral submodules except himself and $(0)$.
– Giuseppe Bargagnati
Dec 22 '17 at 14:35
It is much clearer now, thanks for the edits.
– rschwieb
Dec 22 '17 at 15:12
1
Without loss of generality you can assume $V$ is a faithful $A$ module and think about right primitive rings and apply the Jacobson density theorem.
– rschwieb
Dec 22 '17 at 15:26
2
What is “dimension” “as $A$-module”? Also, what is “countable” - infinite or not necessarily?
– Dap
Dec 29 '17 at 7:37
2
How is $End(V)$ an $A$-module?
– Dap
Dec 30 '17 at 14:09
|
show 7 more comments
I made some mistakes I've just edited; however I mean irreducible module, that is a module with no bilateral submodules except himself and $(0)$.
– Giuseppe Bargagnati
Dec 22 '17 at 14:35
It is much clearer now, thanks for the edits.
– rschwieb
Dec 22 '17 at 15:12
1
Without loss of generality you can assume $V$ is a faithful $A$ module and think about right primitive rings and apply the Jacobson density theorem.
– rschwieb
Dec 22 '17 at 15:26
2
What is “dimension” “as $A$-module”? Also, what is “countable” - infinite or not necessarily?
– Dap
Dec 29 '17 at 7:37
2
How is $End(V)$ an $A$-module?
– Dap
Dec 30 '17 at 14:09
I made some mistakes I've just edited; however I mean irreducible module, that is a module with no bilateral submodules except himself and $(0)$.
– Giuseppe Bargagnati
Dec 22 '17 at 14:35
I made some mistakes I've just edited; however I mean irreducible module, that is a module with no bilateral submodules except himself and $(0)$.
– Giuseppe Bargagnati
Dec 22 '17 at 14:35
It is much clearer now, thanks for the edits.
– rschwieb
Dec 22 '17 at 15:12
It is much clearer now, thanks for the edits.
– rschwieb
Dec 22 '17 at 15:12
1
1
Without loss of generality you can assume $V$ is a faithful $A$ module and think about right primitive rings and apply the Jacobson density theorem.
– rschwieb
Dec 22 '17 at 15:26
Without loss of generality you can assume $V$ is a faithful $A$ module and think about right primitive rings and apply the Jacobson density theorem.
– rschwieb
Dec 22 '17 at 15:26
2
2
What is “dimension” “as $A$-module”? Also, what is “countable” - infinite or not necessarily?
– Dap
Dec 29 '17 at 7:37
What is “dimension” “as $A$-module”? Also, what is “countable” - infinite or not necessarily?
– Dap
Dec 29 '17 at 7:37
2
2
How is $End(V)$ an $A$-module?
– Dap
Dec 30 '17 at 14:09
How is $End(V)$ an $A$-module?
– Dap
Dec 30 '17 at 14:09
|
show 7 more comments
1 Answer
1
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oldest
votes
I have tried to interpret the question as asking about dimension over $mathbb{C}$ , as it is not clear what $A$-module structure $End_A(V)$ has.
If $V$ is a left $A$-module which is irreducible in the sense that it has no proper nontrivial left $A$-modules, then $dim_{mathbb{C}} End_A(V) leq dim_{mathbb C} V$. For if $v in V setminus {0}$, then $V = Av$, and thus the $mathbb{C}$-linear map
begin{align*} End_A(V) &to V \ varphi &mapsto varphi(v) end{align*}
is injective.
Schur's Lemma tells us that every nonzero endomorphism in $End_A(V)$ is in fact invertible, that is, that $End_A(V)$ is a division algebra over $mathbb{C}$ of countable dimension. While it is well known that the only finite-dimensional division algebra over $mathbb{C}$ is $mathbb{C}$ itself, the same result is true of division algebras $D$ of countable dimension, as can be seen by examining for $a in D setminus mathbb{C}$ the set ${z in mathbb{C}: a - z text{ is not invertible}}$.
answered Nov 20 at 5:40
Joshua Mundinger
2,4791026
add a comment |
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1 Answer
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1 Answer
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oldest
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I have tried to interpret the question as asking about dimension over $mathbb{C}$ , as it is not clear what $A$-module structure $End_A(V)$ has.
If $V$ is a left $A$-module which is irreducible in the sense that it has no proper nontrivial left $A$-modules, then $dim_{mathbb{C}} End_A(V) leq dim_{mathbb C} V$. For if $v in V setminus {0}$, then $V = Av$, and thus the $mathbb{C}$-linear map
begin{align*} End_A(V) &to V \ varphi &mapsto varphi(v) end{align*}
is injective.
