Is it possible for an integral domain to have an ideal that cannot be generated by a countable set?
$begingroup$
This question came up when I was working on the following problem
Let $R$ be an integral domain. Prove that if the following two conditions hold then $R$ is a Principal Ideal Domain:
i) any two nonzero elements $a$ and $b$ in $R$ have a greatest common divisor which can be written in the form $ra+sb$ for some $r,s in R$.
ii) if $a_1, a_2, a_3,ldots$ are nonzero elements in $R$ such that $a_{i+1}|a_i$ for all $i$, then there is a positive integer $N$ such that $a_n$ is a unit times $a_N$ for all $n geq N$.
I have managed to prove this by assuming that an arbitrary ideal in an integral domain $R$ can be written as $I = (x_1, x_2, ldots)$ for some $x_i in R$, if however there existed an ideal that could only be generated by an uncountable set in $R$ then my proof would not work.
ring-theory principal-ideal-domains integral-domain
asked Dec 3 '18 at 10:12
DanielDaniel
162
$endgroup$
add a comment |
$begingroup$
This question came up when I was working on the following problem
Let $R$ be an integral domain. Prove that if the following two conditions hold then $R$ is a Principal Ideal Domain:
i) any two nonzero elements $a$ and $b$ in $R$ have a greatest common divisor which can be written in the form $ra+sb$ for some $r,s in R$.
ii) if $a_1, a_2, a_3,ldots$ are nonzero elements in $R$ such that $a_{i+1}|a_i$ for all $i$, then there is a positive integer $N$ such that $a_n$ is a unit times $a_N$ for all $n geq N$.
I have managed to prove this by assuming that an arbitrary ideal in an integral domain $R$ can be written as $I = (x_1, x_2, ldots)$ for some $x_i in R$, if however there existed an ideal that could only be generated by an uncountable set in $R$ then my proof would not work.
ring-theory principal-ideal-domains integral-domain
asked Dec 3 '18 at 10:12
DanielDaniel
162
$endgroup$
add a comment |
$begingroup$
This question came up when I was working on the following problem
Let $R$ be an integral domain. Prove that if the following two conditions hold then $R$ is a Principal Ideal Domain:
i) any two nonzero elements $a$ and $b$ in $R$ have a greatest common divisor which can be written in the form $ra+sb$ for some $r,s in R$.
ii) if $a_1, a_2, a_3,ldots$ are nonzero elements in $R$ such that $a_{i+1}|a_i$ for all $i$, then there is a positive integer $N$ such that $a_n$ is a unit times $a_N$ for all $n geq N$.
I have managed to prove this by assuming that an arbitrary ideal in an integral domain $R$ can be written as $I = (x_1, x_2, ldots)$ for some $x_i in R$, if however there existed an ideal that could only be generated by an uncountable set in $R$ then my proof would not work.
ring-theory principal-ideal-domains integral-domain
asked Dec 3 '18 at 10:12
DanielDaniel
162
$endgroup$
This question came up when I was working on the following problem
Let $R$ be an integral domain. Prove that if the following two conditions hold then $R$ is a Principal Ideal Domain:
i) any two nonzero elements $a$ and $b$ in $R$ have a greatest common divisor which can be written in the form $ra+sb$ for some $r,s in R$.
ii) if $a_1, a_2, a_3,ldots$ are nonzero elements in $R$ such that $a_{i+1}|a_i$ for all $i$, then there is a positive integer $N$ such that $a_n$ is a unit times $a_N$ for all $n geq N$.
I have managed to prove this by assuming that an arbitrary ideal in an integral domain $R$ can be written as $I = (x_1, x_2, ldots)$ for some $x_i in R$, if however there existed an ideal that could only be generated by an uncountable set in $R$ then my proof would not work.
ring-theory principal-ideal-domains integral-domain
ring-theory principal-ideal-domains integral-domain
asked Dec 3 '18 at 10:12
DanielDaniel
162
asked Dec 3 '18 at 10:12
DanielDaniel
162
asked Dec 3 '18 at 10:12
DanielDaniel
162
asked Dec 3 '18 at 10:12
DanielDaniel
162
asked Dec 3 '18 at 10:12
DanielDaniel
162
162
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The obvious counterexample seems to work: that is, $F[{x_imid iin I}]$ for an uncountable index set $I$ and field $F$. The ideal generated by the $x_i$ is not countable generated.
