My instructor wrote in his notes the following example:
"As groups (Z,+) and (nZ,+) are isomorphic. As rings is there any nontrivial homomorphism $\phi$: Z->nZ?
The answer is no and he gives the following justification which I don't quite understand:
Consider a homomorphism $\phi$: Z->nZ with n>1 $1^2=1$ so $\phi(1^2)$=$\phi(1)$ So $\phi(1)$ satisfies $x^2=x$ and this has two solutions 0, and 1 hence $\phi(1)=0$
So what I don't understand is 1. Why did he use the example with $1^2$? and how did he eliminate the possibility that 0 cannot be $\phi(1)$?
It would be great if you can explain this in detail for someone new to the concept of rings.
The argument shows that $\phi(1)\in n\mathbb Z$ satisfies $\phi(1)=\phi(1)^2$. The only element in $n\mathbb Z$ (for $n>1$) that is equal to its square is $0$. So $\phi(1)=0$.
But then $$ \phi(k)=\phi(1+\cdots+1)=k\,\phi(1)=k\cdot0=0. $$ So $\phi=0$, i.e. there is no nontrivial ring homomorphism between $\mathbb Z$ and $n\mathbb Z$.