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$M$ be a complete $R$-module and $N$ be a closed submodule of $M.$ Then $M/N$ is complete.
The number of generators of a submodule over a Principal Ideal Ring.Does every free $R$-module have a maximal proper submodule?Finitely generated submodule of a localisationEvery non-Noetherian module has a submodule maximal with respect to being not finitely generated.Completion of a module is equivalent to Cauchy sequence criterionAnnihilator of a submodule of a moduleEvery $R$-module $M$ contains an indecomposable submodule.projective module which is a submodule of a finitely generated free moduleIs $hatG$ is complete with respect to the induced topology of $G$?Dual of a complete topological group
$begingroup$
Let $M$ be a complete $R$-module with respect to the filteration $M_n.$ Also let $N$ be a closed submodule of $M.$ Then how can I show that $M/N$ is complete with respect to the induced filteration, i.e., $(M/N)_n=(N+M_n)/N.$
We should show that $M/N$ is Hausdorff and every Cauchy sequence in $M/N$ converges. Since $N= cap_n=1^infty(N+M_n)$ clearly $M/N$ is Hausdorff. But How can I show that every Cauchy sequence in $M/N$ is convergent ? I need some help. Thanks.
abstract-algebra commutative-algebra modules topological-groups
asked 2 days ago
user371231user371231
393511
$endgroup$
add a comment |
$begingroup$
Let $M$ be a complete $R$-module with respect to the filteration $M_n.$ Also let $N$ be a closed submodule of $M.$ Then how can I show that $M/N$ is complete with respect to the induced filteration, i.e., $(M/N)_n=(N+M_n)/N.$
We should show that $M/N$ is Hausdorff and every Cauchy sequence in $M/N$ converges. Since $N= cap_n=1^infty(N+M_n)$ clearly $M/N$ is Hausdorff. But How can I show that every Cauchy sequence in $M/N$ is convergent ? I need some help. Thanks.
abstract-algebra commutative-algebra modules topological-groups
asked 2 days ago
user371231user371231
393511
$endgroup$
add a comment |
$begingroup$
Let $M$ be a complete $R$-module with respect to the filteration $M_n.$ Also let $N$ be a closed submodule of $M.$ Then how can I show that $M/N$ is complete with respect to the induced filteration, i.e., $(M/N)_n=(N+M_n)/N.$
We should show that $M/N$ is Hausdorff and every Cauchy sequence in $M/N$ converges. Since $N= cap_n=1^infty(N+M_n)$ clearly $M/N$ is Hausdorff. But How can I show that every Cauchy sequence in $M/N$ is convergent ? I need some help. Thanks.
abstract-algebra commutative-algebra modules topological-groups
asked 2 days ago
user371231user371231
393511
$endgroup$
Let $M$ be a complete $R$-module with respect to the filteration $M_n.$ Also let $N$ be a closed submodule of $M.$ Then how can I show that $M/N$ is complete with respect to the induced filteration, i.e., $(M/N)_n=(N+M_n)/N.$
We should show that $M/N$ is Hausdorff and every Cauchy sequence in $M/N$ converges. Since $N= cap_n=1^infty(N+M_n)$ clearly $M/N$ is Hausdorff. But How can I show that every Cauchy sequence in $M/N$ is convergent ? I need some help. Thanks.
abstract-algebra commutative-algebra modules topological-groups
abstract-algebra commutative-algebra modules topological-groups
asked 2 days ago
user371231user371231
393511
asked 2 days ago
user371231user371231
393511
asked 2 days ago
user371231user371231
393511
asked 2 days ago
user371231user371231
393511
asked 2 days ago
user371231user371231
393511
393511
add a comment |
add a comment |
1 Answer
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$begingroup$
A Cauchy sequence in $M/N$ has the form $x_n+N$, where the $x_nin M$ satisfy relations $x_n+1-x_nin N+M_s(n)$ with $s(n)to infty$. Though $x_n$ might not be a Cauchy sequence in $M$ we can a define a Cauchy sequence $y_n$ in $M$ such that $x_nequiv y_nmod N$. The procedure is as follows. We set $y_1=x_1$. Assume that we have defined $y_n$ so that $y_nequiv x_nmod N$. Note that $x_n+1-x_n-uin M_s(n)$ for some $uin N$. We thus define $y_n+1=x_n+1-(x_n-y_n)-u$. By construction we have $y_n+1equiv x_n+1mod N$ and $y_n+1-y_n=x_n+1-x_n-uin M_s(n)$. It follows that $y_n$ is a Cauchy sequence in $M$ and has therefore a limit, say $yin M$. It is now easy to see that $y+N$ is the limit of $x_n+N=y_n+N$.
