Are two submodules (where one is contained in the other) isomorphic if their quotientmodules are isomorphic? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Error in proof that submodules of f.g. modules are f.g.Pure Submodules and Finitely Presented versus Finitely Generated SubmodulesProving that two quotient modules are isomorphicLet M be an Artinian module over a commutative Artinian unital ring R. Is M necessarily finitely generated?Slick proof of isomorphism of quotients of direct sums of modulesNumber of submodules of a semisimple moduleProduct/coproduct properties: If $N_1simeq N_2$ in some category, then $N_1times N_3simeq N_2times N_3$?A module of a Lie algebra can be decomposed into irreducibles if and only if every submodule has a complement.Confusion over finite generation and noetherian modulesOn showing that $varphi(N)=(N+K)/K$, for modules $N,K<M$, $varphi$ natural map.
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Are two submodules (where one is contained in the other) isomorphic if their quotientmodules are isomorphic?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Error in proof that submodules of f.g. modules are f.g.Pure Submodules and Finitely Presented versus Finitely Generated SubmodulesProving that two quotient modules are isomorphicLet M be an Artinian module over a commutative Artinian unital ring R. Is M necessarily finitely generated?Slick proof of isomorphism of quotients of direct sums of modulesNumber of submodules of a semisimple moduleProduct/coproduct properties: If $N_1simeq N_2$ in some category, then $N_1times N_3simeq N_2times N_3$?A module of a Lie algebra can be decomposed into irreducibles if and only if every submodule has a complement.Confusion over finite generation and noetherian modulesOn showing that $varphi(N)=(N+K)/K$, for modules $N,K<M$, $varphi$ natural map.
$begingroup$
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 9 hours ago
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 9 hours ago
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 9 hours ago
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 9 hours ago
user9620780user9620780
1096
asked 9 hours ago
user9620780user9620780
1096
asked 9 hours ago
user9620780user9620780
1096
asked 9 hours ago
user9620780user9620780
1096
asked 9 hours ago
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 9 hours ago
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
4 hours ago
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 9 hours ago
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
add a comment |
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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 9 hours ago
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
4 hours ago
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 9 hours ago
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
4 hours ago
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 9 hours ago
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 9 hours ago
hunterhunter
15.9k32643
answered 9 hours ago
hunterhunter
15.9k32643
answered 9 hours ago
hunterhunter
15.9k32643
answered 9 hours ago
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
4 hours ago
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
4 hours ago
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
4 hours ago
$begingroup$
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
4 hours ago
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 9 hours ago
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 9 hours ago
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 9 hours ago
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 9 hours ago
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered 9 hours ago
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered 9 hours ago
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered 9 hours ago
Lord Shark the UnknownLord Shark the Unknown
109k1163136
109k1163136
add a comment |
add a comment |
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