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Let $k\ge 2$ and $m_{1},…,m_{k} \in \mathbb{N}$ with $\gcd(m_{i},m_{j}) = 1$ for all $i\ne j$.

Show that $f(x) = (x,…,x)$ defines a ring homomorphism $f: \mathbb{Z}/m\mathbb{Z} \rightarrow \mathbb{Z}/m_{1}\mathbb{Z} \times … \times \mathbb{Z}/m_{k}\mathbb{Z}$ with $m=m_{1}\cdot \cdot \cdot m_{k}$

I am stuck since I don't really see where and how to begin. Therefore I am very thankful for any hints in the right direction.

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    Ring *homomorphism* should not be difficult. You need to show that the mapping is independent of choice of representative, though the wording of the question almost assumes that. Then you need the preservation of addition, multiplication, unit, which should be easy. The mapping is in fact an *isomorphism*. For that part, the Chinese Remainder Theorem is useful. But maybe the question does not ask for proof we have an isomorphism. You should find out, because that's the harder part.2011-10-26
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    Andre is right. Showing this is a homomorphism is quite simple and does not use the hypothesis that $m_i$'s are pairwise relatively prime. You should definitly check and make sure you did not misread the question.2011-10-26
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    Homomorphism: $f(x+y)=(x+y,x+y,…,x+y) = (x,…,x)+(y,…,y)=f(x)+f(y)$ and $f(xy)= (xy,…,xy) = (x,…,x)(y,…,y) = f(x)f(y)$ so homomorphism is shown. I am confused about the Ring property and about how to show isomorphism. Thanks for your efforts2011-10-27

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