infinite upper triangular modules has no simple submodules
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Let $k$ be a field. Let $R$ be the set of infinite upper triangular matrices (columns and row are indexed by $mathbb{Z}$) with only finitely many nonzero entries in any row or column. Then let $V$ be the vector space of infinite column vectors (indexed by $mathbb{Z}$) with finitely many nonzero entries. Then we can view $V$ as an $R$-module with action given by matrix multiplication.
Then I would like to show that $V$ has no simple submodules.
However, to me, this is not intuitively correct. For example, let $W$ consists the vectors with some entries at the first row and zeros elsewhere. Then $W$ has only finitely many nonzero entries (1 or 0 entry) and it is closed under addition. Also for any $Ain R$, $AWsubseteq W$. So it appears to me that $W$ is a submodule, and it is one-dimensional. So it must be simple.
What the role of infinity here?
Because if we consider the finite case (finite upper triangular matrix $R$, finite column vectors space $V$, the submodule $W$ I mentioned before is the only simple submodule of $V$. right?) Why this is not true in infinite case? What I missed?
Thanks!
abstract-algebra modules
asked Nov 13 at 16:17
Abigail
527
|
show 5 more comments
up vote
2
down vote
favorite
Let $k$ be a field. Let $R$ be the set of infinite upper triangular matrices (columns and row are indexed by $mathbb{Z}$) with only finitely many nonzero entries in any row or column. Then let $V$ be the vector space of infinite column vectors (indexed by $mathbb{Z}$) with finitely many nonzero entries. Then we can view $V$ as an $R$-module with action given by matrix multiplication.
Then I would like to show that $V$ has no simple submodules.
However, to me, this is not intuitively correct. For example, let $W$ consists the vectors with some entries at the first row and zeros elsewhere. Then $W$ has only finitely many nonzero entries (1 or 0 entry) and it is closed under addition. Also for any $Ain R$, $AWsubseteq W$. So it appears to me that $W$ is a submodule, and it is one-dimensional. So it must be simple.
What the role of infinity here?
Because if we consider the finite case (finite upper triangular matrix $R$, finite column vectors space $V$, the submodule $W$ I mentioned before is the only simple submodule of $V$. right?) Why this is not true in infinite case? What I missed?
Thanks!
abstract-algebra modules
asked Nov 13 at 16:17
Abigail
527
Could it be that you want the rows and columns of your matrices indexed by $mathbb{Z}$ (that is, bidirectionally infinite matrices)?
– darij grinberg
Nov 14 at 0:05
I just checked. You are right, the rows and columns are indexed by $mathbb{Z}$. But how this helps? Thanks.
– Abigail
Nov 14 at 0:50
1
I don't know how it helps, but it explains why @rschwieb is getting a different answer.
– darij grinberg
Nov 14 at 0:50
2
For each $g in mathbb{Z}$, let $V_g$ be the subspace of $V$ consisting of all vectors $left(ldots, v_{-1}, v_0, v_1, ldotsright)$ such that all $i geq g$ satisfy $v_i = 0$. The trick is to prove that each nontrivial proper $A$-submodule of $V$ is $V_g$ for some $g in mathbb{Z}$. It is clear that none of the $V_g$'s is simple.
– darij grinberg
Nov 14 at 1:07
Ah! I think I know what I am messing up. I was assuming that we start from row 1 and goes to infinity (I was assuming col and rows are indexed by natural number!). But we are indexed by $mathbb{Z}$, so it is not reasonable to start from 1. Thus the submodule I mentioned is not a valid choice! Thanks!!!
– Abigail
Nov 14 at 1:25
|
show 5 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $k$ be a field. Let $R$ be the set of infinite upper triangular matrices (columns and row are indexed by $mathbb{Z}$) with only finitely many nonzero entries in any row or column. Then let $V$ be the vector space of infinite column vectors (indexed by $mathbb{Z}$) with finitely many nonzero entries. Then we can view $V$ as an $R$-module with action given by matrix multiplication.
Then I would like to show that $V$ has no simple submodules.
