TITLE: Chow group of a DVR
QUESTION [6 upvotes]: I need a sanity check on the following: If $A$ is a DVR then $CH_0(A) = 0$.
The proof is simple. A 0-dimensional cycle is of the form $Z= n \cdot [s]$, where s is the closed point of $\operatorname{Spec} A$. Let $\pi$ be a uniformizer of $A$. Then 
$div(\pi^n) = length_A(A/(\pi^n))\cdot [s] = n \cdot [s]$.
Hence $CH_0(A) = Z_0(A)/R_0(A) = 0$. Is this correct?

REPLY [4 votes]: This is a community-wiki answer in order to record the discussion from the comments so that this question may be marked as answered (once this post is upvoted or accepted).

Yes, it is correct. - Mohan

I (KReiser) agree with Mohan's assessment as well. Your work is correct.