Question

Let $\mathit{R}\mathbf{=}\mathbf{\{}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{i}\mathbf{:}\mathbf{a}\mathbf{,}\mathbf{b}\mathbf{\in }\mathbf{\mathbb{Z}}\mathbf{\}}$ is a sub ring of $\mathbf{\mathbb{Q}}$, and $\mathit{K}\mathbf{=}\mathbf{2}\mathbf{+}\mathit{i}$be the principal ideal, then show that $\mathit{R}\mathbf{/}\mathit{K}\mathbf{\cong }{\mathit{\mathbb{Z}}}_{\mathbf{5}}$

Short answer

It is proved that$R/K\cong {\mathbb{Z}}_5$.

Step-by-step solution

Homomorphism of rings

Let $\mathit{R}$and ${\mathit{R}}^{\mathbf{\text{'}}}$be two rings. A mapping $\mathit{f}\mathbf{:}\mathit{R}\mathbf{\to }{\mathit{R}}^{\mathbf{\text{'}}}$is called an homomorphism of rings if,

• $\mathit{f}\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{)}\mathbf{=}\mathit{f}\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{+}\mathit{f}\mathbf{\left(}\mathbf{b}\mathbf{\right)}$
• $\mathit{f}\mathbf{\left(}\mathbf{a}\mathbf{b}\mathbf{\right)}\mathbf{=}\mathit{f}\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathit{f}\mathbf{\left(}\mathbf{b}\mathbf{\right)}$for all $\mathbf{a}\mathbf{,}\mathbf{b}\mathbf{\in }\mathbf{R}$

Proof

Let us consider a mapping $\phi :R\to {\mathbb{Z}}_5$. defined by $\phi (a+bi)=[a+3b]$

We will prove that $\phi$ is a homomorphism. Let $(a+bi),(c+di)\in R$then,

$$\begin{gathered} \phi ((a+bi)+(c+di))=\phi ((a+c)+(b+d)i) \\ \begin{array}{l}=[a+c+3b+3d]\\ =[a+3b]+[c+3d]\\ =\phi (a+bi)+\phi (c+di)\end{array} \end{gathered}$$

$\begin{array}{l}\phi \left((a+bi)(c+di)\right)=\phi ((ac-bd)+(ad+bc)i)\\ =[ac-bd+3ad+3bc]\\ =[ac+9bd+3ad+3bc]\\ =[a+3b][c+3d]\\ =\phi (a+bi)\phi (c+di)\end{array}$

Also,$\left[1\right]$generates $\mathrm{\mathbb{Z}}$so $\phi$represents a surjective homomorphism.

Now, $\phi (2+i)=[2+3]=\left[0\right]$implies that $K\subseteq \ker \phi$.

Also, note that there exists no $(a+bi),(c+di)\in R$such that,

$(a+bi)(c+di)=2+i$, so$K$ is the maximal ideal and hence, $K=\ker \phi$

So, by first theorem of isomorphism,$R/K\cong {\mathbb{Z}}_5$