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Now that I have your attention, here are the pertinent definitions: Let $R$ be an integral domain with field of fractions $K$. A $R$-fractional ideal of $K$ is a $R$-submodule $I$ of $K$ such that $aI\subseteq R$ for some $a\in R\setminus0$. From now on we just say that $I$ is a fractional ideal. In particular every ideal in $R$ is a fractional ideal.

Given fractional ideals $I,J$ we define the set $IJ=\{\sum i_kj_k: i_k\in I, j_k\in J\}$, which in turn is a fractional ideal, and clearly $IR=I$ for all fractional ideal $I$. We say that a fractional ideal $I\ne0$ is invertible if there is some fractional ideal $J$ such that $IJ=R$. There is only a possibility for $J$, namely $J=I^{-1}$, where $I^{-1}$ is the set defined by $I^{-1}=\{z\in K: zI\subseteq R$}. In fact, if $IJ=R$ then from the definition of $I^{-1}$ we have both $II^{-1}\subseteq R$ and $J\subseteq I^{-1}$, and so $II^{-1}J\subseteq RJ$, that is $I^{-1}(IJ)=I^{-1}\subseteq J$. It is not hard to show that $I^{-1}$ is a fractional ideal as well, even if $II^{-1}\subsetneqq R$.

Note then that $I^{-1}$ is just a notation: it denotes the potential inverse of a nonzero fractional ideal $I$.

Now my question is: given an ideal $I$ in $R$ such that $I^{-1}$ is invertible, does it follows that $I$ itself is invertible? I would like to think that this is indeed the case, but who said that life was easy? I think that there must be a counterexample. I am interested in a particular example of this situation, namely if an ideal $J$ of $R$ satisfies $J^{-1}=R$, does it follow that $J=R$? Note that for ideals $J$ in $R$ we always have $R\subseteq J^{-1}$.

UPDATE

I am very grateful to @Youngsu by her/his answer. Now I would like to know if a counterexample exists for the case dim $R=1$ and $R$ Noetherian.

Matemáticos Chibchas
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    Regarding your update, you certainly know the answer if you add in the extra condition that $R$ is normal, so that $R$ is Dedekind. Have you considered moving to the normalization of $R$? – Alex Youcis Nov 29 '13 at 02:11
  • Hi. I believe that a Dedekind domain which is not a PID probably works. For instance, in $R = \mathbb{Z}[\sqrt{-5}]$ take $I = (2, -1 + \sqrt{-5})$ with the field norm $N( a + b \sqrt{-5}) = a^2 + 5b^2$ for $a,b \in \mathbb{Z}$. – Youngsu Dec 10 '13 at 03:41
  • @Youngsu No, a Dedekind domain will never works, because for Dedekind domains every nonzero fractional ideal is invertible. – Matemáticos Chibchas Dec 10 '13 at 04:48
  • @MatemáticosChibchas: Hi. Isn't the question if $I^{-1} = R$, then $I = R$? I think it is stronger than invertible. – Youngsu Dec 10 '13 at 04:59
  • @Youngsu If $I^{-1}=R$, where $R$ is a Dedekind domain, then necessarily $I\ne0$ (otherwise $I^{-1}=0^{-1}=K$, the field of fractions of $R$, and $K\ne R$), hence $II^{-1}=R$, and so $I=R$. – Matemáticos Chibchas Dec 10 '13 at 23:45
  • @MatemáticosChibchas: Hi. Thanks for the comment. I guess then I don't understand the example $R = \mathbb{Z}[\sqrt{-5}]$ with $I = (2, -1 + \sqrt{-5})$. Do you know what $I^{-1}$ is in this example? Here $R$ is a Dedekind domain which is not a PID and $I$ is a maximal ideal which is not principal. – Youngsu Dec 11 '13 at 03:16
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    @Youngsu Let $a,b\in\mathbb Q$ and let $z=a+b\sqrt{-5}$. Then $z\in I^{-1}$ iff both $2z\in R$ and $(-1+\sqrt{-5})z\in R$. Since $R=\mathbb Z[\sqrt{-5}]$, from the first equation we conclude that $2a,2b\in\mathbb Z$, and the second equation is equivalent to both $-a-5b\in\mathbb Z$ and $a-b\in\mathbb Z$. Writing $a=m/2,b=n/2$ with $m,n\in\mathbb Z$, we conclude from the last equation that $m\equiv n(\bmod\ 2)$; conversely, these conditions are sufficient, so $I^{-1}=\bigl{\bigl(m+n\sqrt{-5}\bigr)/2: m,n\in\mathbb Z, m\equiv n(\bmod\ 2)\bigr}$ – Matemáticos Chibchas Dec 11 '13 at 05:19
  • @MatemáticosChibchas: I like your answer very much. This cleared up my confusion and misunderstanding. – Youngsu Dec 11 '13 at 13:04

1 Answers1

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One can take $R = k[x,y]$ and $I = (x,y)$, where $k$ is a field. If $a \in I^{-1}$, then we may write $a = \frac{b}{c}$ where $b,c \in R, c \neq 0$, and $\gcd (b,c) = 1$. Then $bx/c, by/c \in R$ implies that $c$ divides both $x,y$. Since $\gcd(x,y) = 1$, $c$ is a unit in $R$. Therefore, $I^{-1} = R$. However, since $I$ is not principal, $I \neq R$.

Youngsu
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