It’s hard to imagine how your argument relied on countable generation. Could you explain more clearly what you argued? Otherwise it is hard to help.
answered Dec 3 '18 at 11:58
rschwiebrschwieb
107k12102250
$endgroup$
$begingroup$
I start with a field of the form $I=(x_1,x_2,…)$ then define the sequence $a_1 = x_1$, $a_i = gcd(a_{i-1},x_i)$ for $i geq 2$ and then show that $(x_1,ldots,x_i) = (a_i)$ for all $i in mathbb{N}$ using property i). Then I use property ii) to show that there exists an $N$ such that $(a_i) = (a_N)$ for all $i geq N$. Then this shows that $(x_1, x_2, ldots) = (a_N)$. Meaning the ideal $I$ is principal.
$endgroup$
– Daniel
Dec 3 '18 at 12:14
$begingroup$
Sorry I meant Ideal* at the beginning
$endgroup$
– Daniel
Dec 3 '18 at 12:24
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%2f3023872%2fis-it-possible-for-an-integral-domain-to-have-an-ideal-that-cannot-be-generated%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 obvious counterexample seems to work: that is, $F[{x_imid iin I}]$ for an uncountable index set $I$ and field $F$. The ideal generated by the $x_i$ is not countable generated.
It’s hard to imagine how your argument relied on countable generation. Could you explain more clearly what you argued? Otherwise it is hard to help.
answered Dec 3 '18 at 11:58
rschwiebrschwieb
107k12102250
$endgroup$
$begingroup$
I start with a field of the form $I=(x_1,x_2,…)$ then define the sequence $a_1 = x_1$, $a_i = gcd(a_{i-1},x_i)$ for $i geq 2$ and then show that $(x_1,ldots,x_i) = (a_i)$ for all $i in mathbb{N}$ using property i). Then I use property ii) to show that there exists an $N$ such that $(a_i) = (a_N)$ for all $i geq N$. Then this shows that $(x_1, x_2, ldots) = (a_N)$. Meaning the ideal $I$ is principal.
$endgroup$
– Daniel
Dec 3 '18 at 12:14
$begingroup$
Sorry I meant Ideal* at the beginning
$endgroup$
– Daniel
Dec 3 '18 at 12:24
add a comment |
$begingroup$
The obvious counterexample seems to work: that is, $F[{x_imid iin I}]$ for an uncountable index set $I$ and field $F$. The ideal generated by the $x_i$ is not countable generated.
It’s hard to imagine how your argument relied on countable generation. Could you explain more clearly what you argued? Otherwise it is hard to help.
answered Dec 3 '18 at 11:58
rschwiebrschwieb
107k12102250
$endgroup$
$begingroup$
I start with a field of the form $I=(x_1,x_2,…)$ then define the sequence $a_1 = x_1$, $a_i = gcd(a_{i-1},x_i)$ for $i geq 2$ and then show that $(x_1,ldots,x_i) = (a_i)$ for all $i in mathbb{N}$ using property i). Then I use property ii) to show that there exists an $N$ such that $(a_i) = (a_N)$ for all $i geq N$. Then this shows that $(x_1, x_2, ldots) = (a_N)$. Meaning the ideal $I$ is principal.
$endgroup$
– Daniel
Dec 3 '18 at 12:14
$begingroup$
Sorry I meant Ideal* at the beginning
$endgroup$
– Daniel
Dec 3 '18 at 12:24
add a comment |
$begingroup$
The obvious counterexample seems to work: that is, $F[{x_imid iin I}]$ for an uncountable index set $I$ and field $F$. The ideal generated by the $x_i$ is not countable generated.
