# 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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