Example of a ring homomorphism that

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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?
 
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What about Z to Z6 via the natural surjection?
 
What's wrong with the one from Z->Z_n , n not prime?
 
What's the map Z -> Z_6? E.g. what would 7 be mapped to? What's the map for Z -> Z_n?
 
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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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