I have a domain $A$ with field of fractions $K$ and a non-zero fractional $A$-ideal $I$. Let $I^{-1}$ be the fractional ideal $\{a\in K\mid aI\subseteq A\}$.

I assume that $II^{-1}\subseteq \mathfrak p$ for some non zero prime ideal $\mathfrak p$ of $A$.

My question is that if $a(I A_\mathfrak p)\subset A_\mathfrak p$ for some $a\in K$ then do we have $a\in I^{-1}A_\mathfrak p$?


We have $(A:I)A_p=(A_p:IA_p)$, or (if you like) $I^{-1}A_p=(IA_p)^{-1}$. Then from $a(IA_p)\subset A_p$ you get $a\in (A_p:IA_p)=I^{-1}A_p$.

Edit. Let's prove $(A_p:IA_p)\subseteq I^{-1}A_p$: if $xIA_p\subseteq A_p$, then $xa_i\in A_p$ for $a_i\in I$ a system of generators. There is $s_i\in A-p$ such that $s_ixa_i\in A$. For $s=\prod a_i$ we get $sxa_i\in A$. It follows that $sxI\subseteq A$, so $sx\in I^{-1}$, and thus $x\in I^{-1}A_p$.

Remark. The above proof holds for $I$ finitely generated.

  • $\begingroup$ Sorry I don't follow your argument. What is the definition of $(A:I)$ and how do you get the last equality? Regards $\endgroup$ – user149343 Jun 6 '14 at 17:09

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