Kernel of Quadratic Field Norm Let $D > 0$ be squarefree and consider the field $\mathbb{Q}(\sqrt{D})$. Then the field norm $N : \mathbb{Q}(\sqrt{D}) \to \mathbb{Q}^\times$ is given by $N(a + b\sqrt{D}) = a^2 - D b^2$. Let $K$ be the kernel of this map, consisting of elements of norm $+1$. I'm looking for an example where $K$ is not isomorphic to $\mathbb{Z}/2\mathbb{Z} \times \bigoplus_{i = 0}^\infty \mathbb{Z}$ as abelian groups.
The motivation is the Pell conic group $C$ defined on the curve $x^2 - Dy^2 = 1$. If $D$ is a square of a rational number, then $x \mapsto (\frac{x+x^{-1}}{2},\frac{x-x^{-1}}{2\sqrt{D}})$ gives an isomorphism $\mathbb{Q}^{\times} \cong C$, but otherwise we have the isomorphism $K \cong C$ where $K$ is the kernel of the norm map. I am wondering whether the group structure of $C$ alone can detect the irrationality of a square root of a natural number. Below I've included my current thoughts on this question.
Let's consider $D = 2$. Then the group of units of $\mathbb{Z}[\sqrt{2}]$ is generated by powers of $\pm(1 + \sqrt{2})$, so it is isomorphic $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}$. Since $\mathbb{Z}[\sqrt{2}]$ is a UFD, we have that any element of $\mathbb{Q}(\sqrt{2})^\times$ can be written as a unique product of primes and units of $\mathbb{Z}[\sqrt{2}]$, so $\mathbb{Q}(\sqrt{2})^\times \cong (\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}) \times \bigoplus_{p \text{ prime}} \mathbb{Z}$. Now consider the map $h : \mathbb{Q}(\sqrt{2})^\times \to K$ given by $h(a) = \frac{a^*}{a}$ where $a^*$ denotes the conjugate. My understanding is that Hilbert's Theorem 90 implies this map is surjective. Its kernel is $\mathbb{Q}^\times$, so $K$ should be something like $K \cong (\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}) \times \bigoplus_{p \text{ prime, } \notin \mathbb{Q}} \mathbb{Z}$. This paper uses a similar map to obtain the analogous result for $\mathbb{Q}(i)$. It seems like similar reasoning should apply to any UFD.
Now consider a non-UFD, say $D = 10$. Here we have $2 \cdot 5 = \sqrt{10} \cdot \sqrt{10}$ which at first seems like it might change the structure of $\mathbb{Q}(\sqrt{10})^\times$ compared to $\mathbb{Q}(\sqrt{2})^\times$. However, I think you can throw out $5$ as a generator because you can get it from the other irreducibles as as $(\sqrt{10})^2/2$. It seems like $\mathbb{Q}(\sqrt{10})^\times$ has a chance of being isomorphic to $\mathbb{Z}/2\mathbb{Z} \times \bigoplus_{i = 0}^\infty \mathbb{Z}$ despite not being a UFD. Then applying $h$ should then just kill the factors corresponding to primes in $\mathbb{Q}$, so you might get the same thing for $K$ as with $D = 2$.
 A: In what follows, $L$ is any number field.
We have the following theorem.
Thm 1. Let $L$ be a number field, and let $N:L^\times\to \mathbb{Q}^\times$ be the norm map.
Then $\ker(N)\simeq \mathcal{O}_L^\times\cap \ker(N) \times \mathbb{Z}^{(E)}$, for some countable set $E$.
Dirichlet's theorem says that $\mathcal{O}_L^\times $  is finitely generated, thus so is $\mathcal{O}_L^\times\cap \ker(N) $. Note that the torsion part of this last group is the intersection of the torsion group of $\mathcal{O}_L^\times$ , which is the (cyclic) group $\mu_L$ of roots of $1$ contained in $L$,  with $\ker(N)$. Thus:
Corollary. $\ker(N)\simeq  \mu_L\cap \ker(N)\times \mathbb{Z}^{(E')},$ where $E$ is countable and $\mu_L$ is the ( cyclic )group of roots of unity contained in $L$.
Thus, if you stick to subfields of $\mathbb{R}$ (such as real quadratic fields), you won't get any counterexample. You need $L$ to contain at least a root of unity of order $n\geq 3$.
Example. For all $n\geq 3$, you can take $L=\mathbb{Q}(\zeta_n)$. Then norms are always positive (because complex conjugation is a non trivial automorphism), and $\mu_L=\langle \mu_n\rangle\simeq \mathbb{Z}/n\mathbb{Z}$ .
Let's now move to the proof of the theorem.
For $z\in L^\times,$ I will denote by $(z)$ the fractional ideal $z\mathcal{O}_L$. Abusing notation, I will also denote by $N(I)$ the norm of any fractional ideal $I$.
If $z$ has norm $1$, so is $(z)$, since $N((z))=\vert N(z)\vert$.
