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 $infty$ ?empty fiber




algebraic-geometry intersection-theory




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  • I don't also understand what does the notation $mathbbZ[mathbbA^n]$ means ?
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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 $infty$ ?empty fiber




algebraic-geometry intersection-theory




share|cite|improve this question






















  • I don't also understand what does the notation $mathbbZ[mathbbA^n]$ means ?
    – Newbie
    Aug 16 at 5:36












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





I am trying to understand a proof in 3264 and all that. I don't fully understand how do we get empty fiber over $infty$ ?empty fiber




algebraic-geometry intersection-theory




share|cite|improve this question














I am trying to understand a proof in 3264 and all that. I don't fully understand how do we get empty fiber over $infty$ ?empty fiber





algebraic-geometry intersection-theory





share|cite|improve this question













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edited Aug 16 at 5:36









Chaminda Bandara

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asked Aug 16 at 5:01









Newbie

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  • 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




I don't also understand what does the notation $mathbbZ[mathbbA^n]$ means ?
– Newbie
Aug 16 at 5:36










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$Bbb Z[Bbb A^n]$ 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.






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    1 Answer
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    up vote
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    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.






    share|cite|improve this answer
























      up vote
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      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.






      share|cite|improve this answer






















        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.






        share|cite|improve this answer












        $Bbb Z[Bbb A^n]$ 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.







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        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 16 at 11:00









        Nicolas Hemelsoet

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