# How do we get empty fiber over $\infty$?

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

• I don't also understand what does the notation $\mathbb{Z}[\mathbb{A}^n]$ means ? – user329017 Aug 16 '18 at 5:36

$\Bbb Z[\Bbb A^n]$ 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 $F_{|Y} = 0$ and $a \in \overline{Y}$ 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.