Mixing
Finite inner direct sums of simple modules
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Within a proof given in my abstract algebra lecture, we conclude that for any Ring $R$ and any $R$-module $M$ any finite sum of simple submodules $(M_i)_ile n$ is an inner direct sum already, if any two of those submodules intersect trivially, i.e. $forall ineq j:;M_icap M_j=0implies sum^n_i=1M_i=bigoplus_i=1^n M_i$.
I am very unsure about this 'fact' since I cannot prove it. Is this even true? Can anyone outline a proof if it is?
EDIT: My main concern is that for $n>2$ I cannot prove that $M_icapsum^n_oversetj=1ineq jM_j=0$ for every $i$. As far as I know that is the required condition for a sum to be direct.
abstract-algebra ring-theory
asked Jul 24 at 19:02
Sellerie
738
add a comment |Â
up vote
0
down vote
favorite
Within a proof given in my abstract algebra lecture, we conclude that for any Ring $R$ and any $R$-module $M$ any finite sum of simple submodules $(M_i)_ile n$ is an inner direct sum already, if any two of those submodules intersect trivially, i.e. $forall ineq j:;M_icap M_j=0implies sum^n_i=1M_i=bigoplus_i=1^n M_i$.
I am very unsure about this 'fact' since I cannot prove it. Is this even true? Can anyone outline a proof if it is?
EDIT: My main concern is that for $n>2$ I cannot prove that $M_icapsum^n_oversetj=1ineq jM_j=0$ for every $i$. As far as I know that is the required condition for a sum to be direct.
abstract-algebra ring-theory
asked Jul 24 at 19:02
Sellerie
738
2
Is every sum of one-dimensional subspaces of a vector space direct?
– Lord Shark the Unknown
Jul 24 at 19:05
3
It depends on what you mean by "unique." Do you mean non-isomorphic?
– Qiaochu Yuan
Jul 24 at 19:05
Uniqueness in this case is meant as any two of them intersect trivially. What I cannot prove is that, given this intersection property, for any one simple module the intersection of the module with the sum of the remaining (finitely many) modules is also trivial. If $n=2$ this becomes obvious, but I cannot prove it for bigger $n$. Non-isomorphic is definitely not given, since I am looking at isomorphic modules within the proof I'm trying to understand.
– Sellerie
Jul 24 at 19:14
1
In that case the statement is not true, as the vector space example shows.
– Tobias Kildetoft
Jul 24 at 19:22
Sorry for not acknowledging that counterexample sooner, I was stupid enough to only consider the vector space example with 2 subspaces, which is obviously no counter example. Thinking about this further has shown my stupidity, sorry about that. Thanks to everyone, even though this doesn't help me understanding the given proof (which might not even be correct).
– Sellerie
Jul 24 at 19:27
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Within a proof given in my abstract algebra lecture, we conclude that for any Ring $R$ and any $R$-module $M$ any finite sum of simple submodules $(M_i)_ile n$ is an inner direct sum already, if any two of those submodules intersect trivially, i.e. $forall ineq j:;M_icap M_j=0implies sum^n_i=1M_i=bigoplus_i=1^n M_i$.
I am very unsure about this 'fact' since I cannot prove it. Is this even true? Can anyone outline a proof if it is?
EDIT: My main concern is that for $n>2$ I cannot prove that $M_icapsum^n_oversetj=1ineq jM_j=0$ for every $i$. As far as I know that is the required condition for a sum to be direct.
abstract-algebra ring-theory
asked Jul 24 at 19:02
Sellerie
738
Within a proof given in my abstract algebra lecture, we conclude that for any Ring $R$ and any $R$-module $M$ any finite sum of simple submodules $(M_i)_ile n$ is an inner direct sum already, if any two of those submodules intersect trivially, i.e. $forall ineq j:;M_icap M_j=0implies sum^n_i=1M_i=bigoplus_i=1^n M_i$.
I am very unsure about this 'fact' since I cannot prove it. Is this even true? Can anyone outline a proof if it is?
EDIT: My main concern is that for $n>2$ I cannot prove that $M_icapsum^n_oversetj=1ineq jM_j=0$ for every $i$. As far as I know that is the required condition for a sum to be direct.
abstract-algebra ring-theory
asked Jul 24 at 19:02
Sellerie
738
edited Jul 24 at 19:23
asked Jul 24 at 19:02
Sellerie
738
asked Jul 24 at 19:02
Sellerie
738
asked Jul 24 at 19:02
Sellerie
738
738
2
Is every sum of one-dimensional subspaces of a vector space direct?
– Lord Shark the Unknown
Jul 24 at 19:05
3
It depends on what you mean by "unique." Do you mean non-isomorphic?
– Qiaochu Yuan
Jul 24 at 19:05
Uniqueness in this case is meant as any two of them intersect trivially. What I cannot prove is that, given this intersection property, for any one simple module the intersection of the module with the sum of the remaining (finitely many) modules is also trivial. If $n=2$ this becomes obvious, but I cannot prove it for bigger $n$. Non-isomorphic is definitely not given, since I am looking at isomorphic modules within the proof I'm trying to understand.
