Question

Let$\mathit{\omega }$be a complex number such that ${\mathit{\omega }}^{\mathbf{P}}\mathbf{=}\mathbf{1}$. Show that

localid="1659541322947" $\mathbf{\mathbb{Z}}\mathbf{\left[}\mathbf{\omega }\mathbf{\right]}\mathbf{=}\mathbf{\{}{\mathbf{a}}_{\mathbf{0}}\mathbf{+}{\mathbf{a}}_{\mathbf{1}}\mathbf{\omega }\mathbf{+}{\mathbf{a}}_{\mathbf{2}}{\mathbf{\omega }}^{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{\mathbf{a}}_{\mathbf{P}\mathbf{-}\mathbf{1}}{\mathbf{\omega }}^{\mathbf{p}\mathbf{-}\mathbf{1}}\overline{){\mathbf{a}}_{\mathbf{i}}\mathbf{\in }\mathbf{\mathbb{Z}}}\mathbf{\}}$

is an integral domain.$\left[Hint:{\omega }^p=1imples{\omega }^{p+1}=\omega ,{\omega }^{p+2}={\omega }^2,etc.\right]$

Short answer

It is shown that$\mathrm{\mathbb{Z}}\left[\omega \right]$is an integral domain.

Step-by-step solution

Referring to hint

${\omega }^p=1$

Therefore, using ${\omega }^p=1$, we can write${\omega }^{p+2}$ and${\omega }^{p+2}$ as:

$$\begin{gathered} {\omega }^{p+2}=\omega \\ {\omega }^{p+2}={\omega }^2 \end{gathered}$$

Proving that ℤ[ω]  is an integral domain

Given that, $\mathrm{\mathbb{Z}}\left[\omega \right]=\left\{a_0+a_1\omega +a_2{\omega }^2+...+a_{p-1}{\omega }^{p-1}{\overline{)a}}_i\in \mathrm{\mathbb{Z}}\right\}$.

From the above expression of$\mathrm{\mathbb{Z}}\left[\omega \right]$and given hint, we can conclude that$\mathrm{\mathbb{Z}}\left[\omega \right]$is subring of the integral domain of complex number $\mathrm{ℂ}$.

As we know, subrings of the integral domain are also the integral domain.

Hence, it is proved that$\mathrm{\mathbb{Z}}\left[\omega \right]$is an integral domain.