Totally ramified extension of $\mathbb{Q}_{p}$ which is not of a form $\mathbb{Q}_{p}(\sqrt[n]{pu})$ It is known that a finite extension $K/\mathbb{Q}_{p}$ is totally ramified if and only if $K = \mathbb{Q}_{p}(\alpha)$ where $\alpha$ is a root of Eisenstein polynomial. Is there any totally ramified extension that is not of the form $\mathbb{Q}_{p}(\sqrt[n]{pu})$ for some $u\in \mathbb{Z}_{p}^{\times}$? Every degree 2 totally ramified extensions has this form, but I don't know whether this is also true for degree 3 or higher. Thanks in advance. 

There is a general theorem that every tamely totally ramified extension of $\mathbf Q_p$ with degree $n$ has the form $\mathbf Q_p(\sqrt[n]{\pi})$ for some prime $\pi$ in $\mathbf Z_p$, so $\pi = pu$ for a unit $u$ in $\mathbf Z_p$.  (There is a similar theorem over other local fields.) So if you want a totally ramified extension not of that form you need $n$ to be divisible by $p$.
Let's try $n=p$. Something we can say about extensions $\mathbf Q_p(\sqrt[p]{pu})$ for $p>2$ is that they are not Galois over $\mathbf Q_p$: a field $K$ containing a full set of roots of $x^p - pu$ must contain the nontrivial $p$th roots of unity, and those have degree $p-1$ over $\mathbf Q_p$ so $[K:\mathbf Q_p]$ is divisible by $p-1$. Therefore $[K:\mathbf Q_p] \not= p$ when $p>2$. Thus a Galois totally ramified extension of $\mathbf Q_p$ having degree $p$ can't have the form $\mathbf Q_p(\sqrt[p]{pu})$.
Every totally ramified abelian Galois extension of $\mathbf Q_p$ with degree divisible by $p$ contains a subextension with degree $p$ since the Galois group has a subgroup of index $p$: in an abelian group of order $n$ there is a subgroup of each order dividing $n$ and thus also a subgroup of each index dividing $n$ by using a subgroup of order equal to the complementary factor in $n$ of the desired index. Subextensions of totally ramified extensions are totally ramified and subextensions of abelian Galois extensions are abelian Galois extensions. Thus all we need to do now is find a totally ramified abelian Galois extension of $\mathbf Q_p$ with degree divisible by $p$ and inside of it there are extensions of degree $p$, all of which are examples of the kind being sought (not having the form $\mathbf Q_p(\sqrt[n]{pu})$). 
The easiest choice is a cyclotomic extension: $\mathbf Q_p(\zeta_{p^2})$ where $\zeta_{p^2}$ is a root of unity of order $p^2$. This field has degree $p^2-p$ over $\mathbf Q_p$, with cyclic Galois group $(\mathbf Z/p^2\mathbf Z)^\times$, so the field contains a unique subextension with degree $p$ over $\mathbf Q_p$, namely the field fixed by the unique subgroup of the Galois group with order $(p^2-p)/p = p-1$. That subgroup is the solutions to $a^{p-1} \equiv 1 \bmod p^2$, and a  generator of this extension over $\mathbf Q_p$ is $\sum_{a^{p-1} = 1} \zeta_{p^2}^a$ where the sum runs over all solutions of $a^{p-1} \equiv 1 \bmod p^2$.
Example when $p=3$: $a^2 \equiv 1 \bmod 9$ has solutions $\pm 1 \bmod 9$ and $\zeta_{9} + \zeta_9^{-1}$ has minimal polynomial $f(x) = x^3 - 3x + 1$. Then  $f(x-1) = x^3 - 3x^2 + 3$ is Eisenstein at $3$; the polynomial $f(x+1)$ is not. I did my calculation of the minimal polynomial in $\mathbf C$, which is okay since a primitive $p$th-power root of unity has the same degree over $\mathbf Q_p$ as it does over $\mathbf Q$, so the structure of the intermediate fields in a $p$th-power cyclotomic extension over $\mathbf Q_p$ and over $\mathbf Q$ are the same.
Example when $p=5$: solutions to $a^4 \equiv 1 \bmod 25$ are $1, 7, 18$, and $24$, and $\zeta_{25} + \zeta_{25}^7 + \zeta_{25}^{18} + \zeta_{25}^{24}$ has minimal polynomial over $\mathbf Q_5$ equal to $g(x) = x^5 - 10x^3 + 5x^2 + 10x + 1$. (Note $g(x-1) = x^5 - 5x^4 + 25x^2 - 25x + 5$ is Eisenstein at 5; the polynomial $g(x+1)$ is not Eisenstein at $5$.)
