invertible ideal is finitely generated

Theorem.  Let $R$ be a commutative ring containing regular elements.  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.  Let $T$ be the total ring of fractions of $R$ and $e$ the unity of $T$. We first show that the inverse ideal of $\mathfrak{a}$ has the unique quotient presentation (http://planetmath.org/QuotientOfIdeals)  $[R^{\prime}:\mathfrak{a}]$  where  $R^{\prime}:=R+\mathbb{Z}e$.  If $\mathfrak{a}^{-1}$ is an inverse ideal of $\mathfrak{a}$, it means that  $\mathfrak{aa}^{-1}=R^{\prime}$.  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 that  $\mathfrak{aa}^{-1}=\mathfrak{a}[R^{\prime}\!:\!\mathfrak{a}]$,  and because $\mathfrak{a}$ is a cancellation ideal, it must that  $\mathfrak{a}^{-1}=[R^{\prime}\!:\!\mathfrak{a}]$, i.e. $[R^{\prime}\!:\!\mathfrak{a}]$ is the unique inverse of the ideal $\mathfrak{a}$.

Since  $\mathfrak{a}[R^{\prime}\!:\!\mathfrak{a}]=R^{\prime}$,  there exist some elements $a_{1},\,\ldots,\,a_{n}$ of $\mathfrak{a}$ and the elements $b_{1},\,\ldots,\,b_{n}$ of  $[R^{\prime}\!:\!\mathfrak{a}]$  such that  $a_{1}b_{1}\!+\cdots+\!a_{n}b_{n}=e$.  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 $R^{\prime}$.  Accordingly,  $\mathfrak{a}\subseteq(a_{1},\,\ldots,\,a_{n})$.  Since the converse inclusion is apparent, we have seen that  $\{a_{1},\,\ldots,\,a_{n}\}$  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 $R$.  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.  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