If $\omega \not= 1$ is a cube root of unity in $\mathbb{C}$, show that the units in the ring $\mathbb{Z}[\omega]$ are the elements of modulus 1. Hence, or otherwise, show that $U(\mathbb{Z}[\omega]$ is cyclic.
I have absolutely no idea where to begin with this question!!
PLease help, Thanks!
$\mathbb{Z[}x\mathbb{]}$={a+bω:a,b$\in \mathbb{Z}$, ω≠1 and ω³=1}.
Definition of unit tells us the following: if u is a unit then there exists v in the ring such that uv=1 and this v is called as inverse of u. It is fact that the inverse of element is unique. If we find all elements whose inverse also in this ring we are done.
Take a+bω$\in \mathbb{Z}$ it is clear that
Now we need to find all a and b that makes this element in the ring. i.e. the question turns for which a,b$\in \mathbb{Z}$, (a+bω)$^{-1}=\frac{1}{(a+bω )}\in \mathbb{Z[ω]}$
since ω≠1 and $(ω-1)({ω}^{2}-ω-1)=0$, we have $({ω}^{2}-ω-1)=0$ and so we have $(-1-ω)$ instead of ${ω}^{2}$ in the ring and we have $(x+y)^{3}=(x+y)(x^{2}-xy+y^{2})$ so multiply numerator and denominator of $\frac{1}{(a+bω)}$ by $(a^{2}-ab% \mathbb{ω}+(b\mathbb{ω )}^{2})$. We have the following:
$\frac{a^{2}-abω+(bω\mathbb{)}^{2}}{(a+bω)(a^{2}-abω+(bω\mathbb{)}^{2})}$=$\frac{a^{2}-abω+(bω\mathbb{)}^{2}}{a^{3}+b\mathbb{}^{3}}$=$\frac{a^{2}-abω+b^{2}ω^{2}}{a^{3}+b\mathbb{}^{3}}$=$\frac{a^{2}-abω-b^{2}ω-b^{2}}{a^{3}+b\mathbb{}^{3}}$=$\frac{a^{2}-b^{2}-ω(ab+b^{2})}{a^{3}+b\mathbb{}^{3}}$.
therefore, $\frac{a^{2}-b^{2}}{a^{3}+b\mathbb{}^{3}}$ and $\frac{(ab+b^{2})}{a^{3}+b\mathbb{}^{3}}$ has to be integer.
So, we have $\frac{a^{2}-b^{2}}{a^{3}+b\mathbb{}^{3}}$=$\frac{(a-b)(a+b)}{(a+b)(a^{2}-ab+b^{2})}$=$\frac{(a+b)}{(a^{2}-ab+b^{2})}$
and similarly,
$\frac{(ab-b^{2})}{a^{3}+b\mathbb{}^{3}}$=$\frac{(b)(a+b)}{(a+b)(a^{2}-ab+b^{2})}$=$\frac{(b)}{(a^{2}-ab+b^{2})}$
to become this expression integer denominator has to be 1. So, lets determine which a and b makes this expression integer namely inverse element in the ring.
$(a^{2}-ab+b^{2})$=1 so $(a^{2}+b^{2})$=$1+(ab)$. since left hand side always 0 or positive, $ab$ has to be greater then or equal to $-1$ then $((a-b)^{2}+ab)=1$ so $((a-b)^{2})=(1-ab)$ since left hand side always 0 or positiv, $ab$ has to be less than or equal to $1$. So we have following inequalities for $ab$;
$-1\leq ab\leq 1$
since a and b are integer we have following possibilities for this inequality
$a=1$,$b=1$ and $a=1$,$b=-1$ and $a=-1$,$b=1$ and $a=-1$,$b=-1$ and $a=0$,$b=1$ and $a=0$,$b=-1$ and $a=1$,$b=0$ and $a=-1$,$b=0$
as a result we have following elements which are the unit of this ring $1+ω$, $1-ω$, $-1+ω$, $-1-ω$, $ω$, $-ω$, $1$, $-1$