2
$\begingroup$

Let $p>0$ be a prime and let $\omega$ be a primitive $p^{\text{th}}$ root of unity. I am trying to find all units in the ring $\Bbb Z[\omega]$.

Every element of $\Bbb Z[\omega]$ is of the form $z=a_1\omega+a_2\omega^2+\dots+a_{p-1}\omega^{p-1}$ for some $a_1,\dots,a_{p-1}\in\Bbb Z$. Attempting to find the form of the inverse $z$ one could try to make the denominator real by multiplying the complex conjugate. Note that $\overline{\omega}=\omega^{p-1}$ so that \begin{multline} \frac{1}{a_1\omega + a_2\omega^2 + \dots + a_{p-1}\omega^{p-1}} \\[4pt] = \frac{a_1\omega^{p-1} + a_2\omega^{2(p-1)} + \dots + a_{p-1}\omega^{(p-1)(p-1)}} {(a_1\omega + a_2\omega^2 + \dots + a_{p-1}\omega^{p-1})(a_1\omega^{p-1} + a_2\omega^{2(p-1)} + \dots + a_{p-1}\omega^{(p-1)(p-1)})} \end{multline} but note that $$ \omega^{k(p-1)} = \omega^{p-k} $$ so the later fraction is $$ \frac{a_1\omega^{p-1} + a_2\omega^{p-2} + \dots + a_{p-1}\omega} {(a_1\omega + a_2\omega^2 + \dots + a_{p-1}\omega^{p-1}) (a_1\omega^{p-1} + a_2\omega^{p-2} + \dots + a_{p-1}\omega)}. $$

I have trouble simplifying the denominator further, so I took the example where $p=3$ where $\omega$ is then a primitive $3^{\text{rd}}$ root of unity. If $z=a+b\omega$, the fraction above then is easy to calculate as $$ \frac{a + b\omega}{a^2 - ab + b^2}. $$ Since $\{1,\omega\}$ is linearly independent over $\Bbb Q$, $a^2-ab+b^2$ must divide $\gcd(a,b)$ but $\gcd(a,b) = \pm 1$ or else $a+b\omega$ cannot be a unit. We conclude that $$ a^2-ab+b^2 = \pm 1 $$ and an elementary argument shows that the only possibilities are $(a,b)=(0,1)$, $(0,-1)$, $(1,0)$, $(-1,0)$, $(1,1)$, or $(-1, -1)$.

How can I reach a similar conclusion in the general case of the prime $p$?

$\endgroup$
1
  • 1
    $\begingroup$ There will be lots more units as $p$ grows—roughly a $\frac{p-3}2$-parameter family of units, according to Dirichlet's unit theorem. An important class of such units are the cyclotomic units. $\endgroup$ Oct 18, 2023 at 21:41

0

You must log in to answer this question.