Over which commutative rings do we have Smith normal form?
$begingroup$
It is well-known that matrices over a PID
always have Smith normal form as in https://en.m.wikipedia.org/wiki/Smith_normal_form. What about general rings which may not even be noetherian?
abstract-algebra commutative-algebra
asked Dec 4 '18 at 19:02
zzyzzy
2,6111420
$endgroup$
add a comment |
$begingroup$
It is well-known that matrices over a PID
always have Smith normal form as in https://en.m.wikipedia.org/wiki/Smith_normal_form. What about general rings which may not even be noetherian?
abstract-algebra commutative-algebra
asked Dec 4 '18 at 19:02
zzyzzy
2,6111420
$endgroup$
$begingroup$
see here also mathoverflow.net/questions/31275/…
$endgroup$
– Badam Baplan
Dec 5 '18 at 1:33
add a comment |
$begingroup$
It is well-known that matrices over a PID
always have Smith normal form as in https://en.m.wikipedia.org/wiki/Smith_normal_form. What about general rings which may not even be noetherian?
abstract-algebra commutative-algebra
asked Dec 4 '18 at 19:02
zzyzzy
2,6111420
$endgroup$
It is well-known that matrices over a PID
always have Smith normal form as in https://en.m.wikipedia.org/wiki/Smith_normal_form. What about general rings which may not even be noetherian?
abstract-algebra commutative-algebra
abstract-algebra commutative-algebra
asked Dec 4 '18 at 19:02
zzyzzy
2,6111420
asked Dec 4 '18 at 19:02
zzyzzy
2,6111420
asked Dec 4 '18 at 19:02
zzyzzy
2,6111420
asked Dec 4 '18 at 19:02
zzyzzy
2,6111420
asked Dec 4 '18 at 19:02
zzyzzy
2,6111420
2,6111420
$begingroup$
see here also mathoverflow.net/questions/31275/…
$endgroup$
– Badam Baplan
Dec 5 '18 at 1:33
add a comment |
$begingroup$
see here also mathoverflow.net/questions/31275/…
$endgroup$
– Badam Baplan
Dec 5 '18 at 1:33
$begingroup$
see here also mathoverflow.net/questions/31275/…
$endgroup$
– Badam Baplan
Dec 5 '18 at 1:33
$begingroup$
see here also mathoverflow.net/questions/31275/…
$endgroup$
– Badam Baplan
Dec 5 '18 at 1:33
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The algorithm requires at least a Bézout domain to express GCDs as linear combinations, and I think it requires some sort of bound on factorizations to prevent chasing some infinite factorizations forever.
The safest thing to do to control factorizations would be to assume it is a UFD, but a Bézout UFD is already a PID.
answered Dec 4 '18 at 19:20
rschwiebrschwieb
107k12102251
$endgroup$
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%2f3025972%2fover-which-commutative-rings-do-we-have-smith-normal-form%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
$begingroup$
The algorithm requires at least a Bézout domain to express GCDs as linear combinations, and I think it requires some sort of bound on factorizations to prevent chasing some infinite factorizations forever.
The safest thing to do to control factorizations would be to assume it is a UFD, but a Bézout UFD is already a PID.
answered Dec 4 '18 at 19:20
rschwiebrschwieb
107k12102251
$endgroup$
add a comment |
$begingroup$
The algorithm requires at least a Bézout domain to express GCDs as linear combinations, and I think it requires some sort of bound on factorizations to prevent chasing some infinite factorizations forever.
The safest thing to do to control factorizations would be to assume it is a UFD, but a Bézout UFD is already a PID.
answered Dec 4 '18 at 19:20
rschwiebrschwieb
107k12102251
$endgroup$
add a comment |
$begingroup$
The algorithm requires at least a Bézout domain to express GCDs as linear combinations, and I think it requires some sort of bound on factorizations to prevent chasing some infinite factorizations forever.
The safest thing to do to control factorizations would be to assume it is a UFD, but a Bézout UFD is already a PID.
answered Dec 4 '18 at 19:20
rschwiebrschwieb
107k12102251
$endgroup$
The algorithm requires at least a Bézout domain to express GCDs as linear combinations, and I think it requires some sort of bound on factorizations to prevent chasing some infinite factorizations forever.
The safest thing to do to control factorizations would be to assume it is a UFD, but a Bézout UFD is already a PID.
answered Dec 4 '18 at 19:20
rschwiebrschwieb
107k12102251
answered Dec 4 '18 at 19:20
rschwiebrschwieb
107k12102251
answered Dec 4 '18 at 19:20
rschwiebrschwieb
107k12102251
answered Dec 4 '18 at 19:20
rschwiebrschwieb
107k12102251
107k12102251
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%2f3025972%2fover-which-commutative-rings-do-we-have-smith-normal-form%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
$begingroup$
see here also mathoverflow.net/questions/31275/…
$endgroup$
– Badam Baplan
Dec 5 '18 at 1:33