Is $GL\_n(\mathbb{Q}\_p)=GL\_n(\mathbb{Z}\_p)GL\_n(\mathbb{Q})$? Generally, let $R$ be a discrete valuation ring and $K$ its fraction field. Let $\widehat{R}$ be the completion and $\widehat{K}$ the fraction field of $\widehat{R}$. Is $GL\_n(\widehat{K})=GL\_n(\widehat{R})GL\_n(K)$? We know it is true when $n=1$.



Is $GL_n(\mathbb{Q}_p)=GL_n(\mathbb{Z}_p)GL_n(\mathbb{Q})$?
Yes: more generally for every topological group $G$, dense subgroup $H$ and open subgroup $U$, we have $G=UH$.


This applies when $F$ is a valued field, $A$ an open subring (typically, elements of non-negative valuation), and $K$ any dense subfield of $F$, $G=\mathrm{GL}\_n(F)$, $H=\mathrm{GL}\_n(K)$, $U=\mathrm{GL}\_n(A)$.


