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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. – André Nicolas Oct 26 '11 at 18:30
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. – Bill Cook Oct 26 '11 at 18:37
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 efforts – VVV Oct 27 '11 at 3:00

Here's an outline...

Show that $x=y$ mod $m$ if and only if $x=y$ mod $m_i$ for $i=1,\dots,k$.

This will show $x=y$ $\Longleftrightarrow$ $f(x)=f(y)$. The "$\Longrightarrow$" shows $f$ is well defined and "$\Longleftarrow$" shows $f$ is one-to-one.

Showing $f(x+y)=f(x)+f(y)$ should be pretty straightforward as well as $f(xy)=f(x)f(y)$ and $f(1)=(1,\dots,1)$. At this point you'll have established that $f$ is a one-to-one ring homomorphism.

The last step is to show $f$ is onto. For this you'll need to use Chinese remaindering: Suppose $x_i \in \mathbb{Z}/m_i\mathbb{Z}$ for each $i$. You need to find $x$ such that $x=x_i$ mod $m_i$ for each $i$. The hypothesis that the $m_i$'s are pairwise relatively prime guarantees that there is a solution (this is the Chinese remaindering theorem). Thus $f(x)=(x_1,\dots,x_k)$ and so $f$ is also onto.

Edit: I should have read the question more carefully! To show it's a ring homomorphism you just need to verify the homomorphism properties and establish $f$ is well defined.

For showing well defined: Suppose $x=y$ mod $m$. Then $x-y$ is divisible by $m$. You can conclude that $x-y$ is divisible by $m_i$ since each $m_i$ is a divisor of $m$. Thus $x=y$ mod $m_i$.

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The question doesn't actually ask us to prove that $f$ is one-to-one or onto, just that it is a ring homomorphism. – Chris Eagle Oct 26 '11 at 18:26
You will still need to prove the "$\Longrightarrow$" direction of the first statement to establish that $f$ is a (well defined) function. The homomorphism properties are easy: $f(x+y)=(x+y,\dots,x+y)=(x,\dots,x)+(y,\dots,y)=f(x)+f(y)$ etc. – Bill Cook Oct 26 '11 at 18:32

Using the universal property of a product, form the morphism $$ g:\mathbb{Z}\to\frac{\mathbb Z}{m_1\mathbb Z}\times\cdots\times\frac{\mathbb Z}{m_k\mathbb Z} $$ whose $i$ th component is the canonical projection.

Then check that $g$ factors through $\mathbb{Z}/m\mathbb{Z}$ by using the universal property of a quotient.

Variation. Using the universal property of a quotient, check that the canonical projection $\mathbb{Z}\to\mathbb Z/m_i\mathbb Z$ factors through $\mathbb{Z}/m\mathbb{Z}$, and conclude by using the universal property of a product.

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I have now shown the homomorphism properties of f:

additivity: $f(x+y)= (x+y,…,x+y) = (x,…,x)+(y,….,y) = f(x)+f(y)$

multiplicativity: $f(xy) = (xy,…,xy) = (x,…,x)(y,…,y)= f(x)f(y)$

So since $f$ is a homomorphism and $\mathbb{Z} / m \mathbb{Z}$ is a ring, it follows that $\mathbb{Z}/_{m1}\mathbb{Z} \times … \times \mathbb{Z} / m_{k} \mathbb{Z}$ is also a ring and therefore $f$ is a ring homomorphism.

Am I finished?

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