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yesterday comment added Andy Putman @NoahG.Singer: That's correct. You're looking at what is called "unstable" $K_2$, and in his book Milnor computes $K_2(\mathbb{Z})$ by first making an unstable computation.
yesterday history edited Noah G. Singer CC BY-SA 4.0
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2 days ago vote accept Noah G. Singer
2 days ago comment added Noah G. Singer One note: The notation $K_2$ seems to refer to the direct limit of the quotients as $n \to \infty$. I am interested in the case of fixed $n$ (say, $n=3$) so it is not technically sufficient to just show $K_2$ is trivial. However, looks like e.g. Milnor's techniques probably work for the $n=3$ case directly?
Dec 9 at 1:35 comment added Andy Putman If this kind of thing is going to come up in your work repeatedly, it's worth knowing some rudiments of algebraic K-theory. Milnor's short book on the subject starts at the beginning and is basically self-contained (modulo standard graduate algebra stuff). I've had many graduate students read it. Lots of more recent books focus on stuff with a more algebraic topology flavor, so they're probably harder for outsiders to read.
Dec 8 at 3:31 history edited მამუკა ჯიბლაძე CC BY-SA 4.0
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Dec 8 at 3:21 answer added F Zaldivar timeline score: 4
Dec 7 at 20:56 answer added user509184 timeline score: 8
Dec 7 at 19:41 comment added Matthias Wendt Milnor in his paper "Algebraic K-theory and quadratic forms" explains the vanishing of $K_2$ of finite fields in Example 1.5. It's an algebraic consequence of $K_1$ being cyclic. For the polynomial ring case, some more theory is needed.
Dec 7 at 18:35 history edited Noah G. Singer CC BY-SA 4.0
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Dec 7 at 18:31 comment added Noah G. Singer E.g., I found this survey: cs.ox.ac.uk/people/david.mestel/essay.pdf for $\mathbb{F}_q$. I am hoping for a simpler proof in the case where we only care about $K_2$.
Dec 7 at 18:28 history asked Noah G. Singer CC BY-SA 4.0