Since @PaulGarrett has given me the prompt, I’ll say something about higher ramification. This is a way of examining how close to each other the various roots $\lambda,\lambda'$ of the defining equation $f=0$ may be, when you choose $f$ so that one root $\lambda$ generates the ring of integers of the extension field $K$, $\mathcal O_K=\Bbb Z_p[\lambda]$. In the case of an unramified extension, as I recall, the differences between two roots will all be units. It’s the totally ramified extensions where higher ramification theory comes into its own.
Then let $\lambda$ be a uniformizing element of $\mathcal O_K$, $K$ totally ramified over $k$, with $v_k(\lambda)=1/e$ where $e$ is the ramification index (and the degree of the extension). We take $v_k$ normalized so that $v_k(k^\times)=\Bbb Z$. And let $f(x)$ be the minimal polynomial for $\lambda$, necessarily an Eisenstein polynomial over $\mathcal O_k$. Then put $g(x)=f(x+\lambda)$, whose roots are the quantities $\lambda'-\lambda$. We want a good way of measuring the size(s) of the roots of $g$.
Well, you measure the size of the roots of a polynomial with the Newton Polygon, but if you encode the same information in a kind of dual convex body that’s sometimes called the Newton Copolygon, you get something a little more useful. You can stretch this out horizontally by a factor of $e$ and consider the boundary to be the graph of a function, and thereby get the Hassse-Herbrand Transition Function $\psi^K_k$, which also has the wonderful property of being functorial: if $L\supset K\supset k$ is a chain of totally ramified extensions, then $\psi^L_k=\psi^K_k\circ\psi^L_K$.
Note that there are no Galois groups in this story. We can apply it to the two extensions mentioned at the beginning, $\Bbb Q_3(\zeta_9)$ and $\Bbb Q_3((-3)^{1/6})$. The two uniformizers are, respectively, $\tau=\zeta_9-1$ and $\pi=(-3)^{1/6}$. The minimal polynomial for $\zeta_9$ is $\frac{x^9-1}{x^3-1}$, so the minimal polynomial for $\tau$ is $f_1(x)=\frac{(x+1)^9-1}{(x+1)^3-1}$. The numerator and denominator have no constant, while the first-degree coefficient upstairs is $9$, downstairs is $3$, thus the quotient, known to be a polynomial, has constant coefficient $9/3=3$. Looking at the quotient modulo $3$, you see that there, the quotient is $x^6$. Thus $f_1$ is Eisenstein, which we already knew, didn’t we?
Forming $g_1(x)=f_1(x+\tau)$, as above, we need to look closely at $g_1$ and its Newton Polygon (and Copolygon). It’s somewhat of a pain to do the computation this way, but you find that the $v_p$-value of the first coefficient is $3/2$, the value of the third is $1/2$ and of the sixth is zero. This gives vertices of the Polygon at $(1,3/2)$, $(3,1/2)$, and $(6,0)$. The Copolygon (in the Cartesian $(\xi,\eta)$-plane) is described by the corresponding inequalities $\eta\le\xi+3/2$, $\eta\le3\xi+1/2$, and $\eta\le6\xi$. The vertices are where two half-planes’ boundaries intersect, namely at $(1/6,1)$ and $(1/2,2)$. Stretch by a factor of six to get the points $(1,1)$ and $(3,2)$ (the “breaks” in traditional terminology), exactly what you’ll find in Serre’s Corps Locaux, except that my coordinatization of the Cartesian plane is off from Serre’s by one in each direction.
If you do the corresponding computation for $\sqrt[6]{-3}$, you’ll see that it’s much less messy, giving $g_2(x)=(x+\pi)^6+3$ with the Polygon’s vertices at $(1,11/6)$, $(3,1/2)$ and $(6,0)$. The Copolygon is defined by the corresponding inequalities $\eta\le\xi+11/6$, $\eta\le3\xi+1/2$, and $\eta\le6\xi$. Here, the half-planes’ boundaries intersect at $(1/6,1)$
and $(2/3,5/2)$. Stretching by six gives the vertices of the polygonal Hasse-Herbrand function at $(1,1)$ and $(4,5/2)$. Here, the “breaks” are not at integer points, which is fine, because the Galois group of the normal closure of our field is not abelian.
Best Answer
For $K/\Bbb{Q}_p$ a finite extension then $F/K$ is unramified iff $F=K(\zeta_n)=K(\zeta_{q-1})$ with $p\nmid n$ and $q= |O_F/(\pi_F)|$. This is the main application of Hensel lemma.
When $E/K,E'/K$ are ramified then it is not always the case that the residue field of $EE'$ is the smallest field contained those of $E,E'$, try with $E=\Bbb{Q}_2(2^{1/3}),E'=\Bbb{Q}_2(\zeta_3 2^{1/3})$.
When $E'/K$ is unramified then $EE'=E(\zeta_{q-1})$ has residue field $O_E/(\pi_E)(\zeta_{q-1})$.