If $M/Ncong M$, can we conclude that $N=0$? [duplicate]
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Is there a (f.g., free) module isomorphic to a quotient of itself?
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Let $M$ be an $R$-module, where we may assume that $R$ is an integral domain.
Let $N$ be a submodule of $M$.
Suppose that $M/Ncong M$. Can we conclude that $N=0$?
(If no, what are some sufficient conditions that make it true?)
Update: I learn that for "infinite dimensional" cases it can fail. How about when $M$ is finitely generated, does it work?
Thanks a lot.
abstract-algebra modules
asked Nov 21 at 3:51
yoyostein
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marked as duplicate by rschwieb abstract-algebra
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Nov 21 at 15:07
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
Is there a (f.g., free) module isomorphic to a quotient of itself?
4 answers
Let $M$ be an $R$-module, where we may assume that $R$ is an integral domain.
Let $N$ be a submodule of $M$.
Suppose that $M/Ncong M$. Can we conclude that $N=0$?
(If no, what are some sufficient conditions that make it true?)
Update: I learn that for "infinite dimensional" cases it can fail. How about when $M$ is finitely generated, does it work?
Thanks a lot.
abstract-algebra modules
asked Nov 21 at 3:51
yoyostein
7,80483768
marked as duplicate by rschwieb abstract-algebra
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Nov 21 at 15:07
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
This is false, but see statement 3 of Nakayama's Lemma on wikipedia.
– Aaron
Nov 21 at 5:10
how about when $M$ is finitely generated, does it work? Nope. You can get $Roplus Rcong R$ as $R$ modules for some rings. This is clearly finitely generated with a quotient by a nonzero ideal isororphic to $R$. See egreg's answer in the linked duplicate.
– rschwieb
Nov 21 at 15:05
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up vote
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This question already has an answer here:
Is there a (f.g., free) module isomorphic to a quotient of itself?
4 answers
Let $M$ be an $R$-module, where we may assume that $R$ is an integral domain.
Let $N$ be a submodule of $M$.
Suppose that $M/Ncong M$. Can we conclude that $N=0$?
(If no, what are some sufficient conditions that make it true?)
Update: I learn that for "infinite dimensional" cases it can fail. How about when $M$ is finitely generated, does it work?
Thanks a lot.
abstract-algebra modules
asked Nov 21 at 3:51
yoyostein
7,80483768
This question already has an answer here:
Is there a (f.g., free) module isomorphic to a quotient of itself?
4 answers
Let $M$ be an $R$-module, where we may assume that $R$ is an integral domain.
Let $N$ be a submodule of $M$.
Suppose that $M/Ncong M$. Can we conclude that $N=0$?
(If no, what are some sufficient conditions that make it true?)
Update: I learn that for "infinite dimensional" cases it can fail. How about when $M$ is finitely generated, does it work?
Thanks a lot.
This question already has an answer here:
Is there a (f.g., free) module isomorphic to a quotient of itself?
4 answers
abstract-algebra modules
abstract-algebra modules
asked Nov 21 at 3:51
yoyostein
7,80483768
asked Nov 21 at 3:51
yoyostein
7,80483768
edited Nov 21 at 4:00
asked Nov 21 at 3:51
yoyostein
7,80483768
asked Nov 21 at 3:51
yoyostein
7,80483768
asked Nov 21 at 3:51
yoyostein
7,80483768
7,80483768
marked as duplicate by rschwieb abstract-algebra
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Nov 21 at 15:07
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Nov 21 at 15:07
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
This is false, but see statement 3 of Nakayama's Lemma on wikipedia.
– Aaron
Nov 21 at 5:10
how about when $M$ is finitely generated, does it work? Nope. You can get $Roplus Rcong R$ as $R$ modules for some rings. This is clearly finitely generated with a quotient by a nonzero ideal isororphic to $R$. See egreg's answer in the linked duplicate.
