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$\mathit{Q}$, and$\mathit{K}\mathbf{=}\mathbf{2}\mathbf{+}\mathit{i}$be the principal ideal, then show thatlocalid="1657347597760" $\mathit{R}\mathbf{/}\mathit{K}\mathbf{\cong }{\mathit{Z}}_{\mathbf{5}}$

Short answer

It is proved that$\mathit{R}\mathbf{/}\mathit{K}\mathbf{\cong }{\mathit{Z}}_{\mathbf{5}}$

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,

• localid="1657352406725" $\mathit{f}\mathbf{(}\mathit{a}\mathbf{+}\mathit{b}\mathbf{)}\mathbf{=}\mathit{f}\mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{+}\mathit{f}\mathbf{\left(}\mathit{b}\mathbf{\right)}$
• localid="1657353660787" $\mathit{f}\mathbf{\left(}\mathit{a}\mathit{b}\mathbf{\right)}\mathbf{=}\mathit{f}\mathbf{\left(}\mathit{a}\mathbf{\right)}\mathit{f}\mathbf{\left(}\mathit{b}\mathbf{\right)}$for all,localid="1657353669335" $\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{R}$

Proof

Let us consider a mapping.$\mathit{\phi }\mathbf{:}\mathit{R}\mathbf{\to }{\mathit{Z}}_{\mathbf{5}}$defined by $\mathit{\phi }\mathbf{(}\mathit{a}\mathbf{+}\mathit{b}\mathit{i}\mathbf{)}\mathbf{=}\mathbf{[}\mathit{a}\mathbf{+}\mathbf{3}\mathit{b}\mathbf{]}$

We will prove that $\mathit{\phi }$is a homomorphism. Let $\mathbf{(}\mathit{a}\mathbf{+}\mathit{b}\mathit{i}\mathbf{)}\mathbf{,}\mathbf{(}\mathit{c}\mathbf{+}\mathit{d}\mathit{i}\mathbf{)}\mathbf{\in }\mathit{R}$then,

$\mathit{\phi }\mathbf{\left(}\mathbf{\right(}\mathit{a}\mathbf{+}\mathit{b}\mathit{i}\mathbf{)}\mathbf{+}\mathbf{(}\mathit{c}\mathbf{+}\mathit{d}\mathit{i}\mathbf{\left)}\mathbf{\right)}\mathbf{=}\mathit{\phi }\mathbf{\left(}\mathbf{\right(}\mathit{a}\mathbf{+}\mathit{c}\mathbf{)}\mathbf{+}\mathbf{(}\mathit{b}\mathbf{+}\mathit{d}\mathbf{\left)}\mathit{i}\mathbf{\right)}$

$\mathbf{=}\mathbf{[}\mathit{a}\mathbf{+}\mathit{c}\mathbf{+}\mathbf{3}\mathit{b}\mathbf{+}\mathbf{3}\mathit{d}\mathbf{]}$

localid="1657352854458" $\mathbf{=}\mathbf{[}\mathit{a}\mathbf{+}\mathbf{3}\mathit{b}\mathbf{]}\mathbf{+}\mathbf{[}\mathit{c}\mathbf{+}\mathbf{3}\mathit{d}\mathbf{]}$

$\mathbf{=}\mathit{\phi }\mathbf{(}\mathit{a}\mathbf{+}\mathit{b}\mathit{i}\mathbf{)}\mathbf{+}\mathit{\phi }\mathbf{(}\mathit{c}\mathbf{+}\mathit{d}\mathit{i}\mathbf{)}$

$\mathit{\phi }\mathbf{\left(}\mathbf{\right(}\mathit{a}\mathbf{+}\mathit{b}\mathit{i}\mathbf{\left)}\mathbf{\right(}\mathit{c}\mathbf{+}\mathit{d}\mathit{i}\mathbf{\left)}\mathbf{\right)}\mathbf{=}\mathit{\phi }\mathbf{\left(}\mathbf{\right(}\mathit{a}\mathit{c}\mathbf{-}\mathit{b}\mathit{d}\mathbf{)}\mathbf{+}\mathbf{(}\mathit{a}\mathit{d}\mathbf{+}\mathit{b}\mathit{c}\mathbf{\left)}\mathit{i}\right)$

$\mathbf{=}\mathbf{[}\mathit{a}\mathit{c}\mathbf{-}\mathit{b}\mathit{d}\mathbf{+}\mathbf{3}\mathit{a}\mathit{d}\mathbf{+}\mathbf{3}\mathit{b}\mathit{c}\mathbf{]}$

$\mathbf{=}\mathbf{[}\mathit{a}\mathit{c}\mathbf{+}\mathbf{9}\mathit{b}\mathit{d}\mathbf{+}\mathbf{3}\mathit{a}\mathit{d}\mathbf{+}\mathbf{3}\mathit{b}\mathit{c}\mathbf{]}$

$\mathbf{=}\mathbf{[}\mathit{a}\mathbf{+}\mathbf{3}\mathit{b}\mathbf{]}\mathbf{[}\mathit{c}\mathbf{+}\mathbf{3}\mathit{d}\mathbf{]}$

$\mathbf{=}\mathit{\phi }\mathbf{(}\mathit{a}\mathbf{+}\mathit{b}\mathit{i}\mathbf{)}\mathit{\phi }\mathbf{(}\mathit{c}\mathbf{+}\mathit{d}\mathit{i}\mathbf{)}$

Also,$\mathbf{\left[}\mathbf{1}\mathbf{\right]}$generates${\mathit{Z}}_{\mathbf{5}}$so $\mathit{\phi }$represents a surjective homomorphism.

Now,$\mathit{\phi }\mathbf{(}\mathbf{2}\mathbf{+}\mathit{i}\mathbf{)}\mathbf{=}\mathbf{[}\mathbf{2}\mathbf{+}\mathbf{3}\mathbf{]}\mathbf{=}\mathbf{\left[}\mathbf{0}\mathbf{\right]}$implies that $\mathit{K}\mathbf{\subseteq }\mathit{k}\mathit{e}\mathit{r}\mathbf{\hspace{0.17em}}\mathit{\phi }$

Also, note that there exists no $\mathbf{(}\mathit{a}\mathbf{+}\mathit{b}\mathit{i}\mathbf{)}\mathbf{,}\mathbf{(}\mathit{c}\mathbf{+}\mathit{d}\mathit{i}\mathbf{)}\mathbf{\in }\mathit{R}$such that,

$\mathbf{(}\mathit{a}\mathbf{+}\mathit{b}\mathit{i}\mathbf{)}\mathbf{(}\mathit{c}\mathbf{+}\mathit{d}\mathit{i}\mathbf{)}\mathbf{=}\mathbf{2}\mathbf{+}\mathit{i}$,so$\mathit{K}$is the maximal ideal and hence, $\mathit{K}\mathbf{=}\mathit{k}\mathit{e}\mathit{r}\mathbf{\hspace{0.17em}}\mathit{\phi }$

So, by first theorem of isomorphism, $\mathit{R}\mathbf{/}\mathit{K}\mathbf{\cong }{\mathit{Z}}_{\mathbf{5}}$