# Uniformizer for splitting field of p^{1/p^n} over p-adics

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.

• Do you want to do more than calculate the ramification numbers? – Lubin Feb 9 '16 at 15:16
• 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. – Laurent Berger Feb 11 '16 at 8:48

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.
• 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). – David Lampert Nov 3 '19 at 21:39
• 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. – 丢人素学姐 Nov 4 '19 at 5:53