Some counterexamples in basic ring theory
$begingroup$
Give an example if possible, and if not possible explain why not.
a) A subring of a PID that is not PID.
b) A PID that is a subring of a non-PID.
c) A subring of a PID that is not UFD.
My approach:
a) Since $mathbb{Q}$ is field then $mathbb{Q}[x]$ is PID. But it's subring $mathbb{Z}[x]$ is not PID because the ideal $(2,x)$ in $mathbb{Z}[x]$ is not principal.
b) Consider the subring $mathbb{Z}$ of the ring $mathbb{Z}[x]$. We know that $mathbb{Z}$ is euclidean domain and hence is PID.
c) Consider the subring $mathbb{Z}[sqrt{-5}]$ of the ring $mathbb{Q}[sqrt{-5}]$. I know that $mathbb{Z}[sqrt{-5}]$ is not UFD, but $mathbb{Q}[sqrt{-5}]cong mathbb{Q}[x]/(x^2+5)$ and since $x^2+5$ is irreducible in $mathbb{Q}[x]$ then it's field $Rightarrow$ $mathbb{Q}[sqrt{-5}]$ is also field $Rightarrow$ is PID.
Are my examples correct?
Would be very grateful!
abstract-algebra ring-theory principal-ideal-domains unique-factorization-domains
asked Dec 9 '18 at 18:58
ZFRZFR
5,08531340
$endgroup$
add a comment |
$begingroup$
Give an example if possible, and if not possible explain why not.
a) A subring of a PID that is not PID.
b) A PID that is a subring of a non-PID.
c) A subring of a PID that is not UFD.
My approach:
a) Since $mathbb{Q}$ is field then $mathbb{Q}[x]$ is PID. But it's subring $mathbb{Z}[x]$ is not PID because the ideal $(2,x)$ in $mathbb{Z}[x]$ is not principal.
b) Consider the subring $mathbb{Z}$ of the ring $mathbb{Z}[x]$. We know that $mathbb{Z}$ is euclidean domain and hence is PID.
c) Consider the subring $mathbb{Z}[sqrt{-5}]$ of the ring $mathbb{Q}[sqrt{-5}]$. I know that $mathbb{Z}[sqrt{-5}]$ is not UFD, but $mathbb{Q}[sqrt{-5}]cong mathbb{Q}[x]/(x^2+5)$ and since $x^2+5$ is irreducible in $mathbb{Q}[x]$ then it's field $Rightarrow$ $mathbb{Q}[sqrt{-5}]$ is also field $Rightarrow$ is PID.
Are my examples correct?
Would be very grateful!
abstract-algebra ring-theory principal-ideal-domains unique-factorization-domains
asked Dec 9 '18 at 18:58
ZFRZFR
5,08531340
$endgroup$
$begingroup$
looks good to me
$endgroup$
– zoidberg
Dec 9 '18 at 19:04
1
$begingroup$
Each question has been answered at this site already. You could also compare. For example, b) is here. For a), see here. Actually, you have the same example.
$endgroup$
– Dietrich Burde
Dec 9 '18 at 19:17
$begingroup$
Try subrings of a polynomial ring over a field, $k[x]$
$endgroup$
– R.C.Cowsik
Dec 10 '18 at 6:48
add a comment |
$begingroup$
Give an example if possible, and if not possible explain why not.
a) A subring of a PID that is not PID.
b) A PID that is a subring of a non-PID.
c) A subring of a PID that is not UFD.
My approach:
a) Since $mathbb{Q}$ is field then $mathbb{Q}[x]$ is PID. But it's subring $mathbb{Z}[x]$ is not PID because the ideal $(2,x)$ in $mathbb{Z}[x]$ is not principal.
b) Consider the subring $mathbb{Z}$ of the ring $mathbb{Z}[x]$. We know that $mathbb{Z}$ is euclidean domain and hence is PID.
c) Consider the subring $mathbb{Z}[sqrt{-5}]$ of the ring $mathbb{Q}[sqrt{-5}]$. I know that $mathbb{Z}[sqrt{-5}]$ is not UFD, but $mathbb{Q}[sqrt{-5}]cong mathbb{Q}[x]/(x^2+5)$ and since $x^2+5$ is irreducible in $mathbb{Q}[x]$ then it's field $Rightarrow$ $mathbb{Q}[sqrt{-5}]$ is also field $Rightarrow$ is PID.
Are my examples correct?
