If $M$ is a finitely generated torsion module over a PID which is not a field then $M$ is Artinian.
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If $M$ is a finitely generated torsion module over a PID which is not a field then $M$ is Artinian.
Is contradiction the way to go? How does $M$ being torsion help us? Any hints on how to prove?
linear-algebra
edited Nov 22 at 20:00
Bernard
117k637109
asked Nov 22 at 19:15
UserA
496216
add a comment |
up vote
1
down vote
favorite
If $M$ is a finitely generated torsion module over a PID which is not a field then $M$ is Artinian.
Is contradiction the way to go? How does $M$ being torsion help us? Any hints on how to prove?
linear-algebra
edited Nov 22 at 20:00
Bernard
117k637109
asked Nov 22 at 19:15
UserA
496216
@Youngsu Well, $mathbb Z/2mathbb Ztimes mathbb Z/2mathbb Z$ is a counterexample to that... but nevertheless, using the structure theorem is the right way to go...
– rschwieb
Nov 22 at 19:48
You are absolutely right. I should've said product of those. Thank you for pointing it out.
– Youngsu
Nov 22 at 20:24
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If $M$ is a finitely generated torsion module over a PID which is not a field then $M$ is Artinian.
Is contradiction the way to go? How does $M$ being torsion help us? Any hints on how to prove?
linear-algebra
edited Nov 22 at 20:00
Bernard
117k637109
asked Nov 22 at 19:15
UserA
496216
If $M$ is a finitely generated torsion module over a PID which is not a field then $M$ is Artinian.
Is contradiction the way to go? How does $M$ being torsion help us? Any hints on how to prove?
linear-algebra
linear-algebra
edited Nov 22 at 20:00
Bernard
117k637109
asked Nov 22 at 19:15
UserA
496216
edited Nov 22 at 20:00
Bernard
117k637109
asked Nov 22 at 19:15
UserA
496216
edited Nov 22 at 20:00
Bernard
117k637109
edited Nov 22 at 20:00
Bernard
117k637109
edited Nov 22 at 20:00
Bernard
117k637109
117k637109
asked Nov 22 at 19:15
UserA
496216
asked Nov 22 at 19:15
UserA
496216
asked Nov 22 at 19:15
UserA
496216
496216
@Youngsu Well, $mathbb Z/2mathbb Ztimes mathbb Z/2mathbb Z$ is a counterexample to that... but nevertheless, using the structure theorem is the right way to go...
– rschwieb
Nov 22 at 19:48
You are absolutely right. I should've said product of those. Thank you for pointing it out.
– Youngsu
Nov 22 at 20:24
add a comment |
@Youngsu Well, $mathbb Z/2mathbb Ztimes mathbb Z/2mathbb Z$ is a counterexample to that... but nevertheless, using the structure theorem is the right way to go...
– rschwieb
Nov 22 at 19:48
You are absolutely right. I should've said product of those. Thank you for pointing it out.
– Youngsu
Nov 22 at 20:24
@Youngsu Well, $mathbb Z/2mathbb Ztimes mathbb Z/2mathbb Z$ is a counterexample to that... but nevertheless, using the structure theorem is the right way to go...
– rschwieb
Nov 22 at 19:48
@Youngsu Well, $mathbb Z/2mathbb Ztimes mathbb Z/2mathbb Z$ is a counterexample to that... but nevertheless, using the structure theorem is the right way to go...
– rschwieb
Nov 22 at 19:48
You are absolutely right. I should've said product of those. Thank you for pointing it out.
– Youngsu
Nov 22 at 20:24
You are absolutely right. I should've said product of those. Thank you for pointing it out.
– Youngsu
Nov 22 at 20:24
add a comment |
1 Answer
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0
down vote
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How does M being torsion help us?
I assume you are learning about the fundamental structure theorem on f.g. modules over a PID?
The fact that the module is f.g. and torsion implies $Mcong oplus_{i=1}^n R/(p_i^{e_i})$ for some primes $p_i$ in $R$ and exponents $e_i$.
Any hints on how to prove?
You can check that each of these factors has finite composition length, and so the finite product of the factors has finite composition length.
answered Nov 22 at 19:53
rschwieb
104k1299241
add a comment |
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1 Answer
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1 Answer
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up vote
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down vote
accepted
How does M being torsion help us?
I assume you are learning about the fundamental structure theorem on f.g. modules over a PID?
The fact that the module is f.g. and torsion implies $Mcong oplus_{i=1}^n R/(p_i^{e_i})$ for some primes $p_i$ in $R$ and exponents $e_i$.
Any hints on how to prove?
You can check that each of these factors has finite composition length, and so the finite product of the factors has finite composition length.
answered Nov 22 at 19:53
rschwieb
104k1299241
add a comment |
up vote
0
down vote
accepted
How does M being torsion help us?
I assume you are learning about the fundamental structure theorem on f.g. modules over a PID?
The fact that the module is f.g. and torsion implies $Mcong oplus_{i=1}^n R/(p_i^{e_i})$ for some primes $p_i$ in $R$ and exponents $e_i$.
Any hints on how to prove?
You can check that each of these factors has finite composition length, and so the finite product of the factors has finite composition length.
answered Nov 22 at 19:53
rschwieb
104k1299241
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
How does M being torsion help us?
I assume you are learning about the fundamental structure theorem on f.g. modules over a PID?
The fact that the module is f.g. and torsion implies $Mcong oplus_{i=1}^n R/(p_i^{e_i})$ for some primes $p_i$ in $R$ and exponents $e_i$.
Any hints on how to prove?
You can check that each of these factors has finite composition length, and so the finite product of the factors has finite composition length.
answered Nov 22 at 19:53
rschwieb
104k1299241
How does M being torsion help us?
I assume you are learning about the fundamental structure theorem on f.g. modules over a PID?
The fact that the module is f.g. and torsion implies $Mcong oplus_{i=1}^n R/(p_i^{e_i})$ for some primes $p_i$ in $R$ and exponents $e_i$.
Any hints on how to prove?
You can check that each of these factors has finite composition length, and so the finite product of the factors has finite composition length.
answered Nov 22 at 19:53
rschwieb
104k1299241
answered Nov 22 at 19:53
rschwieb
104k1299241
answered Nov 22 at 19:53
rschwieb
104k1299241
answered Nov 22 at 19:53
rschwieb
104k1299241
104k1299241
add a comment |
add a comment |
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@Youngsu Well, $mathbb Z/2mathbb Ztimes mathbb Z/2mathbb Z$ is a counterexample to that... but nevertheless, using the structure theorem is the right way to go...
– rschwieb
Nov 22 at 19:48
You are absolutely right. I should've said product of those. Thank you for pointing it out.
– Youngsu
Nov 22 at 20:24