# applying Chinese Remainder Theorem

Suppose $$N=p_1^{n_1}p_2^{n_2}...p_t^{n_t}$$ where the $$p_i$$ are unique primes .

Now by the Chinese Remainder Theorem

$$SL_2(\mathbb{Z}/N\mathbb{Z})=SL_2(\mathbb{Z}/p_1^{n_1}\mathbb{Z}) \times SL_2(\mathbb{Z}/p_2^{n_2} \mathbb{Z}) \times...\times SL_2(\mathbb{Z}/p_t^{n_t}\mathbb{Z})$$

I only know that $$\mathbb{Z}/N\mathbb{Z}=\mathbb{Z}/p_1^{n_1}\mathbb{Z} \times \mathbb{Z}/p_2^{n_2}\mathbb{Z} \times... \times \mathbb{Z}/p_t^{n_t}\mathbb{Z}$$ by the Chinese Remainder Theorem , but why is it also true for $$SL_2(\mathbb{Z})=\lbrace \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} :a,b,c,d \in \mathbb{Z} \ \ , ad-bc=1 \rbrace$$ ?

Thanks for help .

• What is $SL_2$? Jun 2, 2019 at 11:03
• $D\equiv 1\pmod{N}\iff(\forall i)(D\equiv 1\pmod{p_i^{n_i}})$. Jun 2, 2019 at 11:20

Because, more generally, for any commutative rings $$R$$ and $$S$$, we have an isomorphism of matrix rings $$\operatorname{M}_{n×n} (R × S) \cong \operatorname{M}_{n×n} R × \operatorname{M}_{n×n} S,$$ that restricts to isomorphisms of matrix groups $$\operatorname{GL}_n (R × S) \cong \operatorname{GL}_n R × \operatorname{GL}_n S \quad\text{and}\quad \operatorname{SL}_n (R × S) \cong \operatorname{SL}_n R × \operatorname{SL}_n S.$$ The first isomorphism, the one of the matrix rings, should be fairly obvious: Just split up a matrix of component pairs into a pair of component matrices. No information is lost, addition and multiplication are preserved. Why does it restrict so nicely to these matrix groups?
It’s because $$(R×S)^× = R^× × S^×$$ and $$(1,1)$$ is the unit element in $$R × S$$ and the determinant of a matrix $$A ∈ \operatorname{M}_{n×n} (R×S)$$ is given by the determinants of its component matrices $$A_R$$ and $$A_S$$, that is $$\det A = (\det A_R, \det A_S).$$