Homomorphism ring

May 2010
12
0
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?
 
May 2010
12
0
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
Nov 2009
4,563
1,566
Berkeley, California
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 \(\displaystyle \theta(1)\)
 
May 2010
12
0
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)?
 
May 2010
12
0
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.
 
Last edited:

Drexel28

MHF Hall of Honor
Nov 2009
4,563
1,566
Berkeley, California
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
May 2008
2,295
1,663
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 \(\displaystyle m \mid r^2-r.\)
 
  • Like
Reactions: karlito03
Aug 2009
639
2
i dont get the solution for this question. can someone explain it to me? thanks