4
$\begingroup$

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.)

  • 0
    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.2012-09-04

3 Answers 3

3

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 $JI^{-1} \subset L$

  • 0
    @messi The Noetherian assumption is not necessary.2012-09-05
1

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.

  • 0
    That certainly makes a big difference!2012-09-04
0

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.

  • 0
    Typo. My question is LHS $\supset$RHS.2012-09-03