How do we get empty fiber over $infty$?
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I am trying to understand a proof in 3264 and all that. I don't fully understand how do we get empty fiber over $infty$ ?
algebraic-geometry intersection-theory
edited Aug 16 at 5:36
Chaminda Bandara
1056
asked Aug 16 at 5:01
Newbie
391111
add a comment |Â
up vote
1
down vote
favorite
I am trying to understand a proof in 3264 and all that. I don't fully understand how do we get empty fiber over $infty$ ?
algebraic-geometry intersection-theory
edited Aug 16 at 5:36
Chaminda Bandara
1056
asked Aug 16 at 5:01
Newbie
391111
I don't also understand what does the notation $mathbbZ[mathbbA^n]$ means ?
– Newbie
Aug 16 at 5:36
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am trying to understand a proof in 3264 and all that. I don't fully understand how do we get empty fiber over $infty$ ?
algebraic-geometry intersection-theory
edited Aug 16 at 5:36
Chaminda Bandara
1056
asked Aug 16 at 5:01
Newbie
391111
I am trying to understand a proof in 3264 and all that. I don't fully understand how do we get empty fiber over $infty$ ?
algebraic-geometry intersection-theory
edited Aug 16 at 5:36
Chaminda Bandara
1056
asked Aug 16 at 5:01
Newbie
391111
edited Aug 16 at 5:36
Chaminda Bandara
1056
edited Aug 16 at 5:36
Chaminda Bandara
1056
edited Aug 16 at 5:36
Chaminda Bandara
1056
1056
asked Aug 16 at 5:01
Newbie
391111
asked Aug 16 at 5:01
Newbie
391111
asked Aug 16 at 5:01
Newbie
391111
391111
I don't also understand what does the notation $mathbbZ[mathbbA^n]$ means ?
– Newbie
Aug 16 at 5:36
add a comment |Â
I don't also understand what does the notation $mathbbZ[mathbbA^n]$ means ?
– Newbie
Aug 16 at 5:36
I don't also understand what does the notation $mathbbZ[mathbbA^n]$ means ?
– Newbie
Aug 16 at 5:36
I don't also understand what does the notation $mathbbZ[mathbbA^n]$ means ?
– Newbie
Aug 16 at 5:36
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
$Bbb Z[Bbb A^n]$ is the free abelian group with generator $[Bbb A^n] in A^*(Bbb A^n)$. Recall that the Chow ring $A^*(X)$ is generated by the $Bbb Z$-span of closed subvarieties $Y subset X$.
Recall by definition of the Zariski closure that if $Y$ is an subset of an algebraic variety $X$ and $F$ a regular function on $X$, then $F_ = 0$ and $a in overlineY$ implies $F(a) = 0$.
The proof constructs a function $G$ with $G_infty times Bbb A^n neq 0$ and $G_W = 0$, since $W$ is Zariski closed it implies that $infty times Bbb A^n cap W = emptyset$, i.e the fiber of $W$ over $infty$ is empty as required.
answered Aug 16 at 11:00
Nicolas Hemelsoet
5,040316
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$Bbb Z[Bbb A^n]$ is the free abelian group with generator $[Bbb A^n] in A^*(Bbb A^n)$. Recall that the Chow ring $A^*(X)$ is generated by the $Bbb Z$-span of closed subvarieties $Y subset X$.
Recall by definition of the Zariski closure that if $Y$ is an subset of an algebraic variety $X$ and $F$ a regular function on $X$, then $F_ = 0$ and $a in overlineY$ implies $F(a) = 0$.
The proof constructs a function $G$ with $G_infty times Bbb A^n neq 0$ and $G_W = 0$, since $W$ is Zariski closed it implies that $infty times Bbb A^n cap W = emptyset$, i.e the fiber of $W$ over $infty$ is empty as required.
answered Aug 16 at 11:00
Nicolas Hemelsoet
5,040316
add a comment |Â
up vote
1
down vote
accepted
$Bbb Z[Bbb A^n]$ is the free abelian group with generator $[Bbb A^n] in A^*(Bbb A^n)$. Recall that the Chow ring $A^*(X)$ is generated by the $Bbb Z$-span of closed subvarieties $Y subset X$.
Recall by definition of the Zariski closure that if $Y$ is an subset of an algebraic variety $X$ and $F$ a regular function on $X$, then $F_ = 0$ and $a in overlineY$ implies $F(a) = 0$.
The proof constructs a function $G$ with $G_infty times Bbb A^n neq 0$ and $G_W = 0$, since $W$ is Zariski closed it implies that $infty times Bbb A^n cap W = emptyset$, i.e the fiber of $W$ over $infty$ is empty as required.
answered Aug 16 at 11:00
Nicolas Hemelsoet
5,040316
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$Bbb Z[Bbb A^n]$ is the free abelian group with generator $[Bbb A^n] in A^*(Bbb A^n)$. Recall that the Chow ring $A^*(X)$ is generated by the $Bbb Z$-span of closed subvarieties $Y subset X$.
Recall by definition of the Zariski closure that if $Y$ is an subset of an algebraic variety $X$ and $F$ a regular function on $X$, then $F_ = 0$ and $a in overlineY$ implies $F(a) = 0$.
The proof constructs a function $G$ with $G_infty times Bbb A^n neq 0$ and $G_W = 0$, since $W$ is Zariski closed it implies that $infty times Bbb A^n cap W = emptyset$, i.e the fiber of $W$ over $infty$ is empty as required.
answered Aug 16 at 11:00
Nicolas Hemelsoet
5,040316
$Bbb Z[Bbb A^n]$ is the free abelian group with generator $[Bbb A^n] in A^*(Bbb A^n)$. Recall that the Chow ring $A^*(X)$ is generated by the $Bbb Z$-span of closed subvarieties $Y subset X$.
Recall by definition of the Zariski closure that if $Y$ is an subset of an algebraic variety $X$ and $F$ a regular function on $X$, then $F_ = 0$ and $a in overlineY$ implies $F(a) = 0$.
The proof constructs a function $G$ with $G_infty times Bbb A^n neq 0$ and $G_W = 0$, since $W$ is Zariski closed it implies that $infty times Bbb A^n cap W = emptyset$, i.e the fiber of $W$ over $infty$ is empty as required.
answered Aug 16 at 11:00
Nicolas Hemelsoet
5,040316
answered Aug 16 at 11:00
Nicolas Hemelsoet
5,040316
answered Aug 16 at 11:00
Nicolas Hemelsoet
5,040316
answered Aug 16 at 11:00
Nicolas Hemelsoet
5,040316
5,040316
add a comment |Â
add a comment |Â
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I don't also understand what does the notation $mathbbZ[mathbbA^n]$ means ?
– Newbie
Aug 16 at 5:36