Regulator of a real quadratic field

$K$ has signature $(2,0)$ and $\lambda (x)=(\log \mathrm{\mid }{\sigma }_{1}x\mathrm{\mid },\log \mathrm{\mid }{\sigma }_{2}x\mathrm{\mid })$ for $x\in K^{\times }$: there are two real embeddings and no complex place by [F3], so [F1] has no doubled coordinate, and no logarithm of a nonzero element is undefined.

The fundamental unit generates the unit group: by [F3] the group ${\mathcal{O}}_K^{\times }$ has torsion subgroup $\{\pm 1\}$ and free part of rank $1$, so there is a unit $\gamma >1$ with ${\mathcal{O}}_K^{\times }=\{\pm {\gamma }^n:n\in \mathbb{Z}\}$; every unit $v>1$ is then ${\gamma }^n$ with $n\ge 1$, and ${\gamma }^n=\gamma \cdot {\gamma }^{n-1}\ge \gamma$ because ${\gamma }^{n-1}\ge 1$ for $n\ge 1$, so $\gamma$ is the least unit $>1$ and hence $\gamma =\epsilon$; therefore ${\mathcal{O}}_K^{\times }=\{\pm {\epsilon }^n\}$ and, by [F2] and [F1], $\lambda (-1)=0$ and $\lambda ({\epsilon }^n)=n\lambda (\epsilon )$ for every $n\in \mathbb{Z}$, so $\lambda ({\mathcal{O}}_K^{\times })=\mathbb{Z}\lambda (\epsilon )\mathrm{.}$ Thus $(\epsilon )$ is a system of fundamental units of $K$ in the sense of [F5].

Logarithmic vector: ${\sigma }_1$ may be taken to be the identity, so $\mathrm{\mid }{\sigma }_1\epsilon \mathrm{\mid }=\epsilon >0$; and $\mathrm{\mid }{\sigma }_2\epsilon \mathrm{\mid }=1\mathrm{/}\epsilon$, because $\epsilon$ is a unit and $N_{K\mathrm{/}\mathbb{Q}}(\epsilon )=\epsilon {\sigma }_2(\epsilon )=\pm 1$ by [F4], so ${\sigma }_2(\epsilon )=\pm 1\mathrm{/}\epsilon$. Hence $\lambda (\epsilon )=(\log \epsilon ,\log (1\mathrm{/}\epsilon ))=(\log \epsilon ,-\log \epsilon )$ by [F2], a nonzero vector in the hyperplane $\{({\xi }_1,{\xi }_2):{\xi }_1+{\xi }_2=0\}$.

Regulator: the logarithmic matrix of the system $(\epsilon )$ is the $2\times 1$ matrix $A$ with entries $\log \epsilon$ and $-\log \epsilon$, so deleting row $1$ gives the $1\times 1$ determinant $-\log \epsilon$ and deleting row $2$ gives $\log \epsilon$; by [F6] and step 1.2, $R_K=\mid \det A_k\mid =\log \epsilon ,$ which is positive because $\epsilon >1$ and $\log$ is strictly increasing with $\log 1=0$ by [F2]. In particular the two deleted rows give the same absolute value, and the factor $2$ of the complex coordinates of [F1] is absent.

For $K=\mathbb{Q}(\sqrt{5})$: by [F7] the fundamental unit is $\epsilon =(1+\sqrt{5})\mathrm{/}2\approx 1.6180339887$, so $R_K=\log ((1+\sqrt{5})\mathrm{/}2)\approx \mathrm{0.4812118251.}$

By [F7], the full order unit group is $\mathbb{Z}[\sqrt{5}{]}^{\times }=\pm ⟨{\epsilon }^3⟩$, and its norm-one Pell subgroup is $\pm ⟨{\epsilon }^6⟩=\pm ⟨9+4\sqrt{5}⟩$. The Pell generator has logarithmic coordinate $\log (9+4\sqrt{5})=\log ({\epsilon }^6)=6\log \epsilon =6R_K\approx 2.8872709504,$ so its rank-one deleted-row determinant is six times the field regulator, which is defined using the maximal-order fundamental unit $\epsilon$.

Conclusion and choice accounting: for every real quadratic field the fundamental unit $\epsilon >1$ is a system of fundamental units, its logarithmic vector is $(\log \epsilon ,-\log \epsilon )$, and $R_K=\log \epsilon$; the doubled-complex normalization is vacuous here, and for $K=\mathbb{Q}(\sqrt{5})$ the value is $R_K=\log ((1+\sqrt{5})\mathrm{/}2)\approx 0.4812118251$, six times smaller than the determinant $6R_K$ obtained from the Pell generator $9+4\sqrt{5}$ of the order $\mathbb{Z}[\sqrt{5}]$. Choice enters only through the rank-one unit structure [F3] and the $d=5$ input [F7]; the logarithm computations and the determinant of the $2\times 1$ matrix use no choice.