For fractional ideal $I$ why is $I\cap R \supsetneq \{0\}$?

Question

In the proof in my textbook that a fractional ideal $I$ in a quotient field $K$ of an integral domain $R$ has an inverse $$I^{-1} = \{ x\in K : x I \subseteq R\}\,,$$ it is used that there exists an $r\in R\cap I$ with $r\neq 0$. While it seems very plausible that such an $r$ does exist, I don't think it is obvious. Or am I missing something? How can one show that $I\cap R\supsetneq \{0\}$?

Answer 1

It follows from the very definition: $I$ is a fractional ideal if it is an $R$-module s.t. there is some $0\neq r\in R$ with $rI\subseteq R$. So pick $0\neq a\in I$: then on one hand $ra\in I$ because $I$ is an $R$-module, and on the other hand $ra\subseteq R$ because $I$ is a fractional ideal. Since $ra\neq 0$, you have the claim.