Two independent units in a real cubic field

$\pi >0$, hence $0<\theta <\pi \mathrm{/}3<\pi$ for $\theta =2\pi \mathrm{/}9$.

With $c:=\cos (\pi \mathrm{/}3)$ one has $\cos (2\pi \mathrm{/}3)=\cos (\pi -\pi \mathrm{/}3)=\cos ((-\pi \mathrm{/}3)+\pi )=-\cos (-\pi \mathrm{/}3)=-\cos (\pi \mathrm{/}3)=-c$ by the shift and parity identities, while the double-angle identity gives $\cos (2\pi \mathrm{/}3)=2c^2-1$; hence $2c^2-1=-c$, that is $(2c-1)(c+1)=0$. Since $\pi \mathrm{/}3<\pi$ and cosine is strictly decreasing on $[0,\pi ]$ with $\cos \pi =-1$, one has $c>-1$, so $c=1\mathrm{/}2$.

Values of $f$: $f(2)=3$, $f(1)=-1$, $f(0)=1$, $f(-1)=3$, $f(-2)=-1$.

Hence $\cos (2\pi \mathrm{/}3)=-1\mathrm{/}2$.

Moreover $1<\alpha <2$: from $0<\theta <\pi \mathrm{/}3$ and strict decrease of cosine on $[0,\pi ]$ together with $\cos 0=1$ and $\cos (\pi \mathrm{/}3)=1\mathrm{/}2$ one gets $1\mathrm{/}2<\cos \theta <1$, hence $1<\alpha <2$.

The polynomial $f$ is irreducible over $\mathbb{Q}$: by [F8] a rational root of the monic integer polynomial $f$ must be an integer dividing the constant term $1$, hence equal to $1$ or $-1$; but $f(1)=-1\mathrm{\ne }0$ and $f(-1)=3\mathrm{\ne }0$, so $f$ has no rational root, and since $\mathrm{deg}f=3$ it is irreducible by [F9].

The triple-angle identity gives ${\alpha }^3-3\alpha =(2\cos \theta {)}^3-3(2\cos \theta )=2(4{\cos }^3\theta -3\cos \theta )=2\cos (3\theta )=2\cos (2\pi \mathrm{/}3)=-1$, so $f(\alpha )={\alpha }^3-3\alpha +1=0$.

The polynomial $f$ has exactly one root in $(1,2)$, namely $\alpha$: by step 1.3 and [F10] the sign change from $f(1)<0$ to $f(2)>0$ gives a root in $(1,2)$, and for $1<x<y$ one has $f(y)-f(x)=(y-x)(x^2+xy+y^2-3)>0$, so $f$ is strictly increasing on $(1,\mathrm{\infty })$ and the root is unique; steps 3.1 and 2.2 show that $\alpha$ is such a root.

By step 1.3 and [F10] there are also roots in $(-2,-1)$ and in $(0,1)$, and these three roots are distinct because the intervals are disjoint; a cubic has at most three roots, so these are all the roots of $f$ and all of them are real.

Therefore $f$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}$ (monic, irreducible, with $f(\alpha )=0$ by step 3.1), so $K=\mathbb{Q}(\alpha )$ has degree $3$ over $\mathbb{Q}$; by [F11] the three embeddings $K\to \mathbb{C}$ send $\alpha$ to the three roots of $f$, which are all real by step 5.1, so $K$ is totally real with signature $(3,0)$ and unit rank $r_1+r_2-1=2$ by [F16].

By [F12] and [F13] applied to $f(t)={\prod }_{i=1}^3(t-{\alpha }_i)$, where ${\alpha }_1,{\alpha }_2,{\alpha }_3$ are the three conjugates of $\alpha$ given by the embeddings of step 6.1, one has $N_{K\mathrm{/}\mathbb{Q}}(\alpha )={\alpha }_1{\alpha }_2{\alpha }_3=(-1{)}^3{a}_3=-1$ because the constant coefficient of $f$ is $a_3=1$; and $N_{K\mathrm{/}\mathbb{Q}}(\alpha -1)={\prod }_{i=1}^3({\alpha }_i-1)=(-1{)}^3{\prod }_{i=1}^3(1-{\alpha }_i)=-f(1)=-(-1)=1$, using $f(1)=1-3+1=-1$.

Writing the three real embeddings as ${\sigma }_1=\mathrm{i}\mathrm{d},{\sigma }_2,{\sigma }_3$, the conjugates satisfy ${\sigma }_1(\alpha )=\alpha \in (1,2)$, $0<{\sigma }_2(\alpha )<1$ and $-2<{\sigma }_3(\alpha )<-1$ by steps 2.2 and 5.1; hence $\log \mathrm{\mid }{\sigma }_1\alpha \mathrm{\mid }=\log \alpha >0$, $\log \mathrm{\mid }{\sigma }_1(\alpha -1)\mathrm{\mid }=\log (\alpha -1)<0$, $\log \mathrm{\mid }{\sigma }_2\alpha \mathrm{\mid }<0$, $\log \mathrm{\mid }{\sigma }_2(\alpha -1)\mathrm{\mid }=\log (1-{\sigma }_2\alpha )<0$, $\log \mathrm{\mid }{\sigma }_3\alpha \mathrm{\mid }>0$ and $\log \mathrm{\mid }{\sigma }_3(\alpha -1)\mathrm{\mid }>0$.

