ideal generators in Prüfer ring

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 (http://planetmath.org/IdealGeneratedByASet) $m$ and $n$ elements of $T$ , respectively.

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Then the sum ideal $\mathfrak{a+b}$ may, of course, be generated by $m+n$ elements.
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If $\mathfrak{a}$ or $\mathfrak{b}$ is regular (http://planetmath.org/FractionalIdealOfCommutativeRing), then the product (http://planetmath.org/ProductOfIdeals) ideal $\mathfrak{ab}$ may be generated by $m+n-1$ elements, since in Prüfer rings the

$(a_{1},\,...,\,a_{m})(b_{1},\,...,\,b_{n})=(a_{1}b_{1},\,a_{1}b_{2}+a_{2}b_{1}% ,\,a_{1}b_{3}+a_{2}b_{2}+a_{3}b_{1},\,...,\,a_{m}b_{n})$

holds.
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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.
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If $\mathfrak{a}$ is regular, then it is also invertible (http://planetmath.org/InvertibleIdeal). Its ideal has the expression (http://planetmath.org/QuotientOfIdeals)

$\mathfrak{a}^{-1}=[R:\mathfrak{a}]=\{t\in T|\quad t\mathfrak{a}\subseteq R\}$

and may be generated by $m$ elements of $T$ (see the generators of inverse ideal).

Cf. also the two-generator property.

References

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

Title	ideal generators in Prüfer ring
Canonical name	IdealGeneratorsInPruferRing
Date of creation	2013-03-22 14:33:04
Last modified on	2013-03-22 14:33:04
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	20
Author	pahio (2872)
Entry type	Result
Classification	msc 13C13
Related topic	FractionalIdeal
Related topic	ProductOfFinitelyGeneratedIdeals