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Is $operatornameHom(E,M)$ simple as an $operatornameEnd(M)$-module if $E$ is simple?
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Let $R$ be a ring, let $M$ be an $R$-module and let $R' := operatornameEnd_R(M)$.
Let $E$ be a simple $R$-module with $operatornameHom_R(E,M) neq 0$.
Question:
Is $operatornameHom_R(E,M)$ simple as an $R'$-module?
Background:
While trying to understand (a version of) the double centralizer theorem I learned that the $E$-isotypical component of $M$ can be described as $M_E cong operatornameHom_R(E,M) otimes_operatornameEnd_R(E) E$ as an $(R' otimes_mathbbZ R)$-module.
Because $M_E$ is simple as an $(R' otimes_mathbbZ R)$-module I am wondering about the $R'$-simplicity of the tensor factor $operatornameHom_R(E,M)$.
I think the statement holds if $M$ is semisimple:
We can then assume that $M = M_E cong E^oplus I$ for some index set $I$ (because every $R$-endomorphism of $M_E$ extends to an $R$-endomorphism of $M$).
For the skew field $D := operatornameEnd_R(E)$ we then have that $R' cong operatornameM_I(D)$ (the algebra of column-finite $(I times I)$-matrices) and $operatornameHom_R(E,M) cong D^oplus I$, with $operatornameM_I(D)$ acting transitively on $D^oplus I smallsetminus 0$.
I assume that the statement does not hold for general $M$, but I haven’t found a counterexample.
Any help is welcome.
modules representation-theory noncommutative-algebra
asked Jul 22 at 21:40
Jendrik Stelzner
7,52211037
add a comment |Â
up vote
2
down vote
favorite
Let $R$ be a ring, let $M$ be an $R$-module and let $R' := operatornameEnd_R(M)$.
Let $E$ be a simple $R$-module with $operatornameHom_R(E,M) neq 0$.
Question:
Is $operatornameHom_R(E,M)$ simple as an $R'$-module?
Background:
While trying to understand (a version of) the double centralizer theorem I learned that the $E$-isotypical component of $M$ can be described as $M_E cong operatornameHom_R(E,M) otimes_operatornameEnd_R(E) E$ as an $(R' otimes_mathbbZ R)$-module.
Because $M_E$ is simple as an $(R' otimes_mathbbZ R)$-module I am wondering about the $R'$-simplicity of the tensor factor $operatornameHom_R(E,M)$.
I think the statement holds if $M$ is semisimple:
We can then assume that $M = M_E cong E^oplus I$ for some index set $I$ (because every $R$-endomorphism of $M_E$ extends to an $R$-endomorphism of $M$).
For the skew field $D := operatornameEnd_R(E)$ we then have that $R' cong operatornameM_I(D)$ (the algebra of column-finite $(I times I)$-matrices) and $operatornameHom_R(E,M) cong D^oplus I$, with $operatornameM_I(D)$ acting transitively on $D^oplus I smallsetminus 0$.
I assume that the statement does not hold for general $M$, but I haven’t found a counterexample.
Any help is welcome.
modules representation-theory noncommutative-algebra
asked Jul 22 at 21:40
Jendrik Stelzner
7,52211037
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $R$ be a ring, let $M$ be an $R$-module and let $R' := operatornameEnd_R(M)$.
Let $E$ be a simple $R$-module with $operatornameHom_R(E,M) neq 0$.
Question:
Is $operatornameHom_R(E,M)$ simple as an $R'$-module?
Background:
While trying to understand (a version of) the double centralizer theorem I learned that the $E$-isotypical component of $M$ can be described as $M_E cong operatornameHom_R(E,M) otimes_operatornameEnd_R(E) E$ as an $(R' otimes_mathbbZ R)$-module.
Because $M_E$ is simple as an $(R' otimes_mathbbZ R)$-module I am wondering about the $R'$-simplicity of the tensor factor $operatornameHom_R(E,M)$.
I think the statement holds if $M$ is semisimple:
We can then assume that $M = M_E cong E^oplus I$ for some index set $I$ (because every $R$-endomorphism of $M_E$ extends to an $R$-endomorphism of $M$).
