invertibility of regularly generated ideal

Lemma.   Let $R$ be a commutative ring containing regular elements.  If $𝔞$, $𝔟$ and $𝔠$ are three ideals of $R$ such that  $𝔟+𝔠$,  $𝔠+𝔞$  and  $𝔞+𝔟$  are invertible (http://planetmath.org/FractionalIdealOfCommutativeRing), then also their sum ideal  $𝔞+𝔟+𝔠$  is .

Proof.  We may assume that $R$ has a unity, therefore the product of an ideal and its inverse (http://planetmath.org/FractionalIdealOfCommutativeRing) is always $R$.  Now, the ideals  $𝔟+𝔠$,  $𝔠+𝔞$  and  $𝔞+𝔟$  have the ${𝔣}_{\mathfrak{1}}$, ${𝔣}_{\mathfrak{2}}$ and ${𝔣}_{\mathfrak{3}}$, respectively, so that

$$(𝔟+𝔠){𝔣}_{\mathfrak{1}}=(𝔠+𝔞){𝔣}_{\mathfrak{2}}=(𝔞+𝔟){𝔣}_{\mathfrak{3}}=R.$$

Because  $𝔞{𝔣}_{\mathfrak{2}}\subseteq R$  and  $𝔠{𝔣}_{\mathfrak{1}}\subseteq R$,  we obtain

$(𝔞+𝔟+𝔠)(𝔞{𝔣}_{\mathfrak{2}}{𝔣}_{\mathfrak{3}}+𝔠{𝔣}_{\mathfrak{1}}{𝔣}_{\mathfrak{2}})$$\mathrm{\hspace{0.33em}}=(𝔞+𝔟)𝔞{𝔣}_{\mathfrak{2}}{𝔣}_{\mathfrak{3}}+𝔠(𝔞{𝔣}_{\mathfrak{2}}){𝔣}_{\mathfrak{3}}+𝔞(𝔠{𝔣}_{\mathfrak{1}}){𝔣}_{\mathfrak{2}}+(𝔟+𝔠)𝔠{𝔣}_{\mathfrak{1}}{𝔣}_{\mathfrak{2}}$

$\mathrm{\hspace{0.33em}}=𝔞{𝔣}_{\mathfrak{2}}+𝔠{𝔣}_{\mathfrak{2}}=(𝔠+𝔞){𝔣}_{\mathfrak{2}}$

$\mathrm{\hspace{0.33em}}=R.$

Theorem.  Let $R$ be a commutative ring containing regularelements.  If every ideal of $R$ generated by two regular elements is , then in $R$ also every ideal generated by a finite set of regular elements is .

Proof.  We use induction on $n$, the number of the regular elements of the generating set.  We thus assume that every ideal of $R$ generated by $n$ regular elements  ($n\geqq 2)$  is .  Let  $\{r_1,r_2,\mathrm{\dots },r_{n+1}\}$ be any set of regular elements of $R$.  Denote

$$𝔞=:(r_1),𝔟=:(r_2,\mathrm{\dots },r_n),𝔠=:(r_{n+1}).$$

The sums  $𝔟+𝔠$,  $𝔠+𝔞$  and  $𝔞+𝔟$  are, by the assumptions, .  Then the ideal

$$(r_1,r_2,\mathrm{\dots },r_n,r_{n+1})=𝔞+𝔟+𝔠$$

is, by the lemma, , and the inductionproof is complete.