invertible ideal is finitely generated
Theorem.โ Let be a commutative ring containing regular elements.โ Every invertible (http://planetmath.org/FractionalIdealOfCommutativeRing) fractional ideal of is finitely generated and regular (http://planetmath.org/RegularIdeal), i.e. regular elements.
Proof.โ Let be the total ring of fractions of and the unity of .โWe first show that the inverse ideal of has the unique quotient presentation (http://planetmath.org/QuotientOfIdeals)โ โ whereโ .โ If is an inverse ideal of , it means thatโ .โ Therefore we have
so that
This implies thatโ ,โ and because is a cancellation ideal, it must thatโ , i.e. is the unique inverse of the ideal .
Sinceโ ,โ there exist some elements of and the elements ofโ โ such thatโ .โ Then an arbitrary element of satisfies
because every belongs to the ring .โ Accordingly,โ .โ Since the converse inclusion is apparent, we have seen thatโ โ is a finite of the invertible ideal .
Since the elements belong to the total ring of fractions of , we can choose such a regular element of that each of the products belongs to .โ Then
and thus the fractional ideal contains a regular element of , which obviously is regular in , too.
References
- 1 R. Gilmer: Multiplicative ideal theory.โ 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 |