However, to me, this is not intuitively correct. For example, let $W$ consists the vectors with some entries at the first row and zeros elsewhere. Then $W$ has only finitely many nonzero entries (1 or 0 entry) and it is closed under addition. Also for any $Ain R$, $AWsubseteq W$. So it appears to me that $W$ is a submodule, and it is one-dimensional. So it must be simple.
What the role of infinity here?
Because if we consider the finite case (finite upper triangular matrix $R$, finite column vectors space $V$, the submodule $W$ I mentioned before is the only simple submodule of $V$. right?) Why this is not true in infinite case? What I missed?
Thanks!
abstract-algebra modules
asked Nov 13 at 16:17
Abigail
527
Let $k$ be a field. Let $R$ be the set of infinite upper triangular matrices (columns and row are indexed by $mathbb{Z}$) with only finitely many nonzero entries in any row or column. Then let $V$ be the vector space of infinite column vectors (indexed by $mathbb{Z}$) with finitely many nonzero entries. Then we can view $V$ as an $R$-module with action given by matrix multiplication.
Then I would like to show that $V$ has no simple submodules.
However, to me, this is not intuitively correct. For example, let $W$ consists the vectors with some entries at the first row and zeros elsewhere. Then $W$ has only finitely many nonzero entries (1 or 0 entry) and it is closed under addition. Also for any $Ain R$, $AWsubseteq W$. So it appears to me that $W$ is a submodule, and it is one-dimensional. So it must be simple.
What the role of infinity here?
Because if we consider the finite case (finite upper triangular matrix $R$, finite column vectors space $V$, the submodule $W$ I mentioned before is the only simple submodule of $V$. right?) Why this is not true in infinite case? What I missed?
Thanks!
abstract-algebra modules
abstract-algebra modules
asked Nov 13 at 16:17
Abigail
527
asked Nov 13 at 16:17
Abigail
527
edited Nov 14 at 0:51
asked Nov 13 at 16:17
Abigail
527
asked Nov 13 at 16:17
Abigail
527
asked Nov 13 at 16:17
Abigail
527
527
Could it be that you want the rows and columns of your matrices indexed by $mathbb{Z}$ (that is, bidirectionally infinite matrices)?
– darij grinberg
Nov 14 at 0:05
I just checked. You are right, the rows and columns are indexed by $mathbb{Z}$. But how this helps? Thanks.
– Abigail
Nov 14 at 0:50
1
I don't know how it helps, but it explains why @rschwieb is getting a different answer.
– darij grinberg
Nov 14 at 0:50
2
For each $g in mathbb{Z}$, let $V_g$ be the subspace of $V$ consisting of all vectors $left(ldots, v_{-1}, v_0, v_1, ldotsright)$ such that all $i geq g$ satisfy $v_i = 0$. The trick is to prove that each nontrivial proper $A$-submodule of $V$ is $V_g$ for some $g in mathbb{Z}$. It is clear that none of the $V_g$'s is simple.
– darij grinberg
Nov 14 at 1:07
Ah! I think I know what I am messing up. I was assuming that we start from row 1 and goes to infinity (I was assuming col and rows are indexed by natural number!). But we are indexed by $mathbb{Z}$, so it is not reasonable to start from 1. Thus the submodule I mentioned is not a valid choice! Thanks!!!
– Abigail
Nov 14 at 1:25
|
show 5 more comments
Could it be that you want the rows and columns of your matrices indexed by $mathbb{Z}$ (that is, bidirectionally infinite matrices)?
– darij grinberg
Nov 14 at 0:05
I just checked. You are right, the rows and columns are indexed by $mathbb{Z}$. But how this helps? Thanks.
– Abigail
Nov 14 at 0:50
1
I don't know how it helps, but it explains why @rschwieb is getting a different answer.
– darij grinberg
Nov 14 at 0:50
2
For each $g in mathbb{Z}$, let $V_g$ be the subspace of $V$ consisting of all vectors $left(ldots, v_{-1}, v_0, v_1, ldotsright)$ such that all $i geq g$ satisfy $v_i = 0$. The trick is to prove that each nontrivial proper $A$-submodule of $V$ is $V_g$ for some $g in mathbb{Z}$. It is clear that none of the $V_g$'s is simple.
