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For any field $F$, there is a natural group homomorphism $K_n^{\rm M}(F) \to K_n(F)$ from Milnor's $K$-theory to Quillen's $K$-theory. If $n=2$, it is an isomorphism, by Matsumoto's theorem. It is a well known theorem of Quillen that if $F$ is a number field, then the groups $K_n(F)$ are finitely generated for $n\geq 2$. It is known in particular, that the group $K_2(\mathbb Q)$ is finite, isomorphic to $\mathbb Z/2\mathbb Z$.

I am now confused about the following result of Tate (I got it from Milnors "Introduction to Algebraic K-theory", Theorem 11.6): There is a canonical, split exact sequence of commutative groups $$0 \to \{\pm1\} \to K_2^{\rm M}(\mathbb Q) \to \bigoplus_p (\mathbb Z/p\mathbb Z)^\ast \to 0 $$ and in particular, $K_2^{\rm M}(\mathbb Q)$ is an infinite torsion group. The first term in the sequence should be interpreted as $K_2^{\rm M}(\mathbb Z)$, and each $(\mathbb Z/p\mathbb Z)^\ast$ should be seen as $K_1$ of the finite field with $p$ elements.

What did I misunderstand? Is it not true that $K_2^{\rm M}(\mathbb Q) \cong K_2(\mathbb Q)$ -- or is it not true that $K_2(\mathbb Q)$ is finitely generated --- or did I misquote Tate's theorem?

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up vote 7 down vote accepted

The mistake is your assumption $K_2(\mathbb Q) = \mathbb Z/2$.

For any number field $F$ with ring of integers $\mathcal{O}$, the isomorphism $K_n(F) \cong K_n(\mathcal{O})$ is true only for $n > 1$ odd. If $n > 1$ is even, there is a short exact sequence $$0 \to K_n(\mathcal{O}) \to K_n(F) \to \oplus_P K_{n-1}(\mathcal{O}/P) \to 0$$ where $P$ runs through the primes of $\mathcal{O}$.

Moreover, it's known that $K_n(\mathcal{O})$ is finite for $n>1$ even (for instance $K_2(\mathbb Z) = \mathbb Z/2$). So in this case, $K_n(F)$ is an infinite torsion group.

An excellent survey on the K-theory of number fields can be found in this paper of Weibel.

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thanks- that settles it. – Xandi Tuni Mar 26 '12 at 17:37

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