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Let $R$ be a domain and $K$ be fractional field. Let $I$ an invertible $R$-ideal. Show that $I$ is proper, i.e. $I:I=R$. Deduce that an additive subgroup $I \subset K = Q(R)$ is an invertible ideal for at most one subring of $K$.

By using the property $A:BC=(A:B):C$ then we obtain $I:I=(R:I^{-1}):I=R:R$ but I can not show that $R:R=\{ x \in K : xR \subset R \}=R$.

Edit: As the comment, in particular, $x.1 \in R$, hence $x \in R$.

Desunkid
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    (1) If $x R \subseteq R$ then $x \cdot 1 \in R$, so of course $x \in R$. (2) I don't think you're interpreting the second part of the question correctly. It's not asking about a subring being an invertible ideal. Rather it's asking to you show that if $R_1, R_2$ are two subrings of the quotient field $Q(R)$ of $R$ such that a fixed additive subgroup $I$ of $Q(R)$ is invertible with respect to both $R_1, R_2$, then $R_1 = R_2$. – Badam Baplan Sep 25 '19 at 16:28
  • Thank you @BadamBaplan. As i understand, the question is that given additive subgroup $I \subset K$ then $I$ is invertible $R$-ideal for at most one subring of $K$? – Desunkid Sep 25 '19 at 17:18

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