Is the integral closure of a polynomial ring a UFD?
Let $mathbb{C}(x)$ be the field of rational functions over the complex numbers and $F$ a finite extension of $mathbb{C}(x)$. Suppose $B$ is the integral closure of $mathbb{C}[x]$ (the ring of polynomials over $mathbb{C}$). Is $B$ always a Unique Factorization Domain (UFD)?
commutative-algebra
migrated from mathoverflow.net Dec 26 at 2:14
This question came from our site for professional mathematicians.
|
show 1 more comment
Let $mathbb{C}(x)$ be the field of rational functions over the complex numbers and $F$ a finite extension of $mathbb{C}(x)$. Suppose $B$ is the integral closure of $mathbb{C}[x]$ (the ring of polynomials over $mathbb{C}$). Is $B$ always a Unique Factorization Domain (UFD)?
commutative-algebra
migrated from mathoverflow.net Dec 26 at 2:14
This question came from our site for professional mathematicians.
I meant "B is the integral closure of C[x] in F"
– vassilis papanicolaou
Dec 21 at 16:15
2
Please edit your post rather than amend it in the comments. Please use latex code for math symbols.
– YCor
Dec 21 at 16:19
4
Every integrally closed one dimensional domain, finitely generated over $mathbb{C}$ is the integral closure of $mathbb{C}[x]$ for a suitable $x$ (known as Noether Normalization). Most of these are not UFDs.
– Mohan
Dec 21 at 19:33
3
In fact, $F$ is the field of rational functions of a smooth projective cuve $C$. As soon as $g(C)geq 1$, $B$ is not a UFD.
– abx
Dec 21 at 20:11
I see we can construct rather trivial counterexamples! Even for genus 0:$x^2 - 1 = (x-1)(x+1) = sqrt{x^2 -1} sqrt{x^2 -1}$ (unless I miss something).
– vassilis papanicolaou
Dec 24 at 11:17
|
show 1 more comment
Let $mathbb{C}(x)$ be the field of rational functions over the complex numbers and $F$ a finite extension of $mathbb{C}(x)$. Suppose $B$ is the integral closure of $mathbb{C}[x]$ (the ring of polynomials over $mathbb{C}$). Is $B$ always a Unique Factorization Domain (UFD)?
commutative-algebra
Let $mathbb{C}(x)$ be the field of rational functions over the complex numbers and $F$ a finite extension of $mathbb{C}(x)$. Suppose $B$ is the integral closure of $mathbb{C}[x]$ (the ring of polynomials over $mathbb{C}$). Is $B$ always a Unique Factorization Domain (UFD)?
commutative-algebra
commutative-algebra
asked Dec 21 at 16:14
vassilis papanicolaou
migrated from mathoverflow.net Dec 26 at 2:14
This question came from our site for professional mathematicians.
migrated from mathoverflow.net Dec 26 at 2:14
This question came from our site for professional mathematicians.
I meant "B is the integral closure of C[x] in F"
– vassilis papanicolaou
Dec 21 at 16:15
2
Please edit your post rather than amend it in the comments. Please use latex code for math symbols.
– YCor
Dec 21 at 16:19
4
Every integrally closed one dimensional domain, finitely generated over $mathbb{C}$ is the integral closure of $mathbb{C}[x]$ for a suitable $x$ (known as Noether Normalization). Most of these are not UFDs.
– Mohan
Dec 21 at 19:33
3
In fact, $F$ is the field of rational functions of a smooth projective cuve $C$. As soon as $g(C)geq 1$, $B$ is not a UFD.
– abx
Dec 21 at 20:11
I see we can construct rather trivial counterexamples! Even for genus 0:$x^2 - 1 = (x-1)(x+1) = sqrt{x^2 -1} sqrt{x^2 -1}$ (unless I miss something).
– vassilis papanicolaou
Dec 24 at 11:17
|
show 1 more comment
I meant "B is the integral closure of C[x] in F"
– vassilis papanicolaou
Dec 21 at 16:15
2
Please edit your post rather than amend it in the comments. Please use latex code for math symbols.
