Let $n, m > 1$. The map $\mathbb{Z} \twoheadrightarrow \mathbb{Z}/m\mathbb{Z}$, of reduction mod $m$, induces a group homomorphism $F: \text{SL}_n(\mathbb{Z}) \to \text{SL}_n(\mathbb{Z}/m\mathbb{Z})$. My question is, is $F$ surjective or not?
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1Yes. This is proved in Shimura's book "Introduction to the Arithmetic Theory of Automorphic Functions." I don't have a copy in front of me to find the page. – KCd Aug 25 '15 at 15:39
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5related question: http://math.stackexchange.com/questions/321765/why-is-the-quotient-map-sl-n-mathbbz-to-sl-n-mathbbz-p-mathbb-z-is-s – KCd Aug 25 '15 at 15:42
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Another source for a proof is Lemma 6.3.10 of Henri Cohen's "Number Theory, Volume 1: Tools and Diophantine Equations." – KCd Aug 25 '15 at 15:44
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3As @KCd has said, this is Lemma 1.38 in Shimura's book, where it is proven using elementary divisor theory. I feel that there should be a more elementary proof. – darij grinberg Aug 25 '15 at 15:49
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@KCd Isn't this a duplicate of that related question? – Caleb Stanford Aug 25 '15 at 16:39
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@6005, yes, indeed it is! When I read that related question before I thought the q there was prime, which it need not be. – KCd Aug 25 '15 at 19:22
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This turned out to be a non-example. I'm leaving it up for posterity as a CW answer.
As it turns out, the answer is no. The smallest counterexample I can think of is as follows: we note that $$ \pmatrix{2&1&1\\1&2&1\\1&1&2} \in SL_n(\Bbb Z/3\Bbb Z) $$ However, this matrix is not the reduction mod $3$ of any element of $SL_n(\Bbb Z)$.
Ben Grossmann
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To be sure, I haven't checked all members of the equivalence class of this element, so I'm not certain that this is a counterexample after all. – Ben Grossmann Aug 25 '15 at 15:19
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2I found a matrix in $SL_n(\mathbb Z)$ reducing to this one: $$\pmatrix{-1&1&1\1&-1&-2\-2&1&-1} \in SL_n(\mathbb Z)$$ Unfortunately, I had no more sophisticated method than "check the determinant of a lot of matrices" – Milo Brandt Aug 25 '15 at 15:21
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