# Thread: Homomorphism ring

1. ## Homomorphism ring

This question does not let me sleep.
Fina all Ring homomorphism between Z_n----> Z_m.
I know there should be 3 case:
n>m
n<m and m=n.
the last one is trivial. Anyways, how do you guys think I should proceed or what should I consider?

2. Somehow, I suspect the answer to be gcd(n,m), because I read it somewhere.But I actually want to have an idea of how to get there.

3. Originally Posted by karlito03
Somehow, I suspect the answer to be gcd(n,m), because I read it somewhere.But I actually want to have an idea of how to get there.
Hint:
Spoiler:
If $\theta:\mathbb{Z}_n\to\mathbb{Z}_m$ is a homomorphism then the homomorphism is entirely determined by $\theta(1)$

4. I sure ..
call theta (f).
Then f(1)=1 since you assume it was a ring homomorphism. But how can we get to the point of saying that all homomorphism must be determined by f(1). There must be some cases. For instance m>n, n>m or m=n or Don't m and n have to be such that (n,m)=1. How many homomorphism can one get in each case. Are you saying they will all be determined by f(1)?

5. if n=m then image of any ring Zn-->Zm is determined by the image of 1 mod (n) where m=n.
now if n>m Zn--> Zm need not to be an homomorphism.

6. Originally Posted by karlito03
now if n>m Zn--> Zm need not to be an homomorphism.
I'm not sure what you mean here? Why isn't $f:\mathbb{Z}_n\to\mathbb{Z}_m:z\mapsto 1_m$ not a homomorphism?

7. Hint 1: put $\theta(1+n \mathbb{Z})=r+m\mathbb{Z}$ and extend $\theta$ linearly to all $\mathbb{Z}/n\mathbb{Z}$. now look at the conditions that will make $\theta$ well-defined and multiplicative.

Hint 2: using the above, show that $\theta$ is defined by $\theta(x+n\mathbb{Z})=rx + m\mathbb{Z},$ where $r$ satisfies the following conditions: $\frac{m}{\gcd(n,m)} \mid r$ and $m \mid r^2-r.$

8. i dont get the solution for this question. can someone explain it to me? thanks