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How could I construct a module $M$ that has exactly $n$ composition series?
How could I construct a module $M$ that has exactly $n$ composition series?
I can't seem to find a series of submodules where each have exactly $n in mathbb{N}$ composition series. I don't know if there is something like a classic example of this.
abstract-algebra ring-theory modules noetherian artinian
edited Dec 11 '18 at 23:09
Batominovski
33.8k33292
asked Nov 20 '18 at 17:20
Ben-ZT
1889
add a comment |
How could I construct a module $M$ that has exactly $n$ composition series?
I can't seem to find a series of submodules where each have exactly $n in mathbb{N}$ composition series. I don't know if there is something like a classic example of this.
abstract-algebra ring-theory modules noetherian artinian
edited Dec 11 '18 at 23:09
Batominovski
33.8k33292
asked Nov 20 '18 at 17:20
Ben-ZT
1889
1
If $n=k!$ for some $k$, you can take the direct sum $M=L_1oplus L_2oplus ldots oplus L_k$ where $L_1,L_2,ldots,L_k$ are non-isomorphic simple modules. It would be quite tricky if you require that $M$ be indecomposable.
– Zvi
Nov 20 '18 at 18:16
What do you mean "$n$ composition series"? You don't mean "composition length $n$ for a given $n$", right? You actually want $n$ distinct (in some sense) composition series?
– rschwieb
Nov 20 '18 at 18:55
Ooops, that was wrong. The $mathbb{Z}/2mathbb{Z}oplus mathbb{Z}/4mathbb{Z}$ gives $n=5$, not $n=4$. Old comment deleted.
– Zvi
Nov 21 '18 at 11:27
add a comment |
How could I construct a module $M$ that has exactly $n$ composition series?
I can't seem to find a series of submodules where each have exactly $n in mathbb{N}$ composition series. I don't know if there is something like a classic example of this.
abstract-algebra ring-theory modules noetherian artinian
edited Dec 11 '18 at 23:09
Batominovski
33.8k33292
asked Nov 20 '18 at 17:20
Ben-ZT
1889
How could I construct a module $M$ that has exactly $n$ composition series?
I can't seem to find a series of submodules where each have exactly $n in mathbb{N}$ composition series. I don't know if there is something like a classic example of this.
abstract-algebra ring-theory modules noetherian artinian
abstract-algebra ring-theory modules noetherian artinian
edited Dec 11 '18 at 23:09
Batominovski
33.8k33292
asked Nov 20 '18 at 17:20
Ben-ZT
1889
edited Dec 11 '18 at 23:09
Batominovski
33.8k33292
asked Nov 20 '18 at 17:20
Ben-ZT
1889
edited Dec 11 '18 at 23:09
Batominovski
33.8k33292
edited Dec 11 '18 at 23:09
Batominovski
33.8k33292
edited Dec 11 '18 at 23:09
Batominovski
33.8k33292
33.8k33292
asked Nov 20 '18 at 17:20
Ben-ZT
1889
asked Nov 20 '18 at 17:20
Ben-ZT
1889
asked Nov 20 '18 at 17:20
Ben-ZT
1889
1889
1
If $n=k!$ for some $k$, you can take the direct sum $M=L_1oplus L_2oplus ldots oplus L_k$ where $L_1,L_2,ldots,L_k$ are non-isomorphic simple modules. It would be quite tricky if you require that $M$ be indecomposable.
– Zvi
Nov 20 '18 at 18:16
What do you mean "$n$ composition series"? You don't mean "composition length $n$ for a given $n$", right? You actually want $n$ distinct (in some sense) composition series?
– rschwieb
Nov 20 '18 at 18:55
Ooops, that was wrong. The $mathbb{Z}/2mathbb{Z}oplus mathbb{Z}/4mathbb{Z}$ gives $n=5$, not $n=4$. Old comment deleted.
– Zvi
Nov 21 '18 at 11:27
add a comment |
1
If $n=k!$ for some $k$, you can take the direct sum $M=L_1oplus L_2oplus ldots oplus L_k$ where $L_1,L_2,ldots,L_k$ are non-isomorphic simple modules. It would be quite tricky if you require that $M$ be indecomposable.
