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 13 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 13 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 13 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 13 hours ago
user9620780user9620780
1096
asked 13 hours ago
user9620780user9620780
1096
asked 13 hours ago
user9620780user9620780
1096
asked 13 hours ago
user9620780user9620780
1096
asked 13 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 13 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
8 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 13 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 13 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
8 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 13 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
8 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 13 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 13 hours ago
hunterhunter
15.9k32643
answered 13 hours ago
hunterhunter
15.9k32643
answered 13 hours ago
hunterhunter
15.9k32643
answered 13 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
8 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
8 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
8 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
8 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 13 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 13 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 13 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 13 hours ago
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered 13 hours ago
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered 13 hours ago
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered 13 hours ago
Lord Shark the UnknownLord Shark the Unknown
109k1163136
109k1163136
add a comment |
add a comment |
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-module-isomorphism, modules, vector-space-isomorphism, vector-spaces
