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
Canonical name	InvertibleIdealIsFinitelyGenerated
Date of creation	2015-05-06 14:44:03
Last modified on	2015-05-06 14:44:03
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	10
Author	pahio (2872)
Entry type	Theorem
Classification	msc 13B30
Related topic	InvertibilityOfRegularlyGeneratedIdeal