Let $R$ be a Prüfer ring with total ring of fractions $T$ Let $\mathfrak{a}$ and $\mathfrak{b}$ be fractional ideals of $R$ generated by $m$ and $n$ elements of $T$ respectively.

• Then the sum ideal $\mathfrak{a+b}$ may, of course, be generated by $m+n$ elements.
• If $\mathfrak{a}$ or $\mathfrak{b}$ is regular, then the product ideal $\mathfrak{ab}$ may be generated by $m+n-1$ elements, since in Prüfer rings the formula $$(a_1, \,...,\,a_m)(b_1,\,...,\,b_n) = (a_1b_1,\,a_1b_2+a_2b_1,\,a_1b_3+a_2b_2+a_3b_1,\, ...,\,a_mb_n)$$ holds.
• If both $\mathfrak{a}$ and $\mathfrak{b}$ are regular ideals, then the intersection $\mathfrak{a}\cap\mathfrak{b}$ and the quotient ideal $\mathfrak{a\colon\!b} = \{r \in R| \quad r\mathfrak{b} \subseteq \mathfrak{a}\}$ both may be generated by $m+n$ elements.
• If $\mathfrak{a}$ is regular, then it is also invertible. Its inverse ideal has the expression $$\mathfrak{a}^{-1} = [R:\mathfrak{a}] = \{t\in T|\quad t\mathfrak{a} \subseteq R\}$$ and may be generated by $m$ elements of $T$ (see the generators of inverse ideal).

Cf. also the two-generator property.

Bibliography

J. Pahikkala: ``Some formulae for multiplying and inverting ideals''. $-$ Annales universitatis turkuensis 183. Turun yliopisto (University of Turku) 1982.