I've been trying to see whether following assertion is true in order to give a quick proof of another problem I was doing: if $K$ is a finite dimensional extension of the $p$-adic numbers $\mathbb{Q}_p$, we have the multiplicative field norm $N_{K/\mathbb{Q}_p}: K \rightarrow \mathbb{Q}_p$. While the norm is not guaranteed to be surjective, is it always possible to find some $k \in K$ such that $N_{K/\mathbb{Q}_p}(k)$ has ($p$-adic) absolute value $1/p$?
One way I would imagine to do this is to show that there exist $\alpha, \beta \in K$ such that $|N(\alpha)| = p^A$ and $|N(\beta)| = p^B$ with $A, B$ relatively prime. The assertion then follows because there exist integers $x, y$ such that $xA + yB = 1$, whence $\alpha^{-x}\beta^{-y} \in K$ with $|N(\alpha^{-x}\beta^{-y})| = |N(\alpha^{-x})N(\beta^{-y})| = |a|^{-x}|b|^{-y} = p^{-Ax}p^{-By} = 1/p$.
My claim seems to me altogether reasonable since otherwise there exists some prime $q$ such that the absolute value $p^s$ of every $N(k)$ as $k$ runs through all of $K$ will be such that $s$ is always divisible by $q$, which doesn't seem right.
You ask whether, given a finite extension $K/\mathbf Q_p$, the image of the norm $N_K : K^* \to \mathbf Q_p^*$ contains a uniformiser $\pi \in \mathbf Q_p^*$. In other words, does $\nu(N_K(a)) = 1$ have a solution $a \in K$?
In general, it is not true. For simplicity, suppose $K/\mathbf Q$ is Galois of degree $n$. Since all the conjugates of $a$ have the same $p$-adic valuation, $\nu (N_K(a)) = n\cdot \nu(a)$ for every $a$. Recall that $\nu(K^*) =\frac{1}{e}\mathbf Z$, where $e$ is the ramification index of $K/\mathbf Q_p$. Therefore $\nu(N_K(K^*)) = n\cdot \nu(K^*) = (n/e) \mathbf Z$; so a necessary and sufficient condition for a uniformiser to be in the image of the norm, in the case of a Galois extension, is that $n=e$; i.e. that $K/\mathbf Q_p$ be totally ramified.