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For a field $F$, Let $K_n(F)$ be the Quillen's $n$-th K-group of $F$.Then $K_0(F)\cong \mathbb{Z}$, $K_1(F)\cong F^\times$.

For a finite Galois extension $L/K$, $K_n(L)$ are Galois modules. Then $\mathrm{H}^1(L/K,K_0(L))=0$ trivially holds, and $\mathrm{H}^1(L/K,K_1(L))=0$ by Hilbert 90.

In Srinivas's book Algebraic K-Theory, Page 161, it is proved there if $L/K$ is a cyclic extension, then $\mathrm{H}^1(L/K,K_2(L))=0$.

For general Galois extension $L/K$, do we have $\mathrm{H}^1(L/K,K_2(L))=0$?

Do we have more Hilbert 90 style results on higher K-groups?

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  • $\begingroup$ A simple computation shows that Hilbert 90 holds for the (Quillen) K-theory of finite fields. Does it hold for the K-theory of $\mathbb{C}$ endowed with complex conjugation? $\endgroup$ May 20, 2018 at 20:33
  • $\begingroup$ @FrançoisBrunault: the K-theory of $\mathbb{C}$ can be decomposed into Milnor K-theory and some uniquely divisible stuff. So the inclusion of Milnor K-theory into Quillen K-theory should induce an isomorphism in group cohomology of $\mathbb{Z}/2\mathbb{Z}$ acting by complex conjugation. I think this implies that Hilbert 90 holds in this situation. $\endgroup$ May 25, 2018 at 7:37
  • $\begingroup$ @MatthiasWendt I thought that $K_n^M(\mathbb{C})$ is uniquely divisible for $n \geq 2$ and that $K_n(\mathbb{C})$ has torsion $\mathbb{Q}/\mathbb{Z}$ for odd $n$ (at least this is what I understood from the K-book, see especially VI.1.7.1 and the table VI.3.1). If complex conjugation acts trivially on the torsion then I think we get non-trivialilty of $H^1(\mathbb{Z}/2\mathbb{Z},K_*(\mathbb{C}))$. $\endgroup$ May 25, 2018 at 10:39
  • $\begingroup$ @FrançoisBrunault: of course, you're right. Sorry, I was being stupid. In $K_3$, the torsion is in the indecomposable part. I'm not sure about the action of complex conjugation. I always think of the torsion as coming from roots of unity (in $K_3$ this is more clearly visible). On the roots of unity the complex conjugation wouldn't be trivial, only trivial on $\pm 1$. I'll have to think about the action, but this could be a potential source of counterexample... $\endgroup$ May 28, 2018 at 14:57
  • $\begingroup$ @MatthiasWendt According to VI.1.7.1 in the K-book, complex conjugation acts trivially on $K_3(\mathbb{C})_{\mathrm{tors}}$ so that $H^1(\mathrm{Gal}(\mathbb{C}/\mathbb{R}),K_3(\mathbb{C})) \cong \mathbb{Z}/2\mathbb{Z}$ and similarly for $K_{4i-1}(\mathbb{C})$, $i \geq 1$. This seems to answer the question, but I admit I would be more satisfied with a more concrete description of this 2-torsion element in $K_3(\mathbb{C})$. For example, does it come from the Bloch group of $\mathbb{C}$? $\endgroup$ May 29, 2018 at 4:42

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Another good reference is Weibel's K-book, see section III.6. Voevodsky has proved that Hilbert 90 indeed generalizes to higher Milnor K-theory, see III.7.8.4 in the K-book.

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    $\begingroup$ Note that there is an important difference between Milnor K-theory and K-theory. $\endgroup$ May 20, 2018 at 19:19

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