Show that this submodule is not completely reducible
$begingroup$
Let $M = begin{bmatrix} mathbb{C} \ mathbb{C} end{bmatrix}$.
Show that $M$ is not completely reducible as a left module over $R = begin{bmatrix} mathbb{C} & 0 \ mathbb{C} & mathbb{C} end{bmatrix}$.
I'm not sure how to tackle this? I can spot that $N = begin{bmatrix} 0 \ mathbb{C} end{bmatrix}$ is a submodule of $M$, but how do I show that it has no complement in $M$?
I thought about checking if $begin{bmatrix} mathbb{C} \ 0 end{bmatrix}$ is not a submodule of $M$, but this isn't sufficient as there are plenty of other submodules.
abstract-algebra modules noncommutative-algebra
asked Jan 20 at 19:12
the manthe man
773715
$endgroup$
add a comment |
$begingroup$
Let $M = begin{bmatrix} mathbb{C} \ mathbb{C} end{bmatrix}$.
Show that $M$ is not completely reducible as a left module over $R = begin{bmatrix} mathbb{C} & 0 \ mathbb{C} & mathbb{C} end{bmatrix}$.
I'm not sure how to tackle this? I can spot that $N = begin{bmatrix} 0 \ mathbb{C} end{bmatrix}$ is a submodule of $M$, but how do I show that it has no complement in $M$?
I thought about checking if $begin{bmatrix} mathbb{C} \ 0 end{bmatrix}$ is not a submodule of $M$, but this isn't sufficient as there are plenty of other submodules.
abstract-algebra modules noncommutative-algebra
asked Jan 20 at 19:12
the manthe man
773715
$endgroup$
add a comment |
$begingroup$
Let $M = begin{bmatrix} mathbb{C} \ mathbb{C} end{bmatrix}$.
Show that $M$ is not completely reducible as a left module over $R = begin{bmatrix} mathbb{C} & 0 \ mathbb{C} & mathbb{C} end{bmatrix}$.
I'm not sure how to tackle this? I can spot that $N = begin{bmatrix} 0 \ mathbb{C} end{bmatrix}$ is a submodule of $M$, but how do I show that it has no complement in $M$?
I thought about checking if $begin{bmatrix} mathbb{C} \ 0 end{bmatrix}$ is not a submodule of $M$, but this isn't sufficient as there are plenty of other submodules.
abstract-algebra modules noncommutative-algebra
asked Jan 20 at 19:12
the manthe man
773715
$endgroup$
Let $M = begin{bmatrix} mathbb{C} \ mathbb{C} end{bmatrix}$.
Show that $M$ is not completely reducible as a left module over $R = begin{bmatrix} mathbb{C} & 0 \ mathbb{C} & mathbb{C} end{bmatrix}$.
I'm not sure how to tackle this? I can spot that $N = begin{bmatrix} 0 \ mathbb{C} end{bmatrix}$ is a submodule of $M$, but how do I show that it has no complement in $M$?
I thought about checking if $begin{bmatrix} mathbb{C} \ 0 end{bmatrix}$ is not a submodule of $M$, but this isn't sufficient as there are plenty of other submodules.
abstract-algebra modules noncommutative-algebra
abstract-algebra modules noncommutative-algebra
asked Jan 20 at 19:12
the manthe man
773715
asked Jan 20 at 19:12
the manthe man
773715
asked Jan 20 at 19:12
the manthe man
773715
asked Jan 20 at 19:12
the manthe man
773715
asked Jan 20 at 19:12
the manthe man
773715
773715
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $K$ be a nonzero submodule of $M$ and suppose
$$
x=begin{bmatrix} p \ q end{bmatrix} in K
$$
with $xne0$. Then
$$
begin{bmatrix} 0 & 0 \ beta & gamma end{bmatrix}
begin{bmatrix} p \ q end{bmatrix}=
begin{bmatrix} 0 \ beta p+gamma q end{bmatrix}
$$
and we can obviously choose $beta$ and $gamma$ such that $beta p+gamma q$ is every complex number we fix. This shows $Nsubseteq K$.
answered Jan 20 at 21:38
egregegreg
182k1486204
$endgroup$
$begingroup$
So why does it follow that $N$ doesn’t have a complement in $M$?
