For $k\geq 2$ and $m_1,\ldots,m_k \in \mathbb{N}$ with $\gcd(m_i,m_j) = 1$ for $i\neq j$, show that $f(x)$ is a ring homomorphism 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.
 A: 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 $m_i$.
A: 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.
A: 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?