– rschwieb
Nov 21 at 15:05
add a comment |
This is false, but see statement 3 of Nakayama's Lemma on wikipedia.
– Aaron
Nov 21 at 5:10
how about when $M$ is finitely generated, does it work? Nope. You can get $Roplus Rcong R$ as $R$ modules for some rings. This is clearly finitely generated with a quotient by a nonzero ideal isororphic to $R$. See egreg's answer in the linked duplicate.
– rschwieb
Nov 21 at 15:05
This is false, but see statement 3 of Nakayama's Lemma on wikipedia.
– Aaron
Nov 21 at 5:10
This is false, but see statement 3 of Nakayama's Lemma on wikipedia.
– Aaron
Nov 21 at 5:10
how about when $M$ is finitely generated, does it work? Nope. You can get $Roplus Rcong R$ as $R$ modules for some rings. This is clearly finitely generated with a quotient by a nonzero ideal isororphic to $R$. See egreg's answer in the linked duplicate.
– rschwieb
Nov 21 at 15:05
how about when $M$ is finitely generated, does it work? Nope. You can get $Roplus Rcong R$ as $R$ modules for some rings. This is clearly finitely generated with a quotient by a nonzero ideal isororphic to $R$. See egreg's answer in the linked duplicate.
– rschwieb
Nov 21 at 15:05
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2 Answers
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It fails for vector spaces of infinite dimension.
answered Nov 21 at 3:55
Dante Grevino
90117
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It is not true. Let $M = bigoplus_{n=1}^infty mathbb{Z}$, and let $N = {(z,0,0,0,ldots),|, z in mathbb{Z}}$. Then $M/N cong M$.
answered Nov 21 at 3:56
Monstrous Moonshiner
2,25511337
How about when M is finitely generated, does it work?
– yoyostein
Nov 21 at 4:00
1
If $M$ is a vector space, then yes it does. If $M$ is not a vector space, then I'm not so sure...
– Monstrous Moonshiner
Nov 21 at 4:00
2
It still holds if we let $R = bigoplus^infty mathbb Z$, while $M=R$, and $N$ stays the same. I believe $R$ and $M$ need to be Noetherian for that to hold.
– Quang Hoang
Nov 21 at 4:07
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
It fails for vector spaces of infinite dimension.
answered Nov 21 at 3:55
Dante Grevino
90117
add a comment |
up vote
2
down vote
It fails for vector spaces of infinite dimension.
answered Nov 21 at 3:55
Dante Grevino
90117
add a comment |
up vote
2
down vote
up vote
2
down vote
It fails for vector spaces of infinite dimension.
answered Nov 21 at 3:55
Dante Grevino
90117
It fails for vector spaces of infinite dimension.
answered Nov 21 at 3:55
Dante Grevino
90117
answered Nov 21 at 3:55
Dante Grevino
90117
answered Nov 21 at 3:55
Dante Grevino
90117
answered Nov 21 at 3:55
Dante Grevino
90117
90117
add a comment |
add a comment |
up vote
2
down vote
It is not true. Let $M = bigoplus_{n=1}^infty mathbb{Z}$, and let $N = {(z,0,0,0,ldots),|, z in mathbb{Z}}$. Then $M/N cong M$.
answered Nov 21 at 3:56
Monstrous Moonshiner
2,25511337
How about when M is finitely generated, does it work?
– yoyostein
Nov 21 at 4:00
1
If $M$ is a vector space, then yes it does. If $M$ is not a vector space, then I'm not so sure...
– Monstrous Moonshiner
Nov 21 at 4:00
2
It still holds if we let $R = bigoplus^infty mathbb Z$, while $M=R$, and $N$ stays the same. I believe $R$ and $M$ need to be Noetherian for that to hold.
– Quang Hoang
Nov 21 at 4:07
add a comment |
up vote
2
down vote
It is not true. Let $M = bigoplus_{n=1}^infty mathbb{Z}$, and let $N = {(z,0,0,0,ldots),|, z in mathbb{Z}}$. Then $M/N cong M$.
answered Nov 21 at 3:56
Monstrous Moonshiner
2,25511337
How about when M is finitely generated, does it work?
