Product of fractional ideals 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.)
 A: 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 $JI^{-1} \subset L$
A: 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.
A: 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.
