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To prove the fiber above a codimension 1 point contains a geometrically integral open subscheme
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Suppose $f:Xrightarrow mathbbP_k^n$ is a proper smooth morphism, where $k$ is an algebraically closed field. If $f$ admits a rational section, can we prove that the fiber of $f$ above any codimension 1 point contains a geometrically integral open subscheme?
ag.algebraic-geometry
asked 3 hours ago
Haowen Zhang
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up vote
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Suppose $f:Xrightarrow mathbbP_k^n$ is a proper smooth morphism, where $k$ is an algebraically closed field. If $f$ admits a rational section, can we prove that the fiber of $f$ above any codimension 1 point contains a geometrically integral open subscheme?
ag.algebraic-geometry
asked 3 hours ago
Haowen Zhang
62
New contributor
Haowen Zhang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose $f:Xrightarrow mathbbP_k^n$ is a proper smooth morphism, where $k$ is an algebraically closed field. If $f$ admits a rational section, can we prove that the fiber of $f$ above any codimension 1 point contains a geometrically integral open subscheme?
ag.algebraic-geometry
asked 3 hours ago
Haowen Zhang
62
New contributor
Haowen Zhang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Suppose $f:Xrightarrow mathbbP_k^n$ is a proper smooth morphism, where $k$ is an algebraically closed field. If $f$ admits a rational section, can we prove that the fiber of $f$ above any codimension 1 point contains a geometrically integral open subscheme?
ag.algebraic-geometry
ag.algebraic-geometry
asked 3 hours ago
Haowen Zhang
62
New contributor
Haowen Zhang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Haowen Zhang
62
New contributor
Haowen Zhang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Haowen Zhang
62
New contributor
Haowen Zhang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Haowen Zhang
62
asked 3 hours ago
Haowen Zhang
62
62
New contributor
Haowen Zhang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Haowen Zhang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Haowen Zhang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
1 Answer
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This is true:
Because $f$ is smooth, any fibre $X_y$ is smooth. If $y$ has codimension $1$, then the rational section $sigma colon mathbb P^n_k dashrightarrow X$ is defined at $y$ by the valuative criterion of properness. Thus, $X_y$ has a rational point $sigma(y)$. If $U subseteq X_y$ is the connected component of $sigma(y)$, then $U$ is a smooth connected $kappa(y)$-scheme with a rational point, hence geometrically connected by Tag 04KV.
answered 2 hours ago
R. van Dobben de Bruyn
9,55622959
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
This is true:
Because $f$ is smooth, any fibre $X_y$ is smooth. If $y$ has codimension $1$, then the rational section $sigma colon mathbb P^n_k dashrightarrow X$ is defined at $y$ by the valuative criterion of properness. Thus, $X_y$ has a rational point $sigma(y)$. If $U subseteq X_y$ is the connected component of $sigma(y)$, then $U$ is a smooth connected $kappa(y)$-scheme with a rational point, hence geometrically connected by Tag 04KV.
answered 2 hours ago
R. van Dobben de Bruyn
9,55622959
add a comment |Â
up vote
3
down vote
This is true:
Because $f$ is smooth, any fibre $X_y$ is smooth. If $y$ has codimension $1$, then the rational section $sigma colon mathbb P^n_k dashrightarrow X$ is defined at $y$ by the valuative criterion of properness. Thus, $X_y$ has a rational point $sigma(y)$. If $U subseteq X_y$ is the connected component of $sigma(y)$, then $U$ is a smooth connected $kappa(y)$-scheme with a rational point, hence geometrically connected by Tag 04KV.
answered 2 hours ago
R. van Dobben de Bruyn
9,55622959
add a comment |Â
up vote
3
down vote
up vote
3
down vote
This is true:
Because $f$ is smooth, any fibre $X_y$ is smooth. If $y$ has codimension $1$, then the rational section $sigma colon mathbb P^n_k dashrightarrow X$ is defined at $y$ by the valuative criterion of properness. Thus, $X_y$ has a rational point $sigma(y)$. If $U subseteq X_y$ is the connected component of $sigma(y)$, then $U$ is a smooth connected $kappa(y)$-scheme with a rational point, hence geometrically connected by Tag 04KV.
answered 2 hours ago
R. van Dobben de Bruyn
9,55622959
This is true:
Because $f$ is smooth, any fibre $X_y$ is smooth. If $y$ has codimension $1$, then the rational section $sigma colon mathbb P^n_k dashrightarrow X$ is defined at $y$ by the valuative criterion of properness. Thus, $X_y$ has a rational point $sigma(y)$. If $U subseteq X_y$ is the connected component of $sigma(y)$, then $U$ is a smooth connected $kappa(y)$-scheme with a rational point, hence geometrically connected by Tag 04KV.
answered 2 hours ago
R. van Dobben de Bruyn
9,55622959
answered 2 hours ago
R. van Dobben de Bruyn
9,55622959
answered 2 hours ago
R. van Dobben de Bruyn
9,55622959
answered 2 hours ago
R. van Dobben de Bruyn
9,55622959
9,55622959
add a comment |Â
add a comment |Â
Haowen Zhang is a new contributor. Be nice, and check out our Code of Conduct.
Haowen Zhang is a new contributor. Be nice, and check out our Code of Conduct.
Haowen Zhang is a new contributor. Be nice, and check out our Code of Conduct.
Haowen Zhang is a new contributor. Be nice, and check out our Code of Conduct.
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