ideal generators in Prüfer ring
Let $R$ be a Prüfer ring with total ring of fractions^{} $T$. Let $\U0001d51e$ and $\U0001d51f$ 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 $\U0001d51e+\U0001d51f$ may, of course, be generated by $m+n$ elements.

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If $\U0001d51e$ or $\U0001d51f$ is regular (http://planetmath.org/FractionalIdealOfCommutativeRing), then the product (http://planetmath.org/ProductOfIdeals) ideal $\U0001d51e\U0001d51f$ may be generated by $m+n1$ elements, since in Prüfer rings the
$$({a}_{1},\mathrm{\dots},{a}_{m})({b}_{1},\mathrm{\dots},{b}_{n})=({a}_{1}{b}_{1},{a}_{1}{b}_{2}+{a}_{2}{b}_{1},{a}_{1}{b}_{3}+{a}_{2}{b}_{2}+{a}_{3}{b}_{1},\mathrm{\dots},{a}_{m}{b}_{n})$$ holds.

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If both $\U0001d51e$ and $\U0001d51f$ are regular ideals, then the intersection $\U0001d51e\cap \U0001d51f$ and the quotient ideal $\U0001d51e:\U0001d51f=\{r\in R\mathit{\hspace{1em}}r\U0001d51f\subseteq \U0001d51e\}$ both may be generated by $m+n$ elements.

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If $\U0001d51e$ is regular, then it is also invertible (http://planetmath.org/InvertibleIdeal). Its ideal has the expression (http://planetmath.org/QuotientOfIdeals)
$${\U0001d51e}^{1}=[R:\U0001d51e]=\{t\in T\mathit{\hspace{1em}}t\U0001d51e\subseteq R\}$$ and may be generated by $m$ elements of $T$ (see the generators of inverse ideal).
Cf. also the twogenerator 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  20130322 14:33:04 
Last modified on  20130322 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 