In corollary 4.3.13 of Rosenberg's book Algebraic K-theory and its applications, it's proved that $K_2(F)=0$ if $F$ is a finite field.

The last sentence in the proof says that the central extension $\varphi:St(F)\rightarrow SL(F)$ is trivial over $N(F)$, then it is concluded that p-primary part of $K_2(F)$ vanishes. I could not see why this is true. Can anybody shed some light on this? Thanks!


1 Answer 1


In the following all references refer to Rosenberg's book.

Let $U(n,F) \le Sl(n,F)$ be the group of upper triangular matrices with 1's in the diagonal and let $N(n,F) \le St(n,F)$ be the group in the proof of 4.2.3. Furthermore let $K_2(n,F)$ be the kernel of $\varphi_n: St(n,F) \to Sl(n,F)$ and denote by $E \le St(n,F)$ the subgroup generated by $N(n,F)$ and $K_2(n,F)$. There is a commutative diagramm of central extensions $$ 1 \to K_2(n,F) \to St(n,F) \to Sl(n,F) \to 1 $$ $$ \uparrow \hspace{50pt} \uparrow \hspace{40pt} \uparrow $$ $$ 1 \to K_2(n,F) \to \quad E \quad \to U(n,F) \to 1$$ (the left arrow is the identity and the others are inclusions). Let $e \in H^2(Sl(n,F),K_2(n,F))$ be the extension class from the upper extension. Then $\text{res}^2(e) \in H^2(U(n,F),K_2(n,F))$ corresponds to the lower extension and since it splits by 4.2.3, $\text{res}^2(e) = 0$.

Now the universal coefficient theorem (4.1.13) yields the commutative diagramm $$H^2(Sl(n,F),K_2(n,F)) \to Hom(H_2(Sl(n,F);\mathbb{Z}), K_2(n,F))$$ $$ \downarrow \hspace{60pt} \downarrow \hspace{40pt} $$ $$H^2(U(n,F),K_2(n,F)) \to Hom(H_2(U(n,F);\mathbb{Z}), K_2(n,F))$$ The left arrow is $\text{res}^2$ and the right is the dual $\text{res}_2^\ast$ of $$\text{res}_2: H_2(U(n,F);\mathbb{Z}) \to H_2(Sl(n,F);\mathbb{Z}).$$ Moreover, the upper arrow is an isomorphism, since $Sl(n,F)$ is perfect.

Let $\alpha_n: H_2(Sl(n,F);\mathbb{Z}) \to K_2(n,F)$ correspond to $e$ under this iso. Then $$0 = \text{res}_2^\ast(\alpha_n) = \alpha_n \circ \text{res}_2.$$

By 4.1.19, $\alpha = \text{colim } \alpha_n$ is an isomorphism. So we can assume that $\alpha_n$ is injective, which implies $\text{res}_2 = 0$. Since $U(n,F)$ is a Sylow $p$-subgroup of $Sl(n,F)$, the image of $\text{res}_2 =0$ is the $p$-primary part $H_2(Sl(n,F);\mathbb{Z})_{(p)}$ of $H_2(Sl(n,F);\mathbb{Z})$ by 4.1.24. Thus $H_2(Sl(n,F);\mathbb{Z})_{(p)} =0$ which implies $H_2(Sl(F);\mathbb{Z})_{(p)} =0$. But $H_2(Sl(F);\mathbb{Z}) \cong K_2(F)$ (4.1.19), hence $K_2(F)_{(p)} = 0$, what was to be shown. q.e.d.

I don't know if this is the argument Rosenberg had in mind (perhaps someone from the forum can provide a simpler one). Nevertheless I hope it sheds some light on the subject.

  • $\begingroup$ Thanks! I think it is better if Rosenberg explains what he had in mind in the textbook. $\endgroup$
    – Liu Hang
    Oct 28, 2011 at 3:36

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