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Let $I$ be a fractional ideal of the ring $A$. Now $I$ is called regular if $I$ contains a regular element.

Are regular ideals invertible?

A fractional ideal $I$ is called invertible if $II^{-1} = A$ where $I^{-1} = \{r \in \operatorname{Frac}(A) \mid rI \subseteq A\}$. Now since there is some regular $a\in I$, we know that $a^{-1}\in \operatorname{Frac}(A)$ and thus $1 = aa^{-1} \in II^{-1}$ which implies $II^{-1} = A$ and thus the invertibility of $I$.

Does my proof contain any mistake or is this just simply true?

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Your statement is not true, as there are non-invertible ideals in domains (see Here). The error in your proof would be to assume that if $a\in I$ then $a^{-1}\in I^{-1}$. For example, take $A=k[x,y]$ and $I=(x,y)$, as in the link. Then $x\in (x,y)$ but $x^{-1}$ is not an element of $I^{-1}$ because $x^{-1}y\not\in A$ which means that $x^{-1}I\not\subset A$.

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  • $\begingroup$ Thank you very much! I did not recognize that the elements of $I^{-1}$ need to wedge every element of $I$ into the ground ring. $\endgroup$
    – windsheaf
    Commented May 18, 2018 at 19:56

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