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Let $K$ be a number field which is Galois over $\mathbb{Q}$ with Galois group $G$ and let $p$ be a rational prime. For $\mathfrak{p}$ a prime of $K$ which sits above $p$, we denote by $U_{1,\mathfrak{p}}$ the group of local units in $K_{\mathfrak{p}}$ which are $1$ modulo $\mathfrak{p}$ and let $U_1=\prod\limits_{\mathfrak{p}\mid p} U_{1,\mathfrak{p}}$. It is a standard result in class field theory that $U_1\cong (finite)\times \mathbb{Z}_p^{[K:\mathbb{Q}]}$.

I would like to know if it's true that there exists an element $u \in U_1$ such that $\mathbb{Z}_p[G]u$ has finite index in $U_1$ and when this is the case, whether there is a connection betwen this result and the one I mentioned above.

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I stick to your notations, adding only $G_v=Gal(K_v / \mathbf Q_p)$. Notice that:

1) Because $K/\mathbf Q$ is Galois, $G$ permutes transitively the places $v$ above $p$ and all the $G_v$'s are isomorphic. The semi-local $\mathbf Z_p [G]$-representation $U_1$ is induced from the local $\mathbf Z_p [G_v]$- representation $U_{1,v}$ (for a chosen $v$)

2) Then your question is equivalent to a local one: "Does there exists a cyclic $\mathbf Z_p [G_v]$-submodule $V$ of $U_{1,v}$ with finite index ?". If such a $V$ exists, it must be $\mathbf Z_p [G_v]$-free because of a question of rank

The answer to this local question is "yes", although somewhat hidden in the literature. In Cassels-Fröhlich's book, chap.6, Serre proves the following propos.3 : " $U_{1,v}$ contains an open sub-module which is $G_v$- cohomologically trivial". But if you look carefully at his first proof (he gives two), he actually shows that $V$ is free. I briefly recap the argument: the idea is to compare the additive-multiplicative structures via the exp-log maps. There is a normal (additive) basis of $K_v / \mathbf Q_p$, i.e. a basis consisting of the Galois conjugates of a single element $a$, which we can take in the local ring of integers. Define $A$ = the additive (free) $\mathbf Z_p [G_v]$-module generated by $a$. Take $v(a)$ large enough to ensure that the exp series converges in a neigbourhood of $0$. Then $V=exp(A)$ does the job.

NB: The Galois structure of the global units is much harder, because much deeper, because related, via not yet fully proved conjectures, to special values of L-functions.

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  • $\begingroup$ Thank you very much for this very beautiful and clear answer, it is exactly the kind of argument I was looking for :) $\endgroup$ – Jake Haider Nov 26 '16 at 12:11

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