Is the integral closure of a polynomial ring a UFD?
1
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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
$endgroup$
migrated from mathoverflow.net Dec 26 '18 at 2:14
This question came from our site for professional mathematicians.
|
show 1 more comment
1
$begingroup$
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
$endgroup$
migrated from mathoverflow.net Dec 26 '18 at 2:14
This question came from our site for professional mathematicians.
$begingroup$
I meant "B is the integral closure of C[x] in F"
$endgroup$
– vassilis papanicolaou
Dec 21 '18 at 16:15
2
$begingroup$
Please edit your post rather than amend it in the comments. Please use latex code for math symbols.
$endgroup$
– YCor
Dec 21 '18 at 16:19
5
$begingroup$
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.
$endgroup$
– Mohan
Dec 21 '18 at 19:33
3
$begingroup$
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.
$endgroup$
– abx
Dec 21 '18 at 20:11
$begingroup$
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).
$endgroup$
– vassilis papanicolaou
Dec 24 '18 at 11:17
|
show 1 more comment
1
1
1
$begingroup$
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
$endgroup$
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 '18 at 16:14
vassilis papanicolaou
migrated from mathoverflow.net Dec 26 '18 at 2:14
This question came from our site for professional mathematicians.
migrated from mathoverflow.net Dec 26 '18 at 2:14
This question came from our site for professional mathematicians.
$begingroup$
I meant "B is the integral closure of C[x] in F"
$endgroup$
– vassilis papanicolaou
Dec 21 '18 at 16:15
2
$begingroup$
Please edit your post rather than amend it in the comments. Please use latex code for math symbols.
$endgroup$
– YCor
Dec 21 '18 at 16:19
5
$begingroup$
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.
$endgroup$
– Mohan
Dec 21 '18 at 19:33
3
$begingroup$
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.
$endgroup$
– abx
Dec 21 '18 at 20:11
$begingroup$
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).
$endgroup$
– vassilis papanicolaou
Dec 24 '18 at 11:17
|
show 1 more comment
$begingroup$
I meant "B is the integral closure of C[x] in F"
$endgroup$
– vassilis papanicolaou
Dec 21 '18 at 16:15
2
$begingroup$
Please edit your post rather than amend it in the comments. Please use latex code for math symbols.
$endgroup$
– YCor
Dec 21 '18 at 16:19
5
$begingroup$
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.
$endgroup$
– Mohan
Dec 21 '18 at 19:33
3
$begingroup$
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.
$endgroup$
– abx
Dec 21 '18 at 20:11
$begingroup$
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).
$endgroup$
– vassilis papanicolaou
Dec 24 '18 at 11:17
$begingroup$
I meant "B is the integral closure of C[x] in F"
$endgroup$
– vassilis papanicolaou
Dec 21 '18 at 16:15
$begingroup$
I meant "B is the integral closure of C[x] in F"
$endgroup$
– vassilis papanicolaou
Dec 21 '18 at 16:15
2
2
$begingroup$
Please edit your post rather than amend it in the comments. Please use latex code for math symbols.
$endgroup$
– YCor
Dec 21 '18 at 16:19
$begingroup$
Please edit your post rather than amend it in the comments. Please use latex code for math symbols.
$endgroup$
– YCor
Dec 21 '18 at 16:19
5
5
$begingroup$
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.
$endgroup$
– Mohan
Dec 21 '18 at 19:33
$begingroup$
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.
$endgroup$
– Mohan
Dec 21 '18 at 19:33
3
3
$begingroup$
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.
$endgroup$
– abx
Dec 21 '18 at 20:11
$begingroup$
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.
$endgroup$
– abx
Dec 21 '18 at 20:11
$begingroup$
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).
$endgroup$
– vassilis papanicolaou
Dec 24 '18 at 11:17
$begingroup$
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).
$endgroup$
– vassilis papanicolaou
Dec 24 '18 at 11:17
|
show 1 more comment
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$begingroup$
I meant "B is the integral closure of C[x] in F"
$endgroup$
– vassilis papanicolaou
Dec 21 '18 at 16:15
2
$begingroup$
Please edit your post rather than amend it in the comments. Please use latex code for math symbols.
$endgroup$
– YCor
Dec 21 '18 at 16:19
5
$begingroup$
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.
$endgroup$
– Mohan
Dec 21 '18 at 19:33
3
$begingroup$
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.
$endgroup$
– abx
Dec 21 '18 at 20:11
$begingroup$
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).
$endgroup$
– vassilis papanicolaou
Dec 24 '18 at 11:17