A unimodular change of generators preserves the regulator determinants

The setting is as in [F1]-[F4]: $\lambda (\alpha )=(a,b,c)$ and $\lambda (\alpha -1)$ are vectors in the hyperplane $H$ of ${\mathbb{R}}^3$, where $a=\log x$, $b=\log y$, $c=\log \mathrm{\mid }z\mathrm{\mid }$; these logarithms are defined because $x>1$, $0<y<1$ and $\mathrm{\mid }z\mathrm{\mid }>1$; and $\lambda$ is additive, $\lambda (u_1{u}_2)=\lambda (u_1)+\lambda (u_2)$.

Conjugate relations I: if $r$ is any root of $f$, then $r^2-2$ is again a root: from $r^3=3r-1$ one computes $r^4=3r^2-r$, $r^5=-r^2+9r-3$, $r^6=9r^2-6r+1$, hence $(r^2-2{)}^3=3r^2-7$ and $(r^2-2{)}^3-3(r^2-2)+1=(3r^2-7)-3r^2+6+1=0$. Thus $\phi (r):=r^2-2$ maps the set $\{x,y,z\}$ of the three distinct roots into itself. Now $x>\sqrt{2}$, because $f$ is strictly increasing on $(1,\mathrm{\infty })$ by [F1] and $f(\sqrt{2})=2\sqrt{2}-3\sqrt{2}+1=1-\sqrt{2}<0<f(x)$; hence $x^2-2\in (0,2)$, and the only roots in $(0,2)$ are $y\in (0,1)$ and $x\in (1,2)$, with $x^2-2\mathrm{\ne }x$ because $x^2-x-2=0$ would give $x\in \{-1,2\}$; therefore $\phi (x)=y$. Similarly $\phi (y)=y^2-2\in (-2,-1)$ is the root in $(-2,-1)$, namely $z$. Finally $z\in (-2,-\sqrt{3})$, since for $s<t<-1$ one has $f(t)-f(s)=(t-s)(s^2+st+t^2-3)>0$, so $f$ is strictly increasing on $(-\mathrm{\infty },-1)$, and $f(-2)=-1<0<f(-\sqrt{3})=1$, so $\phi (z)=z^2-2\in (1,2)$, and the only root in $(1,2)$ is $x$; hence $\phi (z)=x$.

Conjugate relations II: for every root $r$ of $f$ one has $r(r^2-3)=-1$, hence $1\mathrm{/}r=3-r^2$. Applying this to $r=x$ and using $x^2-2=y$ gives $1\mathrm{/}x=3-x^2=1-y$; applied to $r=y$ and using $y^2-2=z$ it gives $1\mathrm{/}y=1-z$; applied to $r=z$ and using $z^2-2=x$ it gives $1\mathrm{/}z=1-x$.

Logarithmic coordinates: since $1-y>0$ is the inverse of $x$, $\log (1-y)=-a$, that is $\log \mathrm{\mid }y-1\mathrm{\mid }=-a$; since $1-z>0$ is the inverse of $y$, $\log \mathrm{\mid }z-1\mathrm{\mid }=-b$; and since $1\mathrm{/}z=1-x$ is negative with $x-1>0$, $\log \mathrm{\mid }z\mathrm{\mid }=-\log (x-1)$, that is $\log (x-1)=-c$. Moreover $a+b+c=\log \mathrm{\mid }xyz\mathrm{\mid }=\log \mathrm{\mid }{N}_{K\mathrm{/}\mathbb{Q}}(\alpha )\mathrm{\mid }=\log 1=0$ because $N_{K\mathrm{/}\mathbb{Q}}(\alpha )=-1$. Therefore $\lambda (\alpha )=(a,b,c)$ and $\lambda (\alpha -1)=(\log (x-1),\log \mathrm{\mid }y-1\mathrm{\mid },\log \mathrm{\mid }z-1\mathrm{\mid })=(-c,-a,-b)$ with $a+b+c=0$.

