Question

Question: (a) Show that is ${\mathit{Z}}_{\mathbf{5}}\mathbf{\times }{\mathit{Z}}_{\mathbf{12}}$isomorphic to${\mathit{Z}}_{\mathbf{3}}\mathbf{\times }{\mathit{Z}}_{\mathbf{20}}$ ?

(b) Is ${\mathit{Z}}_{\mathbf{4}}\mathbf{\times }{\mathit{Z}}_{\mathbf{35}}$ isomorphic to ${\mathit{Z}}_{\mathbf{5}}\mathbf{\times }{\mathit{Z}}_{28}$ ?

Short answer

Answer: -

1. The value$Z_5\times Z_{12}$ isomorphic to$Z_3\times Z_{20}$.
2. The value $Z_5\times Z_{28}$ isomorphic to$Z_4\times Z_{35}$ .

Step-by-step solution

Conceptual Introduction

A theorem known as the Chinese remainder theorem provides a singular answer to simultaneous linear congruences with coprime moduli.

Are the values isomorphic?

a. Consider the first ring:

$Z_5\times Z_{12}$, the element $Z_{12}$ is the multiple of $Z_{3\times 4}$ in the ring$Z_3\times Z_{20}$ .

Also, the element $Z_{20}$ is the multiple of $Z_{5\times 4}$, as the Least common multiple (LCM) in between the rings is4 ,the rings $Z_5\times Z_{12}$ and$Z_3\times Z_{20}$ are isomorphic.

Therefore, the $GCD\left(m,n\right)=1$, the ring is isomorphic.

Hence, $Z_5\times Z_{12}$isomorphic to$Z_3\times Z_{20}$ .

Is the value isomorphic?

b. Consider the first ring:

$Z_4\times Z_{35}$, the element $Z_4$ is the multiple of $Z_{7\times 4}$, which is $Z_{28}$ in the ring $Z_5\times Z_{28}$, and the element $Z_5$is the multiple of $Z_{5\times 7}$, which is $Z_{35}$, in the ring $Z_4\times Z_{35}$

It is as the Least common multiple (LCM) in between the rings is 7 , the rings $Z_4\times Z_{35}$and $Z_5\times Z_{28}$are isomorphic.

Therefore, the $GCD\left(m,n\right)=1$, the ring is isomorphic.

Hence, $Z_5\times Z_{28}$ isomorphic to $Z_4\times Z_{35}$.