Question

Let$\mathit{R}\mathbf{=}\mathit{a}\mathbf{+}\mathit{b}\mathit{i}\mathbf{:}\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{Z}$is asub ring of $\mathit{Q}$, and$\mathit{J}\mathbf{=}\mathbf{\{}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{i}\mathbf{:}\mathbf{5}\mathbf{|}\mathbf{a}\mathbf{\hspace{0.33em}}\mathbf{a}\mathbf{n}\mathbf{d}\mathbf{\hspace{0.33em}}\mathbf{5}\mathbf{|}\mathbf{b}\mathbf{\}}$show that . $\mathit{R}\mathbf{/}\mathit{J}\mathbf{\cong }{\mathit{Z}}_{\mathbf{5}}\mathbf{\times }{\mathit{Z}}_{\mathbf{5}}\mathbf{.}$

Short answer

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

Step-by-step solution

First theorem of Isomorphism

Let$\mathit{f}\mathbf{:}\mathit{R}\mathbf{\to }\mathit{S}$be a surjective homomorphism of rings with kernel$\mathit{K}$. Then, the quotient ring $\mathit{R}\mathbf{/}\mathit{K}\mathbf{}$is isomorphic to$\mathit{S}$

Check for surjective

Let us consider a mapping $\mathit{\chi }\mathbf{:}\mathit{R}\mathbf{\to }{\mathit{Z}}_{\mathbf{5}}\mathbf{\times }{\mathit{Z}}_{\mathbf{5}}$defined by $\mathit{\chi }\mathbf{(}\mathit{a}\mathbf{+}\mathit{b}\mathit{i}\mathbf{)}\mathbf{=}\mathbf{\left(}\mathbf{\right[}\mathit{a}\mathbf{+}\mathbf{2}\mathit{b}\mathbf{]}\mathbf{,}\mathbf{[}\mathit{a}\mathbf{+}\mathbf{3}\mathit{b}\mathbf{\left]}\mathbf{\right)}$

Now, we will show that it is a homomorphism. Let$\mathbf{(}\mathit{a}\mathbf{+}\mathit{b}\mathit{i}\mathbf{)}\mathbf{,}\mathbf{(}\mathit{c}\mathbf{+}\mathit{d}\mathit{i}\mathbf{)}\mathbf{\in }\mathit{R}\mathbf{}$

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

$\mathbf{=}\mathit{\chi }\mathbf{\left(}\mathbf{\right(}\mathit{a}\mathbf{+}\mathit{c}\mathbf{)}\mathbf{+}\mathbf{(}\mathit{b}\mathbf{+}\mathit{d}\mathbf{\left)}\mathit{i}\mathbf{\right)}\mathbf{=}\mathbf{\left(}\mathbf{\right[}\mathit{a}\mathbf{+}\mathit{c}\mathbf{+}\mathbf{2}\mathit{b}\mathbf{+}\mathbf{2}\mathit{d}\mathbf{]}\mathbf{,}\mathbf{[}\mathit{a}\mathbf{+}\mathit{c}\mathbf{+}\mathbf{3}\mathit{b}\mathbf{+}\mathbf{3}\mathit{d}\mathbf{\left]}\mathbf{\right)}$

$\mathbf{=}\mathbf{\left(}\mathbf{\right[}\mathit{a}\mathbf{+}\mathbf{2}\mathit{b}\mathbf{]}\mathbf{,}\mathbf{[}\mathit{a}\mathbf{+}\mathbf{3}\mathit{b}\mathbf{\left]}\mathbf{\right)}\mathbf{+}\mathbf{\left(}\mathbf{\right[}\mathit{c}\mathbf{+}\mathbf{2}\mathit{d}\mathbf{]}\mathbf{,}\mathbf{[}\mathit{c}\mathbf{+}\mathbf{3}\mathit{d}\mathbf{\left]}\mathbf{\right)}$

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

And,

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

$\mathbf{=}\mathbf{\left(}\mathbf{\right[}\mathit{a}\mathit{c}\mathbf{-}\mathit{b}\mathit{d}\mathbf{+}\mathbf{2}\mathit{a}\mathit{d}\mathbf{+}\mathbf{2}\mathit{b}\mathit{c}\mathbf{]}\mathbf{,}\mathbf{[}\mathit{a}\mathit{c}\mathbf{-}\mathit{b}\mathit{d}\mathbf{+}\mathbf{3}\mathit{a}\mathit{d}\mathbf{+}\mathbf{3}\mathit{b}\mathit{c}\mathbf{\left]}\mathbf{\right)}$

$\mathbf{=}\mathbf{\left(}\mathbf{\right[}\mathit{a}\mathit{c}\mathbf{+}\mathbf{4}\mathit{b}\mathit{d}\mathbf{+}\mathbf{2}\mathit{a}\mathit{d}\mathbf{+}\mathbf{2}\mathit{b}\mathit{c}\mathbf{]}\mathbf{,}\mathbf{[}\mathit{a}\mathit{c}\mathbf{+}\mathbf{9}\mathit{b}\mathit{d}\mathbf{+}\mathbf{3}\mathit{a}\mathit{d}\mathbf{+}\mathbf{3}\mathit{b}\mathit{c}\mathbf{\left]}\mathbf{\right)}$

