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
Canonical name	ProofOfInvertibleIdealsAreProjective
Date of creation	2013-03-22 18:35:51
Last modified on	2013-03-22 18:35:51
Owner	gel (22282)
Last modified by	gel (22282)
Numerical id	5
Author	gel (22282)
Entry type	Proof
Classification	msc 16D40
Classification	msc 13A15