Question

Let$\mathit{R}\mathbf{=}\mathbf{\{}\mathit{a}\mathbf{+}\mathit{b}\mathit{i}\mathbf{:}\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{Z}\mathbf{\}}$is a sub ring of $\mathit{Q}$ . Show that$\mathit{J}$is not a maximal ideal in$\mathit{R}$,where$\mathit{J}\mathbf{=}\mathit{\{}\mathit{a}\mathbf{+}\mathit{b}\mathit{i}\mathbf{:}\mathbf{5}\mathbf{|}\mathit{a}$and$\mathbf{5}\mathbf{\left|}\mathit{b}\mathbf{\right\}}$[Hint: Consider the principal ideallocalid="1657346860454" $\mathit{K}\mathbf{=}\mathbf{2}\mathbf{+}\mathit{i}$in$\mathit{R}$]

Short answer

It has been proved that $\mathit{J}$is not an ideal in $\mathit{R}$

Step-by-step solution

Maximal Ideal

An ideal$\mathit{M}$in a ringlocalid="1657346611712" $\mathit{R}$is said to be maximal iflocalid="1657344941402" $\mathit{M}\mathbf{\ne }\mathit{R}$and whenever$\mathit{J}$is an ideal

suchlocalid="1657345075315" $\mathit{M}\mathbf{\subseteq }\mathit{J}\mathbf{\subseteq }\mathit{R}$then,$\mathit{M}\mathbf{=}\mathit{J}$orlocalid="1657345204049" $\mathit{J}\mathbf{=}\mathit{R}$.

Conclusion

Let us consider an ideal,$\mathit{K}\mathbf{=}\mathbf{2}\mathbf{+}\mathit{i}$in$\mathit{R}$,then$\mathbf{(}\mathbf{2}\mathbf{+}\mathit{i}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{-}\mathit{i}\mathbf{)}\mathbf{=}\mathbf{5}$

Then we have,$\mathbf{5}\mathbf{\in }\mathit{K}$,hence role="math" localid="1657346051197" $\mathit{J}\mathbf{=}\mathbf{\left(}\mathbf{5}\mathbf{\right)}\mathbf{\subseteq }\mathit{K}\mathbf{.}$

By the definition of $\mathit{J}$given in the question, we must have,$\mathbf{5}\mathbf{|}\mathit{a}$and $\mathbf{5}\mathbf{|}\mathit{b}$for$\mathit{a}\mathbf{+}\mathit{b}\mathit{i}\mathbf{\in }\mathit{J}$

But, $\mathbf{5}\mathbf{\nmid }\mathbf{2}\mathbf{}$and$\mathbf{2}\mathbf{+}\mathit{i}$is not a unit in $\mathit{K}$so, we can say that $\mathit{K}$Is a proper ideal which contains$\mathit{J}$that is,$\mathit{J}\mathbf{\subseteq }\mathit{K}$

So, we can conclude that $\mathit{J}$is not a maximal ideal.