Let $K=\mathbb Q_3$, $K'=\mathbb Q_3(\sqrt[3] 5)$ and $L = \mathbb Q_3(\sqrt[3]{3\cdot \sqrt[3] 5})$. Consider the tower of extensions $K\subseteq K' \subseteq L$. All extensions are totally ramified and $[L:K']=[K':K]=3$. I want to find a prime element in $L$. I tried to use the Norm map $N=N_{L/K}$ and the formula $N(x)=\prod_\sigma \sigma x$, where $\sigma $ goes through all $K$-embeddings from $L$ into a separable closure $\overline K $ of $K$. But I was not able to determine a prime element $\pi_L$ in $L$ such that $v_K(N_{L/K}(\pi_L))=1$.
Does anyone have an idea how to compute a prime element? Perhaps one can use the subextension $K'/K$ as intermediate step?
The key thing to keep in mind is that totally ramified extensions come from eisenstein polynomials, and roots of eisenstein polynomials are uniformizers (i.e. irreducible) so a good place to start is finding eisenstein polynomials that generate the same extension by tweaking your original defining polynomials.
Start with the polynomial $x^3 - 5$ which defines $K'$. Making a change of variable, we get $$(x+2)^3 - 5 = x^3 + 6x^2 + 12x + 2 - 5 = x^3 + 6x^2 + 12x - 3$$
This is eisenstein, so if we let $\alpha$ be any root of it then we know its root is a uniformizer for $\mathcal O_{K'}$. In particular $u\alpha^3 = 3$ for some unit $u$ in $\mathcal O_{K'}$ which is of the form $1 + \alpha(...)$ or $2 + \alpha(...)$ (because the residue field is $\mathbb F_3$). Also $\alpha + 2 = \sqrt[3]5$.
The next extension comes from $$x^3 - 3*(\alpha+2) = x^3 - u\alpha^3(\alpha + 2)$$ which evidently has the same splitting field (substitute $\alpha x$ for $x$) as $$y^3 - u (\alpha+2)$$
This time substitute $y+(2u)$ for $y$ and we get $$y^3 + 3(2u)x^2 + 3(2u)^2 x + (2u)^3 - u(\alpha + 2)$$ If this is eisenstein, then a root of it will be a uniformizer. By inspection we just need to check that the valuation of the constant term is $1$ (in $K'$). The constant term is $$(2u)^3 - u(\alpha + 2) = (2u)^3 - 2u - u\alpha$$
Working with just the first two, try the first case for the unit, $u=1$ mod $\alpha$: $$(2u)^3 - 2u = (2 + O(\alpha))^3 - 2(1 + O(\alpha)) - \alpha + O(\alpha^2)$$ $$= 2 + O(\alpha^3) - 2 + O(\alpha^3) - \alpha + O(\alpha^2)$$ $$= \alpha + o(\alpha^2)$$ which has the right valuation. If you work out the second case you should get $=2\alpha + O(\alpha^2)$
So it is indeed eisenstein, and hence some root of it will be a uniformizer for $\mathcal O_L$.
(for the very careful - if you fix your choices of $\sqrt[3]5$ and $\sqrt[3]{3\sqrt[3]5}$ at the start then you want to make sure to pick the "correct" root corresponding to that one under each substitution. This is easy and not really a problem anyways, since the worst that happens is you pick a conjugate and hence isomorphic field)