$\mathbf{=}\mathbf{\left(}\mathbf{\right[}\mathit{a}\mathbf{+}\mathbf{2}\mathit{b}\mathbf{]}\mathbf{,}\mathbf{[}\mathit{a}\mathbf{+}\mathbf{3}\mathit{b}\mathbf{\left]}\mathbf{\right)}\mathbf{\left(}\mathbf{\right[}\mathit{c}\mathbf{+}\mathbf{2}\mathit{d}\mathbf{]}\mathbf{,}\mathbf{[}\mathit{c}\mathbf{+}\mathbf{3}\mathit{d}\mathbf{\left]}\mathbf{\right)}$

role="math" localid="1657337825893" $\mathbf{=}\mathit{\chi }\mathbf{(}\mathit{a}\mathbf{+}\mathit{b}\mathit{i}\mathbf{)}\mathit{\chi }\mathbf{(}\mathit{c}\mathbf{+}\mathit{d}\mathit{i}\mathbf{)}$

(Since, $\mathbf{-}\mathbf{1}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{+}\mathbf{5}\mathbf{=}\mathbf{4}$and $\mathbf{-}\mathbf{1}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{+}\mathbf{2}\mathbf{\times }\mathbf{5}\mathbf{=}\mathbf{9}\mathit{i}\mathit{n}{\mathit{Z}}_{\mathbf{5}}\mathbf{)}$

Also, $\mathbf{\left(}\mathbf{\right[}\mathbf{1}\mathbf{]}\mathbf{,}\mathbf{[}\mathbf{1}\mathbf{\left]}\mathbf{\right)}\mathbf{}$and $\mathbf{\left(}\mathbf{\right[}\mathbf{2}\mathbf{]}\mathbf{,}\mathbf{[}\mathbf{3}\mathbf{\left]}\mathbf{\right)}\mathbf{}$generate${\mathit{Z}}_{\mathbf{5}}\mathbf{\times }{\mathit{Z}}_{\mathbf{5}}$.So, we can conclude the homomorphism $\mathit{\chi }\mathbf{}$is surjective.

Conclusion

Since, if $\mathbf{(}\mathit{a}\mathbf{+}\mathit{b}\mathit{i}\mathbf{)}\mathbf{\in }\mathit{J}$then, we have $\mathbf{5}\mathbf{|}\mathit{a}\mathbf{\hspace{0.33em}}\mathit{a}\mathit{n}\mathit{d}\mathbf{\hspace{0.33em}}\mathbf{5}\mathbf{|}\mathit{b}$which implies

$\mathbf{5}\mathbf{|}\mathit{a}\mathbf{+}\mathbf{2}\mathit{b}\mathbf{\hspace{0.33em}}\mathit{a}\mathit{n}\mathit{d}\mathbf{\hspace{0.33em}}\mathbf{5}\mathbf{|}\mathit{a}\mathbf{+}\mathbf{3}\mathit{b}\mathbf{}$

localid="1657340302233" role="math" $\mathbf{⇨}\mathbf{(}\mathit{a}\mathbf{+}\mathit{b}\mathit{i}\mathbf{)}\mathbf{\in }\mathit{J}$and hence , $\mathit{J}\mathbf{\subseteq }\mathit{k}\mathit{e}\mathit{r}\mathit{\chi }$

Conversely, if $\mathbf{(}\mathit{a}\mathbf{+}\mathit{b}\mathit{i}\mathbf{)}\mathbf{\in }\mathit{k}\mathit{e}\mathit{r}\mathit{\chi }$then $\mathbf{5}\mathbf{|}\mathit{a}\mathbf{+}\mathbf{2}\mathit{b}\mathbf{}$and $\mathbf{5}\mathbf{|}\mathit{a}\mathbf{+}\mathbf{3}\mathit{b}$. Which implies,

$\mathbf{5}\mathbf{\left|}\mathbf{\right(}\mathit{a}\mathbf{+}\mathbf{3}\mathit{b}\mathbf{)}\mathbf{-}\mathbf{(}\mathit{a}\mathbf{+}\mathbf{2}\mathit{b}\mathbf{)}\mathbf{\Rightarrow }\mathbf{5}\mathbf{|}\mathit{b}\mathbf{}$and, $\mathbf{5}\mathbf{\left|}\mathbf{3}\mathbf{\right(}\mathit{a}\mathbf{+}\mathbf{2}\mathit{b}\mathbf{)}\mathbf{-}\mathbf{2}\mathbf{(}\mathit{a}\mathbf{+}\mathbf{3}\mathit{b}\mathbf{)}\mathbf{\Rightarrow }\mathbf{5}\mathbf{|}\mathit{a}\mathbf{}$

Hence, $\mathbf{(}\mathit{a}\mathbf{+}\mathit{b}\mathit{i}\mathbf{)}\mathbf{\in }\mathit{J}$Which shows that $\mathit{k}\mathit{e}\mathit{r}\mathbf{\hspace{0.17em}}\mathit{\chi }\mathbf{\subseteq }\mathit{J}$

So we can conclude that,$\mathit{J}\mathbf{=}\mathit{k}\mathit{e}\mathit{r}\mathit{\chi }$

Then, by using the first theorem of isomorphism, we came to the conclusion

$\mathit{R}\mathbf{/}\mathit{J}\mathbf{\cong }{\mathit{Z}}_{\mathbf{5}}\mathbf{\times }{\mathit{Z}}_{\mathbf{5}}$

$\mathbf{5}\mathbf{\left|}\mathbf{3}\mathbf{\right(}\mathit{a}\mathbf{+}\mathbf{2}\mathit{b}\mathbf{)}\mathbf{-}\mathbf{2}\mathbf{(}\mathit{a}\mathbf{+}\mathbf{3}\mathit{b}\mathbf{)}\mathbf{\Rightarrow }\mathbf{5}\mathbf{|}\mathit{a}\mathbf{}$