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

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    $\begingroup$ 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
    Commented Sep 17, 2018 at 20:26

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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 $$\mathbb{Z}/(p-1)\mathbb{Z} \oplus \mathbb{Z}/(p-1)\mathbb{Z}$$ 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.

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  • $\begingroup$ That group is cyclic if $p = 2\ldots$ $\endgroup$
    – KConrad
    Commented Sep 17, 2018 at 20:18
  • $\begingroup$ @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. $\endgroup$
    – efs
    Commented Sep 17, 2018 at 20:46
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    $\begingroup$ 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 2-adic case. $\endgroup$
    – KConrad
    Commented Sep 17, 2018 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
    Commented Sep 18, 2018 at 7:12
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    $\begingroup$ $\mathcal{O}_K^*$ is never torsion free, it contains $-1$. $\endgroup$
    – efs
    Commented Sep 18, 2018 at 9:57

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