# The composition of polynomials

## 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, .

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