Real quadratic units and Pell's equation

The ring-of-integers formula gives ${\mathcal{O}}_K=\mathbb{Z}[\sqrt{d}]$ when $d\equiv 2,3(\mathrm{m}\mathrm{o}\mathrm{d}4)$ and ${\mathcal{O}}_K=\mathbb{Z}[(1+\sqrt{d})\mathrm{/}2]⊋\mathbb{Z}[\sqrt{d}]$ when $d\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}4)$.

An element $\alpha =x+y\sqrt{d}\in \mathbb{Z}[\sqrt{d}]$ is a unit of the order $\mathbb{Z}[\sqrt{d}]$ if and only if $N_d(\alpha )=x^2-d{y}^2=\pm 1$: if $N_d(\alpha )=\pm 1$ then ${\alpha }^{-1}=\pm (x-y\sqrt{d})\in \mathbb{Z}[\sqrt{d}]$, while for a unit $\alpha$ one has $N_d(\alpha )N_d({\alpha }^{-1})=N_d(1)=1$ with both factors in $\mathbb{Z}$, so $N_d(\alpha )=\pm 1$.

Seen in $K$, every unit of $\mathbb{Z}[\sqrt{d}]$ has norm $\pm 1$ (its Pell norm is the field norm), and conversely a norm-$\pm 1$ element of $\mathbb{Z}[\sqrt{d}]$ is a unit; thus the units of the order are exactly the integral solutions of $x^2-d{y}^2=\pm 1$, with the norm-one solutions forming $\pm ⟨{\epsilon }_d⟩$.

Take $d=5$ and $\epsilon :=(1+\sqrt{5})\mathrm{/}2\in {\mathcal{O}}_K$. Its norm is $((1+\sqrt{5})\mathrm{/}2)((1-\sqrt{5})\mathrm{/}2)=(1-5)\mathrm{/}4=-1$, so $\epsilon \in {\mathcal{O}}_K^{\times }$; direct multiplication gives ${\epsilon }^2=(3+\sqrt{5})\mathrm{/}2$ and ${\epsilon }^3=2+\sqrt{5}$.

Suppose first that $x^2-d{y}^2=-1$ is solvable, and let $u_d=x_0+y_0\sqrt{d}$ be its least positive solution, whose coordinates are the numerator and denominator of the convergent in [F5]. Then $u_d^2$ is a positive solution of $x^2-d{y}^2=1$, so $u_d^2={\epsilon }_d^n$ for a unique integer $n\ge 1$.

For $d\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}4)$ the elements of ${\mathcal{O}}_K=\mathbb{Z}[(1+\sqrt{d})\mathrm{/}2]$ are the numbers $(x+y\sqrt{d})\mathrm{/}2$ with $x\equiv y(\mathrm{m}\mathrm{o}\mathrm{d}2)$, of norm $(x^2-d{y}^2)\mathrm{/}4$; such an element is a unit exactly when $x^2-d{y}^2=\pm 4$, and it fails to lie in $\mathbb{Z}[\sqrt{d}]$ exactly when $x$ and $y$ are both odd.

The exponent $n$ is odd: if $n=2m$, then $u_d^2=({\epsilon }_d^m{)}^2$ in the domain $\mathbb{Z}[\sqrt{d}]$, so $u_d=\pm {\epsilon }_d^m$ and taking norms gives $N(u_d)=N({\epsilon }_d^m)=+1$, contradicting $N(u_d)=-1$.

The units of ${\mathcal{O}}_K$ for $d=5$ are the elements $(x+y\sqrt{5})\mathrm{/}2$ with $x\equiv y(\mathrm{m}\mathrm{o}\mathrm{d}2)$ and $x^2-5y^2=\pm 4$; for a unit $u>1$ one has $y>0$, and $x=u\pm 1\mathrm{/}u>0$ according as $N(u)=\pm 1$, so $x,y$ are positive integers. Checking the admissible positive pairs in order of $y$: for $y=1$ the equation gives $x^2=5\pm 4$, so $x=1$ (the unit $\epsilon$) or $x=3$ (the unit ${\epsilon }^2$); for $y=2$ it gives $x^2=20\pm 4$, so $x=4$ (the unit $2+\sqrt{5}={\epsilon }^3$); for $y\ge 3$ one has $x^2\ge 5\cdot 9-4=41$, hence $x\ge 7$ and $(x+y\sqrt{5})\mathrm{/}2\ge (7+3\sqrt{5})\mathrm{/}2>{\epsilon }^3>\epsilon$. Therefore $\epsilon$ is the least unit $>1$ of ${\mathcal{O}}_K$.

