# Show their is only one ring morphism from Zn to Zk if k|n

1. Feb 6, 2009

### Daveyboy

1. The problem statement, all variables and given/known data
For the rings Zn and Zk show that if k|n, then the function f: Zn to Zk
s.t [x]n --->[x]k for all x in Zn
is a ring morphism. Show this is the only ring morphism from Zn to Zk.

The attempt at a solution

So I showed it is a ring morphism by just verifying the properties, no big deal. I have no idea how to show that it is the only one though.

I want to start out by contradiction and assume that their is another one... but I don't know what to do with that. Is there a way to show it directly?

2. Feb 6, 2009

### Dick

Once you know what f(1) is, you pretty much know what the morphism is, right? Is there any choice but to make f(1)=1?