Question

If $\left(m,n\right)\mathbf{=}\mathbf{1}$, prove that ${\mathbf{U}}_{\mathbf{mn}}\mathbf{\cong }{\mathbf{U}}_{\mathbf{mn}}\mathbf{}\mathbf{x}\mathbf{}{\mathbf{U}}_{\mathbf{n}}\mathbf{.}$

Short answer

It has been proved that,${\mathrm{U}}_{\mathrm{mn}}\cong {\mathrm{U}}_{\mathrm{m}}\mathrm{x}{\mathrm{U}}_{\mathrm{n}}$.

Step-by-step solution

Define a map

$\mathrm{Define}\mathrm{a}\mathrm{map}\mathrm{f}:{\mathrm{z}}_{\mathrm{mn}}\to {\mathrm{z}}_{\mathrm{m}}\mathrm{x}{\mathrm{z}}_{\mathrm{n}}\mathrm{by}\mathrm{f}\left({\left[\mathrm{a}\right]}_{\mathrm{mn}}\right)=\left({\left[\mathrm{a}\right]}_{\mathrm{m}},{\left[\mathrm{a}\right]}_{\mathrm{n}}\right).$

It is clear that, $\mathrm{f}$is a ring homomorphism.

Given that,$\left(\mathrm{m},\mathrm{n}\right)=1.$

Prove f is an isomorphism

Since $\left(\mathrm{m},\mathrm{n}\right)$= 1 , then kernel is {0}.

Therefore, $\mathrm{f}$is injective.

Since the orders of these rings are equal and it is an homomorphism of rings, $\mathrm{f}$is an isomorphism.

Therefore,${\mathrm{z}}_{\mathrm{mn}}\cong {\mathrm{z}}_{\mathrm{m}}\mathrm{x}{\mathrm{z}}_{\mathrm{n}}$.

Prove Umn≅Um x Un

Since${\mathrm{z}}_{\mathrm{mn}}\cong {\mathrm{z}}_{\mathrm{m}}\mathrm{x}{\mathrm{z}}_{\mathrm{n}}$ and isomorphic rings have isomorphic unit groups,${\mathrm{U}}_{\mathrm{mn}}\cong {\mathrm{U}}_{\mathrm{m}}\mathrm{x}{\mathrm{U}}_{\mathrm{n}}$ .

Therefore, ${\mathrm{U}}_{\mathrm{m}}\cong {\mathrm{U}}_{\mathrm{m}}\mathrm{x}{\mathrm{U}}_{\mathrm{n}}$.