Under What assumptions on $p$, $\mathcal{O}_K^* \simeq \mathbb{Z}_p^{*} \oplus \mathbb{Z}_p^{*}$

Let $p$ be a fixed prime number and $\mathbb{Q}_p$ be the field of $p$-adic numbers and $K$ be an extension of degree $2$ of $\mathbb{Q}_p$. Let $\mathcal{O}_K$ be the ring of integers of $K$ and $\mathcal{O}_K^*$ be the group of units of $\mathcal{O}_K$. Under what conditions on $p$, we can prove that $\mathcal{O}_K^* \simeq \mathbb{Z}_p^{*} \oplus \mathbb{Z}_p^{*}$

For, $\Bbb Z_p^*$ always has a nontrivial torsion subgroup $T_p$, of order $p-1$ if $p\ne2$, and of order two in case $p=2$. Thus $\Bbb Z_p^*\oplus\Bbb Z_p^*$ will always have a finite subgroup $T_p\oplus T_p$, noncyclic. But a finite subgroup of a field must be cyclic, so $\Bbb Z_p^*\oplus\Bbb Z_p^*$ can not be found inside the multiplicative group $\mathcal O^*$.
• If we take the torsion free part of $\mathcal{O}_K^*$, which will be $1+\pi\mathcal{O}_K$ where $\pi$ is the uniformizer of $\mathcal{O}_K$ then under some assumptions on $p$, Is it possible to prove an isomorphism between the $1+\pi\mathcal{O}_K$ and $(1+p\mathbb{Z}_p)\oplus (1+p\mathbb{Z}_p)$. Sep 19, 2018 at 6:34
• Well, for quadratic extensions, in the one case $p=3$, the subgroup $1+\pi\mathcal O_K$ is not torsion-free. In other cases, both should be isomorphic to the additive group $\Bbb Z_p\oplus\Bbb Z_p$. You should be able to use the logarithm on this question. Sep 19, 2018 at 12:57
Let us first recall a few classical results (see e.g. Serre's "Local Fields") on the structure of the group of units $U=U(K)$ of a finite extension $K/\mathbf Q_p$ with residue field $k$ . The valuation $v$ of $K$ produces a natural filtration $U_0=U>U_1>...>U_m>...$, where $x \in U_m$ iff $v(x-1)\ge m$. The successive quotients of the filtration verify $U_0/U_1\cong k^*$ (canonically), and (¤) $U_m/U_{m+1} \cong (k,+)$ via $x\to x-1$ and the choice of a uniformizer. Moreover, if $\phi_p$ denotes the $p$-th power map, $\phi_p (U_m)<U_{m+1}$ for all $m\ge 1$, and if $e$ is the ramification index of $K/\mathbf Q_p$, $\phi_p$ is an isomorphism $U_m \cong U_{m+e}$ for all $m>e':=e/(p-1)$. Assembling these properties together, one can show that, for $m>e'$, the group $U_m$ is a free $\mathbf Z_p$-module of rank $n:=[K:\mathbf Q_p]$, and the group $U_1$ is the product of a free $\mathbf Z_p$-module of rank $n$ and a $p$-primary finite cyclic group (op. cit., chap. XIV, propos. 10). Note that the proof actually shows an isomorphism $\mathbf Z_p^n\cong U_m$, using the step (¤) above and the fact that $U_m/\cong U_{m+e}$ is abelian elementary of order $p^n$ , and choosing a system of generators of $U_m$ which lifts a basis of $U_m/\cong U_{m+e}$ (op. cit.)
Apply this to your quadratic extension $K/\mathbf Q_p$, where $e=1$ or $2$. Leaving it to you to do the particular cases $p=2,3$, I consider only $p>3$, so that $e'<1$, hence $U_0/U_1 \cong \mathbf F_p^*$ or $\mathbf F_{p^2}^*$ and $U_1 \cong (\mathbf Z_p^2,+)$. This shows at the same time that : - your original question has a negative answer (as in @Lubin) - your modified question has an abstract positive answer .
• Yes, but: use the logarithm on $1+\pi\mathcal O_K$ to get a kernel-less map onto a $\Bbb Z_p$-lattice in $\mathcal O_K^+$. Good for $p>3$ only. Also good for $p=3$ and $K\ne\Bbb Q_3(\sqrt{-3}\,)$. Sep 19, 2018 at 13:00