– Zvi
Nov 20 '18 at 18:16
What do you mean "$n$ composition series"? You don't mean "composition length $n$ for a given $n$", right? You actually want $n$ distinct (in some sense) composition series?
– rschwieb
Nov 20 '18 at 18:55
Ooops, that was wrong. The $mathbb{Z}/2mathbb{Z}oplus mathbb{Z}/4mathbb{Z}$ gives $n=5$, not $n=4$. Old comment deleted.
– Zvi
Nov 21 '18 at 11:27
1
1
If $n=k!$ for some $k$, you can take the direct sum $M=L_1oplus L_2oplus ldots oplus L_k$ where $L_1,L_2,ldots,L_k$ are non-isomorphic simple modules. It would be quite tricky if you require that $M$ be indecomposable.
– Zvi
Nov 20 '18 at 18:16
If $n=k!$ for some $k$, you can take the direct sum $M=L_1oplus L_2oplus ldots oplus L_k$ where $L_1,L_2,ldots,L_k$ are non-isomorphic simple modules. It would be quite tricky if you require that $M$ be indecomposable.
– Zvi
Nov 20 '18 at 18:16
What do you mean "$n$ composition series"? You don't mean "composition length $n$ for a given $n$", right? You actually want $n$ distinct (in some sense) composition series?
– rschwieb
Nov 20 '18 at 18:55
What do you mean "$n$ composition series"? You don't mean "composition length $n$ for a given $n$", right? You actually want $n$ distinct (in some sense) composition series?
– rschwieb
Nov 20 '18 at 18:55
Ooops, that was wrong. The $mathbb{Z}/2mathbb{Z}oplus mathbb{Z}/4mathbb{Z}$ gives $n=5$, not $n=4$. Old comment deleted.
– Zvi
Nov 21 '18 at 11:27
Ooops, that was wrong. The $mathbb{Z}/2mathbb{Z}oplus mathbb{Z}/4mathbb{Z}$ gives $n=5$, not $n=4$. Old comment deleted.
– Zvi
Nov 21 '18 at 11:27
add a comment |
1 Answer
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Consider the $mathbb{Z}$-module $M=(mathbb{Z}/2^{n-1}mathbb{Z})oplus(mathbb{Z}/3mathbb{Z})$. There are exactly $n$ distinct composition series for $M$. The composition series are given by
$$0=M_0^isubsetneq M_1^i subsetneq M_2^isubsetneq ldots subsetneq M_{n-1}^isubsetneq M_n^i=M,$$
where $i=1,2,ldots,n$, and
$$M_j^i=begin{cases}biglangle (2^{n-1-j},0)bigrangle&text{for }j=0,1,2,ldots,i-1,\ biglangle (2^{n-j},0),(0,1)bigrangle&text{for }j=i,i+1,i+2,ldots,n.end{cases}$$
Hence, for any integer $ngeq 0$, there exist a ring $R$ and a left $R$-module $M$ of finite length such that $M$ has exactly $n$ distinct composition series.
It would be an interesting question to fix the ring $R$ or demand that $M$ be indecomposable. For example, if $R$ is a finite field of order $q$, then due to this lecture note, the only possible values of $n$ are of the form
$$n=[k]_q!=prod_{i=1}^kfrac{q^i-1}{q-1}=prod_{i=1}^k(1+q+q^2+ldots+q^{i-1}).$$
If $R=mathbb{Z}$ or $R$ is a semisimple ring, and $M$ is an indecomposable left $R$-module of finite length, then $M$ has exactly one composition series.
It is of course possible to have a module of finite length with infinitely many composition series. Take for example $R=mathbb{Q}$ and $M=mathbb{Q}^2$. Then, for every $rinmathbb{Q}$,
$$0subsetneq operatorname{span}big{(1,r)big}subsetneq M$$
is a composition series of $M$ (and along with $0subsetneq operatorname{span}big{(0,1)big}subsetneq M$, these are all composition series of $M$).
answered Dec 6 '18 at 12:01
Zvi
4,810430
add a comment |
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Consider the $mathbb{Z}$-module $M=(mathbb{Z}/2^{n-1}mathbb{Z})oplus(mathbb{Z}/3mathbb{Z})$. There are exactly $n$ distinct composition series for $M$. The composition series are given by
$$0=M_0^isubsetneq M_1^i subsetneq M_2^isubsetneq ldots subsetneq M_{n-1}^isubsetneq M_n^i=M,$$
where $i=1,2,ldots,n$, and
$$M_j^i=begin{cases}biglangle (2^{n-1-j},0)bigrangle&text{for }j=0,1,2,ldots,i-1,\ biglangle (2^{n-j},0),(0,1)bigrangle&text{for }j=i,i+1,i+2,ldots,n.end{cases}$$
Hence, for any integer $ngeq 0$, there exist a ring $R$ and a left $R$-module $M$ of finite length such that $M$ has exactly $n$ distinct composition series.
