Question

Let f be a homomorphism from Z to Z. Prove that either f(x) = 0 for every x in Z, or f(x) = x for every x in Z.

Answer 1

Question 1.

Let $f$ be a homomorphism from $\mathbb{Z}$ to $\mathbb{Z}$. Prove that either $f(x)=0$ for every $x\in\mathbb{Z}$, or $f(x)=x$ for every $x\in\mathbb{Z}$.

Solution.

Consider $f(1)$. Since $f$ is a homomorphism of rings, we have

$f(1)^{2}=f(1^{2})=f(1).$

Therefore, $f(1)(f(1)-1)=0$, i. e. either $f(1)=0$ or $f(1)=1$.

Suppose $f(1)=0$. Then for any $x\in\mathbb{Z}$:

$f(x)=f(x\cdot 1)=f(x)f(1)=f(x)\cdot 0=0,$

so $f$ is the zero homomorphism.

Now let $f(1)=1$. Then for any $x>0$ we have

$f(x)=f(\underbrace{1+\cdots+1}_{x\,terms})=\underbrace{f(1)+\cdots+f(1)}_{x\,terms}=\underbrace{1+\cdots+1}_{x\,terms}=x.$

Furthermore, if $x<0$, then $-x>0$, so

$f(x)=f(-(-x))=-f(-x)=-(-x)=x.$

Finally, $f(0)=0$, because is the zero of $\mathbb{Z}$. Thus, $f$ is the identity homomorphism in this case. ∎