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Let $R$ be a Noetherian commutative ring. Let $I,J\subset K(R)$ be fractional ideals where $K(R)$ is the total quotient ring. Define $I^{-1}:=\{s\in K(R) : sI\subset R\}.$ Further suppose that $I$ is invertible I.e. $I^{-1}I=R$. Then $I^{-1}J=\{s\in K(R) : sI\subset J\}.$ (LHS is a product as ideals.)

Is this true and why?

(If necessary, we can add one more assumption: $J$ is also invertible.)

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Thanks. You are right. My question is the other inclusion part.(Innocent typo: $j(kI )\subset jI => j(kI )\subset jR$) –  Tom Sep 3 '12 at 12:06
What we know now is: (1) $LHS\subset RHS$ is true without invertibility assumption. (2) $LHS\supset RHS$ is fail in general without invertibility assumption, thanks to Jeremy. –  Tom Sep 4 '12 at 0:21

3 Answers 3

up vote 2 down vote accepted

Let $L = \{s\in K(R) : sI\subset J\}.$

Suppose $s \in L$. Then $sII^{-1} \subset JI^{-1}$. Hence $sR \subset JI^{-1}$. Hence $s \in JI^{-1}$. Hence $L \subset JI^{-1}$.

Conversely suppose $s \in JI^{-1}$. Then $sI \subset JI^{-1}I = JR = J$. Hence $s \in L$. Hence $JI^{-1} \subset L$

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Thank you very much. This is what I have wanted. –  Tom Sep 4 '12 at 2:08
was the Noetherian assumption used anywhere? –  messi Sep 5 '12 at 8:25
@messi The Noetherian assumption is not necessary. –  Makoto Kato Sep 5 '12 at 8:49

Take $R=Z[\sqrt{-3}]$, $I=(2,1+\sqrt{-3})$, $J=I$, and $\omega=(1+\sqrt{-3})/2$. Notice that $\omega I=I$. The point is that $I$ is simultaneously an ideal for the ring $R$ and the ring $S=R[\omega]$, but $S$ is a Dedekind domain and its fractional ideals are invertible.

As an $S$-ideal, $I$ is just $2S$ and the set of $x$ for which $xI\subset S$ is $(1/2)S$. Since $R\subset S$, you can use this to see that $I^{-1}$ as an $R$ ideal is just $S$. So $I^{-1}I=SI=I$. Looking at the other side of your proposed equality, we have $SI\subset I$ so that side contains S. Hence the proposed equality fails in general.

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Thank you very much for showing that it is fail in general. I am very sorry I forgot to write a very important assumption I.e. $I$ is invertible. –  Tom Sep 4 '12 at 0:03
That certainly makes a big difference! –  Jeremy Teitelbaum Sep 4 '12 at 10:23

I'm not sure I completely understand even what you're asking, but if your question is about the meaning of that product of ideals, then we have by definition $$I^{-1}J=\{t:=a_1j_1+...+a_nj_n\;\;|\;a_k\in I^{-1}\,,\,j_k\in J\,\,,\,n\in\Bbb N\,\,\text{not fixed}\}$$ But for any $$a\in I^{-1}\,\,,\,\,aI\subset R\Longrightarrow \forall\,\,k\in\Bbb N\,,\,a_kj_kI=\left(a_kI)j_k\subset RJ\subset J\right)\Longrightarrow \forall\,t\in I^{-1}J\,,\,tI\subset J$$ and we're done.

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Thank you. You understand my definition and your proof of $LHS\subset RHS$ is correct. My question is whether $LHS\supset RJS$. –  Tom Sep 3 '12 at 12:44
Typo. My question is LHS $\supset$RHS. –  Tom Sep 3 '12 at 12:45

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