# Decomposition of $\mathbb{C}_p^*$

I'm looking for a topological group decomposition of $\mathbb{C}_p^*$. I know that I can write

$\mathbb{C}_p^*\cong p^\mathbb{Q}\times \mathcal{O}_{\mathbb{C}_p}^* \cong p^\mathbb{Q}\times \mu_\infty^{(p)} \times (1+\mathfrak{m}\mathcal{O}_{\mathbb{C}_p})$,

where $\mu_\infty^{(p)}$ is the group of prime-to-$p$ order roots of unity, $\mathfrak{m}$ is the maximal ideal in $\mathcal{O}_{\mathbb{C}_p}$, and all decompositions are topological. My question is:

Is there a closed subgroup $K$ of $1+\mathfrak{m}\mathcal{O}_{\mathbb{C}_p}$ such that

$1+\mathfrak{m}\mathcal{O}_{\mathbb{C}_p} \cong \mu_{p^\infty}\times K$

as topological groups?

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By the way, it's easy to see that the splitting exists algebraically, since $\mu_{p^\infty}$ is divisible, i.e. injective. –  Kevin Sep 1 '11 at 5:09

Note, just to orient ourselves, that there is a short exact sequence $$0 \to \mu_{p^{\infty}} \to 1 + \mathfrak m\mathcal O_{{\mathbb C}_p} \to \mathbb C_p \to 0,$$ with the second map given by $\log$. Thus if the $K$ you are looking for exists, it is isomorphic to $\mathbb C_p$.
$1+p \mathcal O_{{\mathbb C}_p}$ is a closed subgroup of $1 + \mathfrak m\mathcal O_{{\mathbb C}_p},$ which maps isomorphically via $\log$ to $p\mathcal O_{{\mathbb C}_p}$. (Here I am assuming $p$ odd, otherwise, if $p = 2$, we should replace the factor of $p,$ i.e. of $2$, by a factor of $4$.)
So $\mu_{p^{\infty}} \times 1 + p \mathcal O_{{\mathbb C}_p}$ embeds into $1+\mathfrak m \mathcal O_{{\mathbb C}_p}$, and the projection onto the first factor extends to a morphism $1 + \mathfrak m\mathcal O_{{\mathbb C}_p} \to \mu_{p^{\infty}}$ (using injectivity of the target, as you noted in the original question). If we denote the kernel of this morphism by $K$, then certainly $1 + \mathfrak m \mathcal O_{{\mathbb C}_p} = \mu_{p^{\infty}} \times K.$ Also, $K$ contains the open subgroup $1 + p \mathcal O_{{\mathbb C}_p},$ and so is itself an open (and hence closed) subgroup.
Thus the splitting you ask about does indeed exist. Note though that it is not natural, and e.g. cannot be chosen in an $Aut(\mathbb C_p/\mathbb Q_p)$-equivariant manner.