$mathbb{C}$-valued points and flatness
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Let $X$ be an integral scheme over $mathrm{Spec}(mathbb{Z})$, we denote $X(mathbb{C})$ as the set of $mathbb{C}$-valued points in $X$. Then $Yrightarrow mathrm{Spec}(mathbb{Z})$ is flat if $Y(mathbb{C})not=emptyset$.
Can anyone show me how to prove this?
algebraic-geometry arithmetic-geometry
asked Nov 17 at 12:23
Intoks Liobein
41328
add a comment |
up vote
0
down vote
favorite
Let $X$ be an integral scheme over $mathrm{Spec}(mathbb{Z})$, we denote $X(mathbb{C})$ as the set of $mathbb{C}$-valued points in $X$. Then $Yrightarrow mathrm{Spec}(mathbb{Z})$ is flat if $Y(mathbb{C})not=emptyset$.
Can anyone show me how to prove this?
algebraic-geometry arithmetic-geometry
asked Nov 17 at 12:23
Intoks Liobein
41328
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $X$ be an integral scheme over $mathrm{Spec}(mathbb{Z})$, we denote $X(mathbb{C})$ as the set of $mathbb{C}$-valued points in $X$. Then $Yrightarrow mathrm{Spec}(mathbb{Z})$ is flat if $Y(mathbb{C})not=emptyset$.
Can anyone show me how to prove this?
algebraic-geometry arithmetic-geometry
asked Nov 17 at 12:23
Intoks Liobein
41328
Let $X$ be an integral scheme over $mathrm{Spec}(mathbb{Z})$, we denote $X(mathbb{C})$ as the set of $mathbb{C}$-valued points in $X$. Then $Yrightarrow mathrm{Spec}(mathbb{Z})$ is flat if $Y(mathbb{C})not=emptyset$.
Can anyone show me how to prove this?
algebraic-geometry arithmetic-geometry
algebraic-geometry arithmetic-geometry
asked Nov 17 at 12:23
Intoks Liobein
41328
asked Nov 17 at 12:23
Intoks Liobein
41328
asked Nov 17 at 12:23
Intoks Liobein
41328
asked Nov 17 at 12:23
Intoks Liobein
41328
asked Nov 17 at 12:23
Intoks Liobein
41328
41328
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
Since $mathbb{Z}$ is a pid, flatness is same as torsion free. Since $Y$ is an integral scheme, torsion free just means the map is dominant. Dominance is equivalent to to $Y(mathbb{C})neqemptyset$.
This is just to answer the query by the OP in the comments. $f:Ytomathbb{Z}$ is dominant means the generic point is in the image. That is, there exists a morphism $mathrm{Spec}, Kto Y$ which when composed with $f$ factors through the generic point $mathrm{Spec},mathbb{Q}subsetmathrm{Spec},mathbb{Z}$. Base changing to $mathrm{Spec},mathbb{C}to mathrm{Spec},mathbb{Q}$, we get a morphism $mathrm{Spec},(Kotimes_{mathbb{Q}}mathbb{C})to Ytimes_{mathbb{Z}}mathrm{Spec}, mathbb{C}$. Hope rest is clear.
answered Nov 17 at 14:39
Mohan
11.5k1817
always so prompt and excellent answer
– Intoks Liobein
Nov 17 at 15:11
Can you explain a little bit for your very last step?
– Intoks Liobein
Nov 21 at 0:50
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Since $mathbb{Z}$ is a pid, flatness is same as torsion free. Since $Y$ is an integral scheme, torsion free just means the map is dominant. Dominance is equivalent to to $Y(mathbb{C})neqemptyset$.
This is just to answer the query by the OP in the comments. $f:Ytomathbb{Z}$ is dominant means the generic point is in the image. That is, there exists a morphism $mathrm{Spec}, Kto Y$ which when composed with $f$ factors through the generic point $mathrm{Spec},mathbb{Q}subsetmathrm{Spec},mathbb{Z}$. Base changing to $mathrm{Spec},mathbb{C}to mathrm{Spec},mathbb{Q}$, we get a morphism $mathrm{Spec},(Kotimes_{mathbb{Q}}mathbb{C})to Ytimes_{mathbb{Z}}mathrm{Spec}, mathbb{C}$. Hope rest is clear.
answered Nov 17 at 14:39
Mohan
11.5k1817
always so prompt and excellent answer
– Intoks Liobein
Nov 17 at 15:11
Can you explain a little bit for your very last step?
