# 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.

Title proof of invertible ideals are projective ProofOfInvertibleIdealsAreProjective 2013-03-22 18:35:51 2013-03-22 18:35:51 gel (22282) gel (22282) 5 gel (22282) Proof msc 16D40 msc 13A15