TITLE: Show that $ker(f) = n/k$
QUESTION [2 upvotes]: We suppose that we have integers, $n$ and $k$, such that there exists a ring homomorphism defined as : $$f : \mathbb{Z}/n\mathbb{Z} \rightarrow \mathbb{Z}/k\mathbb{Z}$$ where  $\overline{x} \rightarrow [x]$.
Show that,
(i) k divides n.
(ii) the number of elements in $ker(f)$ is equal to $n/k$.
(iii) $f$ is surjective.
I have been able to show (i), but I am not sure for (ii) and (iii).
For (ii), I have started by using the definition of $ker(f)$. We have $ker(f) = \{\overline{x}, f(\overline{x}) = [0]\} = \{\overline{x}, [x] = [0] = [k]\} = \{\overline{x}, k/x\} = \{\overline{x}, \text{such that x is a multiple of k}\}$. From that point, I don't see how I can conclude that $ker(f) = n/k$, would it be useful to use (i) ?
Also for (iii), we know that $f$ is a ring homomorphism and we know that the homomorphism is entirely determined by it generators. From that the only two generators are $\overline{1}$ and $[1]$ so then we can conclude that $f(\overline{x})=[x]$.
If someone have any ideas for (ii) and (iii), I would appreciate it.

REPLY [3 votes]: Let me denote by $[x]_n$ the residue class of $x$ modulo $n$. Then you have
$$
f([x]_n)=[x]_k
$$
and, in particular,
$$
f([0]_n)=f([n]_n)=[n]_k=[0]_k
$$
Therefore $n\equiv0\pmod{k}$, that is, $k\mid n$.
What's the kernel of $f$? You have $[x]_n\in\ker f$ if and only if $[x]_k=[0]_k$, that is, if and only if $k\mid x$. Hence $\ker f=k\mathbb{Z}/n\mathbb{Z}$, which has precisely $n/k$ elements.
Why? The homomorphism $f$ is clearly surjective, because $[x]_k=f([x]_n)$, and the homomorphism theorem tells you that
$$
(\mathbb{Z}/n\mathbb{Z})/\ker f\cong\mathbb{Z}/k\mathbb{Z}
$$
Therefore
$$
n/\lvert\ker f\rvert=k
$$
by Lagrange's theorem.

You now might be interested in showing that if $k\mid n$, then there does exist a unique ring homomorphism (preserving the unit elements) $\mathbb{Z}/n\mathbb{Z}\to\mathbb{Z}/k\mathbb{Z}$. Which is precisely $[x]_n\mapsto[x]_k$.
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