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,
.
Effect of composition on the evaluation of polynomials
todo
Derivation of the composition of polynomials
todo