TITLE: Kernel of the determinant morphism from the first algebraic K-theory
QUESTION [5 upvotes]: If $A$ is the coordinate ring of a smooth variety over a finite field is it known whether the kernel of the determinant map $K_1(A)\rightarrow A^{\times}$ is torsion or not?

REPLY [3 votes]: I think the following affine deleted quadric is an example where the kernel of the determinant fails to be torsion. Assume we are working in odd characteristic. Then we have the smooth affine quadric $Q_2$ defined by the equation $XY+Z^2=1$. We have $K_0(Q_2)=\mathbb{Z}\oplus\mathbb{Z}$ (either by direct computation or using that $Q_2$ is $\mathbb{A}^1$-equivalent to $\mathbb{P}^1$). Consider the localization sequence involving $Q_2\subseteq \mathbb{A}^3$ with open complement $U=\mathbb{A}^3\setminus Q_2$:
$$
K_1(\mathbb{A}^3)\to K_1(U)\to K_0(Q_2)\to K_0(\mathbb{A}^3)\to K_0(\mathbb{A}^3\setminus Q_2).
$$
The first group is torsion and the last map is an isomorphism, so that $K_1(U)$ has rank 2. On the other hand, the variety $U$ has coordinate ring $\mathbb{F}_q[X,Y,Z][(XY+Z^2-1)^{-1}]$. The element $XY+Z^2-1$ in the polynomial ring (which is a UFD) is irreducible, hence the multiplicative set $\{(XY+Z^2-1)^n\mid n\in \mathbb{N}\}$ is saturated and therefore the group of units of the coordinate ring has rank 1. So the determinant map on $K_1(U)$ must have non-torsion kernel. (This should work more generally whenever we consider the complement of an irreducible smooth hypersurface $Z$ in affine space where the hypersurface $Z$ has non-torsion reduced K-theory.) 
You probably know that the Parshin conjecture would imply that the determinant map $K_1(X)\to \mathbb{F}_q^\times$ has torsion kernel, but that's for smooth projective $X$. I think the above example shows that it's not quite so clear what the right formulation of consequences of Parshin's conjecture for K-theory of smooth affine varieties should be. 
In any case, the Parshin conjecture isn't known at this point, except of course for curves. To emphasize, the simplest case where we don't know what's happening is related to Bloch's conjecture for curves over global function fields. If $X/K$ is a curve over a finite extension $K/\mathbb{F}_q(T)$, then there is a smooth projective curve $C/\mathbb{F}_q$ with function field $K$ and a smooth projective surface $\mathfrak{X}$ with a proper flat map $\mathfrak{X}\to C$ whose generic fiber is $X$. Asking if the determinant map $K_1(\mathfrak{X})\to \mathbb{F}_q^\times$ has torsion kernel is equivalent to asking if the kernel $V(X)$ of the pushforward map $CH^2(X,1)=H^1_{\rm Zar}(X,\mathbf{K}_2)\to K_1(K)$ is torsion. This equivalence is established in 

R. Akhtar. Cycles on curves over global fields of positive characteristic. Trans. Amer. Math. Soc. 357(7), 2557-2569, 2005. 

I guess the best we know about the group $V(X)$ at this point is that the maximal divisible subgroup is uniquely divisible, and the reduced quotient (mod max divisible) is finite. This is proved in 

S. Kondo and S. Yasuda. First and second K-groups of an elliptic curve over a global field of positive characteristic. Ann. Inst. Fourier (Grenoble) 68 (2018), no. 5, 2005–2067. 

So even in the smooth projective case, with appropriate reformulations, the question seems fairly open and it seems we don't know the simplest cases beyond curves.