Is $\Bbb Z_p$ dense in $\Bbb Q_p$ with respect to p-adic topology ?
I couldn't find formal argument of this anywhere.
I know $\Bbb Q_p/ \Bbb Z_p$ is trivial topology, so I guessed $\Bbb Z_p$ is dense in $\Bbb Q_p$, but I’m loosing confident because I have no evidence that $\Bbb Z_p$ is dense in $\Bbb Q_p$.
Any refference are appreciated. Thank you for your help.
To sum up comments: $\mathbb Z_p$ is not dense in $\mathbb Q_p$.
One of many possible ways to see that is that any element $x$ of $\mathbb Q_p$ which is not already in $\mathbb Z_p$ (for example, $1/p$) has valuation $\lvert x \rvert_p > 1$, so for any sequence $(x_n)_n$ where all $x_n \in \mathbb Z_p$ (which means all $\lvert x_n \rvert_p \le 1$) we get $d(x_n,x)=\lvert x_n -x\rvert_p = \max(\lvert x_n\rvert_p, \lvert x\rvert_p) > 1$ for all $n$ by the ultrametric maximum principle, and hence $(x_n)_n$ cannot converge to $x$.
Alternatively, if $\mathbb Z_p$ were dense in $\mathbb Q_p$ then its topological closure $\overline{\mathbb Z_p}$ in $\mathbb Q_p$ would be all of $\mathbb Q_p$; however, it is well-known that $\mathbb Z_p$ is a closed subset of $\mathbb Q_p$ (e.g. because it is compact; or because it is $\{x \in \mathbb Q_p: d(x,0) \le 1 \}$, and such "closed" balls are closed in any metric space), i.e. its topological closure is itself.
Those should be pretty straightforward arguments from the basics of $p$-adics. The reason for your confusion seems to be your statement "I know $\mathbb Q_p/ \mathbb Z_p$ [has the] trivial topology".
Well there is an unfortunate clash of terminologies. On any set $X$, there are two opposite extremes of topologies:
(1) The one where every subset of $X$ is open (and closed).
(2) The one where only the empty set $\emptyset$ and the full set $X$ are open (and closed).
Both are "trivial" in the casual sense of the word (i.e. not particularly interesting), but it is or should be standard terminology to call only (2) the "trivial", but (1) the "discrete" topology. I have seen people refer to (2) as the "indiscrete" topology, which might or might not be a better nomenclature than "trivial". Unfortunately though, I have also seen people refer to both as "trivial", and that is exactly where your confusion seems to come from:
Namely, say, we have a topological group $X$ with a subgroup $A \subseteq X$, and endow the quotient $X/A$ with the quotient topology. Then
But the quotient topology on $\mathbb Q_p/\mathbb Z_p$ is not that trivial topology (2). Actually, it is the other extreme topology (1), the discrete one. Maybe you have seen that one called "trivial" as well, but that one has little to do with density. Or rather, since it is the opposite extreme, it tells us that $\mathbb Z_p$ is "as far from being dense in $\mathbb Q_p$ as possible". Indeed, a third way to see the non-density of $\mathbb Z_p$ would be to say that every $x \in \mathbb Q_p$ which is not in $\mathbb Z_p$ has an open neighbourhood around it which does not intersect $\mathbb Z_p$: indeed, $x +\mathbb Z_p = \{y\in \mathbb Q_p: d(y,x) \le 1\}$ is such an open neighbourhood.