Homomorphism between $\mathbb{Z}_m$ and $\mathbb{Z}_n$ if $m\mid n$

Question

How can I show $\phi:\mathbb{Z}_m \to \mathbb{Z}_n$ defined by

$$ \phi(k) = k\mod{n} $$

satisfies $\phi(k +_m l) = \phi(k)+_n \phi(l)$ when $n\mid m$.

I have tried using remainder terms but the solution gets messy as I have mod n's and m's and cant get rid of them.

You just have to prove that if $k+l\equiv k'+l'\mod m$, then $k+l\equiv k'+l'\mod n$, which is pretty obvious. — Bernard, Nov 05 '20 at 20:56
There are many tens if not hundreds of prior questions on this. Please search first. — Bill Dubuque, Nov 05 '20 at 21:05
@BillDubuque I was after a full computation to cater for an audience with little algebra knowledge. — math111, Nov 05 '20 at 23:00

score 0 · Answer 1 · answered Nov 05 '20 at 21:02

0

You know that $nq=m$ for some integer $q$.

Then you have that $m\in (n)$, in other words $m\equiv_{n}0$.

So if you have $a\equiv_{m}b$, this implies that $a-b\equiv_{m} 0$, and $a-b\equiv_{n} 0$, and therefore $a\equiv_{n}b$.

If you apply this same argument to $k+l\equiv_{m} k'+l'$, you will have that $k+l\equiv_{n} k'+l'$, and you will get your prove.

answered Nov 05 '20 at 21:02

Tio Miserias

1 Answers1