Given a matrix $A\in GL_n(\mathbb{Q})$. Can it be expressed as a product of two matrices $B,C$ with $B\in GL_n(\mathbb{Z}[1/p])$ and $C\in GL_{n}(\mathbb{Z}_{(p)})$, where $ \mathbb{Z_p}$ denotes the localization of $\mathbb{Z}$ at $(p)$ ?
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Yes, there exists such a decomposition. Multiplying $A$ by a suitable integer, we may assume that $A \in M_n(\mathbf{Z})$ and $\det(A) \neq 0$. By the theory of elementary divisors, we have $A=\gamma_1 D \gamma_2$ with $\gamma_1,\gamma_2 \in SL_n(\mathbf{Z})$ and $D$ is a diagonal matrix with nonzero integral entries. Now the result is true for $D$, and letting $D=M_1 M_2$ with $M_1 \in GL_n(\mathbf{Z}[1/p])$ and $M_2 \in GL_n(\mathbf{Z}_{(p)})$, we see that $B=\gamma_1 M_1$ and $C=M_2 \gamma_2$ work.
answered Apr 4, 2011 at 13:15
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$\begingroup$ NOw I am wondering what happens in the more general version where we want to decompose a matrix in $GL_n(R\otimes S)$ in two matrices over $R$ and $S$ like above? I guess in general this is not possible, right? $\endgroup$ Commented Apr 5, 2011 at 8:55
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1$\begingroup$ It is really impossible, as the matrix $(1,xy;0,1)$ over $\mathbb{Z}[X,Y]\cong \mathbb{Z}[X]\otimes \mathbb{Z}[Y]$ shows. $\endgroup$ Commented Apr 5, 2011 at 11:15
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$\begingroup$ @Henrik: indeed, the above proof uses crucially the fact that $\mathbf{Z}$ is principal. I don't know if it still works when the base ring is not principal. $\endgroup$ Commented Apr 5, 2011 at 11:33
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$\begingroup$ Indeed the upper matrix is not even contained in the subgroup spanned by $GL_2(\mathbb{K}[X])$ and $GL_2(\mathbb{K}[Y])$. $\endgroup$ Commented Apr 6, 2011 at 10:23