# Homomorphism ring

#### karlito03

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?

#### 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.

#### Drexel28

MHF Hall of Honor
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:
If $$\displaystyle \theta:\mathbb{Z}_n\to\mathbb{Z}_m$$ is a homomorphism then the homomorphism is entirely determined by $$\displaystyle \theta(1)$$

#### karlito03

I sure (Thinking)..
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)?

#### karlito03

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.

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#### Drexel28

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

#### NonCommAlg

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

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

karlito03

#### alexandrabel90

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