ideal generators in Pr\"ufer ringIdealGeneratorsInPruferRing2004-08-14 17:58:222008-03-11 16:40:55ResultPr\"ufer ring% this is the default PlanetMath preamble. as your knowledge
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% define commands hereLet $R$ be a Pr\"ufer ring with total ring of fractions $T$.
\,Let $\mathfrak{a}$ and $\mathfrak{b}$ be fractional ideals of $R$, \PMlinkname{generated by}{IdealGeneratedByASet} $m$ and $n$ elements of $T$, respectively.
\begin{itemize}
\item\,Then the sum ideal $\mathfrak{a+b}$ may, of course, be generated by $m+n$ elements.
\item\,If $\mathfrak{a}$ or $\mathfrak{b}$ is \PMlinkname{regular}{FractionalIdealOfCommutativeRing}, then the \PMlinkname{product}{ProductOfIdeals} ideal $\mathfrak{ab}$ may be generated by $m+n-1$ elements, since in Pr\"ufer rings the \PMlinkescapetext{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.
\item\,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.
\item\,If $\mathfrak{a}$ is regular, \,then it is also \PMlinkname{invertible}{InvertibleIdeal}.\, Its \PMlinkescapetext{inverse} ideal has the \PMlinkname{expression}{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).
\end{itemize}
Cf. also the two-generator property.
\begin{thebibliography}{9}
J. Pahikkala: \,``Some formulae for multiplying and inverting ideals''. $-$ \emph{Annales universitatis turkuensis} 183. \,Turun yliopisto (University of Turku) 1982.
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