Dimension of $End(V)$ with $V$ countable dimension irreducible module over a complex algebra












5














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







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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
















5














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







share|cite|improve this question
























  • 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














5












5








5


2





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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 22 '17 at 14:33

























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


















  • 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










1 Answer
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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 ${z in mathbb{C}: a - z text{ is not invertible}}$.






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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 ${z in mathbb{C}: a - z text{ is not invertible}}$.






    share|cite|improve this answer


























      2














      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 ${z in mathbb{C}: a - z text{ is not invertible}}$.






      share|cite|improve this answer
























        2












        2








        2






        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 ${z in mathbb{C}: a - z text{ is not invertible}}$.






        share|cite|improve this answer












        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 ${z in mathbb{C}: a - z text{ is not invertible}}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 20 at 5:40









        Joshua Mundinger

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        2,4791026






























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