Chow group of UFD is trivial? In literature I saw the word "Chow group problem" coming up, questioning under what conditions the Chow Group is trivial ($=0$). In general this seems to be quite a difficult question.
What is if the given variety is known well? Let $k$ be a algebraically closed field, we can take e.g. $k=\mathbb{C}$. Given the torus $T_N=\text{Spec } k[x_1^{\pm 1}, ..., x_n^{\pm 1}]$ of a $n$-dimensional toric variety can we conclude that the chow groups $CH_{k}(T_N)=0$ for $0< k < n-1$ (or even for $k=0$ or $k=n$)? Does this hold - if at all - in general for spectra of UFD? I have no clue how to start a proof nor how to construct a counterexample
The case $k=n-1$ is already covered as $CH_{n-1}(T_N)=Cl(T_N) =0$.
 A: Question: "In literature I saw the word "Chow group problem" coming up, questioning under what conditions the Chow Group is trivial (=0)."
Answer: There are some methods available to calculate Chow groups (and grothendieck groups). In the case of  a non-singular curve $C$ over an algebraicallg closed field $k$ there is an isomorphism
$$\phi:K_0(C)\cong Pic(C) \oplus \mathbb{Z} \cong CH^*(C)$$
defined by $\phi([E]):=(det(E), rk(E))$ for a locally trivial rank $r$ sheaf $E$. Here $det(E):=\wedge^r E$.
Here $K_0(C)$ is the grothendieck group of $C$ and $CH^*(C)$ is the Chow group.
There is moreover for any $X$ that is "regular in codimension one" (and noetherian, integral, separated and locally factorial) an isomorphism
$$Pic(X \times_k \mathbb{A}^1_k) \cong Pic(X).$$
Hence
$$Pic(\mathbb{A}^n_k) \cong Pic(Spec(k))\cong \mathbb{Z}$$
hence
$$I1.\text{  }\phi:K_0(\mathbb{A}^1_k) \cong CH^*(\mathbb{A}^1_k)\cong  \mathbb{Z}.$$
It is a classical result that projective modules on polynomial rings are free, hence the map in $I1$ is the following map: Let $E:=k[x]\{e_1,..,e_r\}$ be a free $k[x]$-module. It follows $det(E):=\wedge^r E \cong k[x]e_1\wedge \cdots \wedge e_r$ is the free rank one module on $e_1\wedge \cdots \wedge e_r$.
By definition
$$\phi([E]):= (det(E), r) \in (e)\oplus \mathbb{Z}$$
since $Pic(\mathbb{A}^1_k)\cong (e)$ is the trivial group. Since any finite rank projective module on the polynomial ring $A:=k[x_1,..,x_n]$ is free it follows
$$K_0(\mathbb{A}^n_k) \cong \mathbb{Z}.$$
Similarly
$$ CH^*(\mathbb{A}^n_k)=\mathbb{Z}.$$
You will find some of these questions discussed in Hartshorne, Chapter II.6.
Note: Exercise HH.II.6.10 proves that $K_0(C) \cong Pic(C) \oplus \mathbb{Z}$ for any such $C$. You should consult this exercise.
Question: "Is the Chow group of a UFD is trivial?"
Answer: By the isomorphism in $I1$: No.
