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Let $f:R\rightarrow S$ be a ring homomorphism of commutative rings with identities such that $f(1_R)=1_S$. And let $J$ be an ideal of $S$ such that for every prime ideal $P$ of $R$ and every $n\in \mathbb{N}$, $J\subseteq {\langle f(P)+J \rangle}^{n}$, where $\langle f(P)+J \rangle$ is the ideal generated by $f(P)+J$ in the ring $f(R)+J$, as a subring of $S$. How can we show that $J$ is generated by idempotents, as an ideal of $f (R)+J $ or $ S $?

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This is not true. For instance, let $k$ be a field and let $R=k[x,x^{1/2},x^{1/4},x^{1/8},\dots]$. Then the ideal $J=(x,x^{1/2},x^{1/4},x^{1/8},\dots)$ satisfies $J=J^n$ for all $n>0$, and so will satisfy your condition for any $f:R\to S$ at all. However, $J$ contains no nonzero idempotents (the ring $R$ is a domain and $J$ is a proper ideal), so it is certainly not generated by idempotents.

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  • $\begingroup$ Eric Wofsey: Thank you very mach for your answer. Is this question true when $J $ is finitely generated? $\endgroup$
    – Deroty
    May 6, 2017 at 7:49

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