Question

Show that $\mathit{\mathbb{Z}}\mathbf{\left[}\sqrt{\mathbf{-}\mathbf{2}}\mathbf{\right]}$ is a Euclidean domain with$\mathbf{\delta }\mathbf{(}\mathbf{r}\mathbf{+}\mathbf{s}\sqrt{\mathbf{-}\mathbf{2}}\mathbf{)}\mathbf{=}{\mathbf{r}}^{\mathbf{2}}\mathbf{+}\mathbf{2}{\mathbf{s}}^{\mathbf{2}}$ .

Short answer

It is proved that$\mathrm{\mathbb{Z}}\left[\sqrt{-2}\right]$ is a Euclidean domain.

Step-by-step solution

Definition of Euclidean Domain

An integral domain R is said to be Euclidean domain if there is a function $\mathbf{\delta }$from the non zero elements of R to non-negative integers with these properties:

1. If role="math" localid="1654617178174" $\mathbf{a}\mathbf{}\mathbf{and}\mathbf{}\mathbf{b}$are non-negative elements of R, then $\mathbf{\delta }\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{\le }\mathbf{\delta }\mathbf{\left(}\mathbf{ab}\mathbf{\right)}$.
2. If $\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{R}$and $\mathit{b}\mathbf{\ne }{\mathbf{0}}_{\mathbf{R}}$, then there exist $\mathit{q}\mathbf{,}\mathit{r}\mathbf{\in }\mathit{R}$such that $\mathit{a}\mathbf{=}\mathit{b}\mathit{q}\mathbf{+}\mathit{r}$and either $\mathit{r}\mathbf{=}{\mathbf{0}}_{\mathbf{R}}$or $\mathit{\delta }\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{\le }\mathit{\delta }\mathbf{\left(}\mathbf{b}\mathbf{\right)}$

Here, we know that $\mathrm{\mathbb{Z}}\left[\sqrt{-2}\right]$is an integral domain.

Now, let $a+\sqrt{-2}b,c+\sqrt{-2}d\in \mathbb{Z}\left(\sqrt{-2}\right)$

Now, $\mathrm{\delta }(\mathrm{a}+\sqrt{-2}\mathrm{b})={\mathrm{a}}^2+2{\mathrm{b}}^2$

Also,

role="math" localid="1654617445596" $\begin{array}{c}\delta \left((a+\sqrt{-2}b)(c+\sqrt{-2}d)\right)=\delta ((ac-2bd)+\sqrt{-2}(bc+ad))\\ ={(ac-2bd)}^2+2{(bc+ad)}^2\\ =a^2{c}^2-4acbd+4b^2{d}^2+2b^2{c}^2+4bcad+a^2{d}^2\\ =a^2{c}^2+4b^2{d}^2+2b^2{c}^2+a^2{d}^2\end{array}$

Clearly,$\delta (a+\sqrt{-2}b)\le \delta \left((a+\sqrt{-2}b)(c+\sqrt{-2}d)\right)$

Proving 2nd condition

Now, let by Euclidean algorithm, We can write,

$a,b\in R$and$b\ne 0_R$ , then there exist $q,r\in R$ such that$a=bq+r$ and either $r=0_R$ or $\delta \left(r\right)\le \delta \left(b\right)$

Hence$\mathbb{Z}\left[\sqrt{-2}\right]$ is Euclidean domain.