Understanding the proof of the fact that the Chow group of a scheme $X$ is graded by dimension.

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.

• 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 $(t-t_0)$. – Servaes Apr 1 at 12:26
• @Servaes Why is $\Phi \cap (\mathbb{A}^1 \times X)$ affine? – ciccio Apr 1 at 15:05
• 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 $\mathbb{A}^1$). – Eoin May 28 at 4:31