Schemes in which every affine open subset is principalSchemes covered by finitely-many affine open subsetsdoes every affine closed subscheme have an open affine neighborhood?Is every integral point of an arithmetic scheme contained in an affine open set?Principal open sets of affine schemesLine bundles on affine schemesBirational affine schemesOpen sets in the spectrum of a Dedekind domain are affineAffine Schemes and Basic Open SetsCan scheme be covered by affine schemes like varieties?Strength of “every affine scheme is compact”
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Schemes in which every affine open subset is principal
Schemes covered by finitely-many affine open subsetsdoes every affine closed subscheme have an open affine neighborhood?Is every integral point of an arithmetic scheme contained in an affine open set?Principal open sets of affine schemesLine bundles on affine schemesBirational affine schemesOpen sets in the spectrum of a Dedekind domain are affineAffine Schemes and Basic Open SetsCan scheme be covered by affine schemes like varieties?Strength of “every affine scheme is compact”
$begingroup$
What are some known high dimensional projective schemes in which every affine open subset is principal?
If $X$ is a Dedekind scheme with torsion class group, then the answer is true.
algebraic-geometry commutative-algebra
$endgroup$
|
show 3 more comments
$begingroup$
What are some known high dimensional projective schemes in which every affine open subset is principal?
If $X$ is a Dedekind scheme with torsion class group, then the answer is true.
algebraic-geometry commutative-algebra
$endgroup$
$begingroup$
What is your definition of principal for projective schemes?
$endgroup$
– Mohan
yesterday
$begingroup$
@Mohan in $Proj(S)$ we can call $D_+(f)$ principal open subsets.
$endgroup$
– zzy
yesterday
$begingroup$
If $X$ is smooth (say over an algebraically closed field) and Picard group has rank one, what you say will be true. If Picard group has rank greater than one, it will not be true.
$endgroup$
– Mohan
yesterday
$begingroup$
@Mohan Thank you, how to see that?
$endgroup$
– zzy
yesterday
$begingroup$
By the way, I should have asked for one more clarification. While $S$ determines $X$, the converse is not true. Are you fixing the $S$ or allowing it to be any $S$ such that the proj is $X$?
$endgroup$
– Mohan
yesterday
|
show 3 more comments
$begingroup$
What are some known high dimensional projective schemes in which every affine open subset is principal?
If $X$ is a Dedekind scheme with torsion class group, then the answer is true.
algebraic-geometry commutative-algebra
$endgroup$
What are some known high dimensional projective schemes in which every affine open subset is principal?
If $X$ is a Dedekind scheme with torsion class group, then the answer is true.
algebraic-geometry commutative-algebra
algebraic-geometry commutative-algebra
asked yesterday
zzyzzy
2,6331420
2,6331420
$begingroup$
What is your definition of principal for projective schemes?
$endgroup$
– Mohan
yesterday
$begingroup$
@Mohan in $Proj(S)$ we can call $D_+(f)$ principal open subsets.
$endgroup$
– zzy
yesterday
$begingroup$
If $X$ is smooth (say over an algebraically closed field) and Picard group has rank one, what you say will be true. If Picard group has rank greater than one, it will not be true.
$endgroup$
– Mohan
yesterday
$begingroup$
@Mohan Thank you, how to see that?
$endgroup$
– zzy
yesterday
$begingroup$
By the way, I should have asked for one more clarification. While $S$ determines $X$, the converse is not true. Are you fixing the $S$ or allowing it to be any $S$ such that the proj is $X$?
$endgroup$
– Mohan
yesterday
|
show 3 more comments
$begingroup$
What is your definition of principal for projective schemes?
$endgroup$
– Mohan
yesterday
$begingroup$
@Mohan in $Proj(S)$ we can call $D_+(f)$ principal open subsets.
$endgroup$
– zzy
yesterday
$begingroup$
If $X$ is smooth (say over an algebraically closed field) and Picard group has rank one, what you say will be true. If Picard group has rank greater than one, it will not be true.
$endgroup$
– Mohan
yesterday
$begingroup$
@Mohan Thank you, how to see that?
$endgroup$
– zzy
yesterday
$begingroup$
By the way, I should have asked for one more clarification. While $S$ determines $X$, the converse is not true. Are you fixing the $S$ or allowing it to be any $S$ such that the proj is $X$?
$endgroup$
– Mohan
yesterday
$begingroup$
What is your definition of principal for projective schemes?
$endgroup$
– Mohan
yesterday
$begingroup$
What is your definition of principal for projective schemes?
$endgroup$
– Mohan
yesterday
$begingroup$
@Mohan in $Proj(S)$ we can call $D_+(f)$ principal open subsets.
$endgroup$
– zzy
yesterday
$begingroup$
@Mohan in $Proj(S)$ we can call $D_+(f)$ principal open subsets.
$endgroup$
– zzy
yesterday
$begingroup$
If $X$ is smooth (say over an algebraically closed field) and Picard group has rank one, what you say will be true. If Picard group has rank greater than one, it will not be true.
$endgroup$
– Mohan
yesterday
$begingroup$
If $X$ is smooth (say over an algebraically closed field) and Picard group has rank one, what you say will be true. If Picard group has rank greater than one, it will not be true.
$endgroup$
– Mohan
yesterday
$begingroup$
@Mohan Thank you, how to see that?
$endgroup$
– zzy
yesterday
$begingroup$
@Mohan Thank you, how to see that?
$endgroup$
– zzy
yesterday
$begingroup$
By the way, I should have asked for one more clarification. While $S$ determines $X$, the converse is not true. Are you fixing the $S$ or allowing it to be any $S$ such that the proj is $X$?
$endgroup$
– Mohan
yesterday
$begingroup$
By the way, I should have asked for one more clarification. While $S$ determines $X$, the converse is not true. Are you fixing the $S$ or allowing it to be any $S$ such that the proj is $X$?
$endgroup$
– Mohan
yesterday
|
show 3 more comments
0
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$begingroup$
What is your definition of principal for projective schemes?
$endgroup$
– Mohan
yesterday
$begingroup$
@Mohan in $Proj(S)$ we can call $D_+(f)$ principal open subsets.
$endgroup$
– zzy
yesterday
$begingroup$
If $X$ is smooth (say over an algebraically closed field) and Picard group has rank one, what you say will be true. If Picard group has rank greater than one, it will not be true.
$endgroup$
– Mohan
yesterday
$begingroup$
@Mohan Thank you, how to see that?
$endgroup$
– zzy
yesterday
$begingroup$
By the way, I should have asked for one more clarification. While $S$ determines $X$, the converse is not true. Are you fixing the $S$ or allowing it to be any $S$ such that the proj is $X$?
$endgroup$
– Mohan
yesterday