Would be very grateful!
abstract-algebra ring-theory principal-ideal-domains unique-factorization-domains
asked Dec 9 '18 at 18:58
ZFRZFR
5,08531340
$endgroup$
Give an example if possible, and if not possible explain why not.
a) A subring of a PID that is not PID.
b) A PID that is a subring of a non-PID.
c) A subring of a PID that is not UFD.
My approach:
a) Since $mathbb{Q}$ is field then $mathbb{Q}[x]$ is PID. But it's subring $mathbb{Z}[x]$ is not PID because the ideal $(2,x)$ in $mathbb{Z}[x]$ is not principal.
b) Consider the subring $mathbb{Z}$ of the ring $mathbb{Z}[x]$. We know that $mathbb{Z}$ is euclidean domain and hence is PID.
c) Consider the subring $mathbb{Z}[sqrt{-5}]$ of the ring $mathbb{Q}[sqrt{-5}]$. I know that $mathbb{Z}[sqrt{-5}]$ is not UFD, but $mathbb{Q}[sqrt{-5}]cong mathbb{Q}[x]/(x^2+5)$ and since $x^2+5$ is irreducible in $mathbb{Q}[x]$ then it's field $Rightarrow$ $mathbb{Q}[sqrt{-5}]$ is also field $Rightarrow$ is PID.
Are my examples correct?
Would be very grateful!
abstract-algebra ring-theory principal-ideal-domains unique-factorization-domains
abstract-algebra ring-theory principal-ideal-domains unique-factorization-domains
asked Dec 9 '18 at 18:58
ZFRZFR
5,08531340
asked Dec 9 '18 at 18:58
ZFRZFR
5,08531340
asked Dec 9 '18 at 18:58
ZFRZFR
5,08531340
asked Dec 9 '18 at 18:58
ZFRZFR
5,08531340
asked Dec 9 '18 at 18:58
ZFRZFR
5,08531340
5,08531340
$begingroup$
looks good to me
$endgroup$
– zoidberg
Dec 9 '18 at 19:04
1
$begingroup$
Each question has been answered at this site already. You could also compare. For example, b) is here. For a), see here. Actually, you have the same example.
$endgroup$
– Dietrich Burde
Dec 9 '18 at 19:17
$begingroup$
Try subrings of a polynomial ring over a field, $k[x]$
$endgroup$
– R.C.Cowsik
Dec 10 '18 at 6:48
add a comment |
$begingroup$
looks good to me
$endgroup$
– zoidberg
Dec 9 '18 at 19:04
1
$begingroup$
Each question has been answered at this site already. You could also compare. For example, b) is here. For a), see here. Actually, you have the same example.
$endgroup$
– Dietrich Burde
Dec 9 '18 at 19:17
$begingroup$
Try subrings of a polynomial ring over a field, $k[x]$
$endgroup$
– R.C.Cowsik
Dec 10 '18 at 6:48
$begingroup$
looks good to me
$endgroup$
– zoidberg
Dec 9 '18 at 19:04
$begingroup$
looks good to me
$endgroup$
– zoidberg
Dec 9 '18 at 19:04
1
1
$begingroup$
Each question has been answered at this site already. You could also compare. For example, b) is here. For a), see here. Actually, you have the same example.
$endgroup$
– Dietrich Burde
Dec 9 '18 at 19:17
$begingroup$
Each question has been answered at this site already. You could also compare. For example, b) is here. For a), see here. Actually, you have the same example.
$endgroup$
– Dietrich Burde
Dec 9 '18 at 19:17
$begingroup$
Try subrings of a polynomial ring over a field, $k[x]$
$endgroup$
– R.C.Cowsik
Dec 10 '18 at 6:48
$begingroup$
Try subrings of a polynomial ring over a field, $k[x]$
$endgroup$
– R.C.Cowsik
Dec 10 '18 at 6:48
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%2f3032795%2fsome-counterexamples-in-basic-ring-theory%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%2f3032795%2fsome-counterexamples-in-basic-ring-theory%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
$begingroup$
looks good to me
$endgroup$
– zoidberg
Dec 9 '18 at 19:04
1
$begingroup$
Each question has been answered at this site already. You could also compare. For example, b) is here. For a), see here. Actually, you have the same example.
$endgroup$
– Dietrich Burde
Dec 9 '18 at 19:17
$begingroup$
Try subrings of a polynomial ring over a field, $k[x]$
$endgroup$
– R.C.Cowsik
Dec 10 '18 at 6:48