$endgroup$
– the man
Jan 20 at 21:47
$begingroup$
@theman It follows that $N$ is the unique minimal submodule of $M$. Hence $M$ is not the sum of simple submodules. But the other way is good as well: a complement should not contain $N$, but it can't.
$endgroup$
– egreg
Jan 20 at 21:48
add a comment |
$begingroup$
The subspaces of $M$ are of the form$$left{begin{bmatrix}az\bzend{bmatrix},middle|,zinmathbb{C}right},$$for some complex numbers $a$ and $b$. It is easy to see that, unless $a=0$, none of them is a submodule.
answered Jan 20 at 19:25
José Carlos SantosJosé Carlos Santos
163k22130233
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $K$ be a nonzero submodule of $M$ and suppose
$$
x=begin{bmatrix} p \ q end{bmatrix} in K
$$
with $xne0$. Then
$$
begin{bmatrix} 0 & 0 \ beta & gamma end{bmatrix}
begin{bmatrix} p \ q end{bmatrix}=
begin{bmatrix} 0 \ beta p+gamma q end{bmatrix}
$$
and we can obviously choose $beta$ and $gamma$ such that $beta p+gamma q$ is every complex number we fix. This shows $Nsubseteq K$.
answered Jan 20 at 21:38
egregegreg
182k1486204
$endgroup$
$begingroup$
So why does it follow that $N$ doesn’t have a complement in $M$?
$endgroup$
– the man
Jan 20 at 21:47
$begingroup$
@theman It follows that $N$ is the unique minimal submodule of $M$. Hence $M$ is not the sum of simple submodules. But the other way is good as well: a complement should not contain $N$, but it can't.
$endgroup$
– egreg
Jan 20 at 21:48
add a comment |
$begingroup$
Let $K$ be a nonzero submodule of $M$ and suppose
$$
x=begin{bmatrix} p \ q end{bmatrix} in K
$$
with $xne0$. Then
$$
begin{bmatrix} 0 & 0 \ beta & gamma end{bmatrix}
begin{bmatrix} p \ q end{bmatrix}=
begin{bmatrix} 0 \ beta p+gamma q end{bmatrix}
$$
and we can obviously choose $beta$ and $gamma$ such that $beta p+gamma q$ is every complex number we fix. This shows $Nsubseteq K$.
answered Jan 20 at 21:38
egregegreg
182k1486204
$endgroup$
$begingroup$
So why does it follow that $N$ doesn’t have a complement in $M$?
$endgroup$
– the man
Jan 20 at 21:47
$begingroup$
@theman It follows that $N$ is the unique minimal submodule of $M$. Hence $M$ is not the sum of simple submodules. But the other way is good as well: a complement should not contain $N$, but it can't.
$endgroup$
– egreg
Jan 20 at 21:48
add a comment |
$begingroup$
Let $K$ be a nonzero submodule of $M$ and suppose
$$
x=begin{bmatrix} p \ q end{bmatrix} in K
$$
with $xne0$. Then
$$
begin{bmatrix} 0 & 0 \ beta & gamma end{bmatrix}
begin{bmatrix} p \ q end{bmatrix}=
begin{bmatrix} 0 \ beta p+gamma q end{bmatrix}
$$
and we can obviously choose $beta$ and $gamma$ such that $beta p+gamma q$ is every complex number we fix. This shows $Nsubseteq K$.
answered Jan 20 at 21:38
egregegreg
182k1486204
$endgroup$
Let $K$ be a nonzero submodule of $M$ and suppose
$$
x=begin{bmatrix} p \ q end{bmatrix} in K
$$
with $xne0$. Then
$$
begin{bmatrix} 0 & 0 \ beta & gamma end{bmatrix}
begin{bmatrix} p \ q end{bmatrix}=
begin{bmatrix} 0 \ beta p+gamma q end{bmatrix}
$$
and we can obviously choose $beta$ and $gamma$ such that $beta p+gamma q$ is every complex number we fix. This shows $Nsubseteq K$.
answered Jan 20 at 21:38
egregegreg
182k1486204
answered Jan 20 at 21:38
egregegreg
182k1486204
answered Jan 20 at 21:38
egregegreg
182k1486204
answered Jan 20 at 21:38
egregegreg
182k1486204
182k1486204
$begingroup$
So why does it follow that $N$ doesn’t have a complement in $M$?
