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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 $n=1$.

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  • $\begingroup$ The completion is still a discrete valuation ring. So we may use Smith normal form. $\endgroup$
    – Wilberd van der Kallen
    Commented May 15, 2018 at 7:35

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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 $F$, $G=\mathrm{GL}_n(F)$, $H=\mathrm{GL}_n(K)$, $U=\mathrm{GL}_n(A)$.

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    $\begingroup$ There is a $K$ that should be an $F$. I tried to edit this but for some reason an edit has to be at least 6 characters. $\endgroup$
    – Geoffroy Horel
    Commented May 15, 2018 at 13:14
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    $\begingroup$ @GeoffroyHorel it's now been edited $\endgroup$
    – YCor
    Commented May 15, 2018 at 13:48

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