# 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 (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 GeneratorsOfInverseIdeal 2015-05-06 14:52:30 2015-05-06 14:52:30 pahio (2872) pahio (2872) 18 pahio (2872) Theorem msc 13A15 FractionalIdealOfCommutativeRing IdealGeneratedByASet PruferRing