1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ I don't also understand what does the notation $\mathbb{Z}[\mathbb{A}^n]$ means ? $\endgroup$ – Adeek Aug 16 '18 at 5:36
1
$\begingroup$

$\Bbb Z[\Bbb A^n]$ 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.