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**:

We note that as defined above is trilinear.

Let be a vector space and a trilinear mapping . We must show that there exists a unique linear mapping such that .

For any , the mapping is bilinear; hence there is a unique linear mapping such that for all , we have .

It is easy to check that the mapping is bilinear; hence there is a unique linear mapping such that for all , we have .

Then for any , with :

Hence is such that .

We must now show that if is such that , then .

]]>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 be written with each element of , and being non null for only a finite number of indices .

]]>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 is in . Thus is a filter over .

The cofinite filter over is defined independently of any ordering on .

If is an ultrafilter over , then either it is a principal ultrafilter, or finer than the cofinite filter, but not both.

Proof:

Directed set = *ensemble ordonné filtrant*, a partially ordered set such that for any , there exists with and .

The directed set filter is defined by the filter base ; that is that the elements of are the “segments” of all elements of greater than any .

If has no maximal element (that is, for all elements of , there is another element of that is strictly greater), then the directed set filter is finer than the cofinite filter.

Proof:

Let be an element of the cofinite filter over .

Then is finite. Since is a directed set, there exists an that is greater than all elements of . There exists an that is strictly greater than all elements of . All elements in the segment are strictly greater, hence different from, all elements of , that is, that they belong to . Thus , and since belongs to the filter base , belongs to the filter .

Thus the cofinite filter is a subset of the directed set filter.

]]>filter base over , with a set.

It is always the case that .

**Proof**: Let be an eventual element of , union for all of the sets .

Since is in this union of sets, it is in at least one of them: there exists an such that , that is, such that for all , belongs to .

Now let be any element of . Since and are both in the filter base , their intersection is non-empty and there exists an in this intersection.

Since , we have seen that .

Furthermore, .

Hence .

This being the case for any , we have , that is, .

This being the case for all , we have .

If and are filter bases over a same , with finer than , then

That is, the lower limits increase with the fineness of the filter base, and the upper limits decrease.

**Proof**: is finer than , that is, every is a superset of some .

Let be an eventual element of , that is: . Then there is some such that is in every . Since is finer than , is a superset of some . Then is in every , that is, , with . Hence .

This being the case for all , we have .

Let be an eventual element of , that is, . Let be an element of . Since is finer than , is a superset of some . Since , we have , that is, there is an element such that . Since , we also have , and hence . This being the case for any element of , we have .

This being the case for all , we have .

If two filter bases are equivalent, they have the same upper limits and lower limits.

**Proof**: If the filter bases are equivalent, they are each finer than the other, hence the lower limit of one is both subset and superset of that of the other, that is, are equal; the same for the upper limits.

Let be a sequence of sets, all subsets of a set .

The upper and lower limits and of this sequence are classically defined as:

is the set of elements that are in for an infinity of values of ;

is the set of elements that are in for all but a finite number of .

If , then for any , there is a such that , that is, . Conversely, if for any we have , then . Hence:

If , there must be an such that all the values of for which are less than ; then for all , we have , that is, . Conversely, if there is an such that , then all values of such that must be less than , and hence there are a finite number of them; thus . Hence:

The Fréchet filter is the filter on the elements of which are the cofinite subsets of . Let be defined as the set of subsets for all . It is easily checked that is a filter base and generates on ; we will call it the Fréchet filter base.

Again, let be a sequence of subsets of a set .

We can consider as a mapping ; the Fréchet filter base has a certain image by this mapping, that is a filter base on . An element of corresponds to the element .

]]>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 ; hence is an element of . Similarly ; hence .

Let and be elements of . Then they are both elements of , hence their intersection belongs to . Similarly their intersection belongs to . Hence .

Note:

The filter is also equal to the set of unions of an element of and an element of . In effect, such an is a superset of , hence belongs to , and a superset of , hence belongs to , and so belongs to . Conversely, an element of belongs both to and to , and, being the union of itself with itself, is the union of an element of with an element of .

]]>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 will show that for any concrete category , with faithful functor from to , a sufficient (though not always necessary) condition for a morphism in to be mono is that its image by be itself mono, that is, one-to-one.

More generally, we will show that if categories and are such that there exists a faithful functor from to , a sufficient condition for a morphism in to be mono is that its image in by be itself mono. The same will hold for epimorphisms.

I will first prove the general case last mentioned.

Assume categories and with a faithful functor from the former to the latter.

Let and be two objects in and a morphism from to such that is a monomorphism.

Let be an object in and and two morphisms from to such that .

Since is a functor, we can write , and similarly for ; hence:

Since is a monomorphism, it follows that .

Since is faithful, this implies .

This being the case for any object in and and morphisms from to such that , it follows that is a monomorphism.

We have thus shown that if is mono, then itself is mono.

A similar reasoning shows that if is epi, then itself is epi.

In the case of concrete categories, is the set category, in which injective maps are mono and surjective maps are epi. Typically, the -morphisms are particular mappings between the underlying sets, that is the sets obtained through the functor . In this case, one can say that injective implies mono and surjective implies epi.

]]>If and are such that is separately coprime with each , then is coprime with their product .

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 but one contain as a factor. The one that doesn’t have as a factor is . Hence we obtain an expression of the form:

Hence and are coprime.

The same result holds with polynomials: if and are polynomials over a field , such that is coprime with each , then is coprime with their product .

The proof is identical.

]]>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.