The elements $\alpha$ and $\alpha -1$ lie in ${\mathcal{O}}_K$: $\alpha$ is a root of the monic polynomial $f\in \mathbb{Z}[X]$, hence integral over $\mathbb{Z}$ and in ${\mathcal{O}}_K$ by [F15], and $\alpha -1\in {\mathcal{O}}_K$ because ${\mathcal{O}}_K$ is a subring of $K$; by [F14] with the norms of step 7.1, both are units of ${\mathcal{O}}_K$.

The vectors $\lambda (\alpha )$ and $\lambda (\alpha -1)$ are linearly independent over $\mathbb{R}$: if $a\lambda (\alpha )+b\lambda (\alpha -1)=0$, then reading the first coordinate and dividing by $\log \alpha \mathrm{\ne }0$ gives $a+b{r}_1=0$ with $r_1:=\log (\alpha -1)\mathrm{/}\log \alpha <0$ by step 7.2, while reading the second coordinate and dividing by $\log \mathrm{\mid }{\sigma }_2\alpha \mathrm{\mid }\mathrm{\ne }0$ gives $a+b{r}_2=0$ with $r_2:=\log \mathrm{\mid }{\sigma }_2(\alpha -1)\mathrm{\mid }\mathrm{/}\log \mathrm{\mid }{\sigma }_2\alpha \mathrm{\mid }>0$, a quotient of two negative numbers; subtracting the two equations gives $b(r_1-r_2)=0$, and $r_1\mathrm{\ne }{r}_2$ because their signs differ, so $b=0$ and then $a=0$ from the first equation.

By [F17] the map $\lambda$ turns products into sums, and by [F18] both $\lambda (\alpha )$ and $\lambda (\alpha -1)$ lie in $H$, since $\alpha$ and $\alpha -1$ are units of ${\mathcal{O}}_K$ by step 8.1.

Hence the subgroup $⟨\alpha ,\alpha -1⟩$ is free abelian of rank $2$: if ${\alpha }^m(\alpha -1{)}^n=1$ for integers $m,n$, then $m\lambda (\alpha )+n\lambda (\alpha -1)=\lambda (1)=0$ by [F17] and step 8.2 forces $m=n=0$; thus the homomorphism ${\mathbb{Z}}^2\to {\mathcal{O}}_K^{\times }$, $(m,n)\mapsto {\alpha }^m(\alpha -1{)}^n$, has trivial kernel, and its image is exactly $⟨\alpha ,\alpha -1⟩$.

The image has finite index in $\lambda ({\mathcal{O}}_K^{\times })$: by [F20] fix a system of fundamental units ${\epsilon }_1,{\epsilon }_2$ and write $\alpha =\zeta {\epsilon }_1^{m_1}{\epsilon }_2^{m_2}$ and $\alpha -1={\zeta }^{\mathrm{\prime }}{\epsilon }_1^{n_1}{\epsilon }_2^{n_2}$ with $\zeta ,{\zeta }^{\mathrm{\prime }}\in \mu (K)$ and integers $m_i,n_i$; since $\lambda$ kills $\mu (K)$, $\lambda (\alpha )=m_1\lambda ({\epsilon }_1)+m_2\lambda ({\epsilon }_2)$ and $\lambda (\alpha -1)=n_1\lambda ({\epsilon }_1)+n_2\lambda ({\epsilon }_2)$, so with respect to the $\mathbb{Z}$-basis $(\lambda ({\epsilon }_1),\lambda ({\epsilon }_2))$ of $\lambda ({\mathcal{O}}_K^{\times })$ the two vectors have the integer coordinate columns $(m_1,m_2{)}^{\mathsf{T}}$ and $(n_1,n_2{)}^{\mathsf{T}}$, and the matrix $A=\left(\begin{array}{cc}{m}_1& n_1\\ m_2& n_2\end{array}\right)$ has $\det A\mathrm{\ne }0$ because a zero determinant would make the two coordinate columns, hence $\lambda (\alpha )$ and $\lambda (\alpha -1)$, linearly dependent over $\mathbb{R}$, contradicting step 8.2; therefore the subgroup $\mathbb{Z}\lambda (\alpha )+\mathbb{Z}\lambda (\alpha -1)=A{\mathbb{Z}}^2$ has finite index in $\lambda ({\mathcal{O}}_K^{\times })$ by [F21].

Consequently $⟨\alpha ,\alpha -1⟩$ has finite index in ${\mathcal{O}}_K^{\times }$: the canonical map ${\mathcal{O}}_K^{\times }\mathrm{/}⟨\alpha ,\alpha -1⟩\to \lambda ({\mathcal{O}}_K^{\times })\mathrm{/}(\mathbb{Z}\lambda (\alpha )+\mathbb{Z}\lambda (\alpha -1))$ is surjective onto a finite group by step 10.1, and its kernel $(\mu (K)⟨\alpha ,\alpha -1⟩)\mathrm{/}⟨\alpha ,\alpha -1⟩\cong \mu (K)\mathrm{/}(\mu (K)\cap ⟨\alpha ,\alpha -1⟩)$ is a quotient of the finite group $\mu (K)$ of [F19]; hence the quotient is finite, as claimed.

Choice accounting: AC is used only through the unit theorem [F19], which supplies the finite generation and rank, and through the existence of the system of fundamental units [F20]; the trigonometric, polynomial, norm and logarithm computations, and the independence argument via signs, are elementary and use no further choice.