Show that any epimorphism from the ring $(Z, +, .)$ onto itself is an isomorphism.

Question

Show that any epimorphism from the ring $(Z, +, .)$ onto itself is an isomorphism.

I think by the term epimorphism means an onto homomorphism.

Firstly, I assumed, $\phi:Z\to Z$ such that $\phi$ is an onto homomorphism.

I tried to prove $\phi$ is an isomorphism by showing that $I(\phi)=U=\{0\}.$

Now, $\phi(0)=0$ as $\phi$ is a homomorphism. Now, if $\exists a\in Z$ such that $\phi(a)=0$ then $U=\{$ set of multiples of $a\space \}.$

Next, I claimed that $\phi(x)=0$ for all $x\in Z$.

But I am not able to prove the claim.

If I couldve' proven the claim then I can get a contradiction as $\phi$ is onto. But till now, I haven't been able to do so.

Answer 1

Since $\phi$ is an epimorphism it follows that $\phi(1)=1$. We shall prove that $\ker\phi=\{0\}$. Let $z\in\ker\phi$. Without loss of generality we may assume that $z>0$. Hence, $$0=\phi(z)=\phi(1+\overset{z)}{\ldots}+1)=z\phi(1)=z\cdot 1=z.$$ And we are done.

Answer 2

If the kernel $U$ of $\phi$ isn't $\{0\}$ then the image of $\phi$ is isomorphic to the finite ring $\mathbb{Z}/U$, so $\phi$ isn't surjective. So the kernel is $\{0\}$, so $\phi$ is both injective and surjective, hence an isomorphism.