proof of invertible ideals are projective

We show that a nonzero fractional ideal $\\mathfrak{a}$ of an integral domain $R$ is invertible if and only if it is projective (http://planetmath.org/ProjectiveModule) as an $R$ -module.

Let $\\mathfrak{a}$ be an invertible fractional ideal and $f\\colon M\\rightarrow\\mathfrak{a}$ be an epimorphism of $R$ -modules. We need to show that $f$ has a right inverse.Letting $\\mathfrak{a}^{-1}$ be the inverse ideal of $\\mathfrak{a}$ , there exists $a_{1},\\ldots,a_{n}\\in\\mathfrak{a}$ and $b_{1},\\ldots,b_{n}\\in\\mathfrak{a}^{-1}$ such that

a_{1}b_{1}+\\cdots+a_{n}b_{n}=1

and, as $f$ is onto, there exist $e_{k}\\in M$ such that $f(e_{k})=a_{k}$ . For any $x\\in\\mathfrak{a}$ , $xb_{k}\\in\\mathfrak{a}\\mathfrak{a}^{-1}=R$ , so we can define $g\\colon\\mathfrak{a}\\rightarrow M$ by

g(x)\\equiv(xb_{1})e_{1}+\\cdots+(xb_{n})e_{n}.

Then

f\\circ g(x)=(xb_{1})f(e_{1})+\\cdots+(xb_{n})f(e_{n})=x(b_{1}a_{1}+\\cdots b_{n}%a_{n})=x,

so $g$ is indeed a right inverse of $f$ , and $\\mathfrak{a}$ is projective.

Conversely, suppose that $\\mathfrak{a}$ is projective and let $(a_{i})_{i\\in I}$ generate $\\mathfrak{a}$ (this always exists, as we can let $a_{i}$ include every element of $\\mathfrak{a}$ ). Then let $M$ be a module with free basis $(e_{i})_{i\\in I}$ and define $f\\colon M\\rightarrow\\mathfrak{a}$ by $f(e_{i})=a_{i}$ . As $\\mathfrak{a}$ is projective, $f$ has a right inverse $g\\colon\\mathfrak{a}\\rightarrow M$ . As $e_{i}$ freely generate $M$ , we can uniquely define $g_{i}\\colon\\mathfrak{a}\\rightarrow R$ by

g(x)=\\sum_{i\\in I}g_{i}(x)e_{i},

noting that all but finitely many $g_{i}(x)$ must be zero for any given $x$ . Choosing any fixed nonzero $a\\in\\mathfrak{a}$ , we can set $b_{i}=a^{-1}g_{i}(a)$ so that

g_{i}(x)=a^{-1}g_{i}(ax)=a^{-1}xg_{i}(a)=b_{i}x

for all $x\\in\\mathfrak{a}$ , and $b_{i}$ must equal zero for all but finitely many $i$ . So, we can let $\\mathfrak{b}$ be the fractional ideal generated by the $b_{i}$ and, noting that $xb_{i}=g_{i}(x)\\in R$ we get $\\mathfrak{a}\\mathfrak{b}\\subseteq R$ . Furthermore, for any $x\\in R$ ,

x=a^{-1}f\\circ g(ax)=\\sum_{i}a^{-1}g_{i}(ax)f(e_{i})=\\sum_{i}xb_{i}f(e_{i})\\in%\\mathfrak{b}\\mathfrak{a}

so that $R\\subseteq\\mathfrak{a}\\mathfrak{b}$ , and $\\mathfrak{b}$ is the inverse of $\\mathfrak{a}$ as required.