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Assume $R,S$ are commutative rings, $f:R\to S$ is a surjective ring homomorphism and $I,J$ are coprime ideals in $S$. Must $f^{-1}(I)$ and $f^{-1}(J)$ be coprime in $R$?

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Yes.

If $S=R/K$, then $I=I'/K$ and $J=J'/K$. If $I', J'$ are not coprime, then there is a prime ideal $P$ such that $I'+J'\subseteq P$, and thus $I+J\subseteq P/K$. (Note that $P\supseteq K$ since $I',J'\supseteq K$.)

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Write $1=i+j$ with $i \in I, j \in J$. Find preimages $u,v$ in $f^{-1}(I)$ resp. $f^{-1}(J)$. Then $1 \equiv u+v \bmod \ker(f)$. Since $\ker(f) \subseteq f^{-1}(I)$, it follows that $1 \in f^{-1}(I) + f^{-1}(J)$.

(Notice that the proof by user26857 uses the existence of enough prime ideals, which is independent from ZF.)

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