Projective subvarieties of a quasiprojective varietyWhy are torsion points dense in an abelian variety?Why...
Projective subvarieties of a quasiprojective variety
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Let $X$ be a quasiprojective variety over $mathbf C$. Take the union of all projective subvarieties $W subseteq X$ that have dimension at least $1$. Is the result Zariski closed?
(I was wondering this in the particular setting $X = mathcal M_g$, where the projective subvarieties have been the subject of some study. But the general question seems natural as well.)
ag.algebraic-geometry
edited 9 hours ago
Mark
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$begingroup$
Let $X$ be a quasiprojective variety over $mathbf C$. Take the union of all projective subvarieties $W subseteq X$ that have dimension at least $1$. Is the result Zariski closed?
(I was wondering this in the particular setting $X = mathcal M_g$, where the projective subvarieties have been the subject of some study. But the general question seems natural as well.)
ag.algebraic-geometry
edited 9 hours ago
Mark
1,049714
asked 9 hours ago
user141570user141570
211
New contributor
user141570 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Let $X$ be a quasiprojective variety over $mathbf C$. Take the union of all projective subvarieties $W subseteq X$ that have dimension at least $1$. Is the result Zariski closed?
(I was wondering this in the particular setting $X = mathcal M_g$, where the projective subvarieties have been the subject of some study. But the general question seems natural as well.)
ag.algebraic-geometry
edited 9 hours ago
Mark
1,049714
asked 9 hours ago
user141570user141570
211
New contributor
user141570 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Let $X$ be a quasiprojective variety over $mathbf C$. Take the union of all projective subvarieties $W subseteq X$ that have dimension at least $1$. Is the result Zariski closed?
(I was wondering this in the particular setting $X = mathcal M_g$, where the projective subvarieties have been the subject of some study. But the general question seems natural as well.)
ag.algebraic-geometry
ag.algebraic-geometry
edited 9 hours ago
Mark
1,049714
asked 9 hours ago
user141570user141570
211
New contributor
user141570 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 9 hours ago
Mark
1,049714
asked 9 hours ago
user141570user141570
211
New contributor
user141570 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 9 hours ago
Mark
1,049714
edited 9 hours ago
Mark
1,049714
edited 9 hours ago
Mark
1,049714
1,049714
asked 9 hours ago
user141570user141570
211
New contributor
user141570 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 9 hours ago
user141570user141570
211
asked 9 hours ago
user141570user141570
211
211
New contributor
user141570 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
add a comment |
1 Answer
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No. For example take $X = mathbb{A}^1 times mathbb{P}^1$ minus one point, say $(x,y)$. Then $W = (mathbb{A}^1 setminus x) times mathbb{P}^1$ is not closed.
answered 8 hours ago
SashaSasha
21.8k22859
$endgroup$
$begingroup$
Oops, good point.
$endgroup$
– Mark
6 hours ago
add a comment |
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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 example take $X = mathbb{A}^1 times mathbb{P}^1$ minus one point, say $(x,y)$. Then $W = (mathbb{A}^1 setminus x) times mathbb{P}^1$ is not closed.
answered 8 hours ago
SashaSasha
21.8k22859
$endgroup$
$begingroup$
Oops, good point.
$endgroup$
– Mark
6 hours ago
add a comment |
$begingroup$
No. For example take $X = mathbb{A}^1 times mathbb{P}^1$ minus one point, say $(x,y)$. Then $W = (mathbb{A}^1 setminus x) times mathbb{P}^1$ is not closed.
answered 8 hours ago
SashaSasha
21.8k22859
$endgroup$
$begingroup$
Oops, good point.
$endgroup$
– Mark
6 hours ago
add a comment |
$begingroup$
No. For example take $X = mathbb{A}^1 times mathbb{P}^1$ minus one point, say $(x,y)$. Then $W = (mathbb{A}^1 setminus x) times mathbb{P}^1$ is not closed.
answered 8 hours ago
SashaSasha
21.8k22859
$endgroup$
No. For example take $X = mathbb{A}^1 times mathbb{P}^1$ minus one point, say $(x,y)$. Then $W = (mathbb{A}^1 setminus x) times mathbb{P}^1$ is not closed.
answered 8 hours ago
SashaSasha
21.8k22859
answered 8 hours ago
SashaSasha
21.8k22859
answered 8 hours ago
SashaSasha
21.8k22859
answered 8 hours ago
SashaSasha
21.8k22859
21.8k22859
$begingroup$
Oops, good point.
$endgroup$
– Mark
6 hours ago
add a comment |
$begingroup$
Oops, good point.
$endgroup$
– Mark
6 hours ago
$begingroup$
Oops, good point.
$endgroup$
– Mark
6 hours ago
$begingroup$
Oops, good point.
$endgroup$
– Mark
6 hours ago
add a comment |
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