(Krull) dimension of dense open subset of finite type algebra over a domain
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Let $Dto A$ be a finite type algebra with $D$ a domain. Suppose $Vsubset operatorname{Spec}A$ is open and dense. Is it true that $dim V=dim A$?
I know that if $Xto operatorname{Spec}Bbbk$ is an integral scheme of finite type over a field then for any non-empty open $Usubset X$ we have $dim U=dim X$. Is this possible to globalize to domains?
algebraic-geometry commutative-algebra affine-schemes krull-dimension
asked Jan 18 at 21:26
ArrowArrow
5,21931546
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add a comment |
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Let $Dto A$ be a finite type algebra with $D$ a domain. Suppose $Vsubset operatorname{Spec}A$ is open and dense. Is it true that $dim V=dim A$?
I know that if $Xto operatorname{Spec}Bbbk$ is an integral scheme of finite type over a field then for any non-empty open $Usubset X$ we have $dim U=dim X$. Is this possible to globalize to domains?
algebraic-geometry commutative-algebra affine-schemes krull-dimension
asked Jan 18 at 21:26
ArrowArrow
5,21931546
$endgroup$
add a comment |
$begingroup$
Let $Dto A$ be a finite type algebra with $D$ a domain. Suppose $Vsubset operatorname{Spec}A$ is open and dense. Is it true that $dim V=dim A$?
I know that if $Xto operatorname{Spec}Bbbk$ is an integral scheme of finite type over a field then for any non-empty open $Usubset X$ we have $dim U=dim X$. Is this possible to globalize to domains?
algebraic-geometry commutative-algebra affine-schemes krull-dimension
asked Jan 18 at 21:26
ArrowArrow
5,21931546
$endgroup$
Let $Dto A$ be a finite type algebra with $D$ a domain. Suppose $Vsubset operatorname{Spec}A$ is open and dense. Is it true that $dim V=dim A$?
I know that if $Xto operatorname{Spec}Bbbk$ is an integral scheme of finite type over a field then for any non-empty open $Usubset X$ we have $dim U=dim X$. Is this possible to globalize to domains?
algebraic-geometry commutative-algebra affine-schemes krull-dimension
algebraic-geometry commutative-algebra affine-schemes krull-dimension
asked Jan 18 at 21:26
ArrowArrow
5,21931546
asked Jan 18 at 21:26
ArrowArrow
5,21931546
asked Jan 18 at 21:26
ArrowArrow
5,21931546
asked Jan 18 at 21:26
ArrowArrow
5,21931546
asked Jan 18 at 21:26
ArrowArrow
5,21931546
5,21931546
add a comment |
add a comment |
1 Answer
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No. For instance, let $D=A=mathbb{Z}_p$ (or any other DVR). Then the open set $Vsubsetoperatorname{Spec} A$ where $p$ does not vanish is open and dense, but $V=operatorname{Spec}mathbb{Q}_p$ so $dim V=0neq 1=dim A$.
answered Jan 19 at 0:13
Eric WofseyEric Wofsey
193k14222353
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Ah, the Sierpinksi space! Should have thought of it.
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– Arrow
Jan 19 at 0:15
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
No. For instance, let $D=A=mathbb{Z}_p$ (or any other DVR). Then the open set $Vsubsetoperatorname{Spec} A$ where $p$ does not vanish is open and dense, but $V=operatorname{Spec}mathbb{Q}_p$ so $dim V=0neq 1=dim A$.
answered Jan 19 at 0:13
Eric WofseyEric Wofsey
193k14222353
$endgroup$
$begingroup$
Ah, the Sierpinksi space! Should have thought of it.
$endgroup$
– Arrow
Jan 19 at 0:15
add a comment |
$begingroup$
No. For instance, let $D=A=mathbb{Z}_p$ (or any other DVR). Then the open set $Vsubsetoperatorname{Spec} A$ where $p$ does not vanish is open and dense, but $V=operatorname{Spec}mathbb{Q}_p$ so $dim V=0neq 1=dim A$.
answered Jan 19 at 0:13
Eric WofseyEric Wofsey
193k14222353
$endgroup$
$begingroup$
Ah, the Sierpinksi space! Should have thought of it.
$endgroup$
– Arrow
Jan 19 at 0:15
add a comment |
$begingroup$
No. For instance, let $D=A=mathbb{Z}_p$ (or any other DVR). Then the open set $Vsubsetoperatorname{Spec} A$ where $p$ does not vanish is open and dense, but $V=operatorname{Spec}mathbb{Q}_p$ so $dim V=0neq 1=dim A$.
answered Jan 19 at 0:13
Eric WofseyEric Wofsey
193k14222353
$endgroup$
No. For instance, let $D=A=mathbb{Z}_p$ (or any other DVR). Then the open set $Vsubsetoperatorname{Spec} A$ where $p$ does not vanish is open and dense, but $V=operatorname{Spec}mathbb{Q}_p$ so $dim V=0neq 1=dim A$.
answered Jan 19 at 0:13
Eric WofseyEric Wofsey
193k14222353
answered Jan 19 at 0:13
Eric WofseyEric Wofsey
193k14222353
answered Jan 19 at 0:13
Eric WofseyEric Wofsey
193k14222353
answered Jan 19 at 0:13
Eric WofseyEric Wofsey
193k14222353
193k14222353
$begingroup$
Ah, the Sierpinksi space! Should have thought of it.
$endgroup$
– Arrow
Jan 19 at 0:15
add a comment |
$begingroup$
Ah, the Sierpinksi space! Should have thought of it.
$endgroup$
– Arrow
Jan 19 at 0:15
$begingroup$
Ah, the Sierpinksi space! Should have thought of it.
$endgroup$
– Arrow
Jan 19 at 0:15
$begingroup$
Ah, the Sierpinksi space! Should have thought of it.
$endgroup$
– Arrow
Jan 19 at 0:15
add a comment |
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