If $dim(A/mathfrak p)=dim(A)-1$, then is $mathfrak p$ principal?
$begingroup$
Let $A$ be a commutative ring, and let $mathfrak p$ be a prime ideal in $A$. When is it true that if $dim(A/mathfrak p)=dim(A)-1$ then $mathfrak p$ is a principal ideal in $A$?
I'm pretty sure $A$ must be at least an integral domain, and I'm pretty sure that I have a proof when it's a UFD (in that case $mathfrak p$ must be a minimal nonzero prime, and therefore principal). The question is related to the question of whether or not every codimension $1$ subvariety of an irreducible variety is a complete intersection.
abstract-algebra algebraic-geometry commutative-algebra
edited Jan 15 at 20:48
user26857
39.5k124283
asked Jan 15 at 1:29
leibnewtzleibnewtz
2,6211717
$endgroup$
add a comment |
$begingroup$
Let $A$ be a commutative ring, and let $mathfrak p$ be a prime ideal in $A$. When is it true that if $dim(A/mathfrak p)=dim(A)-1$ then $mathfrak p$ is a principal ideal in $A$?
I'm pretty sure $A$ must be at least an integral domain, and I'm pretty sure that I have a proof when it's a UFD (in that case $mathfrak p$ must be a minimal nonzero prime, and therefore principal). The question is related to the question of whether or not every codimension $1$ subvariety of an irreducible variety is a complete intersection.
abstract-algebra algebraic-geometry commutative-algebra
edited Jan 15 at 20:48
user26857
39.5k124283
asked Jan 15 at 1:29
leibnewtzleibnewtz
2,6211717
$endgroup$
4
$begingroup$
It is not true. Take $A =k[[x,y,z]]/(x^2-yz)$ and $p = (x,y)$. Then $dim A = 2$ and $dim A/p = dim k[[z]] = 1$. However, $p$ is not principal.
$endgroup$
– Youngsu
Jan 15 at 2:57
4
$begingroup$
If all height one primes are principal in a Noetherian domain, then it is a UFD. So, most of the time, you need a UFD.
$endgroup$
– Mohan
Jan 15 at 2:57
add a comment |
$begingroup$
Let $A$ be a commutative ring, and let $mathfrak p$ be a prime ideal in $A$. When is it true that if $dim(A/mathfrak p)=dim(A)-1$ then $mathfrak p$ is a principal ideal in $A$?
I'm pretty sure $A$ must be at least an integral domain, and I'm pretty sure that I have a proof when it's a UFD (in that case $mathfrak p$ must be a minimal nonzero prime, and therefore principal). The question is related to the question of whether or not every codimension $1$ subvariety of an irreducible variety is a complete intersection.
abstract-algebra algebraic-geometry commutative-algebra
edited Jan 15 at 20:48
user26857
39.5k124283
asked Jan 15 at 1:29
leibnewtzleibnewtz
2,6211717
$endgroup$
Let $A$ be a commutative ring, and let $mathfrak p$ be a prime ideal in $A$. When is it true that if $dim(A/mathfrak p)=dim(A)-1$ then $mathfrak p$ is a principal ideal in $A$?
I'm pretty sure $A$ must be at least an integral domain, and I'm pretty sure that I have a proof when it's a UFD (in that case $mathfrak p$ must be a minimal nonzero prime, and therefore principal). The question is related to the question of whether or not every codimension $1$ subvariety of an irreducible variety is a complete intersection.
abstract-algebra algebraic-geometry commutative-algebra
abstract-algebra algebraic-geometry commutative-algebra
edited Jan 15 at 20:48
user26857
39.5k124283
asked Jan 15 at 1:29
leibnewtzleibnewtz
2,6211717
edited Jan 15 at 20:48
user26857
39.5k124283
asked Jan 15 at 1:29
leibnewtzleibnewtz
2,6211717
edited Jan 15 at 20:48
user26857
39.5k124283
edited Jan 15 at 20:48
user26857
39.5k124283
edited Jan 15 at 20:48
user26857
39.5k124283
39.5k124283
asked Jan 15 at 1:29
leibnewtzleibnewtz
2,6211717
asked Jan 15 at 1:29
leibnewtzleibnewtz
2,6211717
asked Jan 15 at 1:29
leibnewtzleibnewtz
2,6211717
2,6211717
4
$begingroup$
It is not true. Take $A =k[[x,y,z]]/(x^2-yz)$ and $p = (x,y)$. Then $dim A = 2$ and $dim A/p = dim k[[z]] = 1$. However, $p$ is not principal.
$endgroup$
– Youngsu
Jan 15 at 2:57
4
$begingroup$
If all height one primes are principal in a Noetherian domain, then it is a UFD. So, most of the time, you need a UFD.
$endgroup$
– Mohan
Jan 15 at 2:57
add a comment |
4
$begingroup$
It is not true. Take $A =k[[x,y,z]]/(x^2-yz)$ and $p = (x,y)$. Then $dim A = 2$ and $dim A/p = dim k[[z]] = 1$. However, $p$ is not principal.
$endgroup$
– Youngsu
Jan 15 at 2:57
4
$begingroup$
If all height one primes are principal in a Noetherian domain, then it is a UFD. So, most of the time, you need a UFD.
$endgroup$
– Mohan
Jan 15 at 2:57
4
4
$begingroup$
It is not true. Take $A =k[[x,y,z]]/(x^2-yz)$ and $p = (x,y)$. Then $dim A = 2$ and $dim A/p = dim k[[z]] = 1$. However, $p$ is not principal.
$endgroup$
– Youngsu
Jan 15 at 2:57
$begingroup$
It is not true. Take $A =k[[x,y,z]]/(x^2-yz)$ and $p = (x,y)$. Then $dim A = 2$ and $dim A/p = dim k[[z]] = 1$. However, $p$ is not principal.
$endgroup$
– Youngsu
Jan 15 at 2:57
4
4
$begingroup$
If all height one primes are principal in a Noetherian domain, then it is a UFD. So, most of the time, you need a UFD.
$endgroup$
– Mohan
Jan 15 at 2:57
$begingroup$
If all height one primes are principal in a Noetherian domain, then it is a UFD. So, most of the time, you need a UFD.
$endgroup$
– Mohan
Jan 15 at 2:57
add a comment |
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$begingroup$
It is not true. Take $A =k[[x,y,z]]/(x^2-yz)$ and $p = (x,y)$. Then $dim A = 2$ and $dim A/p = dim k[[z]] = 1$. However, $p$ is not principal.
$endgroup$
– Youngsu
Jan 15 at 2:57
4
$begingroup$
If all height one primes are principal in a Noetherian domain, then it is a UFD. So, most of the time, you need a UFD.
$endgroup$
– Mohan
Jan 15 at 2:57