For the skew field $D := operatornameEnd_R(E)$ we then have that $R' cong operatornameM_I(D)$ (the algebra of column-finite $(I times I)$-matrices) and $operatornameHom_R(E,M) cong D^oplus I$, with $operatornameM_I(D)$ acting transitively on $D^oplus I smallsetminus 0$.
I assume that the statement does not hold for general $M$, but I haven’t found a counterexample.
Any help is welcome.
modules representation-theory noncommutative-algebra
asked Jul 22 at 21:40
Jendrik Stelzner
7,52211037
Let $R$ be a ring, let $M$ be an $R$-module and let $R' := operatornameEnd_R(M)$.
Let $E$ be a simple $R$-module with $operatornameHom_R(E,M) neq 0$.
Question:
Is $operatornameHom_R(E,M)$ simple as an $R'$-module?
Background:
While trying to understand (a version of) the double centralizer theorem I learned that the $E$-isotypical component of $M$ can be described as $M_E cong operatornameHom_R(E,M) otimes_operatornameEnd_R(E) E$ as an $(R' otimes_mathbbZ R)$-module.
Because $M_E$ is simple as an $(R' otimes_mathbbZ R)$-module I am wondering about the $R'$-simplicity of the tensor factor $operatornameHom_R(E,M)$.
I think the statement holds if $M$ is semisimple:
We can then assume that $M = M_E cong E^oplus I$ for some index set $I$ (because every $R$-endomorphism of $M_E$ extends to an $R$-endomorphism of $M$).
For the skew field $D := operatornameEnd_R(E)$ we then have that $R' cong operatornameM_I(D)$ (the algebra of column-finite $(I times I)$-matrices) and $operatornameHom_R(E,M) cong D^oplus I$, with $operatornameM_I(D)$ acting transitively on $D^oplus I smallsetminus 0$.
I assume that the statement does not hold for general $M$, but I haven’t found a counterexample.
Any help is welcome.
modules representation-theory noncommutative-algebra
asked Jul 22 at 21:40
Jendrik Stelzner
7,52211037
asked Jul 22 at 21:40
Jendrik Stelzner
7,52211037
asked Jul 22 at 21:40
Jendrik Stelzner
7,52211037
asked Jul 22 at 21:40
Jendrik Stelzner
7,52211037
7,52211037
add a comment |Â
add a comment |Â
1 Answer
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Let $k$ be a field, let $R=k[x,y]$, and let $I$ be the maximal ideal $(x,y)subset R$. Let $M=R/I^2$, and let $E=R/I$. Then $R'$ is just the quotient ring $R/I^2$ and $operatornameHom_R(E,M)$ is naturally identified with $I/I^2$. This is not a simple $R'$-module, since $x$ and $y$ act trivially on it so that submodules are the same thing as $k$-subspaces, and $dim_k I/I^2=2$ ($x$ and $y$ are a basis).
(The key point is that the submodule $N$ of $M$ spanned by copies of $E$ is a direct sum of copies of $E$, but not every $operatornameEnd_R(E)$-endomorphism of it extends to an endomorphism of $M$. In this case, $Ncong E^2$ and $operatornameEnd_R(E)cong k$, but $R'$ only contains the diagonal action of $k$ on $E^2$, not the full $M_2(k)$ action, since in $M$ the two copies of $E$ in $N$ are actually both multiples of a single generator over $R$.)
answered Jul 22 at 22:05
Eric Wofsey
162k12189300
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let $k$ be a field, let $R=k[x,y]$, and let $I$ be the maximal ideal $(x,y)subset R$. Let $M=R/I^2$, and let $E=R/I$. Then $R'$ is just the quotient ring $R/I^2$ and $operatornameHom_R(E,M)$ is naturally identified with $I/I^2$. This is not a simple $R'$-module, since $x$ and $y$ act trivially on it so that submodules are the same thing as $k$-subspaces, and $dim_k I/I^2=2$ ($x$ and $y$ are a basis).
