Find all the ring homomorphisms $f$ : $\mathbb{Z}_6\to\mathbb{Z}_3$.
definition of ring homomorphism:
The function f: R → S is a ring homomorphism if:
1) $f(1)$ = $1$
2) $f(a+b)$ = $f(a)$ + $f(b)$ for all a,b, in R
3) $f(ab)$ = $f(a)$ $f(b)$ for all a,b in R
Does it make sense to say that in this case
$f(6) = f(1) + f(1) + f(1) + f(1) +f(1) + f(1) = 1 + 1 + 1 + 1 + 1 + 1 = 0$ in $\mathbb{Z}_3$
Could you explain what do we do to find ring homomorphism in all cases. Not only $\mathbb{Z}_m \to\mathbb{Z}_n$, where $m<n$ .
As I see it, that really depends on the specific rings we are given.
If you have learned module theory, you will know the way to find all the module homomorphisms is to "send the generators" and then "respect the relations". Because the linearity ensures we take care of all the elements as long as we take care of the generators, while the well-definedness requires the relations to be respected.
This is the basic principle to find homomorphisms, and what you are doing is literally the same. In your case, the ring hom is at first place a $\mathbb{Z}$-module hom, and $1$ is the generator. However, for a general ring, things get a bit harder. But you can do your best by getting familiar with some special cases.
For example, to find a ring hom from a polynomial ring $R[x]\rightarrow S$, it suffices to consider where to send $x$, because that gives the destination for every element. And indeed, in many cases, the ring you have is a quotient of a polynomial ring. In that case, you have some constraints: take care of the well-definedness.
Hope this helps.