- #1
jmjlt88
- 96
- 0
I have shown that Z is not isomorphic to 2Z and that 2Z is not isomorphic to 3Z. I need now to generalize this. Thus, to prove that rings kZ and lZ, where l≠k are not isomorphic, I need to define an arbitrary isomorphism, and reach a contradiction. So here's what I am thinking. I let f: kZ->lZ be a isomorphism. Then, f(k)=ln , where n is some integer. This is true since k in kZ has to go to some multiple of l. From here, I have just been tinkering around trying to get k to map to 0 so that f is not injective. This worked for the first two. Is this the right approach? Not really looking for a hint, but just an "okay" to keep trying working with this line of reasoning.