Mixing
Factorisation of a map of modules
$begingroup$
Suppose you have a ring $A$, $B$ with and an $A$-module $M$. Suppose the projection $q:Arightarrow B$ is surjective, and that there's a map $f:Arightarrow M$. Under what assumptions and why does $f$ factor through a map $Brightarrow M$?
$$
newcommand{ra}[1]{!!!!!!!!!!!!xrightarrow{quad#1quad}!!!!!!!!}
newcommand{da}[1]{leftdownarrow{scriptstyle#1}vphantom{displaystyleint_0^1}right.}
%
begin{array}{lllllll}
A & ra{f} & M \
da{q} & & \
B & \
end{array}
$$
Sorry for the shitty diagram, but basically the question is when is there a unique arrow from $B$ to $M$ that makes the diagram commute.
abstract-algebra modules
asked Jan 20 at 12:49
DalamarDalamar
465410
$endgroup$
add a comment |
$begingroup$
Suppose you have a ring $A$, $B$ with and an $A$-module $M$. Suppose the projection $q:Arightarrow B$ is surjective, and that there's a map $f:Arightarrow M$. Under what assumptions and why does $f$ factor through a map $Brightarrow M$?
$$
newcommand{ra}[1]{!!!!!!!!!!!!xrightarrow{quad#1quad}!!!!!!!!}
newcommand{da}[1]{leftdownarrow{scriptstyle#1}vphantom{displaystyleint_0^1}right.}
%
begin{array}{lllllll}
A & ra{f} & M \
da{q} & & \
B & \
end{array}
$$
Sorry for the shitty diagram, but basically the question is when is there a unique arrow from $B$ to $M$ that makes the diagram commute.
abstract-algebra modules
asked Jan 20 at 12:49
DalamarDalamar
465410
$endgroup$
1
$begingroup$
What is $q$ exactly? A ring homomorphism? And what is $B$?
$endgroup$
– Randall
Jan 20 at 12:51
$begingroup$
Yes. You can take even B to be just a quotient ring A/I with I an ideal of A
$endgroup$
– Dalamar
Jan 20 at 13:07
add a comment |
$begingroup$
Suppose you have a ring $A$, $B$ with and an $A$-module $M$. Suppose the projection $q:Arightarrow B$ is surjective, and that there's a map $f:Arightarrow M$. Under what assumptions and why does $f$ factor through a map $Brightarrow M$?
$$
newcommand{ra}[1]{!!!!!!!!!!!!xrightarrow{quad#1quad}!!!!!!!!}
newcommand{da}[1]{leftdownarrow{scriptstyle#1}vphantom{displaystyleint_0^1}right.}
%
begin{array}{lllllll}
A & ra{f} & M \
da{q} & & \
B & \
end{array}
$$
Sorry for the shitty diagram, but basically the question is when is there a unique arrow from $B$ to $M$ that makes the diagram commute.
abstract-algebra modules
asked Jan 20 at 12:49
DalamarDalamar
465410
$endgroup$
Suppose you have a ring $A$, $B$ with and an $A$-module $M$. Suppose the projection $q:Arightarrow B$ is surjective, and that there's a map $f:Arightarrow M$. Under what assumptions and why does $f$ factor through a map $Brightarrow M$?
$$
newcommand{ra}[1]{!!!!!!!!!!!!xrightarrow{quad#1quad}!!!!!!!!}
newcommand{da}[1]{leftdownarrow{scriptstyle#1}vphantom{displaystyleint_0^1}right.}
%
begin{array}{lllllll}
A & ra{f} & M \
da{q} & & \
B & \
end{array}
$$
Sorry for the shitty diagram, but basically the question is when is there a unique arrow from $B$ to $M$ that makes the diagram commute.
abstract-algebra modules
abstract-algebra modules
asked Jan 20 at 12:49
DalamarDalamar
465410
asked Jan 20 at 12:49
DalamarDalamar
465410
asked Jan 20 at 12:49
DalamarDalamar
465410
asked Jan 20 at 12:49
DalamarDalamar
465410
asked Jan 20 at 12:49
DalamarDalamar
465410
465410
1
$begingroup$
What is $q$ exactly? A ring homomorphism? And what is $B$?
$endgroup$
– Randall
Jan 20 at 12:51
$begingroup$
Yes. You can take even B to be just a quotient ring A/I with I an ideal of A
$endgroup$
– Dalamar
Jan 20 at 13:07
add a comment |
1
$begingroup$
What is $q$ exactly? A ring homomorphism? And what is $B$?
$endgroup$
– Randall
Jan 20 at 12:51
$begingroup$
Yes. You can take even B to be just a quotient ring A/I with I an ideal of A
$endgroup$
– Dalamar
Jan 20 at 13:07
1
1
$begingroup$
What is $q$ exactly? A ring homomorphism? And what is $B$?
$endgroup$
– Randall
Jan 20 at 12:51
$begingroup$
What is $q$ exactly? A ring homomorphism? And what is $B$?
$endgroup$
– Randall
Jan 20 at 12:51
$begingroup$
Yes. You can take even B to be just a quotient ring A/I with I an ideal of A
$endgroup$
– Dalamar
Jan 20 at 13:07
$begingroup$
Yes. You can take even B to be just a quotient ring A/I with I an ideal of A
$endgroup$
– Dalamar
Jan 20 at 13:07
add a comment |
0
active
oldest
votes
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
});
}
});
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%2f3080544%2ffactorisation-of-a-map-of-modules%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3080544%2ffactorisation-of-a-map-of-modules%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
1
$begingroup$
What is $q$ exactly? A ring homomorphism? And what is $B$?
$endgroup$
– Randall
Jan 20 at 12:51
$begingroup$
Yes. You can take even B to be just a quotient ring A/I with I an ideal of A
$endgroup$
– Dalamar
Jan 20 at 13:07