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?
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3Chinese Remainder Theorem – May 15 '14 at 14:03
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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 $f(a)=(t,s)$.
Kal S.
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How $f(a)=(s,t)$, I mean how $f(a)=s$ in $\mathbb Z_n$? Isn't it $t$? – Achak0790 Feb 21 '20 at 02:34
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2@Achak0790 $f(a)=(tym \bmod{n}, sxn\bmod{m})=(t\bmod{n},s\bmod{m})$ since $xn\equiv 1\bmod{m},ym\equiv 1\bmod{n}$ – Kal S. Feb 22 '20 at 10:30
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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...
Fredrik Meyer
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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? – The Coding Wombat Oct 31 '18 at 18:19
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@TheCodingWombat The notation $\mathbb Z_6$ is common for the quotient ring $\mathbb Z/(6)$. So it consists of residue classes ${ 0, 1, 2, 3, 4, 5}$. – Fredrik Meyer Nov 01 '18 at 11:07
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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.
Pedro
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