Let $\Gamma$ be the set of principal fractional ideals with norm $1$, and let $f:z\in \ker(N)\mapsto (z)\in\Gamma$.
Then $\Gamma$ is an abelian group, and $f$ is a group morphism.
Lemma. $f$ has  kernel $\mathcal{O}_L^\times\cap\ker(N)$.
Proof. Assume that $z\in\ker(N)$ satisfies $(z)=z\mathcal{O}_L=\mathcal{O}_L$. Then $z\in\mathcal{O}_L$, and since $\mathcal{O}_L$ is a domain, $z$ must be a unit.
Hence  $z\in \mathcal{O}_L^\times \cap\ker(N)$. Conversely, elements of  $\mathcal{O}_L^\times \cap\ker(N)$ lie in $\ker(f)$.
Thm 2. The group $\Gamma$ of principal fractional ideals with norm $1$ is  isomorphic to a subgroup of $ \mathbb{Z}^{(\mathcal{P})}$, where $\mathcal{P}$ is the set of nonzero prime ideals of $\mathcal{O}_L$.
Assume thm 2 is proved, and let us prove thm $1$.
By the lemma, we have an exact sequence of abelian groups  $1\to \mathcal{O}_L^\times\cap\ker(N)\to \ker(N)\to Im(f)\to 1$.
By a theorem of Dedekind, if $S$ is any set, any subgroup of $\mathbb{Z}^{(S)}$ is free, and has a basis indexed by a subset of $S$. By Thm2 and this result, $\Gamma$ is then free, and isomorphic to $\mathbb{Z}^{(X)}$ for some subset $X$ of $\mathcal{P}$. Now, $Im(f)$ is a subgroup of $\Gamma$, so we can apply the same result again to get that $Im(f)\simeq  \mathbb{Z}^{(E)}$, for some countable set $E$.
Now by the lemma,  we have an exact sequence of abelian groups $1\to \mathcal{O}_L^\times\cap \ker(N)\to \ker(N)\to Im(f)\to 1$. Since $Im(f)$ is free, this sequence splits, and we are done.
We now have to prove Thm 2 above.
If $I$ is a  nonzero fractional ideal, write $I=\displaystyle  \prod_{\mathfrak{p}}\mathfrak{p}^{r_{\mathfrak{p}}}$, where the $ r_{\mathfrak{p}}'s$ are integers, which are all zero except for a finite number of them. Note that these integers are unique.
Now we have  $N(\mathfrak{p})=p^{f_{\mathfrak{p}}}$, where $p$ is the unique prime number dividing $\mathfrak{p}$ , so $N(I)=1$ if and only if $\sum_{p\mid\mathfrak{p}}f_{\mathfrak{p}}r_{\mathfrak{p}}=0$ for all prime $p$.
Now, we have to understand on which conditions on the $ r_{\mathfrak{p}}'s$ such an $I$ is principal.
Fix a group isomorphism $\mathbb{Z}/d_1\mathbb{Z}\times \cdots\times\mathbb{Z}/d_k\mathbb{Z}\simeq \mathcal{C}\ell(L)$ (the class group of
$L$
), and let $C_1,\ldots,C_k$ be the images of $(1,0,\ldots,0), (0,1,0,\ldots, 0),\ldots,(0,\ldots,0,1)$ in the class group. Then any element of the class group may be written as $m_1\cdot C_1+\cdots+m_k C_k, m_1,\ldots,m_k\in\mathbb{Z}$, and such a linear combination is zero if and only if $d_i\mid m_i$ for  $i=1,\ldots,k$.
Now for each $\mathfrak{p}$, write $[\mathfrak{p}]=\displaystyle\sum_{i=1}^k a_{i,\mathfrak{p}}\cdot C_i$.
Then $[I]=\displaystyle\sum_{i=1}^k (\sum_{\mathfrak{p}} a_{i,\mathfrak{p}}r_\mathfrak{p})\cdot C_i$ ,and $I$ is principal if and only if $[I]=0$ if and only if $d_i\mid (\sum_{\mathfrak{p}} a_{i,\mathfrak{p}}r_\mathfrak{p})$ for $i=1,\ldots,k$.
Hence, denoting by $\mathcal{P}$ the set of nonzero prime ideals of $\mathcal{O}_L$, we see that $\Gamma$ is canonically isomorphic to the abelian group $$\{(r_\mathfrak{p})_\mathfrak{p}\in\mathbb{Z}^{(\mathcal{P})}\mid \sum_{p\mid\mathfrak{p}}f_{\mathfrak{p}}r_{\mathfrak{p}}=0 \mbox{ for all prime }
 p \mbox{ and } d_i\mid (\sum_{\mathfrak{p}} a_{i,\mathfrak{p}}r_\mathfrak{p})\mbox{ for }i=1,\ldots,k\},$$
which is a subgroup of $\mathbb{Z}^{(\mathcal{P})}$.
