# 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?

• That's an interesting question. We know that $TR_0(\mathcal{O}_K)$ is the Witt ring (which is torsion free), and so if we were going to detect $Pic$ it would have connect around from the cokernel of $1-F$ on $TR_1$ in the fiber sequence $TC \to TR \to TR$. Apr 23, 2015 at 21:46

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.

• If $A$ is the ring of integers in number field, I think it is true more generally that $A_p$ is a finite product of discrete valuation rings and so has trivial class group. I think that your line of argument can therefore be adapted to show that $K_0(A)\to TC_0(A)$ always kills the class group Apr 24, 2015 at 11:22
• Nice! Do you have a reference for the insensitivity of p-adic $TC$ to p-completion? Apr 24, 2015 at 12:48
• It is Addendum 5.2 in the paper you mentioned in the question. Apr 24, 2015 at 13:24
• Look at me, the careful reader. Thanks. Apr 24, 2015 at 14:07