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 $\mathcal{O}_K^* \simeq \mathbb{Z}_p^{*} \oplus \mathbb{Z}_p^{*}$

2$\begingroup$ The structure of the principal units (those congruent to 1 modulo the maximal ideal) is discussed in another MO question: mathoverflow.net/questions/36575/…. See answers and comments. $\endgroup$ – KConrad Sep 17 '18 at 20:26
Let $p\ge3$ be a prime number and $K$ any finite extension of $\mathbb{Q}_p$. The torsion part of $\mathcal{O}_K^*$ is always cyclic. On the other side, the torsion part of $\mathbb{Z}_p^{*} \oplus \mathbb{Z}_p^{*}$ is isomorphic to $$\mathbb{Z}/(p1)\mathbb{Z} \oplus \mathbb{Z}/(p1)\mathbb{Z}$$ which is not cyclic. See Henri Cohen's Number Theory, volume 1, section 4.3. There, he gives an explicit descritption of the group of units of $\mathcal{O}_K$. If I remeber correctly, this also appears in Iwasawa's book on Local Class Field Theory.
Edit: I forgot the condition $p\ge3$, thanks Keith.


$\begingroup$ @KConrad Yes, in the comment that I wrote and deleted, I wrote that condition, and I forgot to put it here. Now it remains the case $p=2$. I'm busy now. I'll check in a little, or please answer the question and I delete this answer. $\endgroup$ – EFinatS Sep 17 '18 at 20:46

2$\begingroup$ Actually, even the case $p=2$ is problematic. The torsion part of $\mathbf Z_2^\times$ is $\{\pm 1\} \cong \mathbf Z/2\mathbf Z$ so the torsion subgroup of $\mathbf Z_2^\times \oplus \mathbf Z_2^\times$ is the noncyclic group $\{\pm 1\}^2$ and thus can't be isomorphic to $\mathcal O_K^\times$ in the 2adic case. $\endgroup$ – KConrad Sep 17 '18 at 21:19

$\begingroup$ If we assume that $\mathcal{O}_K^*$ is torsion free then is it possible $\mathcal{O}_K^*\simeq \mathbb{Z}_p^*\oplus \mathbb{Z}_p^*$. Because I want to know under what assumptions $\mathcal{O}_K^*$ is isomorphic to $\mathbb{Z}_p^*\oplus \mathbb{Z}_p^*$ $\endgroup$ – user89236 Sep 18 '18 at 7:12

1$\begingroup$ $\mathcal{O}_K^*$ is never torsion free, it contains $1$. $\endgroup$ – EFinatS Sep 18 '18 at 9:57