Filtrative Euclidean norm: Difference between revisions
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==Definition== | ==Definition== | ||
A [[Euclidean norm]] on an [[integral domain]] is said to be '''filtrative''' if it satisfies the following conditions: | A [[Euclidean norm]] on an [[integral domain]] is said to be '''filtrative''' if it satisfies the following equivalent conditions: | ||
* For any two elements of the domain, either their sum is zero or the norm of their sum is at most the maximum of the norms | * For any two elements of the domain, either their sum is zero or the norm of their sum is at most the maximum of the norms | ||
* The set of elements of norm <math>r</math>, along with zero, forms an additive subgroup. Thus, the association to each <math>r</math> of the corresponding subgroup forms a filtration of additive subgroups of the integral domain. | * The set of elements of norm at most<math>r</math>, along with zero, forms an additive subgroup. Thus, the association to each <math>r</math> of the corresponding subgroup forms a filtration of additive subgroups of the integral domain. | ||
Clearly, any filtrative Euclidean norm is also uniquely Euclidean. | Clearly, any filtrative Euclidean norm is also uniquely Euclidean. | ||
==Relation with other properties== | |||
==Weaker properties=== | |||
* [[Uniquely Euclidean norm]] | |||
Revision as of 18:32, 5 January 2008
This article defines a property that can be evaluated for a Euclidean norm on a commutative unital ring
Definition
A Euclidean norm on an integral domain is said to be filtrative if it satisfies the following equivalent conditions:
- For any two elements of the domain, either their sum is zero or the norm of their sum is at most the maximum of the norms
- The set of elements of norm at most, along with zero, forms an additive subgroup. Thus, the association to each of the corresponding subgroup forms a filtration of additive subgroups of the integral domain.
Clearly, any filtrative Euclidean norm is also uniquely Euclidean.