# invertible ideal is finitely generated

Theorem.โ Let $R$ be a commutative ring containing regular elements.โ Every invertible (http://planetmath.org/FractionalIdealOfCommutativeRing) fractional ideal $\mathfrak{a}$ of $R$ is finitely generated and regular (http://planetmath.org/RegularIdeal), i.e. regular elements.

Proof.โ Let $T$ be the total ring of fractions of $R$ and $e$ the unity of $T$.โWe first show that the inverse ideal of $\mathfrak{a}$ has the unique quotient presentation (http://planetmath.org/QuotientOfIdeals)โ $[R^{\prime}:\mathfrak{a}]$โ whereโ $R^{\prime}:=R+\mathbb{Z}e$.โ If $\mathfrak{a}^{-1}$ is an inverse ideal of $\mathfrak{a}$, it means thatโ $\mathfrak{aa}^{-1}=R^{\prime}$.โ Therefore we have

 $\mathfrak{a}^{-1}\subseteq\{t\in T\,\vdots\,\,\,t\mathfrak{a}\subseteq R^{% \prime}\}=[R^{\prime}\!:\!\mathfrak{a}],$

so that

 $R^{\prime}=\mathfrak{aa}^{-1}\subseteq\mathfrak{a}[R^{\prime}\!:\!\mathfrak{a}% ]\subseteq R^{\prime}.$

This implies thatโ $\mathfrak{aa}^{-1}=\mathfrak{a}[R^{\prime}\!:\!\mathfrak{a}]$,โ and because $\mathfrak{a}$ is a cancellation ideal, it must thatโ $\mathfrak{a}^{-1}=[R^{\prime}\!:\!\mathfrak{a}]$, i.e. $[R^{\prime}\!:\!\mathfrak{a}]$ is the unique inverse of the ideal $\mathfrak{a}$.

Sinceโ $\mathfrak{a}[R^{\prime}\!:\!\mathfrak{a}]=R^{\prime}$,โ there exist some elements $a_{1},\,\ldots,\,a_{n}$ of $\mathfrak{a}$ and the elements $b_{1},\,\ldots,\,b_{n}$ ofโ $[R^{\prime}\!:\!\mathfrak{a}]$โ such thatโ $a_{1}b_{1}\!+\cdots+\!a_{n}b_{n}=e$.โ Then an arbitrary element $a$ of $\mathfrak{a}$ satisfies

 $a=a_{1}(b_{1}a)\!+\cdots+\!a_{n}(b_{n}a)\in(a_{1},\,\ldots,\,a_{n})$

because every $b_{i}a$ belongs to the ring $R^{\prime}$.โ Accordingly,โ $\mathfrak{a}\subseteq(a_{1},\,\ldots,\,a_{n})$.โ Since the converse inclusion is apparent, we have seen thatโ $\{a_{1},\,\ldots,\,a_{n}\}$โ is a finite of the invertible ideal $\mathfrak{a}$.

Since the elements $b_{i}$ belong to the total ring of fractions of $R$, we can choose such a regular element $d$ of $R$ that each of the products $b_{i}d$ belongs to $R$.โ Then

 $d=a_{1}(b_{1}d)\!+\cdots+\!a_{n}(b_{n}d)\in(a_{1},\,\ldots,\,a_{n})=\mathfrak{% a},$

and thus the fractional ideal $\mathfrak{a}$ contains a regular element of $R$, which obviously is regular in $T$, too.

## References

• 1 R. Gilmer: Multiplicative ideal theory.โ Queens University Press. Kingston, Ontario (1968).
Title invertible ideal is finitely generated InvertibleIdealIsFinitelyGenerated 2015-05-06 14:44:03 2015-05-06 14:44:03 pahio (2872) pahio (2872) 10 pahio (2872) Theorem msc 13B30 InvertibilityOfRegularlyGeneratedIdeal