Voevodsky's proof in any characteristic (for motivic and Chow) Regarding the published version of "Motivic cohomology groups are isomorphic to higher Chow groups in any characterstic" (IMRN) available at:
http://imrn.oxfordjournals.org/content/2002/7/351.full.pdf
I would appreciate any elucidation on the following two points:
(1) a short assertion is made on the second page whose proof is not entirely clear to me 
and
(2) how this assertion factors into the proof of Theorem 1.
For (1) the short assertion is (from page 2, second paragraph from top):
For any smooth scheme $X$ (of finite type) over a field $k$, $x$ a point of $X$ and $k_0$ the subfield of constants of $k$, there exists a smooth variety $X_0$ over $k_0$  and a point $x_0$ on $X_0$ such that the local rings $O_{X, x}$ and $O_{X_0, x_0}$ are isomorphic. 
As noted, the dimension of $X_0$ may be larger than that of $X$. Here is an example I worked out that I think illustrates the principle of the proof:
Example. Let $k  = \mathbb{F}_p(t), X = \mathbb{A}^1_k$ with parameter $T$, and let $x$ correspond to $(T-0)$. Then $k_0 = \mathbb{F}_p$ and put $X_0 = \mathbb{A}^2_{k_0}$ with  parameters $u, v$. Then via $u\mapsto T, v\mapsto t$, and $x_0$ corresponding to $(u-0)$, we have the desired isomorphism of local rings. Note that the residue fields of the local rings are $\mathbb{F}_p(t)$ and $\mathbb{F}_p(v)$, respectively.
In general (assuming $k$ is absolutely finitely generated, i.e. is finitely generated over its prime field): since the question is local, we assume $X$ affine: $X = Spec(A)$ with $A = k[T_1, \ldots, T_n]/I$. Then take a transcendence basis $t_1, \ldots, t_r$ of $k$ over $k_0$. Then it seems that $X_0$ is built from $T_1, \ldots, T_n, t_1, \ldots, t_r$, but I haven't worked this out in general.
Problem: what if $k$ does not have finite transcendence degree over $k_0$? In the absolutely finitely generated case, if $F$ is the prime field of $k$, then $k_0$ is a finite extension of $F$ and $tr.deg_F(k) = tr.deg_{k_0}(k)$. 
As for (2), the part that is not clear to me is when one passes from $X$ to $X_0$ in the course of the proof of the above cited paper (for example, in Prop 4).
NB: the above cited assertion is not included in the preprint http://www.math.uiuc.edu/K-theory/378/allagree.pdf
Cf. the sentence directly after Corollary 2 on page 1 of this preprint
 A: Voevodsky's statement (1) is plainly false if $k$ is not of finite transcendence degree over $k_0$. For then the fraction field of any $O_{X,x}$ is of infinite transcendence degree over $k_0$ whereas the fraction field of $O_{X_0,x_0}$ always has finite transcendence degree over $k_0$.
But anyway here's a more detailed justification of this reduction step. Assume that $k_0$ is a perfect field and that we've already established the equivalence of motivic cohomology and higher Chow groups for schemes in $Sm/k_0$. Let $k_0\subset k$ be any field extension. We'd like to show that
$$H^{n,q}(X)\cong CH^q(X,2q-n)$$
for any $X\in Sm/k$ and any $n,q\in\mathbb Z$.
Any $X\in Sm/k$ can be written as a cofiltered limit of smooth $k_0$-schemes of finite type as follows. It suffices to do it for $X=Spec(k)$ itself, since any smooth scheme of finite type over $S=lim S_\alpha$ comes from a smooth scheme of finite type over some $S_\alpha$ (by EGA4 Prop. 17.7.8). Now $Spec(k)$ is the limit of the schemes $Spec(L)$ where $L$ ranges over the finitely generated subextensions of $k/k_0$, and if $L=k_0(x_1,\dotsc,x_n)$, $Spec(L)$ is the limit of the nonempty open subsets of $Spec(k_0[x_1,\dotsc,x_n])$, which are eventually smooth over $k_0$ since $k_0$ is perfect (see EGA4, Cor. 17.15.13). So in this way we can view any $X\in Sm/k$ as a pro-object $(X_\alpha)$ in $Sm/k_0$.
The key is that both motivic cohomology and higher Chow groups are "continuous" with respect to such inverse limits, i.e., they transform them into colimits:
$$H^{n,q}(X)\cong colim_\alpha H^{n,q}(X_\alpha),\quad CH^q(X,2q-n)\cong colim_\alpha CH^q(X_\alpha,2q-n).$$
The former is Lemma 3.9 in Lecture Notes on Motivic Cohomology. I don't know a reference for the latter, but it should be easy. Since $X_\alpha\in Sm/k_0$ we know that $H^{n,q}(X_\alpha)\cong CH^{q}(X_\alpha,2q-n)$, whence the desired isomorphism in the colimit.
(Of course, if you analyze the proof, you'll find out that this isomorphism between motivic cohomology and higher Chow groups comes from a zig-zag of maps defined at the level of complexes of presheaves, so you've really shown that $\mathbb Z(q)$ and $\mathbb Z^{FS}(q)$ are quasi-isomorphic in the Zariski topology over $Sm/k$.)
