Example of a Noetherian domain which is not equidimensional
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A finite dimensional commutative ring $R$ with unity is called equidimensional if all its minimal prime ideals have same dimension (dimension of a prime ideal $mathfrak p$ is defined to be Krull dimension of the ring $R/mathfrak p$) and every maximal ideal has the height same as dimension of the ring.
I want to have an example of a Noetherian domain which is not equidimensional.
Note that $R$ cannot be a finite type $k$ algebra, neither it can be local.
Thank you.
commutative-algebra noetherian integral-domain
edited Nov 18 at 22:13
user26857
39.2k123882
asked Nov 18 at 8:16
Rtk427
223
add a comment |
up vote
0
down vote
favorite
A finite dimensional commutative ring $R$ with unity is called equidimensional if all its minimal prime ideals have same dimension (dimension of a prime ideal $mathfrak p$ is defined to be Krull dimension of the ring $R/mathfrak p$) and every maximal ideal has the height same as dimension of the ring.
I want to have an example of a Noetherian domain which is not equidimensional.
Note that $R$ cannot be a finite type $k$ algebra, neither it can be local.
Thank you.
commutative-algebra noetherian integral-domain
edited Nov 18 at 22:13
user26857
39.2k123882
asked Nov 18 at 8:16
Rtk427
223
1
"An easy counterexample: consider the polynomial ring B=k[x,y] with coefficients in a field k, let m be a maximal ideal and p a prime ideal of height 1 not contained in m and localize B at the multiplicative subset B∖(m∪p). The ring obtained this way is semi-local, one maximal ideal is generated by m (so has height 2), the other one is generated by p and has height 1."
– user26857
Nov 18 at 22:22
Btw, people uses to distinguish among the two properties calling a ring equidimensional if it satisfies the first property, and equicodimensional if it satisfies the second property.
– user26857
Nov 18 at 22:24
@user26857 Ok but this is the definition given in Eisenbud.
– Rtk427
Nov 19 at 14:07
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A finite dimensional commutative ring $R$ with unity is called equidimensional if all its minimal prime ideals have same dimension (dimension of a prime ideal $mathfrak p$ is defined to be Krull dimension of the ring $R/mathfrak p$) and every maximal ideal has the height same as dimension of the ring.
I want to have an example of a Noetherian domain which is not equidimensional.
Note that $R$ cannot be a finite type $k$ algebra, neither it can be local.
Thank you.
commutative-algebra noetherian integral-domain
edited Nov 18 at 22:13
user26857
39.2k123882
asked Nov 18 at 8:16
Rtk427
223
A finite dimensional commutative ring $R$ with unity is called equidimensional if all its minimal prime ideals have same dimension (dimension of a prime ideal $mathfrak p$ is defined to be Krull dimension of the ring $R/mathfrak p$) and every maximal ideal has the height same as dimension of the ring.
I want to have an example of a Noetherian domain which is not equidimensional.
Note that $R$ cannot be a finite type $k$ algebra, neither it can be local.
Thank you.
commutative-algebra noetherian integral-domain
commutative-algebra noetherian integral-domain
edited Nov 18 at 22:13
user26857
39.2k123882
asked Nov 18 at 8:16
Rtk427
223
edited Nov 18 at 22:13
user26857
39.2k123882
asked Nov 18 at 8:16
Rtk427
223
edited Nov 18 at 22:13
user26857
39.2k123882
edited Nov 18 at 22:13
user26857
39.2k123882
edited Nov 18 at 22:13
user26857
39.2k123882
39.2k123882
asked Nov 18 at 8:16
Rtk427
223
asked Nov 18 at 8:16
Rtk427
223
asked Nov 18 at 8:16
Rtk427
223
223
1
"An easy counterexample: consider the polynomial ring B=k[x,y] with coefficients in a field k, let m be a maximal ideal and p a prime ideal of height 1 not contained in m and localize B at the multiplicative subset B∖(m∪p). The ring obtained this way is semi-local, one maximal ideal is generated by m (so has height 2), the other one is generated by p and has height 1."
– user26857
Nov 18 at 22:22
Btw, people uses to distinguish among the two properties calling a ring equidimensional if it satisfies the first property, and equicodimensional if it satisfies the second property.
– user26857
Nov 18 at 22:24
@user26857 Ok but this is the definition given in Eisenbud.
– Rtk427
Nov 19 at 14:07
add a comment |
1
"An easy counterexample: consider the polynomial ring B=k[x,y] with coefficients in a field k, let m be a maximal ideal and p a prime ideal of height 1 not contained in m and localize B at the multiplicative subset B∖(m∪p). The ring obtained this way is semi-local, one maximal ideal is generated by m (so has height 2), the other one is generated by p and has height 1."
– user26857
Nov 18 at 22:22
Btw, people uses to distinguish among the two properties calling a ring equidimensional if it satisfies the first property, and equicodimensional if it satisfies the second property.
– user26857
Nov 18 at 22:24
@user26857 Ok but this is the definition given in Eisenbud.
– Rtk427
Nov 19 at 14:07
1
1
"An easy counterexample: consider the polynomial ring B=k[x,y] with coefficients in a field k, let m be a maximal ideal and p a prime ideal of height 1 not contained in m and localize B at the multiplicative subset B∖(m∪p). The ring obtained this way is semi-local, one maximal ideal is generated by m (so has height 2), the other one is generated by p and has height 1."
– user26857
Nov 18 at 22:22
"An easy counterexample: consider the polynomial ring B=k[x,y] with coefficients in a field k, let m be a maximal ideal and p a prime ideal of height 1 not contained in m and localize B at the multiplicative subset B∖(m∪p). The ring obtained this way is semi-local, one maximal ideal is generated by m (so has height 2), the other one is generated by p and has height 1."
– user26857
Nov 18 at 22:22
Btw, people uses to distinguish among the two properties calling a ring equidimensional if it satisfies the first property, and equicodimensional if it satisfies the second property.
– user26857
Nov 18 at 22:24
Btw, people uses to distinguish among the two properties calling a ring equidimensional if it satisfies the first property, and equicodimensional if it satisfies the second property.
– user26857
Nov 18 at 22:24
@user26857 Ok but this is the definition given in Eisenbud.
– Rtk427
Nov 19 at 14:07
@user26857 Ok but this is the definition given in Eisenbud.
– Rtk427
Nov 19 at 14:07
add a comment |
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1
"An easy counterexample: consider the polynomial ring B=k[x,y] with coefficients in a field k, let m be a maximal ideal and p a prime ideal of height 1 not contained in m and localize B at the multiplicative subset B∖(m∪p). The ring obtained this way is semi-local, one maximal ideal is generated by m (so has height 2), the other one is generated by p and has height 1."
– user26857
Nov 18 at 22:22
Btw, people uses to distinguish among the two properties calling a ring equidimensional if it satisfies the first property, and equicodimensional if it satisfies the second property.
– user26857
Nov 18 at 22:24
@user26857 Ok but this is the definition given in Eisenbud.
– Rtk427
Nov 19 at 14:07