Given epimorphism $\phi$ and ideal $I$, with $\operatorname{Ker}\phi \subseteq I$, prove $\phi(\sqrt{I}) = \sqrt{\phi(I)}$

Question

I had this question on my exam.

Let $\phi: R\to S$ be epimorphism (Homomorphism and onto). Prove if $I$ is an ideal of $R$ with $\operatorname{Ker}\phi \subseteq I$, then $\phi(\sqrt I) = \sqrt{\phi(I)}$.

I could show $\phi(\sqrt I)⊆ \sqrt{\phi(I)}$. How to show the converse?

Answer 1

Suppose $\;s\in\sqrt{\phi(I)}\;$ , which means that for some $\;n\in\Bbb N\;,\;\;s^n\in\phi(I)\;$ , thus:

$$\exists\,r\in I\,\,s.t.\;\;\phi r=s^n$$ .

Let now $\;t\in R\;$ be such that $\;\phi t=s\implies \phi(t^n)=s^n=\phi r\implies \phi(t^n-r)=0\implies\;$

$$t^n-r\in\ker\phi\subset I\implies \exists i\in I\;\;s.t.\;\;t^n-r=i\implies t^n=r+i\in I+I=I$$

End the proof now.

Answer 2

In general, if $\phi\colon R\to S$ is a ring homomorphism and $J$ is an ideal of $S$, we have an induced injective ring homomorphism $\psi\colon R/\phi^{-1}(J)\to S/J$, defined by $\psi(r+\phi^{-1}(J))=\phi(r)+J$.

If $\phi$ is also surjective, then $\psi$ is an isomorphism.

In your case, taking $J=\phi(I)$, we have $\phi^{-1}(J)=\phi^{-1}(\phi(I))=I+\ker\phi=I$. Therefore $\psi\colon R/I\to S/\phi(I)$ is an isomorphism.

Can you finish? Hint: $r\in\sqrt{I}$ if and only if $r+I$ is nilpotent in $R/I$.