Invertible ideal

Remarks

If $I$ is invertible, its inverse is unique and equals $I^{-1} = (R : I) = \set{x \in K : xI \subseteq R}$. Equivalently, a fractional ideal is invertible exactly when it is a projective $R$-module. For a Noetherian domain, a nonzero ideal is invertible exactly when its localization at every maximal ideal is principal.

Example: let $R = \Z[\sqrt{-5}]$ and $P = (2, 1 + \sqrt{-5})$. Then $P^2 = (4, 2 + 2\sqrt{-5}, -4 + 2\sqrt{-5})$ contains $6$ and $4$, hence $2$, and all its generators are multiples of $2$, so $P^2 = (2)$. Thus $J = \frac{1}{2} P$ is a fractional ideal with $PJ = \frac{1}{2}(2) = R$, and $P$ is invertible.

Non-example: let $R = k[x, y]$ for a field $k$ and $\mathfrak{m} = (x, y)$. If $f \in K$ satisfies $fx \in R$ and $fy \in R$, write $f = a/x = b/y$ with $a, b \in R$; then $ay = bx$, so $x$ divides $a$ and $f \in R$. Hence $(R : \mathfrak{m}) = R$ and $\mathfrak{m} (R : \mathfrak{m}) = \mathfrak{m} \ne R$, so $\mathfrak{m}$ is not invertible.