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

Question

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

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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.

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It would be very nice to hear from you, if my proof is correct.

Best wishes

math12

Answer 1

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.