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I try to prove that theorem.

Let $a,b\in\mathbb N$, $\gcd(a,b)=1$. Then there exist isomorphism between the rings $\mathbb Z_{a} \times \mathbb Z_{b}$ and $\mathbb Z_{ab}$.

My proof:

Let $f:\mathbb Z_{ab} \to \mathbb Z_{a} \times \mathbb Z_{b}$, $x\to (x \bmod a; x \bmod b)$. Then $f$ is a bijection because of Chinese theorem. Prove that $f$ is homomorphism: $f(x)$+$f(y)$=(x mod a; x mod b) + (y mod a; y mod b)=(x+y mod a; x+y mod b)= $f$(x+y mod ab) for x,y elements $\mathbb Z_{ab}$. So $f$ is homomorphism because of definition of homomorphism between rings so finally $\mathbb Z_{ab} $, $ \mathbb Z_{a} \times \mathbb Z_{b}$ are isomorphic.

Do you think it is correct ?

user26857
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joeblack
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  • It looks correct, but I would expect on like a homework assignment that you would have to give more details besides "by chinese remainder theorem." – D_S Mar 14 '16 at 05:11
  • Yes, of course. $f$ is surjectinc function because the range and the domain are equal. It is injective : let x,y from {0,...,ab}. Then a,b|x-y so ab|x-y but |x-y|<ab so x-y=0. Do you agree ? – joeblack Mar 14 '16 at 22:45
  • "$f$ is a surjective function between the range and domain are equal" wouldn't be good enough; you didn't prove anything, you just stated what it means to be surjective. – D_S Mar 15 '16 at 03:48
  • Your argument for injectivity didn't make any sense to me. – D_S Mar 15 '16 at 03:49

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