$endgroup$
– the man
Jan 20 at 21:47
$begingroup$
@theman It follows that $N$ is the unique minimal submodule of $M$. Hence $M$ is not the sum of simple submodules. But the other way is good as well: a complement should not contain $N$, but it can't.
$endgroup$
– egreg
Jan 20 at 21:48
add a comment |
$begingroup$
So why does it follow that $N$ doesn’t have a complement in $M$?
$endgroup$
– the man
Jan 20 at 21:47
$begingroup$
@theman It follows that $N$ is the unique minimal submodule of $M$. Hence $M$ is not the sum of simple submodules. But the other way is good as well: a complement should not contain $N$, but it can't.
$endgroup$
– egreg
Jan 20 at 21:48
$begingroup$
So why does it follow that $N$ doesn’t have a complement in $M$?
$endgroup$
– the man
Jan 20 at 21:47
$begingroup$
So why does it follow that $N$ doesn’t have a complement in $M$?
$endgroup$
– the man
Jan 20 at 21:47
$begingroup$
@theman It follows that $N$ is the unique minimal submodule of $M$. Hence $M$ is not the sum of simple submodules. But the other way is good as well: a complement should not contain $N$, but it can't.
$endgroup$
– egreg
Jan 20 at 21:48
$begingroup$
@theman It follows that $N$ is the unique minimal submodule of $M$. Hence $M$ is not the sum of simple submodules. But the other way is good as well: a complement should not contain $N$, but it can't.
$endgroup$
– egreg
Jan 20 at 21:48
add a comment |
$begingroup$
The subspaces of $M$ are of the form$$left{begin{bmatrix}az\bzend{bmatrix},middle|,zinmathbb{C}right},$$for some complex numbers $a$ and $b$. It is easy to see that, unless $a=0$, none of them is a submodule.
answered Jan 20 at 19:25
José Carlos SantosJosé Carlos Santos
163k22130233
$endgroup$
add a comment |
$begingroup$
The subspaces of $M$ are of the form$$left{begin{bmatrix}az\bzend{bmatrix},middle|,zinmathbb{C}right},$$for some complex numbers $a$ and $b$. It is easy to see that, unless $a=0$, none of them is a submodule.
answered Jan 20 at 19:25
José Carlos SantosJosé Carlos Santos
163k22130233
$endgroup$
add a comment |
$begingroup$
The subspaces of $M$ are of the form$$left{begin{bmatrix}az\bzend{bmatrix},middle|,zinmathbb{C}right},$$for some complex numbers $a$ and $b$. It is easy to see that, unless $a=0$, none of them is a submodule.
answered Jan 20 at 19:25
José Carlos SantosJosé Carlos Santos
163k22130233
$endgroup$
The subspaces of $M$ are of the form$$left{begin{bmatrix}az\bzend{bmatrix},middle|,zinmathbb{C}right},$$for some complex numbers $a$ and $b$. It is easy to see that, unless $a=0$, none of them is a submodule.
answered Jan 20 at 19:25
José Carlos SantosJosé Carlos Santos
163k22130233
answered Jan 20 at 19:25
José Carlos SantosJosé Carlos Santos
163k22130233
answered Jan 20 at 19:25
José Carlos SantosJosé Carlos Santos
163k22130233
answered Jan 20 at 19:25
José Carlos SantosJosé Carlos Santos
163k22130233
163k22130233
add a comment |
add a comment |
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