It would be an interesting question to fix the ring $R$ or demand that $M$ be indecomposable. For example, if $R$ is a finite field of order $q$, then due to this lecture note, the only possible values of $n$ are of the form
$$n=[k]_q!=prod_{i=1}^kfrac{q^i-1}{q-1}=prod_{i=1}^k(1+q+q^2+ldots+q^{i-1}).$$
If $R=mathbb{Z}$ or $R$ is a semisimple ring, and $M$ is an indecomposable left $R$-module of finite length, then $M$ has exactly one composition series.
It is of course possible to have a module of finite length with infinitely many composition series. Take for example $R=mathbb{Q}$ and $M=mathbb{Q}^2$. Then, for every $rinmathbb{Q}$,
$$0subsetneq operatorname{span}big{(1,r)big}subsetneq M$$
is a composition series of $M$ (and along with $0subsetneq operatorname{span}big{(0,1)big}subsetneq M$, these are all composition series of $M$).
answered Dec 6 '18 at 12:01
Zvi
4,810430
add a comment |
Consider the $mathbb{Z}$-module $M=(mathbb{Z}/2^{n-1}mathbb{Z})oplus(mathbb{Z}/3mathbb{Z})$. There are exactly $n$ distinct composition series for $M$. The composition series are given by
$$0=M_0^isubsetneq M_1^i subsetneq M_2^isubsetneq ldots subsetneq M_{n-1}^isubsetneq M_n^i=M,$$
where $i=1,2,ldots,n$, and
$$M_j^i=begin{cases}biglangle (2^{n-1-j},0)bigrangle&text{for }j=0,1,2,ldots,i-1,\ biglangle (2^{n-j},0),(0,1)bigrangle&text{for }j=i,i+1,i+2,ldots,n.end{cases}$$
Hence, for any integer $ngeq 0$, there exist a ring $R$ and a left $R$-module $M$ of finite length such that $M$ has exactly $n$ distinct composition series.
It would be an interesting question to fix the ring $R$ or demand that $M$ be indecomposable. For example, if $R$ is a finite field of order $q$, then due to this lecture note, the only possible values of $n$ are of the form
$$n=[k]_q!=prod_{i=1}^kfrac{q^i-1}{q-1}=prod_{i=1}^k(1+q+q^2+ldots+q^{i-1}).$$
If $R=mathbb{Z}$ or $R$ is a semisimple ring, and $M$ is an indecomposable left $R$-module of finite length, then $M$ has exactly one composition series.
It is of course possible to have a module of finite length with infinitely many composition series. Take for example $R=mathbb{Q}$ and $M=mathbb{Q}^2$. Then, for every $rinmathbb{Q}$,
$$0subsetneq operatorname{span}big{(1,r)big}subsetneq M$$
is a composition series of $M$ (and along with $0subsetneq operatorname{span}big{(0,1)big}subsetneq M$, these are all composition series of $M$).
answered Dec 6 '18 at 12:01
Zvi
4,810430
add a comment |
Consider the $mathbb{Z}$-module $M=(mathbb{Z}/2^{n-1}mathbb{Z})oplus(mathbb{Z}/3mathbb{Z})$. There are exactly $n$ distinct composition series for $M$. The composition series are given by
$$0=M_0^isubsetneq M_1^i subsetneq M_2^isubsetneq ldots subsetneq M_{n-1}^isubsetneq M_n^i=M,$$
where $i=1,2,ldots,n$, and
$$M_j^i=begin{cases}biglangle (2^{n-1-j},0)bigrangle&text{for }j=0,1,2,ldots,i-1,\ biglangle (2^{n-j},0),(0,1)bigrangle&text{for }j=i,i+1,i+2,ldots,n.end{cases}$$
Hence, for any integer $ngeq 0$, there exist a ring $R$ and a left $R$-module $M$ of finite length such that $M$ has exactly $n$ distinct composition series.
