For which rings $R$: If $p,q in R[x,y]$ satisfy $operatorname{Jac}(p,q)=0$, then $p=u(h)$ and $q=v(h)$, for...
Let $R$ be an arbitrary commutative $k$-algebra, where $k$ is a field of characteristic zero.
When $R$ is an algebraically closed field of characteristic zero, Nagata has proved the following nice result in Corollary 1.3 (which does not have a non-commutative analog, see this question):
"Assume that $p,q in R[x,y]$ satisfy $operatorname{Jac}(p,q)=0$. Then $p=u(h)$ and $q=v(h)$, for some $h in R[x,y]$, $u(T),v(T) in R[T]$".
It is known that it is possible to replace $R$ of Nagata's result by an arbitrary field or even by any normal integral domain of characteristic zero.
For which $R$'s Nagata's result is still valid?
What about any integral domain of characteristic zero, not necessarily normal?
What about an arbitrary affine $k$-algebra?
Notice that this question is relevant, but one has to be careful about the eqivalence of being algebraically dependent and having zero Jacobian, when replacing the base field by an arbitrary ring.
Thank you very much!
algebraic-geometry commutative-algebra
asked Nov 29 at 16:01
user237522
2,0891617
add a comment |
Let $R$ be an arbitrary commutative $k$-algebra, where $k$ is a field of characteristic zero.
When $R$ is an algebraically closed field of characteristic zero, Nagata has proved the following nice result in Corollary 1.3 (which does not have a non-commutative analog, see this question):
"Assume that $p,q in R[x,y]$ satisfy $operatorname{Jac}(p,q)=0$. Then $p=u(h)$ and $q=v(h)$, for some $h in R[x,y]$, $u(T),v(T) in R[T]$".
It is known that it is possible to replace $R$ of Nagata's result by an arbitrary field or even by any normal integral domain of characteristic zero.
For which $R$'s Nagata's result is still valid?
What about any integral domain of characteristic zero, not necessarily normal?
What about an arbitrary affine $k$-algebra?
Notice that this question is relevant, but one has to be careful about the eqivalence of being algebraically dependent and having zero Jacobian, when replacing the base field by an arbitrary ring.
Thank you very much!
algebraic-geometry commutative-algebra
asked Nov 29 at 16:01
user237522
2,0891617
add a comment |
Let $R$ be an arbitrary commutative $k$-algebra, where $k$ is a field of characteristic zero.
When $R$ is an algebraically closed field of characteristic zero, Nagata has proved the following nice result in Corollary 1.3 (which does not have a non-commutative analog, see this question):
"Assume that $p,q in R[x,y]$ satisfy $operatorname{Jac}(p,q)=0$. Then $p=u(h)$ and $q=v(h)$, for some $h in R[x,y]$, $u(T),v(T) in R[T]$".
It is known that it is possible to replace $R$ of Nagata's result by an arbitrary field or even by any normal integral domain of characteristic zero.
For which $R$'s Nagata's result is still valid?
What about any integral domain of characteristic zero, not necessarily normal?
What about an arbitrary affine $k$-algebra?
Notice that this question is relevant, but one has to be careful about the eqivalence of being algebraically dependent and having zero Jacobian, when replacing the base field by an arbitrary ring.
Thank you very much!
algebraic-geometry commutative-algebra
asked Nov 29 at 16:01
user237522
2,0891617
Let $R$ be an arbitrary commutative $k$-algebra, where $k$ is a field of characteristic zero.
When $R$ is an algebraically closed field of characteristic zero, Nagata has proved the following nice result in Corollary 1.3 (which does not have a non-commutative analog, see this question):
"Assume that $p,q in R[x,y]$ satisfy $operatorname{Jac}(p,q)=0$. Then $p=u(h)$ and $q=v(h)$, for some $h in R[x,y]$, $u(T),v(T) in R[T]$".
It is known that it is possible to replace $R$ of Nagata's result by an arbitrary field or even by any normal integral domain of characteristic zero.
For which $R$'s Nagata's result is still valid?
What about any integral domain of characteristic zero, not necessarily normal?
What about an arbitrary affine $k$-algebra?
Notice that this question is relevant, but one has to be careful about the eqivalence of being algebraically dependent and having zero Jacobian, when replacing the base field by an arbitrary ring.
Thank you very much!
algebraic-geometry commutative-algebra
algebraic-geometry commutative-algebra
asked Nov 29 at 16:01
user237522
2,0891617
asked Nov 29 at 16:01
user237522
2,0891617
edited Nov 30 at 1:01
asked Nov 29 at 16:01
user237522
2,0891617
asked Nov 29 at 16:01
user237522
2,0891617
asked Nov 29 at 16:01
user237522
2,0891617
2,0891617
add a comment |
add a comment |
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