(The key point is that the submodule $N$ of $M$ spanned by copies of $E$ is a direct sum of copies of $E$, but not every $operatornameEnd_R(E)$-endomorphism of it extends to an endomorphism of $M$. In this case, $Ncong E^2$ and $operatornameEnd_R(E)cong k$, but $R'$ only contains the diagonal action of $k$ on $E^2$, not the full $M_2(k)$ action, since in $M$ the two copies of $E$ in $N$ are actually both multiples of a single generator over $R$.)
answered Jul 22 at 22:05
Eric Wofsey
162k12189300
add a comment |Â
up vote
2
down vote
accepted
Let $k$ be a field, let $R=k[x,y]$, and let $I$ be the maximal ideal $(x,y)subset R$. Let $M=R/I^2$, and let $E=R/I$. Then $R'$ is just the quotient ring $R/I^2$ and $operatornameHom_R(E,M)$ is naturally identified with $I/I^2$. This is not a simple $R'$-module, since $x$ and $y$ act trivially on it so that submodules are the same thing as $k$-subspaces, and $dim_k I/I^2=2$ ($x$ and $y$ are a basis).
(The key point is that the submodule $N$ of $M$ spanned by copies of $E$ is a direct sum of copies of $E$, but not every $operatornameEnd_R(E)$-endomorphism of it extends to an endomorphism of $M$. In this case, $Ncong E^2$ and $operatornameEnd_R(E)cong k$, but $R'$ only contains the diagonal action of $k$ on $E^2$, not the full $M_2(k)$ action, since in $M$ the two copies of $E$ in $N$ are actually both multiples of a single generator over $R$.)
answered Jul 22 at 22:05
Eric Wofsey
162k12189300
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let $k$ be a field, let $R=k[x,y]$, and let $I$ be the maximal ideal $(x,y)subset R$. Let $M=R/I^2$, and let $E=R/I$. Then $R'$ is just the quotient ring $R/I^2$ and $operatornameHom_R(E,M)$ is naturally identified with $I/I^2$. This is not a simple $R'$-module, since $x$ and $y$ act trivially on it so that submodules are the same thing as $k$-subspaces, and $dim_k I/I^2=2$ ($x$ and $y$ are a basis).
(The key point is that the submodule $N$ of $M$ spanned by copies of $E$ is a direct sum of copies of $E$, but not every $operatornameEnd_R(E)$-endomorphism of it extends to an endomorphism of $M$. In this case, $Ncong E^2$ and $operatornameEnd_R(E)cong k$, but $R'$ only contains the diagonal action of $k$ on $E^2$, not the full $M_2(k)$ action, since in $M$ the two copies of $E$ in $N$ are actually both multiples of a single generator over $R$.)
answered Jul 22 at 22:05
Eric Wofsey
162k12189300
Let $k$ be a field, let $R=k[x,y]$, and let $I$ be the maximal ideal $(x,y)subset R$. Let $M=R/I^2$, and let $E=R/I$. Then $R'$ is just the quotient ring $R/I^2$ and $operatornameHom_R(E,M)$ is naturally identified with $I/I^2$. This is not a simple $R'$-module, since $x$ and $y$ act trivially on it so that submodules are the same thing as $k$-subspaces, and $dim_k I/I^2=2$ ($x$ and $y$ are a basis).
(The key point is that the submodule $N$ of $M$ spanned by copies of $E$ is a direct sum of copies of $E$, but not every $operatornameEnd_R(E)$-endomorphism of it extends to an endomorphism of $M$. In this case, $Ncong E^2$ and $operatornameEnd_R(E)cong k$, but $R'$ only contains the diagonal action of $k$ on $E^2$, not the full $M_2(k)$ action, since in $M$ the two copies of $E$ in $N$ are actually both multiples of a single generator over $R$.)
answered Jul 22 at 22:05
Eric Wofsey
162k12189300
edited Jul 22 at 22:14
answered Jul 22 at 22:05
Eric Wofsey
162k12189300
answered Jul 22 at 22:05
Eric Wofsey
162k12189300
answered Jul 22 at 22:05
Eric Wofsey
162k12189300
162k12189300
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