TITLE: A simpler proof that if $m\mid n$ then there is a ring homomorphism from $\mathbb{Z}_n$ onto $\mathbb{Z}_m$?
QUESTION [5 upvotes]: Question 18.I.4 from Pinter's A Book of Abstract Algebra asks for a proof of the following, where $\mathbb{Z}_m$ and $\mathbb{Z}_n$ are treated as rings:

If $n$ is a multiple of $m$, then $\mathbb{Z}_m$ is a homomorphic image of $\mathbb{Z}_n$.

After reading ahead to the next chapter on quotient rings, the method that suggests itself to me is to use cosets. Let $n = md$ and define $f: \mathbb{Z}_n \to \mathbb{Z}_m$ to send $a + n\mathbb{Z} \mapsto a + m\mathbb{Z}$. This is a well-defined map, because if $a + n\mathbb{Z} = b + n\mathbb{Z}$ then $a - b \in n\mathbb{Z} = md\mathbb{Z} \subset m\mathbb{Z}$, and so $a + m\mathbb{Z} = b + m\mathbb{Z}$. Then the homomorphism properties of $f$ are clear from the definitions of coset addition and multiplication.
However, this argument is really about the rings $\mathbb{Z}/(m)$ and $\mathbb{Z}/(n)$, and since the book hasn't defined quotient rings or given the homomorphism theorems up to this point I suspect that a proof using more basic concepts is called for. In fact, the book has not given a formal definition of $\mathbb{Z}_m$ as a ring, so I am not sure how to prove things about it in general.  Should its elements be thought of as equivalence classes of integers, or as the integers $\{0, 1, ... m-1\}$ with addition and multiplication defined by taking remainders modulo $m$?
So I have two questions:

Is the argument above actually a proof of this statement?
Is there a better proof using only the more basic properties of modulo arithmetic?

REPLY [3 votes]: The argument is OK with the caveat that I hope you have a concrete explanation with the part where you say "are clear." You were probably just saving space, but as you know some students often jump the gun on what is clear :)
Both of the "ways of thinking about" the elements of $\Bbb Z_m$ are valid since really they are the same thing. That is, $n$ is equivalent to $k$ if and only if $n-k\in m\Bbb Z$, if and only if $n+m\Bbb Z=k+m\Bbb Z$ as cosets. So, each equivalence class corresponds to one and only one coset.