– Sellerie
Jul 24 at 19:14
1
In that case the statement is not true, as the vector space example shows.
– Tobias Kildetoft
Jul 24 at 19:22
Sorry for not acknowledging that counterexample sooner, I was stupid enough to only consider the vector space example with 2 subspaces, which is obviously no counter example. Thinking about this further has shown my stupidity, sorry about that. Thanks to everyone, even though this doesn't help me understanding the given proof (which might not even be correct).
– Sellerie
Jul 24 at 19:27
add a comment |Â
2
Is every sum of one-dimensional subspaces of a vector space direct?
– Lord Shark the Unknown
Jul 24 at 19:05
3
It depends on what you mean by "unique." Do you mean non-isomorphic?
– Qiaochu Yuan
Jul 24 at 19:05
Uniqueness in this case is meant as any two of them intersect trivially. What I cannot prove is that, given this intersection property, for any one simple module the intersection of the module with the sum of the remaining (finitely many) modules is also trivial. If $n=2$ this becomes obvious, but I cannot prove it for bigger $n$. Non-isomorphic is definitely not given, since I am looking at isomorphic modules within the proof I'm trying to understand.
– Sellerie
Jul 24 at 19:14
1
In that case the statement is not true, as the vector space example shows.
– Tobias Kildetoft
Jul 24 at 19:22
Sorry for not acknowledging that counterexample sooner, I was stupid enough to only consider the vector space example with 2 subspaces, which is obviously no counter example. Thinking about this further has shown my stupidity, sorry about that. Thanks to everyone, even though this doesn't help me understanding the given proof (which might not even be correct).
– Sellerie
Jul 24 at 19:27
2
2
Is every sum of one-dimensional subspaces of a vector space direct?
– Lord Shark the Unknown
Jul 24 at 19:05
Is every sum of one-dimensional subspaces of a vector space direct?
– Lord Shark the Unknown
Jul 24 at 19:05
3
3
It depends on what you mean by "unique." Do you mean non-isomorphic?
– Qiaochu Yuan
Jul 24 at 19:05
It depends on what you mean by "unique." Do you mean non-isomorphic?
– Qiaochu Yuan
Jul 24 at 19:05
Uniqueness in this case is meant as any two of them intersect trivially. What I cannot prove is that, given this intersection property, for any one simple module the intersection of the module with the sum of the remaining (finitely many) modules is also trivial. If $n=2$ this becomes obvious, but I cannot prove it for bigger $n$. Non-isomorphic is definitely not given, since I am looking at isomorphic modules within the proof I'm trying to understand.
– Sellerie
Jul 24 at 19:14
Uniqueness in this case is meant as any two of them intersect trivially. What I cannot prove is that, given this intersection property, for any one simple module the intersection of the module with the sum of the remaining (finitely many) modules is also trivial. If $n=2$ this becomes obvious, but I cannot prove it for bigger $n$. Non-isomorphic is definitely not given, since I am looking at isomorphic modules within the proof I'm trying to understand.
– Sellerie
Jul 24 at 19:14
1
1
In that case the statement is not true, as the vector space example shows.
– Tobias Kildetoft
Jul 24 at 19:22
In that case the statement is not true, as the vector space example shows.
– Tobias Kildetoft
Jul 24 at 19:22
Sorry for not acknowledging that counterexample sooner, I was stupid enough to only consider the vector space example with 2 subspaces, which is obviously no counter example. Thinking about this further has shown my stupidity, sorry about that. Thanks to everyone, even though this doesn't help me understanding the given proof (which might not even be correct).
– Sellerie
Jul 24 at 19:27
Sorry for not acknowledging that counterexample sooner, I was stupid enough to only consider the vector space example with 2 subspaces, which is obviously no counter example. Thinking about this further has shown my stupidity, sorry about that. Thanks to everyone, even though this doesn't help me understanding the given proof (which might not even be correct).
– Sellerie
Jul 24 at 19:27
add a comment |Â
1 Answer
1
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up vote
1
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We can do this inductively in the case where they are pairwise non-isomorphic.
Assume that any family of $n -1$ pairwise non-isomorphic simple sub-modules of $M$ form a direct sum.
Then we may rewrite $M_icapsum^n_oversetj=1ineq jM_j= M_icapbigoplus^n_oversetj = 1jneq i M_j$
Since
$M_i$ is simple it would mean that either $ M_icapbigoplus^n_oversetj = 1jneq i M_j = 0$
or
$M_icapbigoplus^n_oversetj = 1jneq i M_j= M_i$.
Now $M_icapbigoplus^n_oversetj = 1jneq i M_j= M_i$
implies that
$M_i$ has an isomorphic copy in $bigoplus^n_oversetj = 1jneq i M_j$, which is impossible by the uniqueness theorem for semisimple modules.
answered Jul 25 at 4:29
Tam Nguyen
112
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
We can do this inductively in the case where they are pairwise non-isomorphic.
