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Let $R$ and $S$ be commutative rings (with $1$) and $f : R\to S$ be a ring homomorphism. For an ideal $I$ of $R$, set $I^e:=\langle f(I)S\rangle$ (called the extension of $I$ to $S$). When $f$ is surjective, every ideal, $J$, of $S$ is an extension ideal; i.e. $J=I^e$, for some ideal $I$ of $R$.

Question 1. What other (non-trivial) conditions can be posed on rings or the homomorphism to have every ideal of $S$ be an extension ideal?
Is there special rings and homomorphism that we have this property? ( By "special rings and homomorphism", I mean the cases like $f : R\to R[X]$)


Question 2. In the case of "Question 1", (every ideal of $S$ an extension ideal), can one say that

"If $Q$ is a prime ideal of $S$, then there is a prime ideal, $P$, of $R$, such that $P^e=Q$?"

Thank you.

user 1
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  • Cross-posted: http://mathoverflow.net/questions/209267/in-what-conditions-every-ideal-is-an-extension-ideal-is-every-prime-ideal-exten – user26857 May 28 '16 at 06:35

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