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May 16, 2016 at 7:22 comment added naf For a number field $k$, $K_2(k)$ is always torsion and non-zero and has no non-zero divisible subgroups (by a theorem of Garland). If $k$ has no real embeddings, it seems likely that it follows that $K_n(k)$ is zero for $n>2$ (because $k$ has cohomological dimension 2). Moreover, $K_n(k)$ is non-zero for all $n>2$ if $k$ has a real embedding (because $K_n(k)/2$ is non-zero).
May 14, 2016 at 7:08 comment added Jinhyun Park On the other hand, is there a number field $k$ or any characteristic $0$ field for which some $K_n ^M (k)$ could vanish?
May 14, 2016 at 7:06 comment added Jinhyun Park Thank you very much! I wanted to see some concrete examples where the Milnor K-groups could still be zero when $k$ is infinite. It looks I was just given in the above at least two examples where the Milnor K-groups still become zero when $k$ is algebraic over the prime subfield, and when $k$ is not algebraic over the prime subfield.
May 14, 2016 at 5:53 comment added Tyler Lawson Are you asking for nonvanishing of all $K_n^M$, or just if there is some nonvanishing group? If $k$ is a finite field and $k(t)$ is the field of rational functions over it, then there's a Milnor exact sequence which, in particular, shows that $K^M_n(k(t))$ is trivial for $n \geq 3$. (I believe that this exact sequence is discussed at length in Bass and Tate's "The Milnor ring of a global field.")
May 14, 2016 at 4:22 comment added naf Also, if Milnor $K$-theory vanishes for finite fields then it clearly vanishes for any algebraic extension of a finite field (which can be infinite...).
May 14, 2016 at 4:13 comment added naf It follows from Borel's work on the stable cohomology of artihmetic groups that $K_n^M(k)$ is always torsion for $n>1$ if $k$ is a number field.
May 14, 2016 at 1:54 history asked Jinhyun Park CC BY-SA 3.0