Question

Let $\mathbf{R}\mathbf{=}\mathbf{\{}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{i}\mathbf{:}\mathbf{a}\mathbf{,}\mathbf{b}\mathbf{Î}\mathbf{Z}\mathbf{\}}$ is a sub ring of $\mathbf{\mathbb{Q}}$. Show that $\mathit{J}$ is not a maximal ideal in $\mathit{R}$, where $\mathit{J}\mathbf{=}\mathbf{\left\{}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{i}\mathbf{:}\mathbf{5}\mathbf{\left|}\mathbf{a}\mathbf{\text{\hspace{0.17em}and\hspace{0.17em}}}\mathbf{5}\mathbf{\right|}\mathbf{b}\mathbf{\right\}}$.

Short answer

It has been proved that$J$ is not an ideal in $R$.

Step-by-step solution

Maximal Ideal

An ideal $\mathit{M}$ in a ring $\mathit{R}$ is said to be maximal if $\mathit{M}\mathbf{\ne }\mathit{R}$ and whenever $\mathit{J}$ is an ideal such that $\mathit{M}\mathbf{\subseteq }\mathit{J}\mathbf{\subseteq }\mathit{R}$ then, $\mathit{M}\mathbf{=}\mathit{J}$, or $\mathit{J}\mathbf{=}\mathit{R}$ .

Conclusion

Let us consider an ideal, $K=2+i$ in $R$, then $(2+i)(2-i)=5$.

Then we have,$5\in K$, hence, $J=\left(5\right)\subseteq K$.

By the definition of role="math" localid="1654246403816" $J$ given in the question, we must have, role="math" localid="1654246392241" $5|a$and role="math" localid="1654246395574" $5|b$ for role="math" localid="1654246398699" $a+bi\in J$.

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

So, we can conclude that $J$ is not a maximal ideal.