Mixing
Free generators for the localization of a module
$begingroup$
Let $A$ a commutative ring with 1 and $M$ a finitely generated $A$-module. Assume the localization $M_p$ is a free $A_p$-module for some prime ideal $p$. Is it true that there is some set of generators of $M$ whose image in $M_p$ freely generate $M_p$? Feel free to assume $A$ is Noetherian.
Fancy way to ask the same question: is there a surjective map $A^{oplus n} to M$ which induces an isomorphism $A_p^{oplus n} to M_p$?
The reason I'm asking this is: I was able to reduce some Exercise in Hartshorne (II.5.7) to this statement, and all solutions I've found online treat this step as it's obvious but I just don't see it.
algebraic-geometry commutative-algebra
asked Jan 21 at 21:16
Marco FloresMarco Flores
1,768826
$endgroup$
add a comment |
$begingroup$
Let $A$ a commutative ring with 1 and $M$ a finitely generated $A$-module. Assume the localization $M_p$ is a free $A_p$-module for some prime ideal $p$. Is it true that there is some set of generators of $M$ whose image in $M_p$ freely generate $M_p$? Feel free to assume $A$ is Noetherian.
Fancy way to ask the same question: is there a surjective map $A^{oplus n} to M$ which induces an isomorphism $A_p^{oplus n} to M_p$?
The reason I'm asking this is: I was able to reduce some Exercise in Hartshorne (II.5.7) to this statement, and all solutions I've found online treat this step as it's obvious but I just don't see it.
algebraic-geometry commutative-algebra
asked Jan 21 at 21:16
Marco FloresMarco Flores
1,768826
$endgroup$
add a comment |
$begingroup$
Let $A$ a commutative ring with 1 and $M$ a finitely generated $A$-module. Assume the localization $M_p$ is a free $A_p$-module for some prime ideal $p$. Is it true that there is some set of generators of $M$ whose image in $M_p$ freely generate $M_p$? Feel free to assume $A$ is Noetherian.
Fancy way to ask the same question: is there a surjective map $A^{oplus n} to M$ which induces an isomorphism $A_p^{oplus n} to M_p$?
The reason I'm asking this is: I was able to reduce some Exercise in Hartshorne (II.5.7) to this statement, and all solutions I've found online treat this step as it's obvious but I just don't see it.
algebraic-geometry commutative-algebra
asked Jan 21 at 21:16
Marco FloresMarco Flores
1,768826
$endgroup$
Let $A$ a commutative ring with 1 and $M$ a finitely generated $A$-module. Assume the localization $M_p$ is a free $A_p$-module for some prime ideal $p$. Is it true that there is some set of generators of $M$ whose image in $M_p$ freely generate $M_p$? Feel free to assume $A$ is Noetherian.
Fancy way to ask the same question: is there a surjective map $A^{oplus n} to M$ which induces an isomorphism $A_p^{oplus n} to M_p$?
The reason I'm asking this is: I was able to reduce some Exercise in Hartshorne (II.5.7) to this statement, and all solutions I've found online treat this step as it's obvious but I just don't see it.
algebraic-geometry commutative-algebra
algebraic-geometry commutative-algebra
asked Jan 21 at 21:16
Marco FloresMarco Flores
1,768826
asked Jan 21 at 21:16
Marco FloresMarco Flores
1,768826
asked Jan 21 at 21:16
Marco FloresMarco Flores
1,768826
asked Jan 21 at 21:16
Marco FloresMarco Flores
1,768826
asked Jan 21 at 21:16
Marco FloresMarco Flores
1,768826
1,768826
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No. For instance, if $p$ and $q$ are two different maximal ideals of $A$ and $M=A/q$, then $M_p=0$ so we would have to have $n=0$ but no map $A^0to M$ can be surjective.
This result is not needed to prove Exercise II.5.7, though, since you only need such a map $A^nto M$ locally (i.e., after localizing at some element $fin Asetminus p$.). As a sketch of a proof, you can take a basis of $M_p$ and lift each element of this basis to $M_f$ for some $fin Asetminus p$. This gives a homomorphism $A_f^nto M_f$ which is an isomorphism after localizing at $p$. To make this homomorphism surjective, note that its cokernel is a finitely generated module whose localization at $p$ is $0$, so you can find some $gin Asetminus p$ which annihilates the cokernel, and then the map $A_{fg}^nto M_{fg}$ obtained by also localizing at $g$ will be surjective. (Assuming $A$ is Noetherian, you can also do the same thing for the kernel and find a localization at an element of $Asetminus p$ which makes the map an isomorphism.)
answered Jan 21 at 22:05
Eric WofseyEric Wofsey
188k14216346
$endgroup$
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
});
}
});
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%2f3082425%2ffree-generators-for-the-localization-of-a-module%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No. For instance, if $p$ and $q$ are two different maximal ideals of $A$ and $M=A/q$, then $M_p=0$ so we would have to have $n=0$ but no map $A^0to M$ can be surjective.
