# olivierd

## Associativity of the tensor product

All vector spaces are by default over or . We consider three vector spaces , and . Let be a tensor product of and , and a tensor product of and . The aim is first to show that , together with the mapping:     is a tensor product of , and . Proof:…

## Free associative algebra on a vector space

Vector spaces and associative algebras will be implicitely all on the same field or . Let be a vector space. We wish to build a free associative algebra on . For any natural number , we will note (tensor product times). Each is a vector space. We may form:     An element of can…

## Cofinite filters and directed set filters

We assume that is an infinite set. Cofinite filters The cofinite filter over is the set of subsets of such that is finite. Since is infinite, is not in . It is trivial to check that any superset in of an element of is in , and that the intersection of any two elements of…

## Lower and upper limits of a filter base

Two cases, that I have been unable to unify: a filter base over a power set with a set, or over . First case: filter bases over a power set filter base over , with a set. Definitions         Property It is always the case that . Proof: Let be an eventual…

## Intersection of filters

Let and be filters on a same set . Then is also a filter on . Proof: belongs to both and , hence it belongs to , and . Since is not an element of , it is not an element of . Let be an element of and a superset of . Then and…

## Concrete categories: injective/surjective implies mono/epi; functor generalization

A category is by definition concrete if there exists a faithful functor from to the category of sets . (A functor is faithful if it is one-to-one (injective) regarding morphisms.) It happens to be the case that in the category , monomorphisms and one-to-one mappings are the same thing; similarly for epimorphisms and surjections. We…

## Coprimality with a product of integers

The theorem If and are such that is separately coprime with each , then is coprime with their product . The proof Proof: Through Bézout. being coprime with each , we can write, for each : If we take the product of each of these expressions, we get: Among the terms of this product, all…

## Degree and Euclidean division of polynomials

For the definition and characterization of the polynomials over a field , see Polynomials over a field. Definition of the degree of a polynomial The degree of a polynomial P, written , is an element of . The symbol is taken to have the usual properties of order and of addition relative to natural numbers….