$a,b \in \mathbb Z$ we have that $a+ b\phi \in \mathbb Z[\phi]^* \iff a^2 +ab-b^2 = \pm 1$

Question

I am thinking about the following question:

Let $\phi= \frac{1+ \sqrt 5}{2}$ be the golden ratio, and define: $$ \mathbb Z [\phi]=\{a+ b\phi : \quad a, b \in \mathbb Z\} $$ We wish to prove that for $a,b \in \mathbb Z$ we have that $a+ b\phi \in \mathbb Z[\phi]^* \iff a^2 +ab-b^2 = \pm 1$. Note: the star indicates the multiplicative group of our ring.

This exercise comes with a hint: let $\bar{\phi}=\frac{1-\sqrt 5}{2}$, You may use without proof that $(a + b \phi)(c + d\phi) = 1$ if and only if $(a + b\bar{\phi})(c + d\bar{\phi}) = 1$.

I have trouble interpreting how the hint plays out.

Answer 1

For $z = a + b \phi$, we write $\overline{z} = a + b \overline{\phi}$.

Define the function $N: \mathbb{Z}[\phi] \rightarrow \mathbb{Z}$ by $N(z)=z \overline{z}$. It's easy to check that $N(a+b \phi) = a^2 +ab -b^2 \in \mathbb{Z}.$

Proposition: $z \in \mathbb{Z}[\phi]$ is a unit if, and only if, $N(z) \in \mathbb{Z}$ is a unit.

Proof:

($\Rightarrow$) Suppose $(a+b\phi)(c+d\phi)=1$. By the hint we have $(a+b\overline{\phi})(c+d\overline{\phi})=1$, and so $$N(a+b\phi)N(c+d\phi) = (a+b\phi)(a+b\overline{\phi})(c+d\phi)(c+d\overline{\phi}) = 1.$$ Since $N(a+b\phi)$, $N(c+d\phi) \in \mathbb{Z}$, we must have either $$N(a+b\phi) = N(c+d\phi) = 1 \ \ \text{or} \ \ N(a+b\phi) = N(c+d\phi) = -1.$$

($\Leftarrow$) Suppose that $N(a+b\phi) = (a+b\phi)(a+b\overline{\phi}) = (-1)^k, k\in \{ 0,1 \}$.

We want to find $c,d \in \mathbb{Z}$ such that $c+d\phi = a+b\overline{\phi}$. So write $$ c+d\phi = c+ \frac{1}{2} d + \frac{\sqrt{5}}{2} \equiv a + \frac{1}{2}b -\frac{\sqrt{5}}{2} = a+b\overline{\phi},$$ and equate coefficients in $\mathbb{Z} + \sqrt{5}\mathbb{Z}$ to get $c = a+b$ and $d=-b$. That is, we have $$(a+b\phi)((-1)^k (a+b) + (-1)^{k+1} b\phi) = 1.$$

$\hspace{18cm} \square$

Therefore, given $a+b\phi \in \mathbb{Z}[\phi]$, $$a+b\phi \in \mathbb{Z}[\phi]^* \Leftrightarrow a^2 +ab -b^2 = N(a+b\phi ) = \pm 1.$$

Answer 2

I do not see how the hint is of any use in answering the question. There are many different approaches, depending how much theory you are familiar with. I'll give a very basic approach, assuming very little theory.

Note that for $a,b,c,d\in\Bbb{Z}$ we have $a+b\phi=c+d\phi$ if and only if $a=c$ and $b=d$. Also note that $\phi+\bar\phi=1$ and $\phi\bar\phi=-1$, meaning that both are roots of $X^2-X-1$. This shows that $$\Bbb{Z}[\phi]\ \longrightarrow\ \Bbb{Z}[\phi]:\ a+b\phi\ \longmapsto\ a+b\bar\phi,$$ is a ring automorphism of $\Bbb{Z}[\phi]$, from which the hint follows. I do not see how the hint is relevant though.

We also see that for every $x\in\Bbb{Z}[\phi]$ there exist unique $a,b,a',b'\in\Bbb{Z}$ such that $$x=a+b\phi=a'+b'\bar\phi.$$ Now if $a,b\in\Bbb{Z}$ are such that $a^2+ab-b^2=\pm1$ then $$(a+b\phi)(a+b\bar\phi)=a^2+ab(\phi+\bar\phi)+b^2\phi\bar\phi=a^2+ab-b^2=\pm1,$$ which shows that $a+b\phi\in\Bbb{Z}[\phi]^{\times}$ with $(a+b\phi)^{-1}=\pm(a+b\bar\phi)$.

Conversely, if $a,b\in\Bbb{Z}$ are such that $a+b\phi\in\Bbb{Z}[\phi]^{\times}$ then there exist unique $c,d\in\Bbb{Z}[\phi]$ such that $$(a+b\phi)(c+d\phi)=1.$$ Then $\gcd(a,b)=1$, and there exist unique $c',d'\in\Bbb{Z}$ such that \begin{eqnarray*} 1&=&(a+b\phi)(c+d\phi)=(a+b\phi)(c'+d'\bar\phi) =ac'+ad'\bar\phi+bc'\phi+bd\phi\bar\phi. \end{eqnarray*} Because $\phi\bar\phi=-1$ and $\phi+\bar\phi=1$ it follows that $$1=ac'+ad'\bar\phi+bc'\phi+bd'\phi\bar\phi=(ac'+ad'-bd')+(bc'-ad')\phi,$$ which implies that $bc'=ad'$. Because $\gcd(a,b)=1$ this shows that $c'=\pm a$ and $d'=\pm b$, where the two signs agree. Hence $$1=(a+b\phi)(c+d\phi)=\pm(a+b\phi)(a+b\bar\phi)=\pm(a^2+ab-b^2).$$ This proves that for all $a,b\in\Bbb{Z}$ we have $$a+b\phi\in\Bbb{Z}[\phi]^{\times}\qquad\iff\qquad a^2+ab-b^2=\pm1.$$