Schur's Lemma tells us that every nonzero endomorphism in $End_A(V)$ is in fact invertible, that is, that $End_A(V)$ is a division algebra over $mathbb{C}$ of countable dimension. While it is well known that the only finite-dimensional division algebra over $mathbb{C}$ is $mathbb{C}$ itself, the same result is true of division algebras $D$ of countable dimension, as can be seen by examining for $a in D setminus mathbb{C}$ the set ${z in mathbb{C}: a - z text{ is not invertible}}$.
answered Nov 20 at 5:40
Joshua Mundinger
2,4791026
add a comment |
I have tried to interpret the question as asking about dimension over $mathbb{C}$ , as it is not clear what $A$-module structure $End_A(V)$ has.
If $V$ is a left $A$-module which is irreducible in the sense that it has no proper nontrivial left $A$-modules, then $dim_{mathbb{C}} End_A(V) leq dim_{mathbb C} V$. For if $v in V setminus {0}$, then $V = Av$, and thus the $mathbb{C}$-linear map
begin{align*} End_A(V) &to V \ varphi &mapsto varphi(v) end{align*}
is injective.
Schur's Lemma tells us that every nonzero endomorphism in $End_A(V)$ is in fact invertible, that is, that $End_A(V)$ is a division algebra over $mathbb{C}$ of countable dimension. While it is well known that the only finite-dimensional division algebra over $mathbb{C}$ is $mathbb{C}$ itself, the same result is true of division algebras $D$ of countable dimension, as can be seen by examining for $a in D setminus mathbb{C}$ the set ${z in mathbb{C}: a - z text{ is not invertible}}$.
answered Nov 20 at 5:40
Joshua Mundinger
2,4791026
add a comment |
I have tried to interpret the question as asking about dimension over $mathbb{C}$ , as it is not clear what $A$-module structure $End_A(V)$ has.
If $V$ is a left $A$-module which is irreducible in the sense that it has no proper nontrivial left $A$-modules, then $dim_{mathbb{C}} End_A(V) leq dim_{mathbb C} V$. For if $v in V setminus {0}$, then $V = Av$, and thus the $mathbb{C}$-linear map
begin{align*} End_A(V) &to V \ varphi &mapsto varphi(v) end{align*}
is injective.
Schur's Lemma tells us that every nonzero endomorphism in $End_A(V)$ is in fact invertible, that is, that $End_A(V)$ is a division algebra over $mathbb{C}$ of countable dimension. While it is well known that the only finite-dimensional division algebra over $mathbb{C}$ is $mathbb{C}$ itself, the same result is true of division algebras $D$ of countable dimension, as can be seen by examining for $a in D setminus mathbb{C}$ the set ${z in mathbb{C}: a - z text{ is not invertible}}$.
answered Nov 20 at 5:40
Joshua Mundinger
2,4791026
I have tried to interpret the question as asking about dimension over $mathbb{C}$ , as it is not clear what $A$-module structure $End_A(V)$ has.
If $V$ is a left $A$-module which is irreducible in the sense that it has no proper nontrivial left $A$-modules, then $dim_{mathbb{C}} End_A(V) leq dim_{mathbb C} V$. For if $v in V setminus {0}$, then $V = Av$, and thus the $mathbb{C}$-linear map
begin{align*} End_A(V) &to V \ varphi &mapsto varphi(v) end{align*}
is injective.
Schur's Lemma tells us that every nonzero endomorphism in $End_A(V)$ is in fact invertible, that is, that $End_A(V)$ is a division algebra over $mathbb{C}$ of countable dimension. While it is well known that the only finite-dimensional division algebra over $mathbb{C}$ is $mathbb{C}$ itself, the same result is true of division algebras $D$ of countable dimension, as can be seen by examining for $a in D setminus mathbb{C}$ the set ${z in mathbb{C}: a - z text{ is not invertible}}$.
answered Nov 20 at 5:40
Joshua Mundinger
2,4791026
answered Nov 20 at 5:40
Joshua Mundinger
2,4791026
answered Nov 20 at 5:40
Joshua Mundinger
2,4791026
answered Nov 20 at 5:40
Joshua Mundinger
2,4791026
2,4791026
add a comment |
add a comment |
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I made some mistakes I've just edited; however I mean irreducible module, that is a module with no bilateral submodules except himself and $(0)$.
– Giuseppe Bargagnati
Dec 22 '17 at 14:35
It is much clearer now, thanks for the edits.
– rschwieb
Dec 22 '17 at 15:12
1
Without loss of generality you can assume $V$ is a faithful $A$ module and think about right primitive rings and apply the Jacobson density theorem.
– rschwieb
Dec 22 '17 at 15:26
2
What is “dimension” “as $A$-module”? Also, what is “countable” - infinite or not necessarily?
– Dap
Dec 29 '17 at 7:37
2
How is $End(V)$ an $A$-module?
– Dap
Dec 30 '17 at 14:09