Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
    
I removed the newly created tag 'algebraic' since it seems to be created in error and in any case would be too unspecific as a tag. –  quid Feb 19 '13 at 13:50
2  
Voevodsky's statement (1) is plainly false if $k$ is not of finite transcendence degree over $k_0$ ($k$ itself then cannot be of the form $O_{X,x}$). However, it is nonetheless true that if $\mathbb{Z}(q)\simeq\mathbb{Z}^{FS}(q)$ over a perfect field $k_0$, then this automatically holds over any field extension $k$ of $k_0$. This is because these sheaves over $Sm/k$ (and the canonical map between them) are pulled back from $Sm/k_0$. Ultimately this is because any cycle on an inverse limit of schemes comes from one of the schemes in the diagram. I'll try to write a more detailed answer later. –  Marc Hoyois Feb 19 '13 at 18:38
    
Marc: Thanks for your comment. What do you intend to show by saying a field is not of the form of a local ring in your first sentence? Perhaps you meant something about the residue fields? In general, residue fields of local rings can have infinite transcendence degree over the prime field. – Ivan 2 mins ago –  Ivan Feb 19 '13 at 19:34
    
@Ivan: Take $X=Spec(k)$ so that $O_{X,x}=k$. According to Voevodsky's statement there should exist $X_0$ smooth of finite type over $k_0$ whose function field is $k$. So the transcendence degree of $k$ must be finite. –  Marc Hoyois Feb 19 '13 at 20:16

1 Answer 1

up vote 4 down vote accepted

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 $Sm/k$.)

share|improve this answer
    
Thank you for the clear explanation! –  Ivan Feb 20 '13 at 12:54

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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