Question

If $\mathbf{r}\mathbf{+}\mathbf{s}\sqrt{\mathbf{-}\mathbf{5}}\mathbf{\in }\mathbf{\mathbb{Z}}\mathbf{\left[}\sqrt{\mathbf{-}\mathbf{5}}\mathbf{\right]}$with $\mathit{s}\mathbf{\ne }\mathbf{0}$, then prove that 2 is not in the principal ideal $\mathbf{(}\mathbf{r}\mathbf{+}\mathbf{s}\sqrt{\mathbf{-}\mathbf{5}}\mathbf{)}$

Short answer

It is proved that$2$ is not in the principal ideal $\mathrm{r}+\mathrm{s}\sqrt{-5}$.

Step-by-step solution

Principal Ideal 

An ideal $\mathit{I}$of a ring R is called principal if there is an elementrole="math" localid="1654763347888" $\mathit{a}$ of$\mathit{R}$ such that $\mathit{I}\mathbf{=}\mathit{a}\mathit{R}\mathbf{=}\mathbf{\{}\mathbf{a}\mathbf{r}\mathbf{:}\mathbf{r}\mathbf{\in }\mathbf{R}\mathbf{\}}$.

Prove that 2 is not in the principal ideal (r+s−5)

Let assume that $2$is the principal ideal $r+s\sqrt{-5}$then there exists

$a=u+v\sqrt{-5}$ such that $2\in I$.So,

$\begin{array}{c}2=(r+s\sqrt{-5})(u+v\sqrt{-5})\\ =(ru-5sv)+(rv+su)r\sqrt{-5}\\ =m+n\sqrt{-5}\end{array}$

Where$m=2$ and$n=0$ .

It is given$(\mathrm{r}+\mathrm{s}\sqrt{-5})\in \mathrm{\mathbb{Z}}\sqrt{-5}$ with$s\ne 0$ . So, $n$cannot be equal to zero. This is a contradiction to the assumption.

Therefore,$2$ is not in the principal ideal $r+s\sqrt{-5}$.