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Let $f: R \rightarrow S$ be an injective ring homomorphism. Suppose $S/J$ is flat over $R$ for every ideal $J$ of $S$. Show that for any ideal $I$ of $R$ for any ideal $J$ of $S$, we have $(IS) \cap J = IJ$.

Here is my approach:we have injective and surjective $IS \hookrightarrow S \twoheadrightarrow S/J$ and the kenel($IS\to S/J$) is $IS\cap J$ so we get injective $IS/(IS\cap J) \hookrightarrow S/J$ and by $S/J$ is flat over R, we have the injective $(IS/(IS\cap J))\otimes_R S/J \hookrightarrow S/J\otimes_R S/J$.

It is also obvious the $IJ \subset IS\cap J $ How can we show the other side. thanks a lot.

lee
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  • The assumption holds for all ideals $J$ of $S$, but in the claim there is only one ideal $J$. Maybe we can weaken the assumption that $S/J$ is flat only for this specific $J$? Can you please check the original form of the claim? – Martin Brandenburg Dec 04 '23 at 19:03
  • @MartinBrandenburg I think it is arbitrary J, and I edited, thanks. – lee Dec 05 '23 at 02:06
  • @lee Sorry, on second glance I realized the thing I wanted doesn't work. – Shrugs Dec 05 '23 at 05:39

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