Find units in $\mathbb{Z}[(\sqrt{13}+1)/2]$

Question

I need to show that the units in the ring $\mathbb{Z}[x] =\{a + bx : a,b \in \mathbb Z\}$ where $x = \frac{1+\sqrt{13}}{2}$ (just considering the positive root of $13$), are those which $a^2 +ab -3b^2$ is a unit in $\mathbb{Z}$, that is, those which $a^2 +ab -3b^2= \pm 1$.

I have tried it by finding the solutions to $(a + bx)(a'+b'x)=1$, but this leads me to the following equation with 4 variables and many possible restrictions that I am stucked with:

$ \displaystyle (a + bx)(a'+b'x)= a a ' +ab'x + a'bx + bb'x^2 = aa' + \frac{ab'}{2}+ \frac{a'b}{2} +\frac{ab'}{2}\sqrt{13} + \frac{a'b}{2}\sqrt{13} + \frac{bb'}{2}\sqrt{13} + \frac{7}{2}bb'=1$

where the terms with $\sqrt{13}$ are necessarily $0$, that is $ab'+a'b+bb'=0$

I cannot transform the last condition into anything more easy to handle, and I don't know if there is a more efficient way tho prove this. (I do not know much about Number Theory nor advanced Group or Ring Theory, which I have seen that is often useful in similar problems).

Answer 1

Let handle your case first:

In $\Bbb{Z}[\frac{1 + \sqrt{13}}{2}]$, we can define a function $N:\Bbb{Z}[\frac{1 + \sqrt{13}}{2}] \to \mathbb{Z}$ by $(a + b\frac{1 + \sqrt{13}}{2}) \mapsto (a + b\frac{1 + \sqrt{13}}{2})(a + b\frac{1 - \sqrt{13}}{2}) = (a+\frac{b}{2})^2 - \frac{13}{4}b^2= a^2 +ab - 3b^2$.

Claim: $\alpha\in \Bbb{Z}[\frac{1 + \sqrt{13}}{2}]$ be unit if and only if $N(\alpha) = \pm 1$.

Observation: $(a + b\frac{1 - \sqrt{13}}{2}) = (a+b) + (-b)\frac{1 + \sqrt{13}}{2}$.

Proof of the claim:

If $N(\alpha) = \pm 1$, clearly $\alpha$ is an unit from definition of $N$. Conversely, if $\alpha$ is a unit, then there exist a $\beta \in \Bbb{Z}[\frac{1 + \sqrt{13}}{2}]$ such that $\alpha \beta = 1$. Now, $N(\alpha \beta) = N(1) = 1$. If we can show that $N$ is multiplicative, i.e., $N(\alpha \beta)=N(\alpha)N(\beta)$, then we are done because $N(\alpha)N(\beta) = 1$ means $|N(\alpha)| = 1$. Hence, $N(\alpha)= \pm 1$.

Proof of $N(\alpha \beta) = N(\alpha)N(\beta)$: Left to be check. Do brutal computation.

Remarks: We can see in the proof all we need is properties of $N$ only. So is there a way to generalize this thing?

Answer is yes!

Let $d$ be a squarefree integer (maybe negeative) and we define

$\mathcal{O}(d) = \cases{\mathbb{Z}[\sqrt{d}] & $\text{if } d \equiv 2, 3 \, (\text{mod } 4)$ \\ \mathbb{Z}[\frac{1+\sqrt{d}}{2}] & $\text{if } d \equiv 1 \, (\text{mod } 4)$}$

We can define $N$ similarly for $\mathcal{O}(d)$ and our claim is also true here.

$N$ is called norm and $\mathcal{O}(d)$ is called ring of integer in $\mathbb{Q}(\sqrt{d})$.

Answer 2

A multiplicative norm can be defined on $\mathbb Z[x]$ as $N(a+bx)=a^2+ab-3b^2$. This norm is always an integer and $N(1)=1$, so $N(u)=\pm1$ iff $u$ is a unit in $\mathbb Z[x]$.

Answer 3

The group of units of the ring of integers $\mathcal{O}_K=\Bbb Z\Large[ \frac{1+\sqrt{13}}{2}\Large]$ is isomorphic to $\Bbb{Z}$ by Dirichlet's unit theorem, and a fundamental unit is given by $$\epsilon=\frac{3+\sqrt{13}}{2}.$$ This gives a complete description. Of course, the units are also the elements of norm $\pm 1$ here.

Answer 4

You were on the right track, but must continue.

Having derived the equation

$ba'+(a+b)b'=0$ (note the rearrangement to get terms with the same primed variable together)

Now equate the rational parts to get a second relation

$(2a+b)a'+(a+7b)b'=2$

You now consider these two equations as a linear system, with $a'$ and $b'$ your "unknowns" for which you solve in terms of $a$ and $b$. By standard methods for solving linear systems obtain the results

$a'=\dfrac{a+b}{a^2+ab-3b^2}$

$b'=\dfrac{-b}{a^2+ab-3b^2}$

For a unit you must have $a'$ and $b'$ both integers. From the above solutions it follows that the common denominator $a^2+ab-3b^2$ must be a common factor of $a+b$ and $-b$, therefore also a common factor of $a=(a+b)-b$ and $b$.

Now suppose that $f$ is any such common factor. Then $f^2|(a^2+ab-3b^2)$ because each term in $a^2+ab-3b^2$ is of degree $2$. Then $a+b,b$ must both be multiples of $f^2$ and thus $a,b$ must be multiples of $f^2$. Then $f^4=(f^2)^2$ must also divide both $a$ and $b$, $f^8$ must also be a common factor and there will be a contradiction (the intended divisor grows absolutely larger than $a$ and $b$) unless the sequence $f,f^2,f^4,f^8,...,f^{2^{n-1}},...$ is bounded. This means $a,b$ can have no common factors other than $\pm1$ forcing $a^2+ab-3b^2$ to also be one of these values.