6
$\begingroup$

Let $\mathbf{Q}^{\mathrm{ab}}$ be the maximal abelian extension of the field of rational numbers $\mathbf{Q}$. I'm interested in the following question:

Is it true that $K^{M}_{2}(\mathbf{Q}^{\mathrm{ab}})/pK^{M}_{2}(\mathbf{Q}^{\mathrm{ab}})=0$ for any prime $p$ ?

where $K^{M}_{\ast}$ is Milnor $K$-theory.

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
13
$\begingroup$

By the Milnor-Bloch-Kato conjecture, this is equivalent to $\mathrm{Br}(\mathbf{Q}^{\mathrm{ab}})[p] = 0$ (by the Kummer sequence $1 \to \mu_p \to \mathbf{G}_m \to \mathbf{G}_m \to 1$, Hilbert 90 $H^1(K,\mathbf{G}_m) = 0$ and $H^2(K,\mathbf{G}_m) = \mathrm{Br}(K)$). This follows from [Neukirch-Schmidt-Wingberg, Cohomology of Number Fields https://www.mathi.uni-heidelberg.de/~schmidt/NSW2e/NSW2.2.pdf ], Proposition (8.1.14) (ii).

Improve this answer
$\endgroup$
1
  • 9
    $\begingroup$ I would only add that for the second Milnor K-group and algebraic extensions of global fields, the Milnor-Bloch-Kato conjecture is due to Tate (Relations between K_2 and Galois cohomology, Inv. math 36 (1976), 257-274). $\endgroup$
    – user05811
    Commented Dec 8, 2016 at 18:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .