How to show that $\mathbb Z_{mn}$ is isomorphic to $\mathbb Z_m \times\mathbb Z_n$ when $m$ and $n$ are coprime? It is easy to show that the natural map from $\mathbb Z_{mn}$ to $\mathbb Z_m \times\mathbb Z_n$ is a ring homomorphism. How to show that it is bijective?

  • 2
    $\begingroup$ Chinese Remainder Theorem $\endgroup$ – user14972 May 15 '14 at 14:03

You define $f:\mathbb{Z}\rightarrow \mathbb{Z}/n\mathbb{Z}\times\mathbb{Z}/m\mathbb{Z}$ by $f(a)=(a\bmod{n},a\bmod{m})$ which is a ring homomorphism.

Then you can verify that $\ker{f}=mn\mathbb{Z}$ and to show that $f$ surjective:

$\gcd{(m,n)}=1$ so there exists $x,y$ such that $xn+ym=1$, so for $(s,t)\in\mathbb{Z}/n\mathbb{Z}\times\mathbb{Z}/m\mathbb{Z}$ consider $a=sxn+tym$. Then since: $$ xn\equiv 1\bmod{m}\text{ and } ym\equiv 1\bmod{n} $$ you have $f(a)=(s,t)$.

  • $\begingroup$ How $f(a)=(s,t)$, I mean how $f(a)=s$ in $\mathbb Z_n$? Isn't it $t$? $\endgroup$ – Achak0790 2 days ago
  • $\begingroup$ @Achak0790 $f(a)=(tym \bmod{n}, sxn\bmod{m})=(t\bmod{n},s\bmod{m})$ since $xn\equiv 1\bmod{m},ym\equiv 1\bmod{n}$ $\endgroup$ – Kal S. yesterday

The two rings have the same cardinality, so it is enough to show that the homomorphism is injective.

Now, what is the kernel? The kernel consists precisely of those $a \in \mathbb Z_{mn}$ such that $a \equiv 0 \pmod{n}$ and $a \equiv 0 \pmod{m}$. The gcd is 1, so...

  • $\begingroup$ Does $\mathbb{Z}_6$ mean the set $\{0, 6, -6, 12, -12, ...\}$ and are you using that the kernel of f only contains the number $0$, and that that implies that the function is injective? $\endgroup$ – The Coding Wombat Oct 31 '18 at 18:19
  • $\begingroup$ @TheCodingWombat The notation $\mathbb Z_6$ is common for the quotient ring $\mathbb Z/(6)$. So it consists of residue classes $\{ 0, 1, 2, 3, 4, 5\}$. $\endgroup$ – Fredrik Meyer Nov 1 '18 at 11:07

The element $(1,1)$ in the direct product has order $mn$ (why?). Then you know the direct product is cyclic, and a cyclic group is uniquely determined (up to isomorphism) by its order.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.