In fact $n=1$. If $n\ge 3$, write $n=2m+1$ and put $w:=u_d{\epsilon }_d^{-m}$. Both $u_d$ and ${\epsilon }_d$ are units of $\mathbb{Z}[\sqrt{d}]$ by [F3], so $w$ is a unit of that order as well; in particular $w=x+y\sqrt{d}$ for integers $x,y$. From step 2.1, $u_d^2={\epsilon }_d^n$, and therefore $w^2=u_d^2{\epsilon }_d^{-2m}={\epsilon }_d$ and $u_d=w{\epsilon }_d^m=w(w^2{)}^m=w^n$. Since $u_d>0$ and ${\epsilon }_d>1$, the definition of $w$ gives $w>0$, and $w^2={\epsilon }_d>1$ gives $w>1$. By [F3] the order norm $N_d(w)$ is $+1$ or $-1$. If $N_d(w)=+1$, multiplicativity in [F3] and $u_d=w^n$ give $N_d(u_d)=N_d(w{)}^n=+1$, contradicting $N_d(u_d)=-1$; hence $N_d(w)=-1$. Thus $w=x+y\sqrt{d}$ is a positive integral solution of $x^2-d{y}^2=-1$: its conjugate is $-1\mathrm{/}w$, so $x=(w-1\mathrm{/}w)\mathrm{/}2>0$ and $y=(w+1\mathrm{/}w)\mathrm{/}(2\sqrt{d})>0$. As $n\ge 3$ and $w>1$, $u_d=w^n>w$; moreover $x=(w-1\mathrm{/}w)\mathrm{/}2<(u_d-1\mathrm{/}{u}_d)\mathrm{/}2$, so this solution has smaller first coordinate than the least positive solution $u_d$, a contradiction. Therefore $n=1$ and ${\epsilon }_d=u_d^2$.

Since ${\mathcal{O}}_K^{\times }=\{\pm {\gamma }^n\}$ for some $\gamma >1$ and the least unit $>1$ in such a group is $\gamma$, step 3.2 gives $\gamma =\epsilon$, so ${\mathcal{O}}_K^{\times }=\{\pm {\epsilon }^n:n\in \mathbb{Z}\}$.

Consequently, in the solvable case every norm-one unit is $\pm {\epsilon }_d^m=\pm u_d^{2m}$ and every norm-$(-1)$ unit is $\pm u_d^{2m+1}$, because multiplying it by $u_d^{-1}$ gives norm $1$; thus $\mathbb{Z}[\sqrt{d}{]}^{\times }=\pm ⟨u_d⟩$, and the norm-one subgroup $\pm ⟨{\epsilon }_d⟩=\pm ⟨u_d^2⟩$ consists of the even powers of $u_d$, of index $2$.

If instead $x^2-d{y}^2=-1$ is unsolvable, every unit of $\mathbb{Z}[\sqrt{d}]$ has norm $+1$, so $\mathbb{Z}[\sqrt{d}{]}^{\times }=\pm ⟨{\epsilon }_d⟩$; setting $u_d:={\epsilon }_d$ gives $\mathbb{Z}[\sqrt{d}{]}^{\times }=\pm ⟨u_d⟩$ in both cases.

The order $\mathbb{Z}[\sqrt{d}]$ is a subring of ${\mathcal{O}}_K$, so its unit group is a subgroup of ${\mathcal{O}}_K^{\times }$ and equals the units computed in steps 5.1 and 6.1; if the fundamental unit of ${\mathcal{O}}_K$ (its least unit $>1$) is an element with $x,y$ both odd, then it is not in $\mathbb{Z}[\sqrt{d}]$, so $\mathbb{Z}[\sqrt{d}{]}^{\times }⊊{\mathcal{O}}_K^{\times }$, and since $\pm ⟨{\epsilon }_d⟩\subseteq \mathbb{Z}[\sqrt{d}{]}^{\times }$ also $\pm ⟨{\epsilon }_d⟩⊊{\mathcal{O}}_K^{\times }$.

In the order, [F5] applied with the period length $\mathrm{\ell }=1$ computed in [F7] makes the pair $(p_0,q_0)=(2,1)$ the least positive solution of $x^2-5y^2=-1$ and the pair $(p_1,q_1)=(9,4)$ the least positive solution of $x^2-5y^2=1$, whose associated element is the fundamental Pell unit ${\epsilon }_5=9+4\sqrt{5}$; by steps 5.1 and 6.1 applied to $d=5$, $\mathbb{Z}[\sqrt{5}{]}^{\times }=\pm ⟨2+\sqrt{5}⟩$, and by step 1.4 this is $\pm ⟨{\epsilon }^3⟩$, while the identity $(2+\sqrt{5}{)}^2=9+4\sqrt{5}$ of [F7] gives $9+4\sqrt{5}={\epsilon }^6={\epsilon }_5$.

Finally $\mathbb{Z}[\sqrt{5}{]}^{\times }=\pm ⟨{\epsilon }^3⟩\subseteq \pm ⟨\epsilon ⟩={\mathcal{O}}_K^{\times }$ with index $[⟨\epsilon ⟩:⟨{\epsilon }^3⟩]=3$, because the multiples of $3$ in $\mathbb{Z}$ have index $3$; a generator of the larger group, for instance $\epsilon$, is not in the smaller order, so the two unit groups are not equal and the order's norm-one Pell subgroup is proper in ${\mathcal{O}}_K^{\times }$.

Scope and choice accounting: the general statements of the example are steps 5.1, 6.1 and 7.1, and the failure of equality is witnessed by $d=5$ in steps 7.2 and 8.1; AC is used only through the rank-one structure [F6], all computations here being elementary arithmetic in $\mathbb{Z}[\sqrt{5}]$.