Lemma. Let $R$ be a commutative ring containing regular elements. If $\mathfrak{a}$ $\mathfrak{b}$ and $\mathfrak{c}$ are three ideals of $R$ such that $\mathfrak{b\!+\!c}$ $\mathfrak{c\!+\!a}$ , and $\mathfrak{a\!+\!b}$ , are invertible, then also their sum ideal $\mathfrak{a\!+\!b\!+\!c}$ , is invertible.

Proof. We may assume that $R$ has a unity, therefore the product of an ideal and its inverse is always $R$ Now, the ideals $\mathfrak{b+c}$ $\mathfrak{c+a}$ , and $\mathfrak{a+b}$ , have the inverses $\mathfrak{f_1}$ $\mathfrak{f_2}$ and $\mathfrak{f_3}$ respectively, so that $$\mathfrak{(b+c)f_1 = (c+a)f_2 = (a+b)f_3} = R.$$ Because $\mathfrak{af_2} \subseteq R$ , and $\mathfrak{cf_1} \subseteq R$ we obtain $$\mathfrak{(a+b+c)(af_2f_3+cf_1f_2) = (a+b)af_2f_3+c(af_2)f_3+a(cf_1)f_2+(b+c)cf_1f_2 = af_2+cf_2 = (c+a)f_2} = R.$$

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 invertible. Let $\{r_1,\,r_2,\,\ldots,\,r_{n+1}\}$ be any set of regular elements of $R$ Denote $$\mathfrak{a} := (r_1),\quad \mathfrak{b} := (r_2,\,\ldots,\,r_n), \quad \mathfrak{c} := (r_{n+1}).$$ The sums $\mathfrak{b+c}$ $\mathfrak{c+a}$ , and $\mathfrak{a+b}$ , are, by the assumptions, invertible. Then the ideal $$(r_1,\,r_2,\,\ldots,\,r_n,\,r_{n+1}) = \mathfrak{a+b+c}$$ is, by the lemma, invertible, and the induction proof is ready.

Bibliography

1

R. GILMER: Multiplicative ideal theory. Queens University Press. Kingston, Ontario (1968).