– darij grinberg
Nov 14 at 1:07
Ah! I think I know what I am messing up. I was assuming that we start from row 1 and goes to infinity (I was assuming col and rows are indexed by natural number!). But we are indexed by $mathbb{Z}$, so it is not reasonable to start from 1. Thus the submodule I mentioned is not a valid choice! Thanks!!!
– Abigail
Nov 14 at 1:25
Could it be that you want the rows and columns of your matrices indexed by $mathbb{Z}$ (that is, bidirectionally infinite matrices)?
– darij grinberg
Nov 14 at 0:05
Could it be that you want the rows and columns of your matrices indexed by $mathbb{Z}$ (that is, bidirectionally infinite matrices)?
– darij grinberg
Nov 14 at 0:05
I just checked. You are right, the rows and columns are indexed by $mathbb{Z}$. But how this helps? Thanks.
– Abigail
Nov 14 at 0:50
I just checked. You are right, the rows and columns are indexed by $mathbb{Z}$. But how this helps? Thanks.
– Abigail
Nov 14 at 0:50
1
1
I don't know how it helps, but it explains why @rschwieb is getting a different answer.
– darij grinberg
Nov 14 at 0:50
I don't know how it helps, but it explains why @rschwieb is getting a different answer.
– darij grinberg
Nov 14 at 0:50
2
2
For each $g in mathbb{Z}$, let $V_g$ be the subspace of $V$ consisting of all vectors $left(ldots, v_{-1}, v_0, v_1, ldotsright)$ such that all $i geq g$ satisfy $v_i = 0$. The trick is to prove that each nontrivial proper $A$-submodule of $V$ is $V_g$ for some $g in mathbb{Z}$. It is clear that none of the $V_g$'s is simple.
– darij grinberg
Nov 14 at 1:07
For each $g in mathbb{Z}$, let $V_g$ be the subspace of $V$ consisting of all vectors $left(ldots, v_{-1}, v_0, v_1, ldotsright)$ such that all $i geq g$ satisfy $v_i = 0$. The trick is to prove that each nontrivial proper $A$-submodule of $V$ is $V_g$ for some $g in mathbb{Z}$. It is clear that none of the $V_g$'s is simple.
– darij grinberg
Nov 14 at 1:07
Ah! I think I know what I am messing up. I was assuming that we start from row 1 and goes to infinity (I was assuming col and rows are indexed by natural number!). But we are indexed by $mathbb{Z}$, so it is not reasonable to start from 1. Thus the submodule I mentioned is not a valid choice! Thanks!!!
– Abigail
Nov 14 at 1:25
Ah! I think I know what I am messing up. I was assuming that we start from row 1 and goes to infinity (I was assuming col and rows are indexed by natural number!). But we are indexed by $mathbb{Z}$, so it is not reasonable to start from 1. Thus the submodule I mentioned is not a valid choice! Thanks!!!
– Abigail
Nov 14 at 1:25
|
show 5 more comments
1 Answer
1
active
oldest
votes
up vote
2
down vote
I'm doing everything assuming that the sides of $R$ are indexed by $mathbb N$, and so are the positions of $V$, and that is how matrix multiplication is defined.
The subspace ${(lambda, 0, 0ldots)^Tmid lambda in k}$ appears to be a simple submodule of $_RV$.
(N.B. At the time of this answer, the indexing by $mathbb Z$ was not mentioned.)
answered Nov 13 at 19:11
rschwieb
103k1299238
Maybe there is a mixup in the problem setup? or maybe I am doing something silly...
– rschwieb
Nov 13 at 19:12
Hi, in fact, the subspace you mentioned is exactly I was guessing to be the simple submodule. But it appears that the rows and columns are indexed by $mathbb{Z}$, does this condition gives us a different answer? (sorry I did not mention this earlier...I didn't expect this to be important)
– Abigail
Nov 14 at 0:56
And I suppose that if we indexed by $mathbb{N}$, then ${(lambda,0,0,ldots)^T}$ should be the unique simple submodule?
