generators of inverse ideal

Theorem.  Let $R$ be a commutative ring with non-zerounity and let $T$ be the total ring of fractions of $R$.  If $𝔞=(a_1,\mathrm{\dots },a_n)$  is an invertiblefractional ideal (http://planetmath.org/FractionalIdealOfCommutativeRing) of $R$ with  $𝔞𝔟=R$,  then also the inverseideal $𝔟$ can be generated by $n$ elements of $T$.

Proof.  The equation  $𝔞𝔟=(1)$  implies the existence of the elements $a_i^{\prime }$ of $𝔞$ and $b_i^{\prime }$ of $𝔟$  $(i=1,\mathrm{\dots },m)$ such that  $a_1^{\prime }{b}_1^{\prime }+\mathrm{\cdots }+a_m^{\prime }{b}_m^{\prime }=1$.  Because the $a_i^{\prime }$’s arein $𝔞$, they may be expressed as

$$a_i^{\prime }=\sum _{j=1}^n{r}_{ij}{a}_j\mathit{\hspace{1em}\hspace{1em}}(i=1,\mathrm{\dots },m),$$

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

$$1=\sum _{i=1}^m{a}_i^{\prime }{b}_i^{\prime }=\sum _{i=1}^m\sum _{j=1}^n{r}_{ij}{a}_j{b}_i^{\prime }=\sum _{j=1}^n{a}_j\sum _{i=1}^m{r}_{ij}{b}_i^{\prime }=\sum _{j=1}^n{a}_j{b}_j,$$

in which

$$b_j=\sum _{i=1}^m{r}_{ij}{b}_i^{\prime }\in R𝔟=𝔟\mathit{\hspace{1em}\hspace{1em}}(j=1,\mathrm{\dots },n).$$

Thus an arbitrary element $b$ of the $𝔟$ satisfies the condition

$$b=b\cdot 1=\sum _{j=1}^n(a_{j}b)b_j\in R{b}_1+\mathrm{\cdots }+R{b}_n=(b_1,\mathrm{\dots },b_n).$$

Consequently,  $𝔟\subseteq (b_1,\mathrm{\dots },b_n)$.  Since the inverse inclusion is apparent, we have the equality

$${𝔞}^{-1}=𝔟=(b_1,\mathrm{\dots },b_n).$$