– Intoks Liobein
Nov 21 at 0:50
add a comment |
up vote
3
down vote
accepted
Since $mathbb{Z}$ is a pid, flatness is same as torsion free. Since $Y$ is an integral scheme, torsion free just means the map is dominant. Dominance is equivalent to to $Y(mathbb{C})neqemptyset$.
This is just to answer the query by the OP in the comments. $f:Ytomathbb{Z}$ is dominant means the generic point is in the image. That is, there exists a morphism $mathrm{Spec}, Kto Y$ which when composed with $f$ factors through the generic point $mathrm{Spec},mathbb{Q}subsetmathrm{Spec},mathbb{Z}$. Base changing to $mathrm{Spec},mathbb{C}to mathrm{Spec},mathbb{Q}$, we get a morphism $mathrm{Spec},(Kotimes_{mathbb{Q}}mathbb{C})to Ytimes_{mathbb{Z}}mathrm{Spec}, mathbb{C}$. Hope rest is clear.
answered Nov 17 at 14:39
Mohan
11.5k1817
always so prompt and excellent answer
– Intoks Liobein
Nov 17 at 15:11
Can you explain a little bit for your very last step?
– Intoks Liobein
Nov 21 at 0:50
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Since $mathbb{Z}$ is a pid, flatness is same as torsion free. Since $Y$ is an integral scheme, torsion free just means the map is dominant. Dominance is equivalent to to $Y(mathbb{C})neqemptyset$.
This is just to answer the query by the OP in the comments. $f:Ytomathbb{Z}$ is dominant means the generic point is in the image. That is, there exists a morphism $mathrm{Spec}, Kto Y$ which when composed with $f$ factors through the generic point $mathrm{Spec},mathbb{Q}subsetmathrm{Spec},mathbb{Z}$. Base changing to $mathrm{Spec},mathbb{C}to mathrm{Spec},mathbb{Q}$, we get a morphism $mathrm{Spec},(Kotimes_{mathbb{Q}}mathbb{C})to Ytimes_{mathbb{Z}}mathrm{Spec}, mathbb{C}$. Hope rest is clear.
answered Nov 17 at 14:39
Mohan
11.5k1817
Since $mathbb{Z}$ is a pid, flatness is same as torsion free. Since $Y$ is an integral scheme, torsion free just means the map is dominant. Dominance is equivalent to to $Y(mathbb{C})neqemptyset$.
This is just to answer the query by the OP in the comments. $f:Ytomathbb{Z}$ is dominant means the generic point is in the image. That is, there exists a morphism $mathrm{Spec}, Kto Y$ which when composed with $f$ factors through the generic point $mathrm{Spec},mathbb{Q}subsetmathrm{Spec},mathbb{Z}$. Base changing to $mathrm{Spec},mathbb{C}to mathrm{Spec},mathbb{Q}$, we get a morphism $mathrm{Spec},(Kotimes_{mathbb{Q}}mathbb{C})to Ytimes_{mathbb{Z}}mathrm{Spec}, mathbb{C}$. Hope rest is clear.
answered Nov 17 at 14:39
Mohan
11.5k1817
edited Nov 21 at 14:28
answered Nov 17 at 14:39
Mohan
11.5k1817
answered Nov 17 at 14:39
Mohan
11.5k1817
answered Nov 17 at 14:39
Mohan
11.5k1817
11.5k1817
always so prompt and excellent answer
– Intoks Liobein
Nov 17 at 15:11
Can you explain a little bit for your very last step?
– Intoks Liobein
Nov 21 at 0:50
add a comment |
always so prompt and excellent answer
– Intoks Liobein
Nov 17 at 15:11
Can you explain a little bit for your very last step?
– Intoks Liobein
Nov 21 at 0:50
always so prompt and excellent answer
– Intoks Liobein
Nov 17 at 15:11
always so prompt and excellent answer
– Intoks Liobein
Nov 17 at 15:11
Can you explain a little bit for your very last step?
– Intoks Liobein
Nov 21 at 0:50
Can you explain a little bit for your very last step?
– Intoks Liobein
Nov 21 at 0:50
add a comment |
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