– Abigail
Nov 14 at 2:09
@Abigail the indexing issue would be good to clarify. I think the last claim you mentioned is correct.
– rschwieb
Nov 14 at 4:48
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
I'm doing everything assuming that the sides of $R$ are indexed by $mathbb N$, and so are the positions of $V$, and that is how matrix multiplication is defined.
The subspace ${(lambda, 0, 0ldots)^Tmid lambda in k}$ appears to be a simple submodule of $_RV$.
(N.B. At the time of this answer, the indexing by $mathbb Z$ was not mentioned.)
answered Nov 13 at 19:11
rschwieb
103k1299238
Maybe there is a mixup in the problem setup? or maybe I am doing something silly...
– rschwieb
Nov 13 at 19:12
Hi, in fact, the subspace you mentioned is exactly I was guessing to be the simple submodule. But it appears that the rows and columns are indexed by $mathbb{Z}$, does this condition gives us a different answer? (sorry I did not mention this earlier...I didn't expect this to be important)
– Abigail
Nov 14 at 0:56
And I suppose that if we indexed by $mathbb{N}$, then ${(lambda,0,0,ldots)^T}$ should be the unique simple submodule?
– Abigail
Nov 14 at 2:09
@Abigail the indexing issue would be good to clarify. I think the last claim you mentioned is correct.
– rschwieb
Nov 14 at 4:48
add a comment |
up vote
2
down vote
I'm doing everything assuming that the sides of $R$ are indexed by $mathbb N$, and so are the positions of $V$, and that is how matrix multiplication is defined.
The subspace ${(lambda, 0, 0ldots)^Tmid lambda in k}$ appears to be a simple submodule of $_RV$.
(N.B. At the time of this answer, the indexing by $mathbb Z$ was not mentioned.)
answered Nov 13 at 19:11
rschwieb
103k1299238
Maybe there is a mixup in the problem setup? or maybe I am doing something silly...
– rschwieb
Nov 13 at 19:12
Hi, in fact, the subspace you mentioned is exactly I was guessing to be the simple submodule. But it appears that the rows and columns are indexed by $mathbb{Z}$, does this condition gives us a different answer? (sorry I did not mention this earlier...I didn't expect this to be important)
– Abigail
Nov 14 at 0:56
And I suppose that if we indexed by $mathbb{N}$, then ${(lambda,0,0,ldots)^T}$ should be the unique simple submodule?
– Abigail
Nov 14 at 2:09
@Abigail the indexing issue would be good to clarify. I think the last claim you mentioned is correct.
– rschwieb
Nov 14 at 4:48
add a comment |
up vote
2
down vote
up vote
2
down vote
I'm doing everything assuming that the sides of $R$ are indexed by $mathbb N$, and so are the positions of $V$, and that is how matrix multiplication is defined.
The subspace ${(lambda, 0, 0ldots)^Tmid lambda in k}$ appears to be a simple submodule of $_RV$.
(N.B. At the time of this answer, the indexing by $mathbb Z$ was not mentioned.)
answered Nov 13 at 19:11
rschwieb
103k1299238
I'm doing everything assuming that the sides of $R$ are indexed by $mathbb N$, and so are the positions of $V$, and that is how matrix multiplication is defined.
The subspace ${(lambda, 0, 0ldots)^Tmid lambda in k}$ appears to be a simple submodule of $_RV$.
(N.B. At the time of this answer, the indexing by $mathbb Z$ was not mentioned.)
answered Nov 13 at 19:11
rschwieb
103k1299238
edited Nov 14 at 14:42
answered Nov 13 at 19:11
rschwieb
103k1299238
answered Nov 13 at 19:11
rschwieb
103k1299238
answered Nov 13 at 19:11
rschwieb
103k1299238
103k1299238
Maybe there is a mixup in the problem setup? or maybe I am doing something silly...
