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