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Bryan Shih
Bryan Shih
3:52 AM
Gabber rigidity: if (R,I) is henselian pair, n is invertible, then K(R)/n->K(R/I)/n is an equivalence.

Is the same true with K replaced by TC?

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Why I asked: Some form of this statement seem to be implied in the paper by Clausen, Mathew, Morrow, https://arxiv.org/pdf/1803.10897.pdf
1. page 3 says when n is iinvertible Theorem A reduces to a computation of finite field.
2. Last step of Theorem 4.36 (in char k\not= p case)
 
 
3 hours later…
Dustin Clausen
Dustin Clausen
6:52 AM
Hi Bryan, yes, it's true for 'trivial' reasons: if R is a ring in which n is invertible, then R/n =0, so THH(R)/n=0, so TC(R)/n=0. As the same applies to R/I this means the answer is yes because both terms vanish. We should have probably made this remark in the introduction to the paper.
You can be a bit more specific about this, and I think we mention this at some point in the paper: the map R --> R^_p to the derived p-completion is an equivalence on mod p TC. Mod p TC really is a theory for p-adic rings. It doesn't give any information when p is invertible.
In some sense K(1)-local K-theory is a complementary theory, since R --> R[1/p] is an equivalence in K(1)-local K-theory.
Indeed TC and K(1)-local K-theory play similar roles in their respective settings since they are very close to etale K-theory
You can see my paper with Akhil on etale hyperdescent for more on this
They may seem different because TC/p is basically connective (only possible exception is pi_{-1}) whereas K(1)-local K-theory is basically periodic, but this is a red herring
In fact they're both "periodic" in the most relevant sense, that for a base scheme S their value on P^1_S rel \infty_S is the same as their value on S
i.e. they're periodic w.r.t. an algebraic sphere
 
 
2 hours later…
Bryan Shih
Bryan Shih
8:47 AM
@DustinClausen Thanks, this is interesting :O
 

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