I am trying to find all homomorphisms from $\mathbb{Z}$ to $\mathbb{Z}$. I am looking at a solution, but i do not understand it. Here it is:
Let $\Phi\colon \mathbb{Z}\to\mathbb{Z}$ be a ring homomorphism. Because $1^2 = 1$, we see that $\Phi(1)$ must be an integer whose square is itself, namely either $0$ or $1$. If $\Phi(1) = 1$ then $\Phi(n)= \Phi(n · 1)= n$, so $\Phi$ is the identity map of $\mathbb{Z}$ onto itself which is a homomorphism. If $\Phi(1)= 0$, then $\Phi(n)=\Phi(n · 1)0$, so maps everthing onto $0$, which also yields a homomorphism.
I do not understand here what "Because $1^2= 1$, we see that $\Phi(1)$ must be an integer whose square is itself, namely either $0$ or $1$.". All proof is built upon this and I do not understand what it means. Can someone help me with this?
Thank you
Best Answer
Remember the basic properties of a ring homomorphism: $\Phi$ must be such that
$$\Phi(m+n)=\Phi(m)+\Phi(n)$$
and
$$\Phi(mn)=\Phi(m)\Phi(n)$$
for all $m,n\in\Bbb Z$. Since $1\cdot1=1$, we must have
$$\Phi(1)=\Phi(1\cdot1)=\Phi(1)\Phi(1)\;,$$
i.e., $\Phi(1)^2=\Phi(1)$. The only two integers with the property that $n^2=n$ are $0$ and $1$, so $\Phi(1)$ must be one of them.