Valuation ring, of infinite global dimension, with principal maximal ideal The 2019 Stack...
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Valuation ring, of infinite global dimension, with principal maximal ideal
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Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Weak Global Dimension and Global DimensionProjective dimension of the residue field of a noetherian local ring.Dimension of quotients of a discrete valuation domainLocal Noetherian domain of dimension one with principal maximal idealGlobal Dimension of a Ring and its LocalizationsLocalization of a valuation ring at a prime is abstractly isomorphic to the original ringValuation ring and localization.Almost a discrete valuation ringGlobal dimension of $prod k$Valuation ring of finite Krull dimension whose every non-maximal ideal is principal
$begingroup$
Does there exist a Valuation ring $(R, mathfrak m)$ , with principal maximal ideal, of infinite global dimension ?
Corollary 2 of the following paper by Osofsky has an example of Valuation ring of infinite global dimension. But I'm not sure whether we can also construct such example with principal maximal ideal.
Please help
commutative-algebra homological-algebra valuation-theory global-dimension
edited Mar 23 at 13:16
Andrés E. Caicedo
66k8160252
asked Mar 22 at 22:23
user521337user521337
1,2201417
$endgroup$
add a comment |
$begingroup$
Does there exist a Valuation ring $(R, mathfrak m)$ , with principal maximal ideal, of infinite global dimension ?
Corollary 2 of the following paper by Osofsky has an example of Valuation ring of infinite global dimension. But I'm not sure whether we can also construct such example with principal maximal ideal.
Please help
commutative-algebra homological-algebra valuation-theory global-dimension
edited Mar 23 at 13:16
Andrés E. Caicedo
66k8160252
asked Mar 22 at 22:23
user521337user521337
1,2201417
$endgroup$
add a comment |
$begingroup$
Does there exist a Valuation ring $(R, mathfrak m)$ , with principal maximal ideal, of infinite global dimension ?
Corollary 2 of the following paper by Osofsky has an example of Valuation ring of infinite global dimension. But I'm not sure whether we can also construct such example with principal maximal ideal.
Please help
commutative-algebra homological-algebra valuation-theory global-dimension
edited Mar 23 at 13:16
Andrés E. Caicedo
66k8160252
asked Mar 22 at 22:23
user521337user521337
1,2201417
$endgroup$
Does there exist a Valuation ring $(R, mathfrak m)$ , with principal maximal ideal, of infinite global dimension ?
Corollary 2 of the following paper by Osofsky has an example of Valuation ring of infinite global dimension. But I'm not sure whether we can also construct such example with principal maximal ideal.
Please help
commutative-algebra homological-algebra valuation-theory global-dimension
commutative-algebra homological-algebra valuation-theory global-dimension
edited Mar 23 at 13:16
Andrés E. Caicedo
66k8160252
asked Mar 22 at 22:23
user521337user521337
1,2201417
edited Mar 23 at 13:16
Andrés E. Caicedo
66k8160252
asked Mar 22 at 22:23
user521337user521337
1,2201417
edited Mar 23 at 13:16
Andrés E. Caicedo
66k8160252
edited Mar 23 at 13:16
Andrés E. Caicedo
66k8160252
edited Mar 23 at 13:16
Andrés E. Caicedo
66k8160252
66k8160252
asked Mar 22 at 22:23
user521337user521337
1,2201417
asked Mar 22 at 22:23
user521337user521337
1,2201417
asked Mar 22 at 22:23
user521337user521337
1,2201417
1,2201417
add a comment |
add a comment |
1 Answer
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The example of a valuation domain of infinite global dimension Barbara Osofsky gives is a generalized power series ring $k[[Gamma]]$ with a well-chosen ordered abelian group $Gamma$. But as far as I see the residue field $k$ can be chosen arbitrarily. Hence one can take any field $k$, that carries a discrete valuation ring $overline{O}subset k$. Let $Osubset k[[Gamma]]$ be the preimage of $overline{O}$ under the natural homomorphism $k[[Gamma]]rightarrow k$. Then $O$ is a valuation ring of the fraction field of $k[[Gamma]]$ and $k[[Gamma]]$ is a localization of $O$. Consequently $O$ must have infinite global dimension, since global dimension decreases under localization. On the other side the maximal ideal of $O$ is generated by any preimage of a generator of the maximal ideal of $overline{O}$.
answered Mar 23 at 3:33
Hagen KnafHagen Knaf
6,9521318
$endgroup$
$begingroup$
I'm quite not sure why $O$ is a Valuation ring ... why are the ideals linearly ordered ?
$endgroup$
– user521337
Mar 24 at 0:40
$begingroup$
To check that $O$ is a valuation ring one verifies that either $xin O$ or $x^{-1}in O$ for every $xin k((Gamma))$. So assume $xnotin O$. Case 1: $xnotin k[[Gamma]]$. Then $x^{-1}$ is in the maximal ideal of $k[[Gamma]]$, which is the preimage of $0$ under the natural map. Hence $x^{-1}in O$. Case 2: $xin k[[Gamma]]$. Then the image $overline{x}$ of $x$ is not in $overline{O}$ and thus $overline{x}^{-1}inoverline{O}$. Consequently $x^{-1}in O$. (Remark: in valuation theory $O$ is called the composite valuation ring of $overline{O}$ and $k[[Gamma]]$).