The family is a -vector space basis of . For any , there exists an unique , only finitely many of which are nonzero, such that . If all the are zero, that is, if is the zero element of , then is defined as . If not all the are zero, since only finitely many are nonzero, there is a greatest value of such that . Then is defined as this greatest value.

The following are easily checked to hold for any polynomials and (even when one or both are zero):

if , then

The notion of the degree of a polynomial allows us to formulate the following property of Euclidean division:

For all polynomials and , with nonzero, there exists exactly one pair of polynomials such that with .

For the proof, we fix the value of nonzero polynomial , and recurse over the degree of .

More specifically, being given, we consider the following proposition dependent on :

iff for all polynomials such that , there exist polynomials , with , such that .

is trivially true (for means that is zero, and satisfies with ).

Let us suppose true for some , and consider a polynomial the degree of which is less or equal to the successor of , that is to if , or otherwise.

If the degree of is less or equal to , then applies and there exist the wanted and .

If the degree of is less than that of , then we can directly write with and .

Remains the case in which is the successor of .

]]>itself, as a vector space over itself and the usual multiplication in as the third law, is a -aua,

Morphisms between two -auas are linear mappings that conserve the third law and map the unit of the first algebra onto the unit of the second one.

All that will be said of the polynomials over will be based on the universal property given below. The polynomials will thus be defined uniquely up to an isomorphism.

The polynomials over is any -aua together with a distinguished element of such that the following universal property holds:

For any -aua and any , there exists exactly one morphism such that .

Let and be two sets of polynomials over . The universal property implies the existence of a morphism such that , and of a morphism such that .

We then have . Hence is a morphism that maps to . Now is another such morphism. By virtue of the universal property of , it follows that , that is, is the -aua category identity on .

The same reasoning shows that is the -aua category identity on .

Hence is an isomorphism that maps the formal variable of to that of .

Let be an associative unital algebra over , and an element of . Then represents the polynomials over if and only if the family , with and , is a family-base of the -vector space .

Let us first suppose that the latter condition holds on , and prove the universal property.

Let be a -aua and an element of .

Let us suppose that is a morphism such that . Then for all , we have . Since as a -aua morphism, it is in particular a linear mapping. Since a linear mapping is defined by the image of a basis, and is a basis, necessarily is the unique linear mapping such that . Now it is easy to check that this linear mapping does preserve the multiplication in , and is hence a -aua morphism.

Thus the universal property is proven for .

Let us now suppose the universal property for , and show that is a family-base of the -vector space .

Let us first show that is linearly independent.

Let be a free -vector space on set . This entails that is a family-base of . To make into an associative unital algebra, we must define a multiplication law that is bilinear, and also associative.

A bilinear law can be defined by the images of all the ordered pairs of base elements. Hence we may define the multiplication in by:

We then have .

Hence the multiplication is associative with regard to the elements of the base; it is easy to check that this follow through to arbitrary elements, which makes an associative algebra.

is a unit; hence is an associative unital algebra.

Applying the universal property of , we find that there exists exactly one -aua morphism such that .

We then have ; ; and, for any , . Hence for all , .

Let us consider a linear combination of the family that vanishes; that is, a finite sequence of values of , , such that . Taking the image of this linear combination by , we obtain . Since the form a basis, all the must be zero.

Thus the family is linearly independent.

Lastly, we must show that this family generates the -vector space .

Let be the sub-aua of generated by .

There exists a unique aua morphism such that .

Let then be , but with all of as codomain; in other words, . It is clear that too is an aua morphism, and it is such that .

But there is exactly one aua morphism that maps to ; and is such.

Hence .

The image of is that of , and is a subset of . But the image of is itself. Hence .

Thus the sub-aua of generated by is itself. This sub-aua is the set of linear combinations of all the powers of ; hence the sub-vector space generated by is .

Thus is a family-base of the -vector space .

]]>Let be a commutative field. The polynomials form a commutative unital algebra over containing an element such that the following universal property holds:

For any unital algebra (commutative or otherwise) and any element , there exists exactly one unital algebra morphism such that .

If and are elements of , let be the unique morphism such that . Then the composition of and is defined by .

In terms of polynomial functions, the composition of polynomials is the equivalent to the composition (via “”) of the associated functions.

Let us first define the function that is associated with a polynomial.

The set of functions is made into a commutative and unital associative algebra by the usual operations of summing (, multiplication by a scalar ( and internal multiplication ().

Hence, following the universal property of the algebra of polynomials, there exists a unique morphism such that .

By definition, the function associated with a polynomial is simply .

We wish to show that if and are polynomials and and the respective associated functions, the function associated with is .

Let be a polynomial and . Let us consider the mapping .

It is easy to check that is a morphism.

Furthermore, .

On the other hand, is also a morphism from to , and .

Thus both and are morphisms and they agree on the value . The universal property of polynomials entails that they are equal.

Hence for any , .

That is, .

If and , the evaluation of at is, by definition, the value at of the unique morphism such that .

Given polynomials and and , we wish to evaluate at , and, specifically, show that .

Since , we have .

Both and are morphisms . Both evaluate at to . Hence they are equal:

which entails

hence: , as announced.

In particular: if and , then .

If , and are polynomials in , we wish to show that:

Using the definition of composition, this translates into:

That is:

We need to show that for any and in , the two mappings and are identical.

Both are endomorphisms of .

We have:

- by definition of .
- by definition of .

Hence both morphisms agree on the image of . By virtue of the universal property of polynomials, they are equal.

Hence for any , we indeed have , that is, .

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