Prove that the only ring homomorphism from $\mathbb Z$ to $\mathbb Z$ is Identity mapping

Question

Suppose there exists some other ring homomorphism than the identity mapping, $f$ from $\mathbb Z$ to $\mathbb Z$. First I observe $f$ sends 0 to 0, and 1 to 1, and permutes some other elements in $\mathbb Z$. Then, I ponder my life decisions, and consider trying to find help on Stack Exchange. Could someone hint me in the right direction?

Answer 1

Let $f : \mathbb{Z} \to \mathbb{Z}$ be a function such that it verifies every condition of a ring morphism, except $f(1) = 1$ (i.e. a non-unital ring morphism). Now, since $f(a+b) = f(a) + f(b)$, using Robert Shore's hint, for $n \geq 1$ we get

$$ f(n) = f(1 + \cdots+ 1) = f(1) + \cdots + f(1) = nf(1) $$

and $f(-n) = f( (-1) + \cdots + (-1)) = nf(-1)$. Moreover,

$$ f(-1) + f(1) = f(-1+1) = f(0) = 0 $$

and thus, by definition of additive inverse, we see that $f(-1) = -f(1)$. Hence, $f$ is completely determined by $f(1)$ via

$$ f(k) = kf(1) \ (\forall k \in \mathbb{Z}). $$

Also, one should have $f(1) = f(1^2) = f(1)^2$ which in $\mathbb{Z}$ means that either $f(1) = 1$ or $f(1) = 0$. If $f$ ought to be unital, then necessarily $f \equiv id$.