Does anyone know an explicit uniformizer for $\mathbf{Q}_p(\zeta_{p^n}, p^{\frac{1}{p^n}}) / \mathbf{Q}_p$? I was reading the question "adding an n-th root to Q_p" where dke mentions this question but it remains unanswered and I would like to use this for some calculations.
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2$\begingroup$ Do you want to do more than calculate the ramification numbers? $\endgroup$– LubinCommented Feb 9, 2016 at 15:16
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1$\begingroup$ I am pretty sure that the answer is "no, it's too messy". If however somebody does have an answer, then we could use it to study the field of norms of the union of all those fields, and I'd be very interested in that. $\endgroup$– Laurent BergerCommented Feb 11, 2016 at 8:48
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For n=2 use the p-adic expansion of $\zeta$ to express an explicit uniformizer: combinations of $(\zeta-1)$ and $p^{1/p^2}$ continue canceling terms in the expansion until the first term with p-exponent having denominator divisible by $p^3$. For example with p=2: $(\zeta-1)2^{-3/4}+2^{-1/4}+1 = 2^{1/8}+...$ is a uniformizer.
Edit: With p=3: $\zeta_9 = 1 + \zeta_4\times3^{1/6} + 3^{1/3} - 3^{4/9} + 3^{13/27} ...$ and so $3^{2/3}\times((\zeta_9 - 1 - 3^{1/3} + 3^{4/9})^2 + 3^{1/3})^{-1} = -\zeta_4/2\times3^{1/54}+ ...$ is a uniformizer.
answered Feb 9, 2016 at 14:59
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1$\begingroup$ Could you please explain more? 1. How do you calculate the p-adic expansion of the unit root? 2. Why the element you find is a uniformizer? $\endgroup$ Commented Nov 3, 2019 at 11:23
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$\begingroup$ Algebraic P-adic Expansions explains how to calculate p-adic expansions with some results about roots of unity (I think the statements there about $p^2$-roots of unity maybe technically incorrect and the "Edit: With p=3" is correct). The element identified is a uniformizer because its valuation is 1/(ramification index). $\endgroup$ Commented Nov 3, 2019 at 21:39
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$\begingroup$ Well, I calculated the case of p=3 yesterday, but it seems to me that the element(let's call it $\pi$) you give has valuation 6. With the help of Mathematica, its minimal polynomial is $729+\cdots+627342091483343791624192x^{54}$, where this large constant is coprime to 3. Thus, $v(\pi)=v(729/627...192)/54=v(729)/54=6v(3)=v(3)/9=6$. Tell me if I'm wrong. $\endgroup$ Commented Nov 4, 2019 at 5:53
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$\begingroup$ Maybe my p=3 calculations are wrong (I did them years ago and without computer). Anyway the method (for any p) and the p=2 result are correct. $\endgroup$ Commented Nov 4, 2019 at 19:15