Show: $\varphi\colon\mathbb{Z}_{mn}\to\mathbb{Z}_m\times\mathbb{Z}_n, k\mapsto (k\% m,k\% n)$ is a ring isomorphism for $m$ and $n$ relatively prim Let $m\in\mathbb{Z}, n\in\mathbb{N}$. Then there exist unique elements $q\in\mathbb{Z}, r\in\mathbb{N}$ with $0\leq r<n$ and $m=qn+r$. We write $r:=m\% n$.

Let $m,n\in\mathbb{N}$ be relatively prim. Show by explicit calculation that 
  $$
\varphi\colon\mathbb{Z}_{mn}\to\mathbb{Z}_m\times\mathbb{Z}_n,~~~ k\mapsto (k\% m,k\% n)
$$ is a ring isomorphism


First some notation stuff: I have two rings, namely $(\mathbb{Z}_{mn},\oplus,\odot)$ and $(\mathbb{Z}_m\times\mathbb{Z}_n,\boxplus,\boxdot)$.
I have two show two things: (i) $\varphi$ is a ring homomorphism, i.e.
$$
\varphi(x+mn\mathbb{Z}\oplus y+mn\mathbb{Z})=\varphi(x+mn\mathbb{Z})\boxplus\varphi(y+mn\mathbb{Z}),\\\varphi(x+mn\mathbb{Z}\odot y+mn\mathbb{Z})=\varphi(x+mn\mathbb{Z})\boxdot\varphi(y+mn\mathbb{Z})
$$
and (ii) that $\varphi$ is bijective.
Proof:
(i) 
$\begin{align}
\varphi(k+mn\mathbb{Z}\oplus l+mn\mathbb{Z})&=\varphi((k+l)\% mn+mn\mathbb{Z})\\
&=(((k+l)\%mn)\%m+m\mathbb{Z},((k+l)\%mn)\% n+n\mathbb{Z})\\
&=((k\% mn+l\% mn)\% m+m\mathbb{Z},(k\% mn+l\% mn)\% n+n\mathbb{Z})\\
&=((k\% mn)\% m+m\mathbb{Z}\oplus (l\% mn)\%m+m\mathbb{Z},(k\% mn)\%n+n\mathbb{Z}\oplus (l\% mn)\% n+n\mathbb{Z})\\
&=(k\% m+m\mathbb{Z}\oplus l\% m+m\mathbb{Z},k\% n+n\mathbb{Z}\oplus l\% n+n\mathbb{Z})\\
&=(k\% m+m\mathbb{Z},k\% n+n\mathbb{Z})\boxplus(l\% m+m\mathbb{Z},l\% n+n\mathbb{Z})\\
&=\varphi(k+mn\mathbb{Z})\boxplus\varphi(l+mn\mathbb{Z})
\end{align}$
(and analog for $\varphi(x+mn\mathbb{Z}\odot y+mn\mathbb{Z})=\varphi(x+mn\mathbb{Z})\boxdot\varphi(y+mn\mathbb{Z})$)
(ii)
Surjectivity is clear: Consider any $z:=(k\% m+m\mathbb{Z},k\% n+n\mathbb{Z})\in\mathbb{Z}_m\times\mathbb{Z}_n$, then $z=\varphi(k+mn\mathbb{Z})$.
Consider
$$
\mbox{ker}\varphi=\left\{k+mn\mathbb{Z}: (k\%m+m\mathbb{Z},k\% n+n\mathbb{Z})=(0+m\mathbb{Z},0+n\mathbb{Z})\right\}\\=\left\{k+mn\mathbb{Z}: \text{k is multiple of m and n}\right\}.
$$
Because $m$ and $n$ are relatively prim, $k$ can be a multiple of $m$ and a multiple of $n$ only if $k=0$. So
$$
\left\{k+mn\mathbb{Z}: \text{k is multiple of m and n}\right\}=\left\{0+mn\mathbb{Z}\right\}.
$$
So $\varphi$ is injective.

It would be very nice to hear from you, if my proof is correct.
Best wishes
math12
 A: Your proof of $\varphi$ respecting $\oplus$ and $\odot$ is correct but a bit cumbersome. Alternatively, you could use the theorem

If $\phi:R\to S$ is a ring homomorphism, and $I$ is an ideal in $R$ with $I\le\text{ker}(\phi)$, then there is a unique ring homomorphism $φ:R/I→S$ so that commutes
  $$\begin{array}{}
\ \ R & \large\longrightarrow & S\\
\small q\large\downarrow & \nearrow φ\\
R/I
\end{array}$$ where $q$ is the canonical ring epimorphism $r\mapsto r+I$.

Take $\phi:\Bbb Z→\Bbb Z_m×\Bbb Z_n,\phi(k)=(k\%m,k\%n)$ and 
$I=mn\Bbb Z$. Then 
$$\begin{eqnarray}
ϕ(k+l)& = &((k+l)\%m,(k+l)\%n) \\
& = &((k\%m+l\%m)\%m,(k\%n+l\%n)\%n)\\
&  = &ϕ (k)+ϕ(l)
\end{eqnarray}$$
Or you argue that $ϕ$ is just the composition
$\Bbb Z→\Bbb Z×\Bbb Z→\Bbb Z_m×\Bbb Z_n$, then you get the linearity almost for free. It follows that $φ(k+mn\Bbb Z)=(k\%m,k\%n)$ is a ring homomorphism.
You also showed successfully that $φ$ is injective. Since both $\Bbb Z_{mn}$ and $\Bbb Z_m×\Bbb Z_n$ have $mn$ elements, $φ$ must be bijective.
