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Let us fix a square free positive integer $n\in\mathbb{N}$ and consider the number field $\mathbb{Q}(\sqrt n)$ with ring of integers $K=\mathbb{Z}[\sqrt n]$. Let us denote the Galois norm of elements in this ring by $\mathcal{N}$. Every invertible element in $K$ has to satisfy $$\mathcal{N}(a+b\sqrt n)=a^2-b^2n=1.$$ Moreover, these elements form the group of units $K^\times$, which thanks to Dirichlet's unit theorem can be expressed as $(\mathbb{Z}[\sqrt n])^\times = \{\pm1\}\times \langle\epsilon\rangle$ for a fundamental unit $\epsilon\in\mathbb Z[\sqrt n]$.

Let us now fix some prime $p$ and $k\in\mathbb N$. I wonder if one can find an analogous description of elements of norm 1 in $(K/p^kK)^\times$: more formally, the Galois norm descends mod $p^k$ to a map $\mathcal{N}_{p^k}:(K/p^kK)^\times\to (\mathbb{Z}/p^k\mathbb{Z})^*$ and we consider the group $$\mathcal{I}_{p^k}=\{a+b\sqrt n \in(K/p^kK)^\times:\mathcal{N}_{p^k}(a+b\sqrt n)=1\pmod p^k\}.$$ Can one describe it explicitly? In fact, can it be equal to $$\{1,p^k-1\}\times \langle\epsilon_{p^k}\rangle$$ for $\epsilon\equiv \epsilon_{p^k}\pmod{p^k}$?

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  • $\begingroup$ Things for finite commutative rings are usually much simpler: a finite commutative ring is a product $\prod_{i=1}^n A_i$ and its unit ring is $\prod_{i=1}^nA_i^\times$, where $A_i^\times$ is the complement of the maximal ideal of $A_i$. Of course this depends on understanding this product decomposition; which essentially means, for $K/p^kK$, understanding the corresponding decomposition for the ring $K/pK$. In the case at hand, this shouldn't be hard. $\endgroup$
    – YCor
    Commented Oct 10, 2021 at 18:04
  • $\begingroup$ @YCor Thank you. What is $A_i$? A finite local ring? So your suggestion is to look at $K/p^kK$ and divide to cases based on the ramification index? $\endgroup$ Commented Oct 10, 2021 at 18:47
  • $\begingroup$ Sorry, yes I meant to say $A_i$ is local. $\endgroup$
    – YCor
    Commented Oct 10, 2021 at 19:40
  • $\begingroup$ @YCor alright but then unless I'm missing something $A_i^\times$ might include invertible elements with $\mathcal{N}_{p^k}-$norm different than 1. My question is about the group of invertible elements of unit norm. $\endgroup$ Commented Oct 10, 2021 at 20:08
  • $\begingroup$ To clarify your 1st paragraph, the ring of integers of $\mathbf Q(\sqrt{n})$ does not have to be $\mathbf Z[\sqrt{n}]$ (try $n = 12$ and $n = 5$) and it is pretty unusual to write the ring of integers of a number field as $K$ when that letter is used so often in number theory to refer to number fields. The units in $\mathbf Z[\sqrt{n}]$ might include elements of norm $-1$, not only norm $1$ (such as $n = 2$ and $n = 5$). For all $n$ that are not a perfect square, the unit group of $\mathbf Z[\sqrt{n}]$ (even if it's not the integers of $\mathbf Q(\sqrt{n})$) is $\pm\varepsilon^{\mathbf Z}$. $\endgroup$
    – KConrad
    Commented Oct 11, 2021 at 7:26

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