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 $n=1$.
Asked
Active
Viewed 658 times
17
-
The completion is still a discrete valuation ring. So we may use Smith normal form. – Wilberd van der Kallen May 15 '18 at 07:35
1 Answers
26
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 $F$, $G=\mathrm{GL}_n(F)$, $H=\mathrm{GL}_n(K)$, $U=\mathrm{GL}_n(A)$.
Desiderius Severus
- 5,324
YCor
- 60,149
-
1There 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. – Geoffroy Horel May 15 '18 at 13:14
-
1