Assume that any family of $n -1$ pairwise non-isomorphic simple sub-modules of $M$ form a direct sum.
Then we may rewrite $M_icapsum^n_oversetj=1ineq jM_j= M_icapbigoplus^n_oversetj = 1jneq i M_j$
Since
$M_i$ is simple it would mean that either $ M_icapbigoplus^n_oversetj = 1jneq i M_j = 0$
or
$M_icapbigoplus^n_oversetj = 1jneq i M_j= M_i$.
Now $M_icapbigoplus^n_oversetj = 1jneq i M_j= M_i$
implies that
$M_i$ has an isomorphic copy in $bigoplus^n_oversetj = 1jneq i M_j$, which is impossible by the uniqueness theorem for semisimple modules.
answered Jul 25 at 4:29
Tam Nguyen
112
add a comment |Â
up vote
1
down vote
We can do this inductively in the case where they are pairwise non-isomorphic.
Assume that any family of $n -1$ pairwise non-isomorphic simple sub-modules of $M$ form a direct sum.
Then we may rewrite $M_icapsum^n_oversetj=1ineq jM_j= M_icapbigoplus^n_oversetj = 1jneq i M_j$
Since
$M_i$ is simple it would mean that either $ M_icapbigoplus^n_oversetj = 1jneq i M_j = 0$
or
$M_icapbigoplus^n_oversetj = 1jneq i M_j= M_i$.
Now $M_icapbigoplus^n_oversetj = 1jneq i M_j= M_i$
implies that
$M_i$ has an isomorphic copy in $bigoplus^n_oversetj = 1jneq i M_j$, which is impossible by the uniqueness theorem for semisimple modules.
answered Jul 25 at 4:29
Tam Nguyen
112
add a comment |Â
up vote
1
down vote
up vote
1
down vote
We can do this inductively in the case where they are pairwise non-isomorphic.
Assume that any family of $n -1$ pairwise non-isomorphic simple sub-modules of $M$ form a direct sum.
Then we may rewrite $M_icapsum^n_oversetj=1ineq jM_j= M_icapbigoplus^n_oversetj = 1jneq i M_j$
Since
$M_i$ is simple it would mean that either $ M_icapbigoplus^n_oversetj = 1jneq i M_j = 0$
or
$M_icapbigoplus^n_oversetj = 1jneq i M_j= M_i$.
Now $M_icapbigoplus^n_oversetj = 1jneq i M_j= M_i$
implies that
$M_i$ has an isomorphic copy in $bigoplus^n_oversetj = 1jneq i M_j$, which is impossible by the uniqueness theorem for semisimple modules.
answered Jul 25 at 4:29
Tam Nguyen
112
We can do this inductively in the case where they are pairwise non-isomorphic.
Assume that any family of $n -1$ pairwise non-isomorphic simple sub-modules of $M$ form a direct sum.
Then we may rewrite $M_icapsum^n_oversetj=1ineq jM_j= M_icapbigoplus^n_oversetj = 1jneq i M_j$
Since
$M_i$ is simple it would mean that either $ M_icapbigoplus^n_oversetj = 1jneq i M_j = 0$
or
$M_icapbigoplus^n_oversetj = 1jneq i M_j= M_i$.
Now $M_icapbigoplus^n_oversetj = 1jneq i M_j= M_i$
implies that
$M_i$ has an isomorphic copy in $bigoplus^n_oversetj = 1jneq i M_j$, which is impossible by the uniqueness theorem for semisimple modules.
answered Jul 25 at 4:29
Tam Nguyen
112
answered Jul 25 at 4:29
Tam Nguyen
112
answered Jul 25 at 4:29
Tam Nguyen
112
answered Jul 25 at 4:29
Tam Nguyen
112
112
add a comment |Â
add a comment |Â
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2
Is every sum of one-dimensional subspaces of a vector space direct?
– Lord Shark the Unknown
Jul 24 at 19:05
3
It depends on what you mean by "unique." Do you mean non-isomorphic?
– Qiaochu Yuan
Jul 24 at 19:05
Uniqueness in this case is meant as any two of them intersect trivially. What I cannot prove is that, given this intersection property, for any one simple module the intersection of the module with the sum of the remaining (finitely many) modules is also trivial. If $n=2$ this becomes obvious, but I cannot prove it for bigger $n$. Non-isomorphic is definitely not given, since I am looking at isomorphic modules within the proof I'm trying to understand.
– Sellerie
Jul 24 at 19:14
1
In that case the statement is not true, as the vector space example shows.
– Tobias Kildetoft
Jul 24 at 19:22
Sorry for not acknowledging that counterexample sooner, I was stupid enough to only consider the vector space example with 2 subspaces, which is obviously no counter example. Thinking about this further has shown my stupidity, sorry about that. Thanks to everyone, even though this doesn't help me understanding the given proof (which might not even be correct).
– Sellerie
Jul 24 at 19:27