answered yesterday
DiegoDiego
2,007715
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1 Answer
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1 Answer
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$begingroup$
A Cauchy sequence in $M/N$ has the form $x_n+N$, where the $x_nin M$ satisfy relations $x_n+1-x_nin N+M_s(n)$ with $s(n)to infty$. Though $x_n$ might not be a Cauchy sequence in $M$ we can a define a Cauchy sequence $y_n$ in $M$ such that $x_nequiv y_nmod N$. The procedure is as follows. We set $y_1=x_1$. Assume that we have defined $y_n$ so that $y_nequiv x_nmod N$. Note that $x_n+1-x_n-uin M_s(n)$ for some $uin N$. We thus define $y_n+1=x_n+1-(x_n-y_n)-u$. By construction we have $y_n+1equiv x_n+1mod N$ and $y_n+1-y_n=x_n+1-x_n-uin M_s(n)$. It follows that $y_n$ is a Cauchy sequence in $M$ and has therefore a limit, say $yin M$. It is now easy to see that $y+N$ is the limit of $x_n+N=y_n+N$.
answered yesterday
DiegoDiego
2,007715
$endgroup$
add a comment |
$begingroup$
A Cauchy sequence in $M/N$ has the form $x_n+N$, where the $x_nin M$ satisfy relations $x_n+1-x_nin N+M_s(n)$ with $s(n)to infty$. Though $x_n$ might not be a Cauchy sequence in $M$ we can a define a Cauchy sequence $y_n$ in $M$ such that $x_nequiv y_nmod N$. The procedure is as follows. We set $y_1=x_1$. Assume that we have defined $y_n$ so that $y_nequiv x_nmod N$. Note that $x_n+1-x_n-uin M_s(n)$ for some $uin N$. We thus define $y_n+1=x_n+1-(x_n-y_n)-u$. By construction we have $y_n+1equiv x_n+1mod N$ and $y_n+1-y_n=x_n+1-x_n-uin M_s(n)$. It follows that $y_n$ is a Cauchy sequence in $M$ and has therefore a limit, say $yin M$. It is now easy to see that $y+N$ is the limit of $x_n+N=y_n+N$.
answered yesterday
DiegoDiego
2,007715
$endgroup$
add a comment |
$begingroup$
A Cauchy sequence in $M/N$ has the form $x_n+N$, where the $x_nin M$ satisfy relations $x_n+1-x_nin N+M_s(n)$ with $s(n)to infty$. Though $x_n$ might not be a Cauchy sequence in $M$ we can a define a Cauchy sequence $y_n$ in $M$ such that $x_nequiv y_nmod N$. The procedure is as follows. We set $y_1=x_1$. Assume that we have defined $y_n$ so that $y_nequiv x_nmod N$. Note that $x_n+1-x_n-uin M_s(n)$ for some $uin N$. We thus define $y_n+1=x_n+1-(x_n-y_n)-u$. By construction we have $y_n+1equiv x_n+1mod N$ and $y_n+1-y_n=x_n+1-x_n-uin M_s(n)$. It follows that $y_n$ is a Cauchy sequence in $M$ and has therefore a limit, say $yin M$. It is now easy to see that $y+N$ is the limit of $x_n+N=y_n+N$.
answered yesterday
DiegoDiego
2,007715
$endgroup$
A Cauchy sequence in $M/N$ has the form $x_n+N$, where the $x_nin M$ satisfy relations $x_n+1-x_nin N+M_s(n)$ with $s(n)to infty$. Though $x_n$ might not be a Cauchy sequence in $M$ we can a define a Cauchy sequence $y_n$ in $M$ such that $x_nequiv y_nmod N$. The procedure is as follows. We set $y_1=x_1$. Assume that we have defined $y_n$ so that $y_nequiv x_nmod N$. Note that $x_n+1-x_n-uin M_s(n)$ for some $uin N$. We thus define $y_n+1=x_n+1-(x_n-y_n)-u$. By construction we have $y_n+1equiv x_n+1mod N$ and $y_n+1-y_n=x_n+1-x_n-uin M_s(n)$. It follows that $y_n$ is a Cauchy sequence in $M$ and has therefore a limit, say $yin M$. It is now easy to see that $y+N$ is the limit of $x_n+N=y_n+N$.
answered yesterday
DiegoDiego
2,007715
answered yesterday
DiegoDiego
2,007715
answered yesterday
DiegoDiego
2,007715
answered yesterday
DiegoDiego
2,007715
2,007715
add a comment |
add a comment |
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