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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?

ougao
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  • Sorry ougao, actually my post was not sensical (I had been thinking of the general linear group). Anyhow, it is odd of you to start eliminating cases ("n>1"). If it's true for $n>1$, chances are it's going to be true for $n=1$ also. – rschwieb Mar 05 '13 at 20:11

3 Answers3

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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.

Julien
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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.

Sean Eberhard
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  • You cannot simply make it a $1$ by row or column operations. In the end you have to lift matrices of the form $\begin{pmatrix} a & 0 \ 0 & a^{-1} \end{pmatrix}$. And then one uses this mysterious factorization http://math.stackexchange.com/questions/111650 – Martin Brandenburg Apr 11 '13 at 15:41
  • @MartinBrandenburg Granted my outline was very rough, I think my algorithm should not stumble on that matrix. See my new answer to your linked question for instance. – Sean Eberhard Apr 11 '13 at 16:29
  • So what do you mean by "and making it a $1$"? It seems to me that this cannot be done in one step using a row or column operations (but rather in three steps and it is quite tricky). – Martin Brandenburg Apr 11 '13 at 16:39
  • I agree, you can't do it one step, but you can do it by messing about. – Sean Eberhard Apr 11 '13 at 16:48
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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!

Vatsa Srinivas
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