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Problem. Show that the natural map $\mathrm{SL}_{n}(\mathbb{Z}) \to \mathrm{SL}_{n}(\mathbb{Z}/m\mathbb{Z})$ is surjective, for all $m$ and $n$. Denoting its kernel by $K_{n}(m)$, show that $$\varprojlim(\mathrm{SL}_{n}(\mathbb{Z})/K_{n}(p^{i}))_{i \in \mathbb{N}} \simeq \mathrm{SL}_{n}(\mathbb{Z}_{p}),$$ $$\varprojlim(\mathrm{SL}_{n}(\mathbb{Z})/K_{n}(m))_{m \in \mathbb{N}} \simeq \mathrm{SL}_{n}(\hat{\mathbb{Z}})$$

For the first part, I can prove that $\mathrm{SL}_{n}(\mathbb{Z}) = \langle I + e_{ij}\rangle$. In this paper by Keith Conrad (theorem 3.2), he does a proof using the Chinese Remainder Theorem and the generators of $\mathrm{SL}_{2}(\mathbb{Z})$. I think that the same ideia works for $n > 2$.

Now, since $K_{n}(m)$ is the kernel of a surjective map, so $$\mathrm{SL}_{n}(\mathbb{Z})/K_{n}(p^{i}) \simeq \mathrm{SL}_{n}(\mathbb{Z}/p^{i}\mathbb{Z})$$ and $$\mathrm{SL}_{n}(\mathbb{Z})/K_{n}(m) \simeq \mathrm{SL}_{n}(\mathbb{Z}/m\mathbb{Z}).$$

I know that $$\varprojlim \mathbb{Z}/p^{i}\mathbb{Z} = \mathbb{Z}_{p}$$ and $$\varprojlim \mathbb{Z}/m\mathbb{Z} = \hat{\mathbb{Z}}.$$

Now, its enough to show that

$$\varprojlim \mathrm{SL}_{n}(\mathbb{Z}/p^{i}\mathbb{Z}) \simeq \mathrm{SL}_{n}(\mathbb{Z}_{p})\tag{i}$$

and

$$\varprojlim \mathrm{SL}_{n}(\mathbb{Z}/m\mathbb{Z}) \simeq \mathrm{SL}_{n}(\hat{\mathbb{Z}}).\tag{ii}$$


How can I show (i) and (ii)? PS. I can use that $\mathrm{SL}_{n}(\hat{\mathbb{Z}}) \simeq \hat{\mathrm{SL}_{n}(\mathbb{Z})}$ (sorry for the horrible symbol for completion).

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    $\begingroup$ The reduction modulo every $p^i$ makes $\mathrm{SL}_{n}(\mathbb{Z}_{p})\subset \varprojlim \mathrm{SL}_{n}(\Bbb{Z}_p/p^{i}\Bbb{Z}_p)=\varprojlim \mathrm{SL}_{n}(\mathbb{Z}/p^{i}\mathbb{Z})$, then for $(a_i) \in \varprojlim \mathrm{SL}_{n}(\mathbb{Z}/p^{i}\mathbb{Z})$, for each index $u,v$ let $A_{uv}= \lim_{i \to \infty} (a_i)_{uv} \in \Bbb{Z}_p$, the obtained matrix $A \in SL_n(\Bbb{Z}_p)$. Otherwise send each $a_i$ to some $A_i\in SL_{n}(\mathbb{Z}), \equiv a_i \bmod p^i$ then $\lim_{m \to \infty} A_i$ converges in $\mathrm{SL}_{n}(\mathbb{Z}_{p})$. It works the same way for $\hat{\Bbb{Z}}$. $\endgroup$ – reuns May 23 at 23:53
  • $\begingroup$ @reuns, can you explain a little more about $A_{uv}$ (I cannot see what is the role of $u,v$) and who is $A \in SL_n(\mathbb{Z}_p)$? My book doesn't often use sequence arguments. So I'm not very accustomed to this kind of proof. $\endgroup$ – Lucas Corrêa May 24 at 23:30
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    $\begingroup$ $a_i$ is a matrix $(a_i)_{uv}$ is its $u,v$ entry. $\endgroup$ – reuns May 25 at 21:36

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