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



Voevodsky's proof in any characteristic (for motivic and Chow)
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](http://www.claymath.org/library/monographs/cmim02.pdf). 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 $Sm/k$.)


