It is a well-known fact that the group homomorphism $\mathrm{SL}(2,\mathbb{Z}) \rightarrow \mathrm{SL}(2,\mathbb{Z}/N\mathbb{Z})$ is surjective.

What I want is a proof by method of algebraic geometry. For example, Let $G= \mathrm{SL}(n,\mathbb{Z})$ be the corresponding group scheme over $\mathbb{Z}$. Then $\mathrm{Spec}(\mathbb{Z}/N) \hookrightarrow \mathrm{Spec}(\mathbb{Z})$ gives $G(\mathbb{Z}) \rightarrow G(\mathbb{Z}/N)$. Then use algebro-geometric method to show that this map is surjective.

  • 7
    $\begingroup$ Are you asking about "strong approximation" for simply connected, semisimple algebraic groups? This was proved over number fields by Kneser and Platonov. If Borovoi sees this post, he can give you far more detail than I can. $\endgroup$ May 12, 2021 at 13:42
  • 8
    $\begingroup$ I doubt that what you are looking for exists since the surjectivity boils down to properties of simply connected Chevalley group schemes over $\mathbb{Z}$ (and fails for things like $\mathrm{GL}_n$). There is a story that Demazure asked Serre and Grothendieck something similar to your question and the answer that eventually emerged is strong approximation and the theory in SGA 3. $\endgroup$ May 13, 2021 at 10:18
  • 6
    $\begingroup$ Then the OP is asking about whether the fact that the map of groups $\mathrm{SL}(2,\mathbb{Z}) \to \mathrm{SL}(2,\mathbb{Z}/N\mathbb{Z})$ is surjective can be proved using tools from algebraic geometry. Your answer and resulting comments do not address this question. Since this is a fact from number theory some number theory should play a role in proving this, but it seems reasonable to suspect there may be a proof using tools from algebraic geometry over $\mathbb{Z}$, which is what the OP's question is about. $\endgroup$ May 14, 2021 at 8:41
  • 7
    $\begingroup$ @hm2020: This is bad notation which should be avoided. For a scheme $X$ over a ring $A$ it is standard notation in algebraic geometry to write $X(A)$ for the set of $A$-points of $X$. This is part of Grothendieck's functorial view of algebraic geometry. I think the notation of the OP is perfectly clear to anyone who knows this result in the context of strong approximation and no further clarification is necessary. $\endgroup$ May 14, 2021 at 8:42
  • 13
    $\begingroup$ @hm2020 I must strongly disagree with the asserion that $SL(n,R)$ usually means spectrum of some ring. In fact I have never seen this notation used to denote such a thing, even in algebraic geometry. That notation to me always indicated the mere group of matrices. $\endgroup$
    – Wojowu
    May 14, 2021 at 10:13


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.