– rschwieb
Nov 13 at 19:12
Hi, in fact, the subspace you mentioned is exactly I was guessing to be the simple submodule. But it appears that the rows and columns are indexed by $mathbb{Z}$, does this condition gives us a different answer? (sorry I did not mention this earlier...I didn't expect this to be important)
– Abigail
Nov 14 at 0:56
And I suppose that if we indexed by $mathbb{N}$, then ${(lambda,0,0,ldots)^T}$ should be the unique simple submodule?
– Abigail
Nov 14 at 2:09
@Abigail the indexing issue would be good to clarify. I think the last claim you mentioned is correct.
– rschwieb
Nov 14 at 4:48
add a comment |
Maybe there is a mixup in the problem setup? or maybe I am doing something silly...
– rschwieb
Nov 13 at 19:12
Hi, in fact, the subspace you mentioned is exactly I was guessing to be the simple submodule. But it appears that the rows and columns are indexed by $mathbb{Z}$, does this condition gives us a different answer? (sorry I did not mention this earlier...I didn't expect this to be important)
– Abigail
Nov 14 at 0:56
And I suppose that if we indexed by $mathbb{N}$, then ${(lambda,0,0,ldots)^T}$ should be the unique simple submodule?
– Abigail
Nov 14 at 2:09
@Abigail the indexing issue would be good to clarify. I think the last claim you mentioned is correct.
– rschwieb
Nov 14 at 4:48
Maybe there is a mixup in the problem setup? or maybe I am doing something silly...
– rschwieb
Nov 13 at 19:12
Maybe there is a mixup in the problem setup? or maybe I am doing something silly...
– rschwieb
Nov 13 at 19:12
Hi, in fact, the subspace you mentioned is exactly I was guessing to be the simple submodule. But it appears that the rows and columns are indexed by $mathbb{Z}$, does this condition gives us a different answer? (sorry I did not mention this earlier...I didn't expect this to be important)
– Abigail
Nov 14 at 0:56
Hi, in fact, the subspace you mentioned is exactly I was guessing to be the simple submodule. But it appears that the rows and columns are indexed by $mathbb{Z}$, does this condition gives us a different answer? (sorry I did not mention this earlier...I didn't expect this to be important)
– Abigail
Nov 14 at 0:56
And I suppose that if we indexed by $mathbb{N}$, then ${(lambda,0,0,ldots)^T}$ should be the unique simple submodule?
– Abigail
Nov 14 at 2:09
And I suppose that if we indexed by $mathbb{N}$, then ${(lambda,0,0,ldots)^T}$ should be the unique simple submodule?
– Abigail
Nov 14 at 2:09
@Abigail the indexing issue would be good to clarify. I think the last claim you mentioned is correct.
– rschwieb
Nov 14 at 4:48
@Abigail the indexing issue would be good to clarify. I think the last claim you mentioned is correct.
– rschwieb
Nov 14 at 4:48
add a comment |
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Could it be that you want the rows and columns of your matrices indexed by $mathbb{Z}$ (that is, bidirectionally infinite matrices)?
– darij grinberg
Nov 14 at 0:05
I just checked. You are right, the rows and columns are indexed by $mathbb{Z}$. But how this helps? Thanks.
– Abigail
Nov 14 at 0:50
1
I don't know how it helps, but it explains why @rschwieb is getting a different answer.
– darij grinberg
Nov 14 at 0:50
2
For each $g in mathbb{Z}$, let $V_g$ be the subspace of $V$ consisting of all vectors $left(ldots, v_{-1}, v_0, v_1, ldotsright)$ such that all $i geq g$ satisfy $v_i = 0$. The trick is to prove that each nontrivial proper $A$-submodule of $V$ is $V_g$ for some $g in mathbb{Z}$. It is clear that none of the $V_g$'s is simple.
– darij grinberg
Nov 14 at 1:07
Ah! I think I know what I am messing up. I was assuming that we start from row 1 and goes to infinity (I was assuming col and rows are indexed by natural number!). But we are indexed by $mathbb{Z}$, so it is not reasonable to start from 1. Thus the submodule I mentioned is not a valid choice! Thanks!!!
– Abigail
Nov 14 at 1:25