It would be an interesting question to fix the ring $R$ or demand that $M$ be indecomposable. For example, if $R$ is a finite field of order $q$, then due to this lecture note, the only possible values of $n$ are of the form
$$n=[k]_q!=prod_{i=1}^kfrac{q^i-1}{q-1}=prod_{i=1}^k(1+q+q^2+ldots+q^{i-1}).$$
If $R=mathbb{Z}$ or $R$ is a semisimple ring, and $M$ is an indecomposable left $R$-module of finite length, then $M$ has exactly one composition series.
It is of course possible to have a module of finite length with infinitely many composition series. Take for example $R=mathbb{Q}$ and $M=mathbb{Q}^2$. Then, for every $rinmathbb{Q}$,
$$0subsetneq operatorname{span}big{(1,r)big}subsetneq M$$
is a composition series of $M$ (and along with $0subsetneq operatorname{span}big{(0,1)big}subsetneq M$, these are all composition series of $M$).
answered Dec 6 '18 at 12:01
Zvi
4,810430
Consider the $mathbb{Z}$-module $M=(mathbb{Z}/2^{n-1}mathbb{Z})oplus(mathbb{Z}/3mathbb{Z})$. There are exactly $n$ distinct composition series for $M$. The composition series are given by
$$0=M_0^isubsetneq M_1^i subsetneq M_2^isubsetneq ldots subsetneq M_{n-1}^isubsetneq M_n^i=M,$$
where $i=1,2,ldots,n$, and
$$M_j^i=begin{cases}biglangle (2^{n-1-j},0)bigrangle&text{for }j=0,1,2,ldots,i-1,\ biglangle (2^{n-j},0),(0,1)bigrangle&text{for }j=i,i+1,i+2,ldots,n.end{cases}$$
Hence, for any integer $ngeq 0$, there exist a ring $R$ and a left $R$-module $M$ of finite length such that $M$ has exactly $n$ distinct composition series.
It would be an interesting question to fix the ring $R$ or demand that $M$ be indecomposable. For example, if $R$ is a finite field of order $q$, then due to this lecture note, the only possible values of $n$ are of the form
$$n=[k]_q!=prod_{i=1}^kfrac{q^i-1}{q-1}=prod_{i=1}^k(1+q+q^2+ldots+q^{i-1}).$$
If $R=mathbb{Z}$ or $R$ is a semisimple ring, and $M$ is an indecomposable left $R$-module of finite length, then $M$ has exactly one composition series.
It is of course possible to have a module of finite length with infinitely many composition series. Take for example $R=mathbb{Q}$ and $M=mathbb{Q}^2$. Then, for every $rinmathbb{Q}$,
$$0subsetneq operatorname{span}big{(1,r)big}subsetneq M$$
is a composition series of $M$ (and along with $0subsetneq operatorname{span}big{(0,1)big}subsetneq M$, these are all composition series of $M$).
answered Dec 6 '18 at 12:01
Zvi
4,810430
edited Dec 6 '18 at 12:31
answered Dec 6 '18 at 12:01
Zvi
4,810430
answered Dec 6 '18 at 12:01
Zvi
4,810430
answered Dec 6 '18 at 12:01
Zvi
4,810430
4,810430
add a comment |
add a comment |
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If $n=k!$ for some $k$, you can take the direct sum $M=L_1oplus L_2oplus ldots oplus L_k$ where $L_1,L_2,ldots,L_k$ are non-isomorphic simple modules. It would be quite tricky if you require that $M$ be indecomposable.
– Zvi
Nov 20 '18 at 18:16
What do you mean "$n$ composition series"? You don't mean "composition length $n$ for a given $n$", right? You actually want $n$ distinct (in some sense) composition series?
– rschwieb
Nov 20 '18 at 18:55
Ooops, that was wrong. The $mathbb{Z}/2mathbb{Z}oplus mathbb{Z}/4mathbb{Z}$ gives $n=5$, not $n=4$. Old comment deleted.
– Zvi
Nov 21 '18 at 11:27