# Homework Help: Example of a ring homomorphism that

1. Jun 10, 2010

### logarithmic

Can anyone give an example of a ring homomorphism f : R -> R', such that R is a integral domain but the Image(f) is not an integral domain.

I was thinking that since we want two non zero elements of Image(f) multiply to 0, we require: f(xy) = f(x)f(y) = 0, with f(x), f(y) not 0. Now f(xy) = 0, we don't want xy = 0, because then either x = 0 or y = 0 as R is a domain, meaning f(x) = 0 or f(y) = 0, so we want xy not 0, but still gets mapped to 0. So f can't be injective.

But apart from that I can't think of anything else. I can't think of any function f that would make it work for simple domains like Z or Z_p p prime, or simple non domains like Z_n where n not prime, so can anyone think of a integral domain R and function f to make this work?

2. Jun 10, 2010

### Tedjn

What about Z to Z6 via the natural surjection?

3. Jun 10, 2010

### boboYO

What's wrong with the one from Z->Z_n , n not prime?

4. Jun 10, 2010

### logarithmic

What's the map Z -> Z_6? E.g. what would 7 be mapped to? What's the map for Z -> Z_n?

5. Jun 10, 2010

### Tedjn

The integers mod 6 or mod n. 7 would map to 1 in my case. You do have to check that this is a ring homomorphism but in fact it turns out to be so. More generally, since nZ is an ideal, the surjective homomorphism from Z to Z/nZ is always a ring homomorphism.