Why is the quotient map $SL_n(\mathbb{Z})$ to $SL_n(\mathbb{Z}/p\mathbb Z)$ is surjective?

Question

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?

Answer 1

The result is true for $n\geq 1$ and any integer $q\geq 1$.

The group $SL_n(\mathbb{Z}/q\mathbb{Z})$ is generated by the elementary (transvection) matrices. It is easily seen that every elementary matrix is in the image of $\pi$, as the image of an elementary matrix in $SL_n(\mathbb{Z})$. So $\pi$ is surjective.

Answer 2

It suffices to prove that $SL_n(\mathbb{Z}/q)$ is generated by the elementary matrices (the matrices with ones along the diagonal and exactly one other nonzero entry). Equivalently, given any matrix in $SL_n(\mathbb{Z}/q)$, you can reduce it to $I$ using elementary row and column operations (add a multiple of a row to another row, or add a multiple of a column to another column).

Let $A\in SL_n(\mathbb{Z}/q)$, and consider its bottom row. By performing elementary column operations you can effectively perform the Euclidean algorithm, reducing the entries in the bottom row to all zero except for one entry which must be invertible modulo $q$. Moving this entry to the bottom-right position and making it a $1$, we reduce to the $n-1$ case.

Answer 3

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!