It’s hard to imagine how your argument relied on countable generation. Could you explain more clearly what you argued? Otherwise it is hard to help.
answered Dec 3 '18 at 11:58
rschwiebrschwieb
107k12102250
$endgroup$
The obvious counterexample seems to work: that is, $F[{x_imid iin I}]$ for an uncountable index set $I$ and field $F$. The ideal generated by the $x_i$ is not countable generated.
It’s hard to imagine how your argument relied on countable generation. Could you explain more clearly what you argued? Otherwise it is hard to help.
answered Dec 3 '18 at 11:58
rschwiebrschwieb
107k12102250
answered Dec 3 '18 at 11:58
rschwiebrschwieb
107k12102250
answered Dec 3 '18 at 11:58
rschwiebrschwieb
107k12102250
answered Dec 3 '18 at 11:58
rschwiebrschwieb
107k12102250
107k12102250
$begingroup$
I start with a field of the form $I=(x_1,x_2,…)$ then define the sequence $a_1 = x_1$, $a_i = gcd(a_{i-1},x_i)$ for $i geq 2$ and then show that $(x_1,ldots,x_i) = (a_i)$ for all $i in mathbb{N}$ using property i). Then I use property ii) to show that there exists an $N$ such that $(a_i) = (a_N)$ for all $i geq N$. Then this shows that $(x_1, x_2, ldots) = (a_N)$. Meaning the ideal $I$ is principal.
$endgroup$
– Daniel
Dec 3 '18 at 12:14
$begingroup$
Sorry I meant Ideal* at the beginning
$endgroup$
– Daniel
Dec 3 '18 at 12:24
add a comment |
$begingroup$
I start with a field of the form $I=(x_1,x_2,…)$ then define the sequence $a_1 = x_1$, $a_i = gcd(a_{i-1},x_i)$ for $i geq 2$ and then show that $(x_1,ldots,x_i) = (a_i)$ for all $i in mathbb{N}$ using property i). Then I use property ii) to show that there exists an $N$ such that $(a_i) = (a_N)$ for all $i geq N$. Then this shows that $(x_1, x_2, ldots) = (a_N)$. Meaning the ideal $I$ is principal.
$endgroup$
– Daniel
Dec 3 '18 at 12:14
$begingroup$
Sorry I meant Ideal* at the beginning
$endgroup$
– Daniel
Dec 3 '18 at 12:24
$begingroup$
I start with a field of the form $I=(x_1,x_2,…)$ then define the sequence $a_1 = x_1$, $a_i = gcd(a_{i-1},x_i)$ for $i geq 2$ and then show that $(x_1,ldots,x_i) = (a_i)$ for all $i in mathbb{N}$ using property i). Then I use property ii) to show that there exists an $N$ such that $(a_i) = (a_N)$ for all $i geq N$. Then this shows that $(x_1, x_2, ldots) = (a_N)$. Meaning the ideal $I$ is principal.
$endgroup$
– Daniel
Dec 3 '18 at 12:14
$begingroup$
I start with a field of the form $I=(x_1,x_2,…)$ then define the sequence $a_1 = x_1$, $a_i = gcd(a_{i-1},x_i)$ for $i geq 2$ and then show that $(x_1,ldots,x_i) = (a_i)$ for all $i in mathbb{N}$ using property i). Then I use property ii) to show that there exists an $N$ such that $(a_i) = (a_N)$ for all $i geq N$. Then this shows that $(x_1, x_2, ldots) = (a_N)$. Meaning the ideal $I$ is principal.
$endgroup$
– Daniel
Dec 3 '18 at 12:14
$begingroup$
Sorry I meant Ideal* at the beginning
$endgroup$
– Daniel
Dec 3 '18 at 12:24
$begingroup$
Sorry I meant Ideal* at the beginning
$endgroup$
– Daniel
Dec 3 '18 at 12:24
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%2f3023872%2fis-it-possible-for-an-integral-domain-to-have-an-ideal-that-cannot-be-generated%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