– yoyostein
Nov 21 at 4:00
1
If $M$ is a vector space, then yes it does. If $M$ is not a vector space, then I'm not so sure...
– Monstrous Moonshiner
Nov 21 at 4:00
2
It still holds if we let $R = bigoplus^infty mathbb Z$, while $M=R$, and $N$ stays the same. I believe $R$ and $M$ need to be Noetherian for that to hold.
– Quang Hoang
Nov 21 at 4:07
add a comment |
up vote
2
down vote
up vote
2
down vote
It is not true. Let $M = bigoplus_{n=1}^infty mathbb{Z}$, and let $N = {(z,0,0,0,ldots),|, z in mathbb{Z}}$. Then $M/N cong M$.
answered Nov 21 at 3:56
Monstrous Moonshiner
2,25511337
It is not true. Let $M = bigoplus_{n=1}^infty mathbb{Z}$, and let $N = {(z,0,0,0,ldots),|, z in mathbb{Z}}$. Then $M/N cong M$.
answered Nov 21 at 3:56
Monstrous Moonshiner
2,25511337
answered Nov 21 at 3:56
Monstrous Moonshiner
2,25511337
answered Nov 21 at 3:56
Monstrous Moonshiner
2,25511337
answered Nov 21 at 3:56
Monstrous Moonshiner
2,25511337
2,25511337
How about when M is finitely generated, does it work?
– yoyostein
Nov 21 at 4:00
1
If $M$ is a vector space, then yes it does. If $M$ is not a vector space, then I'm not so sure...
– Monstrous Moonshiner
Nov 21 at 4:00
2
It still holds if we let $R = bigoplus^infty mathbb Z$, while $M=R$, and $N$ stays the same. I believe $R$ and $M$ need to be Noetherian for that to hold.
– Quang Hoang
Nov 21 at 4:07
add a comment |
How about when M is finitely generated, does it work?
– yoyostein
Nov 21 at 4:00
1
If $M$ is a vector space, then yes it does. If $M$ is not a vector space, then I'm not so sure...
– Monstrous Moonshiner
Nov 21 at 4:00
2
It still holds if we let $R = bigoplus^infty mathbb Z$, while $M=R$, and $N$ stays the same. I believe $R$ and $M$ need to be Noetherian for that to hold.
– Quang Hoang
Nov 21 at 4:07
How about when M is finitely generated, does it work?
– yoyostein
Nov 21 at 4:00
How about when M is finitely generated, does it work?
– yoyostein
Nov 21 at 4:00
1
1
If $M$ is a vector space, then yes it does. If $M$ is not a vector space, then I'm not so sure...
– Monstrous Moonshiner
Nov 21 at 4:00
If $M$ is a vector space, then yes it does. If $M$ is not a vector space, then I'm not so sure...
– Monstrous Moonshiner
Nov 21 at 4:00
2
2
It still holds if we let $R = bigoplus^infty mathbb Z$, while $M=R$, and $N$ stays the same. I believe $R$ and $M$ need to be Noetherian for that to hold.
– Quang Hoang
Nov 21 at 4:07
It still holds if we let $R = bigoplus^infty mathbb Z$, while $M=R$, and $N$ stays the same. I believe $R$ and $M$ need to be Noetherian for that to hold.
– Quang Hoang
Nov 21 at 4:07
add a comment |
This is false, but see statement 3 of Nakayama's Lemma on wikipedia.
– Aaron
Nov 21 at 5:10
how about when $M$ is finitely generated, does it work? Nope. You can get $Roplus Rcong R$ as $R$ modules for some rings. This is clearly finitely generated with a quotient by a nonzero ideal isororphic to $R$. See egreg's answer in the linked duplicate.
– rschwieb
Nov 21 at 15:05