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invertibility of regularly generated ideal

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\textbf{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 \PMlinkname{invertible}{FractionalIdealOfCommutativeRing}, then also their sum ideal\, $\mathfrak{a\!+\!b\!+\!c}$\, is \PMlinkescapetext{invertible}.

{\em Proof.} \,We may assume that $R$ has a unity, therefore the product of an ideal and its \PMlinkname{inverse}{FractionalIdealOfCommutativeRing} is always $R$.\, Now, the ideals\, $\mathfrak{b+c}$,\, $\mathfrak{c+a}$\, and\, $\mathfrak{a+b}$\, have the \PMlinkescapetext{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)} &\;=\; \mathfrak{(a+b)af_2f_3+c(af_2)f_3+a(cf_1)f_2+(b+c)cf_1f_2}\\ 
                                   &\;=\; \mathfrak{af_2+cf_2 \;=\; (c+a)f_2}\\ 
                                   &\;=\; R.

\textbf{Theorem.}\, Let $R$ be a commutative ring containing regular 
elements.\, If every ideal of $R$ generated by two regular elements is \PMlinkescapetext{invertible}, then in $R$ also every ideal generated by a finite set of regular elements is \PMlinkescapetext{invertible}.\\

{\em 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 \PMlinkescapetext{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, \PMlinkescapetext{invertible}.\, Then the ideal
    $$(r_1,\,r_2,\,\ldots,\,r_n,\,r_{n+1}) \;=\; \mathfrak{a+b+c}$$
is, by the lemma, \PMlinkescapetext{invertible}, and the induction 
proof is complete.

\bibitem{RG}{\sc R. Gilmer:} {\em Multiplicative ideal theory}. \,Queens University Press. Kingston, Ontario (1968).