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Let $p$ be a prime, $L$ be a finite, totally ramified-extension of $Q_p,$ and $U$ be the group of principal units in the ring of integers of $L.$ Then $U$ is a finitely generated $\mathbb{Z}_p$-module under exponentiation. I'm looking for an explicit set of generators for $U$ under this action; is there a resource where I can find this information?

Thanks Robin but am I afraid I need more control on the set of generators. In particular, I'm interested if G is a set of generators of the value m = max {$v_L(g-1):g\in G$}

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    $\begingroup$ See Fesenko & Vostokov's Local Fields and Their Extensions. In the first part of the book they discuss the structure of the unit group as a Z_p-module. It gets tricky. $\endgroup$
    – KConrad
    Commented Aug 24, 2010 at 18:52

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Well you can lift any set of generators for the finite group $U/U^p$ and they will be a set of generators of $U$. If $n$ is the degree of $L$ over $\mathbb{Q}_p$ then $U/U^p$ will have order $p^{n+1}$ or $p^n$ according to whether or not $L$ has the $p$-th roots of unity. Certainly there's a system of generators of the form $1+\pi^j$ for a certain set of positive integers $j$ where $\pi$ is a uniformizer. You choose each $j$ to be the next one such that $1+\pi^j$ is not in the group generated by $U^p$ and the previous generators.

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Note that the $p$-adic logarithm gives a $\mathbb{Z}_p$-module homomorphism of $1+\mathfrak{m}_K$ into $\mathfrak{m}_K^{-e}$, where $e=[K:\mathbb{Q}_p]$ which has finite kernel (by compactness and discreteness). Since $\mathbb{Z}_p$ is Noetherian, this shows that $1+\mathfrak{m}_K$ is finitely generated. To find explicit generators, use the complex exponential to map generators of the image of $1+\mathfrak{m}_K$ back to $1+\mathfrak{m}_K$.

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  • $\begingroup$ I've thought of this and it doesn't help me answer the second part of the question; This is because $v_L(Log(1+\pi_L^j))$ is difficult to deduce if $v_L(\pi_L^j) = v_L(1-r)$ for some $r\in\mu_{p^\infty}.$ (in all other cases this value can be calculated using the newton copolygon). This makes finding gererators for Log(1+m_K) hard for an arbitary extension L. $\endgroup$
    – Joel
    Commented Aug 24, 2010 at 19:02

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