This result is not needed to prove Exercise II.5.7, though, since you only need such a map $A^nto M$ locally (i.e., after localizing at some element $fin Asetminus p$.). As a sketch of a proof, you can take a basis of $M_p$ and lift each element of this basis to $M_f$ for some $fin Asetminus p$. This gives a homomorphism $A_f^nto M_f$ which is an isomorphism after localizing at $p$. To make this homomorphism surjective, note that its cokernel is a finitely generated module whose localization at $p$ is $0$, so you can find some $gin Asetminus p$ which annihilates the cokernel, and then the map $A_{fg}^nto M_{fg}$ obtained by also localizing at $g$ will be surjective. (Assuming $A$ is Noetherian, you can also do the same thing for the kernel and find a localization at an element of $Asetminus p$ which makes the map an isomorphism.)
answered Jan 21 at 22:05
Eric WofseyEric Wofsey
188k14216346
$endgroup$
add a comment |
$begingroup$
No. For instance, if $p$ and $q$ are two different maximal ideals of $A$ and $M=A/q$, then $M_p=0$ so we would have to have $n=0$ but no map $A^0to M$ can be surjective.
This result is not needed to prove Exercise II.5.7, though, since you only need such a map $A^nto M$ locally (i.e., after localizing at some element $fin Asetminus p$.). As a sketch of a proof, you can take a basis of $M_p$ and lift each element of this basis to $M_f$ for some $fin Asetminus p$. This gives a homomorphism $A_f^nto M_f$ which is an isomorphism after localizing at $p$. To make this homomorphism surjective, note that its cokernel is a finitely generated module whose localization at $p$ is $0$, so you can find some $gin Asetminus p$ which annihilates the cokernel, and then the map $A_{fg}^nto M_{fg}$ obtained by also localizing at $g$ will be surjective. (Assuming $A$ is Noetherian, you can also do the same thing for the kernel and find a localization at an element of $Asetminus p$ which makes the map an isomorphism.)
answered Jan 21 at 22:05
Eric WofseyEric Wofsey
188k14216346
$endgroup$
add a comment |
$begingroup$
No. For instance, if $p$ and $q$ are two different maximal ideals of $A$ and $M=A/q$, then $M_p=0$ so we would have to have $n=0$ but no map $A^0to M$ can be surjective.
This result is not needed to prove Exercise II.5.7, though, since you only need such a map $A^nto M$ locally (i.e., after localizing at some element $fin Asetminus p$.). As a sketch of a proof, you can take a basis of $M_p$ and lift each element of this basis to $M_f$ for some $fin Asetminus p$. This gives a homomorphism $A_f^nto M_f$ which is an isomorphism after localizing at $p$. To make this homomorphism surjective, note that its cokernel is a finitely generated module whose localization at $p$ is $0$, so you can find some $gin Asetminus p$ which annihilates the cokernel, and then the map $A_{fg}^nto M_{fg}$ obtained by also localizing at $g$ will be surjective. (Assuming $A$ is Noetherian, you can also do the same thing for the kernel and find a localization at an element of $Asetminus p$ which makes the map an isomorphism.)
answered Jan 21 at 22:05
Eric WofseyEric Wofsey
188k14216346
$endgroup$
No. For instance, if $p$ and $q$ are two different maximal ideals of $A$ and $M=A/q$, then $M_p=0$ so we would have to have $n=0$ but no map $A^0to M$ can be surjective.
This result is not needed to prove Exercise II.5.7, though, since you only need such a map $A^nto M$ locally (i.e., after localizing at some element $fin Asetminus p$.). As a sketch of a proof, you can take a basis of $M_p$ and lift each element of this basis to $M_f$ for some $fin Asetminus p$. This gives a homomorphism $A_f^nto M_f$ which is an isomorphism after localizing at $p$. To make this homomorphism surjective, note that its cokernel is a finitely generated module whose localization at $p$ is $0$, so you can find some $gin Asetminus p$ which annihilates the cokernel, and then the map $A_{fg}^nto M_{fg}$ obtained by also localizing at $g$ will be surjective. (Assuming $A$ is Noetherian, you can also do the same thing for the kernel and find a localization at an element of $Asetminus p$ which makes the map an isomorphism.)
answered Jan 21 at 22:05
Eric WofseyEric Wofsey
188k14216346
answered Jan 21 at 22:05
Eric WofseyEric Wofsey
188k14216346
answered Jan 21 at 22:05
Eric WofseyEric Wofsey
188k14216346
answered Jan 21 at 22:05
Eric WofseyEric Wofsey
188k14216346
188k14216346
add a comment |
add a comment |
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%2f3082425%2ffree-generators-for-the-localization-of-a-module%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
