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I want to show in homomorphic image of Jacobson ring Nilradical is equal to the Jacobson radical. By Jacobson ring we know that every prime ideal is intersection of maximal ideals.

Where I stuck is the following. Given $f:A \to B$ surjective ring homomorphism and $A$ be a Jacobson ring. Consider $y \in Jac(B)$ then want to show that for any prime ideal $p$ of $B,$ $y \in p.$ Now if $f^{-1}(y) \cap f^{-1}(p)= \phi $ I cannot draw any contradiction. Please help me to prove this way.

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Hint: We know that $B\cong A/I$ for some ideal $I\subset A$. Thus, what you want to show is that $\operatorname{nil}(A/I)=\mathcal{J}(A/I)$. Use the correspondence between prime ideals of $A$ and those of $A/I$.

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  • $\begingroup$ Enough to show $rad(I)= \cap m_{\alpha}$ where $m_{\alpha}$ the collection of all maximal ideals of $A$ containing $I.$ But since every prime ideal is intersection of maximal ideals $A$ and $rad(I)$ is the intersection of all prime ideals containing $p$..it follows. Am I correct ? $\endgroup$ – user371231 Feb 9 '18 at 10:40
  • $\begingroup$ @user371231 Yes, try to write down all the details! (There is a typo in what you wrote. It should say $I$ instead of $p$) $\endgroup$ – user347489 Feb 9 '18 at 10:52
  • $\begingroup$ yes $rad(I)$ is intersection of all prime ideals containing $I.$ $\endgroup$ – user371231 Feb 11 '18 at 6:35

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