Why is the quotient map $SL_n(\mathbb{Z})$ to $SL_n(\mathbb{Z}/p\mathbb Z)$ is surjective? Recall that $SL_n(\mathbb{Z})$ is the special linear group, $n\geq 2$, and let $q\geq 2$ be any integer. We have a natural quotient map $$\pi: SL_n(\mathbb{Z})\to SL_n(\mathbb{Z}/q).$$ I remember that this map is surjective (is it correct?). It seems the Chinese Remainder Theorem might be helpful, but I forgot how to prove it.
Can anyone give some tips?

Here is a proof that $SL_n(\mathbb{Z}/q\mathbb{Z})$ is indeed generated by matrices of the form
$$ A_{ij}(t) =
\begin{cases}
t & \text{at the $ij^{th}$ position} \\
0 & \text{everywhere else}
\end{cases} $$
We shall call these transvections. Note that $SL_n(R \prod S) \simeq  SL_n(R) \prod SL_n(S)$ with the natural map; which sends transvections to a tuple of transvections and whose inverse sends a tuple of transvections to a transvection. Thus if
$$ R = \prod_{i \in I} R_i$$ it suffices to check that $SL_n(R_i)$ is generated by transvections for all $i \in I$ in order to say that $SL_n(R)$ is generated by transvections.
Note that by the CRT we have that
$$ \mathbb{Z}/q\mathbb{Z} = \prod_{i \in I} \mathbb{Z}/q_i\mathbb{Z} $$
with $q_i = p_i^{k_i}$ for some prime $p_i$.
We now have to prove that for $q = p^k$, $SL_n(\mathbb{Z}/q\mathbb{Z})$ is generated by transvections. In order to prove this note that every element in $\mathbb{Z}/q\mathbb{Z}$ is a unit or is annihilated by $p^{k-1}$. We thus conclude that if a matrix $M$ is in $SL_n(\mathbb{Z}/q\mathbb{Z})$ then every row must contain a unit (else we would have that the determinant would be annihilated by $p^{k-1}$ and hence not a unit). Thus the first row of $M$ must contain a unit. Now one can proceed as if the problem was stated over a field!
