Explicitly computing uniformisers of local fields Consider the field tower $L/K'/K$ where $L=\mathbb{Q_3}(\xi,2^{1/3})$, $K'=\mathbb{Q_3}(\xi) $ and $K=\mathbb{Q_3}$. Here, $\xi$ is a primitive cube root of unity, and $\mathbb{Q_3}$ is the 3-adics.
I know that a uniformiser for $\mathbb{Q_3}$ is the rational prime $3 \in \mathbb{Z}$. Am I correct in thinking that, because $K'/K$ is unramified, 3 is still a uniformiser for $K'$? How would I compute a uniformiser for $L$?
Relevant facts I know are: 


*

*$O_L$ is the integral closure of $O_K$ in $L$

*If $3O_K$ factorises into prime ideals in $O_{K'}$ as $3O_K=p^e$ then anything in $p-p^2$ works as a uniformiser.
 A: Here’s the method that I (try to) use for questions like this. First you must remember that $\xi=(-1+\sqrt{-3}\,)/2$, so that of course its field is ramified over $\Bbb Q_3$.
Let’s name $2^{1/3}=\rho$ for convenience, and similarly $\tau=\sqrt{-3}$. Your field is $\Bbb Q_3(\rho,\tau)$. Now, using the additive valuation $v=v_3$ normalized so that $v(3)=1$, we have $v(\tau)=1/2$, and I need also a prime element in $\Bbb Q_3(\rho)$. Since $\text{Irr}\bigr(\rho,\Bbb Q_3[X]\bigr)=X^3-2=f(X)$, the irreducible polynomial for $\rho+1$ is $f(X-1)=X^3=3X^3+3X-3=g(X)$, and there’s your prime element for the cubic field.
Now you get something of valuation $1/6$ simply by writing down $\tau/(\rho+1)=\sigma$. For completeness, I’ll show you how to find its minimal polynomial.
Form $h(X)=\frac1{\tau^3}g(\tau X)=X^3+\tau X^2-X-1/\tau$, you see that this is $\text{Irr}\bigl((\rho+1)/\tau,\Bbb Q_3(\tau)[X]\bigr)$. Reverse it to get $H(X)=-X^3/\tau-X^2+\tau X+1$, a polynomial vanishing at $\sigma$. Multiply $H$ by $-\tau$ to get $X^3+\tau X^2 +3X-\tau$, the irreducible polynomial for $\sigma$ over $\Bbb Q_3(\tau)$, a pleasingly Eisenstein polynomial. If you want the polynomial for $\sigma$ over $\Bbb Q_3$, just multiply the last polynomial by its conjugate (replacing $\tau$ by $-\tau$).