$endgroup$
– Hagen Knaf
Mar 24 at 9:41
add a comment |
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1 Answer
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$begingroup$
The example of a valuation domain of infinite global dimension Barbara Osofsky gives is a generalized power series ring $k[[Gamma]]$ with a well-chosen ordered abelian group $Gamma$. But as far as I see the residue field $k$ can be chosen arbitrarily. Hence one can take any field $k$, that carries a discrete valuation ring $overline{O}subset k$. Let $Osubset k[[Gamma]]$ be the preimage of $overline{O}$ under the natural homomorphism $k[[Gamma]]rightarrow k$. Then $O$ is a valuation ring of the fraction field of $k[[Gamma]]$ and $k[[Gamma]]$ is a localization of $O$. Consequently $O$ must have infinite global dimension, since global dimension decreases under localization. On the other side the maximal ideal of $O$ is generated by any preimage of a generator of the maximal ideal of $overline{O}$.
answered Mar 23 at 3:33
Hagen KnafHagen Knaf
6,9521318
$endgroup$
$begingroup$
I'm quite not sure why $O$ is a Valuation ring ... why are the ideals linearly ordered ?
$endgroup$
– user521337
Mar 24 at 0:40
$begingroup$
To check that $O$ is a valuation ring one verifies that either $xin O$ or $x^{-1}in O$ for every $xin k((Gamma))$. So assume $xnotin O$. Case 1: $xnotin k[[Gamma]]$. Then $x^{-1}$ is in the maximal ideal of $k[[Gamma]]$, which is the preimage of $0$ under the natural map. Hence $x^{-1}in O$. Case 2: $xin k[[Gamma]]$. Then the image $overline{x}$ of $x$ is not in $overline{O}$ and thus $overline{x}^{-1}inoverline{O}$. Consequently $x^{-1}in O$. (Remark: in valuation theory $O$ is called the composite valuation ring of $overline{O}$ and $k[[Gamma]]$).
$endgroup$
– Hagen Knaf
Mar 24 at 9:41
add a comment |
$begingroup$
The example of a valuation domain of infinite global dimension Barbara Osofsky gives is a generalized power series ring $k[[Gamma]]$ with a well-chosen ordered abelian group $Gamma$. But as far as I see the residue field $k$ can be chosen arbitrarily. Hence one can take any field $k$, that carries a discrete valuation ring $overline{O}subset k$. Let $Osubset k[[Gamma]]$ be the preimage of $overline{O}$ under the natural homomorphism $k[[Gamma]]rightarrow k$. Then $O$ is a valuation ring of the fraction field of $k[[Gamma]]$ and $k[[Gamma]]$ is a localization of $O$. Consequently $O$ must have infinite global dimension, since global dimension decreases under localization. On the other side the maximal ideal of $O$ is generated by any preimage of a generator of the maximal ideal of $overline{O}$.
answered Mar 23 at 3:33
Hagen KnafHagen Knaf
6,9521318
$endgroup$
$begingroup$
I'm quite not sure why $O$ is a Valuation ring ... why are the ideals linearly ordered ?
$endgroup$
– user521337
Mar 24 at 0:40
$begingroup$
To check that $O$ is a valuation ring one verifies that either $xin O$ or $x^{-1}in O$ for every $xin k((Gamma))$. So assume $xnotin O$. Case 1: $xnotin k[[Gamma]]$. Then $x^{-1}$ is in the maximal ideal of $k[[Gamma]]$, which is the preimage of $0$ under the natural map. Hence $x^{-1}in O$. Case 2: $xin k[[Gamma]]$. Then the image $overline{x}$ of $x$ is not in $overline{O}$ and thus $overline{x}^{-1}inoverline{O}$. Consequently $x^{-1}in O$. (Remark: in valuation theory $O$ is called the composite valuation ring of $overline{O}$ and $k[[Gamma]]$).
$endgroup$
– Hagen Knaf
Mar 24 at 9:41
add a comment |
$begingroup$
The example of a valuation domain of infinite global dimension Barbara Osofsky gives is a generalized power series ring $k[[Gamma]]$ with a well-chosen ordered abelian group $Gamma$. But as far as I see the residue field $k$ can be chosen arbitrarily. Hence one can take any field $k$, that carries a discrete valuation ring $overline{O}subset k$. Let $Osubset k[[Gamma]]$ be the preimage of $overline{O}$ under the natural homomorphism $k[[Gamma]]rightarrow k$. Then $O$ is a valuation ring of the fraction field of $k[[Gamma]]$ and $k[[Gamma]]$ is a localization of $O$. Consequently $O$ must have infinite global dimension, since global dimension decreases under localization. On the other side the maximal ideal of $O$ is generated by any preimage of a generator of the maximal ideal of $overline{O}$.
