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Let $R$ be an integral domain with fraction field $K$. For a fractional ideal $I$ (i.e. $I$ is an $R$-submodule of $K$ such that $\exists 0\ne r\in R $ with $rI \subseteq R$) of $R$, define $(R:I)=\{x\in K : xI \subseteq R\}$.

Is it true that $(R:I\cap J)=(R:I)+(R:J)$ for $I,J$ fractional ideals?

I can easily see that $(R:I)+(R:J) \subseteq (R:I\cap J)$, however I am having difficulty proving $(R:I\cap J) \subseteq (R:I)+(R:J)$.

user26857
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user521337
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  • I don't think so. If one denotes $(R:I)$ by $I^{-1}$, and suppose that your claim is true, then $((I\cap J)^{-1})^{-1}=(I^{-1}+J^{-1})^{-1}$, that is, $((I\cap J)^{-1})^{-1}=(I^{-1})^{-1}\cap (J^{-1})^{-1}$, which doesn't hold in general; see here. – user26857 Jan 25 '19 at 22:17
  • @user26857: I see ... is it at least true when $R$ is Noetherian / normal ? – user521337 Jan 25 '19 at 22:20
  • @user26857: I don't think the counterexample you refer to doesn't quite apply here ... $I,J$ in my case are fractional ideals ... – user521337 Jan 26 '19 at 05:43
  • The characterization given in the linked thread involves fractional ideals. – user26857 Jan 26 '19 at 08:25
  • According to the paper in the link on v-domains, R would be a v-domain (and I believe your identity would then hold) if R were noetherian and integrally closed. – vacant Jan 26 '19 at 23:27

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