Theorem.โ Let $R$ be a commutative ring containing regular elements.โ Every invertible fractional ideal $\mathfrak{a}$ of $R$ is finitely generated and regular, i.e. contains regular elements.

Proof.โ Let $T$ be the total ring of fractions of $R$ and $e$ the unity of $T$.โWe first show that the inverse ideal of $\mathfrak{a}$ has the unique quotient presentationโ $[R^{\prime}:\mathfrak{a}]$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 mean 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 generator system 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).

## Comments

## invertible ideals can be principal in a local ring

Can anyone show this?

An invertible fractional ideal in an integral domain that is a local ring is principal.

## Re: invertible ideals can be principal in a local ring

Let I be an invertible fractional ideal of an integral domain D. Let J be its inverse. So IJ=D.

Write 1=sum of (a_k * b_k) , where a_k are in I and b_k are in J, k ranges from 1 to n.

If r is any element in I, multiply r on both sides of the equation in the previous paragraph, and rearrange the terms so that r = sum of (a_k *(r*b_k)). Since b_k are in J, r*b_k are in D. So I is generated by a_k, k ranging from 1 to n. I is finitely generated.

If D is local, it has a unique maximal ideal M, which contains all the non-units. If all of a_k are non-units, then 1, which is the sum of a_k*b_k, must also be a non-unit. This is impossible. So one of the a_k must be a unit. Being a unit, it must generate all of I. So I is principal.

## Re: invertible ideals can be principal in a local ring

What I mean in the last paragraph of the proof is that if all of a_k are non-units, they must all be in M. But then, 1, being the sum of a_k*b_k, must be in M as well. This means that M=D, an impossibility.