If $M/Ncong M$, can we conclude that $N=0$? [duplicate]
Multi tool use
up vote
1
down vote
favorite
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
7,75383667
marked as duplicate by rschwieb abstract-algebra
Users with the abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
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.
add a comment |
up vote
1
down vote
favorite
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
7,75383667
marked as duplicate by rschwieb abstract-algebra
Users with the abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
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 |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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
7,75383667
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
7,75383667
asked Nov 21 at 3:51
yoyostein
7,75383667
edited Nov 21 at 4:00
asked Nov 21 at 3:51
yoyostein
7,75383667
asked Nov 21 at 3:51
yoyostein
7,75383667
asked Nov 21 at 3:51
yoyostein
7,75383667
7,75383667
marked as duplicate by rschwieb abstract-algebra
Users with the abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
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.
marked as duplicate by rschwieb abstract-algebra
Users with the abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.
StackExchange.ready(function() {
if (StackExchange.options.isMobile) return;
$('.dupe-hammer-message-hover:not(.hover-bound)').each(function() {
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');
$hover.hover(
function() {
$hover.showInfoMessage('', {
messageElement: $msg.clone().show(),
transient: false,
position: { my: 'bottom left', at: 'top center', offsetTop: -7 },
dismissable: false,
relativeToBody: true
});
},
function() {
StackExchange.helpers.removeMessages();
}
);
});
});
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
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
It fails for vector spaces of infinite dimension.
answered Nov 21 at 3:55
Dante Grevino
4616
New contributor
Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
2,25911337
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 |
2 Answers
2
active
oldest
votes
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
4616
New contributor
Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
2
down vote
It fails for vector spaces of infinite dimension.
answered Nov 21 at 3:55
Dante Grevino
4616
New contributor
Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
4616
New contributor
Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
It fails for vector spaces of infinite dimension.
answered Nov 21 at 3:55
Dante Grevino
4616
New contributor
Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Nov 21 at 3:55
Dante Grevino
4616
New contributor
Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Nov 21 at 3:55
Dante Grevino
4616
answered Nov 21 at 3:55
Dante Grevino
4616
4616
New contributor
Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
2,25911337
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
2,25911337
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
2,25911337
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
2,25911337
answered Nov 21 at 3:56
Monstrous Moonshiner
2,25911337
answered Nov 21 at 3:56
Monstrous Moonshiner
2,25911337
answered Nov 21 at 3:56
Monstrous Moonshiner
2,25911337
2,25911337
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 |
b0mr0HNCWQ Vyl,Z kluzwAAAUk,EShQs4Mo
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