# 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.

• Then the sum ideal $\mathfrak{a+b}$ may, of course, be generated by $m+n$ elements.

• 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.

• 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 (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 IdealGeneratorsInPruferRing 2013-03-22 14:33:04 2013-03-22 14:33:04 pahio (2872) pahio (2872) 20 pahio (2872) Result msc 13C13 FractionalIdeal ProductOfFinitelyGeneratedIdeals