– YCor
Dec 21 at 16:19
4
Every integrally closed one dimensional domain, finitely generated over $mathbb{C}$ is the integral closure of $mathbb{C}[x]$ for a suitable $x$ (known as Noether Normalization). Most of these are not UFDs.
– Mohan
Dec 21 at 19:33
3
In fact, $F$ is the field of rational functions of a smooth projective cuve $C$. As soon as $g(C)geq 1$, $B$ is not a UFD.
– abx
Dec 21 at 20:11
I see we can construct rather trivial counterexamples! Even for genus 0:$x^2 - 1 = (x-1)(x+1) = sqrt{x^2 -1} sqrt{x^2 -1}$ (unless I miss something).
– vassilis papanicolaou
Dec 24 at 11:17
I meant "B is the integral closure of C[x] in F"
– vassilis papanicolaou
Dec 21 at 16:15
I meant "B is the integral closure of C[x] in F"
– vassilis papanicolaou
Dec 21 at 16:15
2
2
Please edit your post rather than amend it in the comments. Please use latex code for math symbols.
– YCor
Dec 21 at 16:19
Please edit your post rather than amend it in the comments. Please use latex code for math symbols.
– YCor
Dec 21 at 16:19
4
4
Every integrally closed one dimensional domain, finitely generated over $mathbb{C}$ is the integral closure of $mathbb{C}[x]$ for a suitable $x$ (known as Noether Normalization). Most of these are not UFDs.
– Mohan
Dec 21 at 19:33
Every integrally closed one dimensional domain, finitely generated over $mathbb{C}$ is the integral closure of $mathbb{C}[x]$ for a suitable $x$ (known as Noether Normalization). Most of these are not UFDs.
– Mohan
Dec 21 at 19:33
3
3
In fact, $F$ is the field of rational functions of a smooth projective cuve $C$. As soon as $g(C)geq 1$, $B$ is not a UFD.
– abx
Dec 21 at 20:11
In fact, $F$ is the field of rational functions of a smooth projective cuve $C$. As soon as $g(C)geq 1$, $B$ is not a UFD.
– abx
Dec 21 at 20:11
I see we can construct rather trivial counterexamples! Even for genus 0:$x^2 - 1 = (x-1)(x+1) = sqrt{x^2 -1} sqrt{x^2 -1}$ (unless I miss something).
– vassilis papanicolaou
Dec 24 at 11:17
I see we can construct rather trivial counterexamples! Even for genus 0:$x^2 - 1 = (x-1)(x+1) = sqrt{x^2 -1} sqrt{x^2 -1}$ (unless I miss something).
– vassilis papanicolaou
Dec 24 at 11:17
|
show 1 more comment
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%2f3052583%2fis-the-integral-closure-of-a-polynomial-ring-a-ufd%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
active
oldest
votes
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3052583%2fis-the-integral-closure-of-a-polynomial-ring-a-ufd%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
I meant "B is the integral closure of C[x] in F"
– vassilis papanicolaou
Dec 21 at 16:15
2
Please edit your post rather than amend it in the comments. Please use latex code for math symbols.
– YCor
Dec 21 at 16:19
4
Every integrally closed one dimensional domain, finitely generated over $mathbb{C}$ is the integral closure of $mathbb{C}[x]$ for a suitable $x$ (known as Noether Normalization). Most of these are not UFDs.
– Mohan
Dec 21 at 19:33
3
In fact, $F$ is the field of rational functions of a smooth projective cuve $C$. As soon as $g(C)geq 1$, $B$ is not a UFD.
– abx
Dec 21 at 20:11
I see we can construct rather trivial counterexamples! Even for genus 0:$x^2 - 1 = (x-1)(x+1) = sqrt{x^2 -1} sqrt{x^2 -1}$ (unless I miss something).
– vassilis papanicolaou
Dec 24 at 11:17