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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?

user26857
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MathGeek1998
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  • Either $\phi(u)=0$ or $\phi(u)^{n-2}=0$. Repeat now - either $\phi(u)=0$ or $\phi(u)^{n-3}=0$. Repeat now. Either $\phi(u)=0$ or .... or $\phi(u)^1=0$. – Dietrich Burde Apr 02 '20 at 13:34
  • It suffices to prove that the only nilpotent in a domain is zero, because ring homomorphisms obviously preserve nilpotents. – Geoffrey Trang Apr 02 '20 at 14:35
  • For a nontrivial application see this proof that a polynomial $p(x)$ is a unit (invertible) $\iff p(0)$ is a unit and all other coef's are nilpotent. – Bill Dubuque Apr 02 '20 at 16:33

2 Answers2

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Hint

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

Surb
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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 $\phi(u)=0$.

Delong
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