generators of inverse ideal

Theorem. Let $R$ be a commutative ring with non-zero unity and let $T$ be the total ring of fractions of $R$ . If $\mathfrak{a}=(a_{1},\,\ldots,\,a_{n})$ is an invertible fractional ideal (http://planetmath.org/FractionalIdealOfCommutativeRing) of $R$ with $\mathfrak{ab}=R$ , then also the inverse ideal $\mathfrak{b}$ can be generated by $n$ elements of $T$ .

Proof. The equation $\mathfrak{ab}=(1)$ implies the existence of the elements $a_{i}^{\prime}$ of $\mathfrak{a}$ and $b_{i}^{\prime}$ of $\mathfrak{b}$ $(i=1,\ldots,\,m)$ such that $a_{1}^{\prime}b_{1}^{\prime}\!+\cdots+\!a_{m}^{\prime}b_{m}^{\prime}=1$ . Because the $a_{i}^{\prime}$ ’s are in $\mathfrak{a}$ , they may be expressed as

$a_{i}^{\prime}=\sum_{j=1}^{n}r_{ij}a_{j}\qquad(i=1,\ldots,m),$

where the $r_{ij}$ ’s are some elements of $R$ . Now the unity acquires the form

$1=\sum_{i=1}^{m}a_{i}^{\prime}b_{i}^{\prime}=\sum_{i=1}^{m}\sum_{j=1}^{n}r_{ij% }a_{j}b_{i}^{\prime}=\sum_{j=1}^{n}a_{j}\sum_{i=1}^{m}r_{ij}b_{i}^{\prime}=% \sum_{j=1}^{n}a_{j}b_{j},$

in which

$b_{j}=\sum_{i=1}^{m}r_{ij}b_{i}^{\prime}\,\in R\mathfrak{b}=\mathfrak{b}\qquad% (j=1,\ldots,n).$

Thus an arbitrary element $b$ of the $\mathfrak{b}$ satisfies the condition

$b=b\!\cdot\!1=\sum_{j=1}^{n}(a_{j}b)b_{j}\,\in Rb_{1}\!+\cdots+\!Rb_{n}=(b_{1}% ,\,\ldots,\,b_{n}).$

Consequently, $\mathfrak{b}\subseteq(b_{1},\,\ldots,\,b_{n})$ . Since the inverse inclusion is apparent, we have the equality

$\mathfrak{a}^{-1}=\mathfrak{b}=(b_{1},\,\ldots,\,b_{n}).$

Title	generators of inverse ideal
Canonical name	GeneratorsOfInverseIdeal
Date of creation	2015-05-06 14:52:30
Last modified on	2015-05-06 14:52:30
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	18
Author	pahio (2872)
Entry type	Theorem
Classification	msc 13A15
Related topic	FractionalIdealOfCommutativeRing
Related topic	IdealGeneratedByASet
Related topic	PruferRing