Fdtxjr

Movie: boy escapes the real world and goes to a fantasy world with big furry trolls

Help! My Character is too much for her story!

Logistic regression BIC: what's the right N?

Rationale to prefer local variables over instance variables?

Can the Witch Sight warlock invocation see through the Mirror Image spell?

How to copy the rest of lines of a file to another file

Has a sovereign Communist government ever run, and conceded loss, on a fair election?

How do you make a gun that shoots melee weapons and/or swords?

Too soon for a plot twist?

PTIJ: Who was the sixth set of priestly clothes for?

Cycles on the torus

ESPP--any reason not to go all in?

Why does Central Limit Theorem break down in my simulation?

Use Mercury as quenching liquid for swords?

Under what conditions can the right to remain silent be revoked in the USA?

The (Easy) Road to Code

(Codewars) Linked Lists-Sorted Insert

Is "cogitate" used appropriately in "I cogitate that success relies on hard work"?

Sampling from Gaussian mixture models, when are the sampled data independent?

How do spaceships determine each other's mass in space?

School performs periodic password audits. Is my password compromised?

What does the Digital Threat scope actually do?

Can I negotiate a patent idea for a raise, under French law?

Writing text next to a table

$M$ 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

1

$begingroup$

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

share|cite|improve this question

asked 2 days ago

user371231user371231

393511

$endgroup$

add a comment | 

1

$begingroup$

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

share|cite|improve this question

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

share|cite|improve this question

asked 2 days ago

user371231user371231

393511

$endgroup$

share|cite|improve this question

asked 2 days ago

user371231user371231

393511

share|cite|improve this question

asked 2 days ago

user371231user371231

393511

asked 2 days ago

user371231user371231

393511

asked 2 days ago

user371231user371231

393511

1 Answer
1

active

oldest

votes

2

$begingroup$

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 $x_n+N=y_n+N$.

share|cite|improve this answer

answered yesterday

DiegoDiego

2,007715

$endgroup$

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

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
);



);

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3138684%2fm-be-a-complete-r-module-and-n-be-a-closed-submodule-of-m-then-m-n-is%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

2

$begingroup$

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 $x_n+N=y_n+N$.

share|cite|improve this answer

answered yesterday

DiegoDiego

2,007715

$endgroup$

add a comment | 

2

$begingroup$

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 $x_n+N=y_n+N$.

share|cite|improve this answer

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 $x_n+N=y_n+N$.

share|cite|improve this answer

answered yesterday

DiegoDiego

2,007715

$endgroup$

share|cite|improve this answer

answered yesterday

DiegoDiego

2,007715

answered yesterday

DiegoDiego

2,007715

answered yesterday

DiegoDiego

2,007715