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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:

  1. $O_L$ is the integral closure of $O_K$ in $L$

  2. 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.

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    $\begingroup$ I don't think $K'/K$ is unramified. Let $\pi=1-\xi$. Then $$\pi^3=1-3\xi+3\xi^2-\xi^3=3\xi(\xi-1)=-3\xi\pi.$$ Therefore $\pi^2=-3\xi$. As $-\xi$ is a unit, this implies that $\pi$ is a uniformizer in $K'$. Compare this with the situation in the number field side. If $\omega$ is a primitive root of unity of order $p$, $p$ a prime, then the prime $p$ is totally ramified in $\Bbb{Q}(\omega)/\Bbb{Q}$. More precisely, $$(p)=(\lambda)^{p-1}$$ with $\lambda=1-\omega$. $\endgroup$ – Jyrki Lahtonen Jan 12 '15 at 16:34
  • $\begingroup$ Not an exact match, but related. $\endgroup$ – Jyrki Lahtonen Jan 12 '15 at 16:54
  • $\begingroup$ Thanks for commenting, Jyrki. If I understand you correctly, you've shown $3O_k=\pi^2 O_{K'}$ so the ramification index $e=2$. Using the formula $[K':K]=ef$, this gives $f=1$. This confuses me, however, because I have a result in my lecture notes which says that any finite extension $M/\mathbb{Q}_3$ can be 'split' into an unramified bit and a totally ramified bit. We constructed the unramified bit by taking a generator $g\in M$ for the image of $\mathbb{F}_3$ under the Teichmuller map $\mathbb{F}_3 \rightarrow M$, and then my notes say $\mathbb{Q_3}(g)$ is unramified. What's my mistake? $\endgroup$ – user3131035 Jan 12 '15 at 16:57
  • $\begingroup$ IIRC the Teichmüller map sends $\Bbb{F}_p^*$ into the group of roots of unity. So those roots of unity are of order $p-1$. See professor Lubin's comment. In other words here $\Bbb{F}_3$ mapsto $\{0,1,-1\}\subset K$. I also think that in general (that is, $f>1$) the Teichmüller map sends $\Bbb{F}_{3^f}^*$ to a group of roots of unity (of order $3^f-1$). $\endgroup$ – Jyrki Lahtonen Jan 12 '15 at 17:05
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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 $-\tau$).

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