Are two submodules (where one is contained in the other) isomorphic if their quotientmodules are isomorphic?
$begingroup$
Let $M$ be an $R$ module and $N_1 subset N_2$ be submodules of $M$ such that $M / N_1 cong M / N_2$. Can I know conclude $N_1 cong N_2$ or even $N_1 = N_2$? I know that a proper submodule can be isomorphic to the original module (I know some examples when considering vectorspaces) which is why I am almost certain that $N_1 neq N_2$ in the general case, but $N_1 cong N_2$ seems intuitive enough.
I would appreciate any help, since I am relatively new to modules.
vector-spaces modules vector-space-isomorphism module-isomorphism
asked yesterday
user9620780user9620780
1096
$endgroup$
add a comment |
$begingroup$
Let $M$ be an $R$ module and $N_1 subset N_2$ be submodules of $M$ such that $M / N_1 cong M / N_2$. Can I know conclude $N_1 cong N_2$ or even $N_1 = N_2$? I know that a proper submodule can be isomorphic to the original module (I know some examples when considering vectorspaces) which is why I am almost certain that $N_1 neq N_2$ in the general case, but $N_1 cong N_2$ seems intuitive enough.
I would appreciate any help, since I am relatively new to modules.
vector-spaces modules vector-space-isomorphism module-isomorphism
asked yesterday
user9620780user9620780
1096
$endgroup$
add a comment |
$begingroup$
Let $M$ be an $R$ module and $N_1 subset N_2$ be submodules of $M$ such that $M / N_1 cong M / N_2$. Can I know conclude $N_1 cong N_2$ or even $N_1 = N_2$? I know that a proper submodule can be isomorphic to the original module (I know some examples when considering vectorspaces) which is why I am almost certain that $N_1 neq N_2$ in the general case, but $N_1 cong N_2$ seems intuitive enough.
I would appreciate any help, since I am relatively new to modules.
vector-spaces modules vector-space-isomorphism module-isomorphism
asked yesterday
user9620780user9620780
1096
$endgroup$
Let $M$ be an $R$ module and $N_1 subset N_2$ be submodules of $M$ such that $M / N_1 cong M / N_2$. Can I know conclude $N_1 cong N_2$ or even $N_1 = N_2$? I know that a proper submodule can be isomorphic to the original module (I know some examples when considering vectorspaces) which is why I am almost certain that $N_1 neq N_2$ in the general case, but $N_1 cong N_2$ seems intuitive enough.
I would appreciate any help, since I am relatively new to modules.
vector-spaces modules vector-space-isomorphism module-isomorphism
vector-spaces modules vector-space-isomorphism module-isomorphism
asked yesterday
user9620780user9620780
1096
asked yesterday
user9620780user9620780
1096
asked yesterday
user9620780user9620780
1096
asked yesterday
user9620780user9620780
1096
asked yesterday
user9620780user9620780
1096
1096
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Nice question. No, this is not true in general, even if $R$ is a field.
Say the ring is a field, and let $V$ be a vector space over it with uncountable dimension. We can pick two subspaces $W_f$ and $W_infty$ of finite, respectively countable, dimension. Then the quotients will both have the same dimension as $V$ and thus be isomorphic to each other (we are using the axiom of choice here), but the subspaces are not isomorphic to each other.
answered yesterday
hunterhunter
15.9k32643
$endgroup$
2
$begingroup$
You don't need the axiom of choice to make a concrete example out of this idea. Just declare that $V$ is, for example, the space of functions $mathbb Rto F$ with finite support, and let $W_f$ be the trivial subspace and $W_infty$ be the subspace of functions that vanish on $(-infty,0]$. Then an isomorphism between the quotients can easily be written down expicitly.
$endgroup$
– Henning Makholm
yesterday
add a comment |
$begingroup$
Let $R$ be a nontrivial (necessarily
noncommutative) ring with the property that $Rcong Roplus R$ as a left $R$-module. (This is the case if $R$ is the endomorphism
ring of an infinite-dimensional vector space.)
Then $R$ has a proper submodule $N$ with $Ncong R$ and $R/N
cong R$ as $R$-modules. Take $M=R$, $N_1=0$ and $N_2=N$.
Clearly $N_1notcong N_2$.
answered yesterday
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3189898%2fare-two-submodules-where-one-is-contained-in-the-other-isomorphic-if-their-quo%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Nice question. No, this is not true in general, even if $R$ is a field.
Say the ring is a field, and let $V$ be a vector space over it with uncountable dimension. We can pick two subspaces $W_f$ and $W_infty$ of finite, respectively countable, dimension. Then the quotients will both have the same dimension as $V$ and thus be isomorphic to each other (we are using the axiom of choice here), but the subspaces are not isomorphic to each other.
answered yesterday
hunterhunter
15.9k32643
$endgroup$
2
$begingroup$
You don't need the axiom of choice to make a concrete example out of this idea. Just declare that $V$ is, for example, the space of functions $mathbb Rto F$ with finite support, and let $W_f$ be the trivial subspace and $W_infty$ be the subspace of functions that vanish on $(-infty,0]$. Then an isomorphism between the quotients can easily be written down expicitly.
$endgroup$
– Henning Makholm
yesterday
add a comment |
$begingroup$
Nice question. No, this is not true in general, even if $R$ is a field.