The logarithmic matrix of the tuple $(u_1,u_2)$ is $A=\left(\begin{array}{cc}a& -c\\ b& -a\\ c& -b\end{array}\right),a+b+c=0.$ Deleting row $1$ gives ${\mathrm{\Delta }}_1=b(-b)-(-a)c=-b^2+ac=-(a^2+ab+b^2)$; deleting row $2$ gives ${\mathrm{\Delta }}_2=a(-b)-(-c)c=-ab+c^2=a^2+ab+b^2$; and deleting row $3$ gives ${\mathrm{\Delta }}_3=a(-a)-(-c)b=-a^2+bc=-(a^2+ab+b^2)$, where $c=-a-b$ is used in each reduction. In particular all three deleted-row determinants are nonzero and have absolute value $Q:=a^2+ab+b^2$.

The second tuple has the same lattice and the same determinants: by additivity $\lambda (u_1{u}_2)=\lambda (u_1)+\lambda (u_2)$, so $\mathbb{Z}\lambda (u_1)+\mathbb{Z}\lambda (u_2)=\mathbb{Z}(\lambda (u_1)+\lambda (u_2))+\mathbb{Z}\lambda (u_2)$, an equality of subgroups of $H$. In coordinates, the logarithmic matrix of $(u_1{u}_2,u_2)$ is $A^{\mathrm{\prime }}=AC$ with $C=\left(\begin{array}{cc}1& 0\\ 1& 1\end{array}\right)$ acting on columns, whose inverse $C^{-1}=\left(\begin{array}{cc}1& 0\\ -1& 1\end{array}\right)$ is integral and whose determinant is $1$; deleting row $k$ commutes with right multiplication, so $A_k^{\mathrm{\prime }}=A_{k}C$ and $\det A_k^{\mathrm{\prime }}=\det (A_k)\det (C)=\det A_k$ by [F7]. Thus the determinants of the two tuples are equal, not merely equal in absolute value, and $\mid \det A_k^{\mathrm{\prime }}\mid =Q$ as well.

Swapping the two units, that is passing to $(u_2,u_1)$, replaces $A$ by $A^{\mathrm{\prime }\mathrm{\prime }}=AP$ with $P=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, whose inverse is itself and whose determinant is $-1$; then $\det A_k^{\mathrm{\prime }\mathrm{\prime }}=\det (A_k)\det (P)=-\det A_k$, so the sign of every deleted-row determinant is reversed and the absolute value $Q$ is unchanged. Both $C$ and $P$ are invertible over $\mathbb{Z}$, so by [F8] their determinants are units of $\mathbb{Z}$, that is $\pm 1$, in agreement with the direct computations $\det C=1$ and $\det P=-1$.

Numerical value: evaluating cosine gives $\cos (2\pi \mathrm{/}9)\approx 0.7660444431$, so $x=\alpha =2\cos (2\pi \mathrm{/}9)\approx 1.5320888862$, $y=x^2-2\approx 0.3472963553$, and $z=y^2-2$ satisfies $\mathrm{\mid }z\mathrm{\mid }\approx 1.8793852416$ because $z\in (-2,-1)$ by step 1.2; hence $a=\log x\approx 0.4266320894$, $b=\log y$ satisfies $\mathrm{\mid }b\mathrm{\mid }\approx 1.0575768136$, and $Q=a^2+ab+b^2\approx \mathrm{0.8492874506.}$ So every deleted-row determinant of the two tuples has absolute value the number $Q$ of the display, the signs for the tuple $(u_1,u_2)$ being $-,+,-$ as computed in step 4.1.

Relation to the regulator definition and choice accounting: the computation is the explicit ${\mathrm{GL}}_2(\mathbb{Z})$ step that [F5] and [F9] single out — right multiplication by an integral matrix of determinant $\pm 1$ changes the deleted-row determinants by that same factor, so the absolute value is the invariant and the regulator is defined from it. Nothing here asserts that $(u_1,u_2)$ or $(u_1{u}_2,u_2)$ is a system of fundamental units: the common number $Q$ is the absolute deleted-row determinant of the rank-two subgroup lattice generated by the tuple, and it coincides with the field regulator only when the tuple generates all of $\lambda ({\mathcal{O}}_K^{\times })$. AC enters only through the unit-theoretic inputs quoted in [F1] and [F9] and the hyperplane input [F4]; the algebraic relations among conjugates, the logarithmic matrix computations and the determinant identities use no choice.