Let $ϕ: R → S$ be a ring homomorphism, where $S$ is a domain. Prove that if $u ∈ R$ is nilpotent then $u ∈ \ker ϕ$.

Question

Let $ϕ: R → S$ be a ring homomorphism, where $S$ is a domain. Prove that if $u ∈ R$ is nilpotent then $u ∈ ker ϕ$.

Solution: Suppose that $n ∈ N$ and $u^n = 0_R$. Then $0_S = ϕ(0_R) = ϕ(u^n) = ϕ(u)^n = ϕ(u)^{n−1}ϕ(u)$, so either $ϕ(u) = 0_S$ or $ϕ(u)^{n−1} = 0_S$ since $S$ is a domain. Repeating this, we eventually find that $ϕ(u) = 0_S$ and so $u ∈ \ker ϕ$.

I don't understand how "Repeating this, we eventually find that $ϕ(u) = 0_S$ and so $u ∈ \ker ϕ$." Could someone show me how you repeat it to get the final answer?

Answer 1

Hint

$$0=\varphi (u)^{n-1}=\varphi (u)^{n-2}\varphi (u).$$

Answer 2

Claim. Let $S$ be a domain. Let $r\in S$ and $n\geq 1$. If $r^{n}=0$, then $r=0$.

Proof. We prove by induction on $n$. If $n=1$, then we are done. Let $n\geq 2$. Then $r^{n-1}r=r^{n}=0$. So $r=0$ or $r^{n-1}=0$. If $r=0$, we are done. If $r^{n-1}=0$, then by induction hypothesis, we have $r=0$.

Using this claim, since $\phi(u)^{n}=0$, we have $\phi(u)=0$.