Say the ring is a field, and let $V$ be a vector space over it with uncountable dimension. We can pick two subspaces $W_f$ and $W_infty$ of finite, respectively countable, dimension. Then the quotients will both have the same dimension as $V$ and thus be isomorphic to each other (we are using the axiom of choice here), but the subspaces are not isomorphic to each other.
answered yesterday
hunterhunter
15.9k32643
$endgroup$
2
$begingroup$
You don't need the axiom of choice to make a concrete example out of this idea. Just declare that $V$ is, for example, the space of functions $mathbb Rto F$ with finite support, and let $W_f$ be the trivial subspace and $W_infty$ be the subspace of functions that vanish on $(-infty,0]$. Then an isomorphism between the quotients can easily be written down expicitly.
$endgroup$
– Henning Makholm
yesterday
add a comment |
$begingroup$
Nice question. No, this is not true in general, even if $R$ is a field.
Say the ring is a field, and let $V$ be a vector space over it with uncountable dimension. We can pick two subspaces $W_f$ and $W_infty$ of finite, respectively countable, dimension. Then the quotients will both have the same dimension as $V$ and thus be isomorphic to each other (we are using the axiom of choice here), but the subspaces are not isomorphic to each other.
answered yesterday
hunterhunter
15.9k32643
$endgroup$
Nice question. No, this is not true in general, even if $R$ is a field.
Say the ring is a field, and let $V$ be a vector space over it with uncountable dimension. We can pick two subspaces $W_f$ and $W_infty$ of finite, respectively countable, dimension. Then the quotients will both have the same dimension as $V$ and thus be isomorphic to each other (we are using the axiom of choice here), but the subspaces are not isomorphic to each other.
answered yesterday
hunterhunter
15.9k32643
answered yesterday
hunterhunter
15.9k32643
answered yesterday
hunterhunter
15.9k32643
answered yesterday
hunterhunter
15.9k32643
15.9k32643
2
$begingroup$
You don't need the axiom of choice to make a concrete example out of this idea. Just declare that $V$ is, for example, the space of functions $mathbb Rto F$ with finite support, and let $W_f$ be the trivial subspace and $W_infty$ be the subspace of functions that vanish on $(-infty,0]$. Then an isomorphism between the quotients can easily be written down expicitly.
$endgroup$
– Henning Makholm
yesterday
add a comment |
2
$begingroup$
You don't need the axiom of choice to make a concrete example out of this idea. Just declare that $V$ is, for example, the space of functions $mathbb Rto F$ with finite support, and let $W_f$ be the trivial subspace and $W_infty$ be the subspace of functions that vanish on $(-infty,0]$. Then an isomorphism between the quotients can easily be written down expicitly.
$endgroup$
– Henning Makholm
yesterday
2
2
$begingroup$
You don't need the axiom of choice to make a concrete example out of this idea. Just declare that $V$ is, for example, the space of functions $mathbb Rto F$ with finite support, and let $W_f$ be the trivial subspace and $W_infty$ be the subspace of functions that vanish on $(-infty,0]$. Then an isomorphism between the quotients can easily be written down expicitly.
$endgroup$
– Henning Makholm
yesterday
$begingroup$
You don't need the axiom of choice to make a concrete example out of this idea. Just declare that $V$ is, for example, the space of functions $mathbb Rto F$ with finite support, and let $W_f$ be the trivial subspace and $W_infty$ be the subspace of functions that vanish on $(-infty,0]$. Then an isomorphism between the quotients can easily be written down expicitly.
$endgroup$
– Henning Makholm
yesterday
add a comment |
$begingroup$
Let $R$ be a nontrivial (necessarily
noncommutative) ring with the property that $Rcong Roplus R$ as a left $R$-module. (This is the case if $R$ is the endomorphism
ring of an infinite-dimensional vector space.)
Then $R$ has a proper submodule $N$ with $Ncong R$ and $R/N
cong R$ as $R$-modules. Take $M=R$, $N_1=0$ and $N_2=N$.
Clearly $N_1notcong N_2$.
answered yesterday
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
add a comment |
$begingroup$
Let $R$ be a nontrivial (necessarily
noncommutative) ring with the property that $Rcong Roplus R$ as a left $R$-module. (This is the case if $R$ is the endomorphism
ring of an infinite-dimensional vector space.)
Then $R$ has a proper submodule $N$ with $Ncong R$ and $R/N
cong R$ as $R$-modules. Take $M=R$, $N_1=0$ and $N_2=N$.
Clearly $N_1notcong N_2$.
answered yesterday
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
add a comment |
$begingroup$
Let $R$ be a nontrivial (necessarily
noncommutative) ring with the property that $Rcong Roplus R$ as a left $R$-module. (This is the case if $R$ is the endomorphism
ring of an infinite-dimensional vector space.)
Then $R$ has a proper submodule $N$ with $Ncong R$ and $R/N
cong R$ as $R$-modules. Take $M=R$, $N_1=0$ and $N_2=N$.
Clearly $N_1notcong N_2$.
answered yesterday
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
Let $R$ be a nontrivial (necessarily
noncommutative) ring with the property that $Rcong Roplus R$ as a left $R$-module. (This is the case if $R$ is the endomorphism
ring of an infinite-dimensional vector space.)
Then $R$ has a proper submodule $N$ with $Ncong R$ and $R/N
cong R$ as $R$-modules. Take $M=R$, $N_1=0$ and $N_2=N$.
Clearly $N_1notcong N_2$.
answered yesterday
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered yesterday
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered yesterday
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered yesterday
Lord Shark the UnknownLord Shark the Unknown
109k1163136
109k1163136
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3189898%2fare-two-submodules-where-one-is-contained-in-the-other-isomorphic-if-their-quo%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
