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Let $X$ be an scheme. Fix a prime $l$ which is invertible in $X$. Consider the $K(1)$-localization at prime $l$ of algebraic K theory $L_1K(X)$ and $l$-adic completion of etale K theory $K^{et}(X)$.

Is there an map from one to the other?

If $X$ is an Noetherian scheme of finite Krull dimension, then $L_1K(X)$ and $K^{et}(X)$ are equivalent as spectra?

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Yes, there is a map: by Thomason's work, $L_{K(1)}K$ satisfies étale descent and the canonical map $K\to L_{K(1)}K$ therefore induces a canonical map $K^{ét}\to L_{K(1)}K$.

This paper by Clausen and Mathew shows that if you tweak $L_{K(1)}K$ a little bit, using $TC$, then you get something which looks a lot like $K^{ét}$, namely the canonical map from $K^{ét}$ incudes an isomorphism on $\pi_*, *\geq -1$.

I don't know about the second question beyond this, but this was a little too long for a comment.

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  • $\begingroup$ The OP seems to be interested in $\ell$-adic $K$-theory. If I remember correctly, it satisfies étale (hyper)descent, and would be much easier to analyze (e.g. via Gabber–Suslin rigidity)? $\endgroup$
    – Z. M
    Commented Jul 31, 2022 at 9:29
  • $\begingroup$ @Z.M : as I said, I don't know a lot about this - if you have an answer to add along the lines of your comment, I'm sure that would be very helpful $\endgroup$
    – Maxime Ramzi
    Commented Jul 31, 2022 at 14:38
  • $\begingroup$ Basically if $\ell$ is invertible on $X$, it is invertible on $TC$, and therefore $L_{K(1)}TC$ should vanish. It follows that the $\ell$-adic completion of $K^{ét}$ should just be $L_{K(1)}K$ in degrees $\geq 0$ , if I'm not saying anything silly $\endgroup$
    – Maxime Ramzi
    Commented Jul 31, 2022 at 16:02
  • $\begingroup$ Just to add a remark that what I said was incorrect: even $\ell$-completely, $K$-theory does not seem to satisfy étale descent, but it seems to be controlled by the norm-residue isomorphism (I only heard these from talks so incompetent to say anything seriously). $\endgroup$
    – Z. M
    Commented Jul 31, 2022 at 18:03

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