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? 



About Tate's computation of $K_2^{\rm M}(\mathbb Q)$
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](http://www.math.uiuc.edu/K-theory/0691/KZsurvey.pdf) of Weibel. 


