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I would like to understand the proof of this fact:

If $X$ is a scheme (separated, of finite type over $k=\overline{k}$) then the Chow group of $X$ is graded by dimension; that is, \begin{equation} A(X)=\oplus A_k(x) \end{equation} with $A_k(x)$ the group of rational equivalence classes of $k$-cycles. In the proof it is said that if $\Phi \subset \mathbb{P}^1 \times_{k} X$ is an irreducible variety (not contained in a fiber over $X$) then, in an appropriate affine open set $\Phi \cap (\mathbb{A}^1\ \times X) \subset \Phi$, the scheme $\Phi \cap(\{t_0\} \times X)$ is defined by the vanishing of the single nonzerodivisor $t-t_0$. It follows that the components of this intersection are all of codimension exactly $1$ in $\Phi$, and similarly for $\Phi \cap (\{t_1\} \times X)$.

I don't understand why $\Phi \cap (\mathbb{A}^1 \times X)$ is affine, and i cannot see the fact "the scheme $\Phi \cap(\{t_0\} \times X)$ is defined by the vanishing of the single nonzerodivisor $t-t_0$": it should mean that the intersection above is defined by an ideal generated by one element (i.e. $t-t_0$), but why?

In the final part i would use the Corollary 2.5.26 of Liu's Algebraic geometry and Arithmetic Curves.

Maybe the question seems stupid, but i cannot see those things.

EDIT: the proof is given on "3264 and all that" of Eisenbud and Harris.

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  • $\begingroup$ The second part suggests that, since $\Phi \cap (\mathbb{A}^1 \times X)$ is affine, for $t_0\in\Bbb{A}^1$ the subscheme $\Phi \cap(\{t_0\} \times X)$ is defined by $(t-t_0)$. $\endgroup$ – Servaes Apr 1 at 12:26
  • $\begingroup$ @Servaes Why is $\Phi \cap (\mathbb{A}^1 \times X)$ affine? $\endgroup$ – ciccio Apr 1 at 15:05
  • $\begingroup$ If $X=\mathbb{P}^1$ and $\Phi$ is all of $\mathbb{P}^1\times \mathbb{P}^1$ then it certainly doesn't need to be affine (either as a variety or over $\mathbb{A}^1$). $\endgroup$ – Eoin May 28 at 4:31

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