answered Mar 23 at 3:33
Hagen KnafHagen Knaf
6,9521318
$endgroup$
The example of a valuation domain of infinite global dimension Barbara Osofsky gives is a generalized power series ring $k[[Gamma]]$ with a well-chosen ordered abelian group $Gamma$. But as far as I see the residue field $k$ can be chosen arbitrarily. Hence one can take any field $k$, that carries a discrete valuation ring $overline{O}subset k$. Let $Osubset k[[Gamma]]$ be the preimage of $overline{O}$ under the natural homomorphism $k[[Gamma]]rightarrow k$. Then $O$ is a valuation ring of the fraction field of $k[[Gamma]]$ and $k[[Gamma]]$ is a localization of $O$. Consequently $O$ must have infinite global dimension, since global dimension decreases under localization. On the other side the maximal ideal of $O$ is generated by any preimage of a generator of the maximal ideal of $overline{O}$.
answered Mar 23 at 3:33
Hagen KnafHagen Knaf
6,9521318
answered Mar 23 at 3:33
Hagen KnafHagen Knaf
6,9521318
answered Mar 23 at 3:33
Hagen KnafHagen Knaf
6,9521318
answered Mar 23 at 3:33
Hagen KnafHagen Knaf
6,9521318
6,9521318
$begingroup$
I'm quite not sure why $O$ is a Valuation ring ... why are the ideals linearly ordered ?
$endgroup$
– user521337
Mar 24 at 0:40
$begingroup$
To check that $O$ is a valuation ring one verifies that either $xin O$ or $x^{-1}in O$ for every $xin k((Gamma))$. So assume $xnotin O$. Case 1: $xnotin k[[Gamma]]$. Then $x^{-1}$ is in the maximal ideal of $k[[Gamma]]$, which is the preimage of $0$ under the natural map. Hence $x^{-1}in O$. Case 2: $xin k[[Gamma]]$. Then the image $overline{x}$ of $x$ is not in $overline{O}$ and thus $overline{x}^{-1}inoverline{O}$. Consequently $x^{-1}in O$. (Remark: in valuation theory $O$ is called the composite valuation ring of $overline{O}$ and $k[[Gamma]]$).
$endgroup$
– Hagen Knaf
Mar 24 at 9:41
add a comment |
$begingroup$
I'm quite not sure why $O$ is a Valuation ring ... why are the ideals linearly ordered ?
$endgroup$
– user521337
Mar 24 at 0:40
$begingroup$
To check that $O$ is a valuation ring one verifies that either $xin O$ or $x^{-1}in O$ for every $xin k((Gamma))$. So assume $xnotin O$. Case 1: $xnotin k[[Gamma]]$. Then $x^{-1}$ is in the maximal ideal of $k[[Gamma]]$, which is the preimage of $0$ under the natural map. Hence $x^{-1}in O$. Case 2: $xin k[[Gamma]]$. Then the image $overline{x}$ of $x$ is not in $overline{O}$ and thus $overline{x}^{-1}inoverline{O}$. Consequently $x^{-1}in O$. (Remark: in valuation theory $O$ is called the composite valuation ring of $overline{O}$ and $k[[Gamma]]$).
$endgroup$
– Hagen Knaf
Mar 24 at 9:41
$begingroup$
I'm quite not sure why $O$ is a Valuation ring ... why are the ideals linearly ordered ?
$endgroup$
– user521337
Mar 24 at 0:40
$begingroup$
I'm quite not sure why $O$ is a Valuation ring ... why are the ideals linearly ordered ?
$endgroup$
– user521337
Mar 24 at 0:40
$begingroup$
To check that $O$ is a valuation ring one verifies that either $xin O$ or $x^{-1}in O$ for every $xin k((Gamma))$. So assume $xnotin O$. Case 1: $xnotin k[[Gamma]]$. Then $x^{-1}$ is in the maximal ideal of $k[[Gamma]]$, which is the preimage of $0$ under the natural map. Hence $x^{-1}in O$. Case 2: $xin k[[Gamma]]$. Then the image $overline{x}$ of $x$ is not in $overline{O}$ and thus $overline{x}^{-1}inoverline{O}$. Consequently $x^{-1}in O$. (Remark: in valuation theory $O$ is called the composite valuation ring of $overline{O}$ and $k[[Gamma]]$).
$endgroup$
– Hagen Knaf
Mar 24 at 9:41
$begingroup$
To check that $O$ is a valuation ring one verifies that either $xin O$ or $x^{-1}in O$ for every $xin k((Gamma))$. So assume $xnotin O$. Case 1: $xnotin k[[Gamma]]$. Then $x^{-1}$ is in the maximal ideal of $k[[Gamma]]$, which is the preimage of $0$ under the natural map. Hence $x^{-1}in O$. Case 2: $xin k[[Gamma]]$. Then the image $overline{x}$ of $x$ is not in $overline{O}$ and thus $overline{x}^{-1}inoverline{O}$. Consequently $x^{-1}in O$. (Remark: in valuation theory $O$ is called the composite valuation ring of $overline{O}$ and $k[[Gamma]]$).
$endgroup$
– Hagen Knaf
Mar 24 at 9:41
add a comment |
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