Is $\phi :SL_2(Z) \to SL_2(Z/NZ)$ still surjective if we replace Z with some ring of integers?

Question

It is well known that the natural map $\phi :SL_2(Z) \to SL_2(Z/NZ)$ is surjective. So that the kernel, i.e. the principal congruence subgroup is of finite index.

But what if we replace Z with some ring of integers? Is this map still surjective?

Answer 1

The map $SL_2(O_K) \to SL_2(O_K / I)$ is surjective for any ideal $I \triangleleft O_K$, where $K$ is a number field. The "usual" proof for $SL_2(Z)$ works without change for $SL_2(O_K)$ if $O_K$ is Euclidean, see here. The general case is a little more advanced; there is a very slick proof in these notes of Brian Conrad if you're confident working with adeles.

Answer 2

Let $R$ be a commutative ring with unit and let $I$ be an ideal of $R$. Then the natural map $SL(n,R) \rightarrow SL(n,R/I)$ is surjective for all $n$, provided $SL(n,R/I)$ is generated by elementary matrices. In particular this is true for $R$ being the ring of integers $\mathcal{O}_K$ of a number field $K$.

The question has been discussed in more generality here. In fact, there are rings $R$ such that the above map is not surjective, e.g., for $R=\mathbb{Z}[a,b,c,d]$ and ideal $I$ generated by $ad-bc-1$.