Equivalent characterizations of invertible fractional ideals

Assume (1). Choose a finite relation $1={\sum }_{i=1}^n{x}_i{y}_i$ with $x_i\in I$ and $y_i\in (R:I)$. For any $x\in I$ one has $x={\sum }_i(x{y}_i)x_i$, so $x_1,\dots ,x_n$ generate $I$. Define $\varphi :R^n\to I$ by $e_i\mapsto x_i$ and $\psi :I\to R^n$ by $x\mapsto (x{y}_1,\dots ,x{y}_n)$. Then $\varphi \psi ={\mathrm{id}}_I$, so $I$ is a direct summand of a finite free module and hence finite projective by [L4].

Assume (3), and choose generators $x_1,\dots ,x_n$ of $I$. For a maximal ideal $\mathfrak{m}$, write $I_{\mathfrak{m}}=a{R}_{\mathfrak{m}}$ with $a\in K^{\times }$. For each $i$ there is $u_i\in R_{\mathfrak{m}}$ such that $x_i\mathrm{/}1=a{u}_i$, and we may choose $s_i\mathrm{\notin }\mathfrak{m}$ with $s_i{u}_i\in R$. Putting $t:=s_1\cdots s_n$, we get $(t{a}^{-1})x_i\in R$ for every $i$, so $t{a}^{-1}\in (R:I)$ and $a^{-1}\mathrm{/}1=(t{a}^{-1})\mathrm{/}t\in (R:I{)}_{\mathfrak{m}}$. Hence $1=(a\mathrm{/}1)(a^{-1}\mathrm{/}1)\in I_{\mathfrak{m}}(R:I{)}_{\mathfrak{m}}\subseteq R_{\mathfrak{m}}$, so in fact $I_{\mathfrak{m}}(R:I{)}_{\mathfrak{m}}=R_{\mathfrak{m}}$. By [L1], localising the quotient $R\mathrm{/}I(R:I)$ and using [L2] gives $(R\mathrm{/}I(R:I){)}_{\mathfrak{m}}=0$ for every maximal ideal $\mathfrak{m}$. Therefore [L3] gives $I(R:I)=R$, so $I$ is invertible.

Still under (1), localise at a maximal ideal $\mathfrak{m}$. From $1=\sum x_i{y}_i$, one summand $x_j{y}_j$ is a unit in the local ring $R_{\mathfrak{m}}$, so $x_j$ generates $I_{\mathfrak{m}}$. Thus (1) implies (3), and also yields the local rank-one part of (2).

Condition (2) implies (3) because finite projective modules are finitely generated, and a free rank-one module over $R_{\mathfrak{m}}$ is principal. Steps 1.1 and 2.1 prove $(1)\Rightarrow (2)$ and $(1)\Rightarrow (3)$, step 1.2 proves $(3)\Rightarrow (1)$, and the present step proves $(2)\Rightarrow (3)$. Therefore the three conditions are equivalent.