## 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 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.

## In the case of polynomials

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.

## 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.

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.

## Elementary properties of the degree

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

if , then

## Euclidean division

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 .

## Polynomials over a field

In this context, we consider a commutative field (simply: field) . The polynomials will be constructed as a certain associative unital algebra over (“-aua”), together with a distinguished element called the formal variable.

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.

## Definition

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 .

## Unicity up to an isomorphism

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 .

## Characterization

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 .

## Definition

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 .

## Effect of composition of polynomials on the corresponding functions

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, .

## Effect on the evaluation of a polynomial

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 .

## Associativity

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, .

todo

todo