Suppose that $G$ is an absolutely quasi-simple algebraic group over a non-archimedean local field $k$ (of either zero or positive characteristic). Is it known whether or not it is necessarily the case that the group $G(k)$ with the strong $k$-topology is locally topologically finitely generated and locally hereditarily just infinite?



query about quasi-simple algebraic groups over local fields
The answer to your question is yes. Let $G(k)$ be the group of rational points of $G$. Choose a global field $K$ contained in $k$ such that $G$ is defined over $K$ (this can be done, by e.g. classification of $G$ over $k$). Let $S$ be a large finite set of places of $K$ such that $S$ contains all the Archimedean ones, and such that $G(O\_S)$ has "higher rank" and such that $S$ contains a non-archimedean place different from the completion $v:K\subset k$, and so that $S$ does not contain the place $v$. The rank of $G(O\_S)$ is by definition the sum
$$\sum \_{v\in S} K\_v-rank (G);$$ one says that $G(O\_S)$ has higher rank if the rank of $G(O\_S)$ is $\geq 2$. By strong approximation, $G(O\_S)$ is dense in the compact open subgroup $G(O\_k)$. 


Suppose $U\subset G(k)$ is a compact open subgroup. Then the intersection $\Gamma =G(O\_S)\cap U$ has finite index in $G(O\_S)$; moreover, since $G(O\_S)$ has higher rank, it is finitely generated (it always is f.g in char zero - a result of Borel-Harish-Chandra, but in char p, you need higher rank). Hence $\Gamma $ is f.g and dense in $U$. Thus $G(k)$ is topologically finitely generated. 


We now turn to proving that $G(k)$ is "locally hereditarily just infinite" (in OP's terminology). If $G$ is isotropic over $k$, then we can choose a $K$-form of $G$ (again using classification) so that $G$ is isotropic over $K$; now suppose $N$ is a normal subgroup of an open $U$ in $G(O\_k)$. If N'= $N\cap \Gamma$ is non-trivial, then 
by the Margulis normal subgroup theorem, $N'$ has finite index in $N$ and again $\Gamma /N'$ is finite hence, by density of $\Gamma $ in $U$, $U/N$ is finite. 


If $N'$ is trivial, the argument (that I am thinking of) is more involved; the quotient $U/N$ is a profinite group in which the $S$-arithmetic group $\Gamma$ sits densely; but the congruence subgroup property (proved in this generality by Raghunathan) shows that such a quotient $U/N$ cannot exist; i.e. $U/N$ is finite. Hence $G(k)$ is hereditarily just infinite. 


Finally, the anisotropic case can be handled directly. This involves looking at $G=SL\_1(D)$ or $SU(D)$ for suitable division algebras; the argument uses congruence subgroup property for "semi-local" rings, i.e. groups like $G(K)$ or $G(O\_T)$ where $T$ is the $complement$ of a finite set.


By the way, all this is unnecessary if $char~k$ is zero; every normal subgroup $N$ of $U$ is automatically open and hence of finite index, by p-adic Lie algebra considerations 


