Can topological cyclic homology compute Picard groups? Let $K$ be a number field, and $\mathcal{O}_K$ its ring of integers.  Then there is an isomorphism
$$K_0(\mathcal{O}_K) \cong \mathbb{Z} \oplus Pic(\mathcal{O}_K)$$
where $Pic(\mathcal{O}_K)$ is the Picard group (or ideal class group) of $\mathcal{O}_K$.
The cyclotomic trace defines a map 
$$trc: K(\mathcal{O}_K) \to TC(\mathcal{O}_K)$$
to the topological cyclic homology of $\mathcal{O}_K$.  
My question is: to what extent is this map an equivalence in degree 0?  That is, does $TC_0(\mathcal{O}_K)$ compute $Pic(\mathcal{O}_K)$?
I know a complete, local statement of this sort: Hesselholt and Madsen proved that $trc$ is an equivalence on $(-1)$-connected covers of $p$-completions if $\mathcal{O}_K$ is replaced with the ring of Witt vectors of a perfect field of characteristic $p$ (in "On the K-theory of finite algebras over Witt vectors of a perfect field").
Is there anything of this sort in the number field setting?  
 A: Warning: I know nothing about this subject, but found the question interesting so decided to learn something about it. Approach the following with caution.
Consider the ring of integers $\mathbb{Z}[\tfrac{1+\sqrt{-15}}{2}] \subset \mathbb{Q}[\sqrt{-15}]$. This has class number 2, and the ideal
$$I:=(2, \tfrac{1+\sqrt{-15}}{2})$$
generates the ideal class group: in other words this is a projective module of rank 1, and is not free.
Writing $\zeta = \tfrac{1+\sqrt{-15}}{2}$ we have
$$\mathbb{Z}[\tfrac{1+\sqrt{-15}}{2}] = \mathbb{Z}[\zeta]/(\zeta^2-\zeta+4),$$
and so its 2-adic completion is
$$(\mathbb{Z}[\tfrac{1+\sqrt{-15}}{2}])^\hat{}_2 = \mathbb{Z}^\hat{}_2[\zeta]/(\zeta^2-\zeta+4).$$
Using [http://www.numbertheory.org/php/2adic.html] we find that $-15$ has a 2-adic square root $\eta$ starting $(1,0,0,1,1,0,1,1,0,0,...)$. Thus $1+\eta = (0,1,0,1,1,0,1,1,0,0,...)$ is divisible by 2, and we set $\bar{\eta} = \tfrac{1+\eta}{2}=(1,0,1,1,0,1,1,0,0,...)$. Hence we have
$$\zeta^2-\zeta+4 = (\zeta - \bar{\eta})(\zeta -\tfrac{4}{\bar{\eta}}) \in \mathbb{Z}^\hat{}_2[{\zeta}],$$
and so
$$\mathbb{Z}^\hat{}_2[\zeta]/(\zeta^2-\zeta+4) = \mathbb{Z}^\hat{}_2[\zeta]/((\zeta - \bar{\eta})(\zeta -\tfrac{4}{\bar{\eta}})) \cong \mathbb{Z}^\hat{}_2[\zeta]/(\zeta - \bar{\eta}) \times \mathbb{Z}^\hat{}_2[\zeta]/(\zeta -\tfrac{4}{\bar{\eta}})$$
which is isomorphic to $\mathbb{Z}^\hat{}_2 \times \mathbb{Z}^\hat{}_2$ as a ring. As $K_0(\mathbb{Z}^\hat{}_2 \times \mathbb{Z}^\hat{}_2) = K_0(\mathbb{Z}^\hat{}_2) \oplus K_0(\mathbb{Z}^\hat{}_2) = \mathbb{Z} \oplus \mathbb{Z}$ is torsion-free, the conclusion is that 
$$K_0(\mathbb{Z}[\tfrac{1+\sqrt{-15}}{2}]) \to K_0((\mathbb{Z}[\tfrac{1+\sqrt{-15}}{2}])^\hat{}_2)$$
fails to be injective: in fact, its kernel is precisely $\mathbb{Z}/2$.
The relevance of this to the question is that, as I understand it, 2-adic $TC$ is insensitive to 2-adic completion of the ring (at least for rings which are finitely generated as abelian groups), that is,
$$TC(\mathbb{Z}[\tfrac{1+\sqrt{-15}}{2}])^\hat{}_2 \to TC((\mathbb{Z}[\tfrac{1+\sqrt{-15}}{2}])^\hat{}_2)^\hat{}_2$$
is an equivalence. It follows from the evident commutative square that
$$K_0(\mathbb{Z}[\tfrac{1+\sqrt{-15}}{2}])^\hat{}_2 \to TC_0(\mathbb{Z}[\tfrac{1+\sqrt{-15}}{2}])^\hat{}_2$$
is not injective.
